The traditional double-six domino set (28 pieces with zero to six spots) is used. The specific rules used for placing the dominoes—e.g., whether doubles serve as spinners—are up to the player. Draw 14 tiles to form two rows of seven tiles with each row having five tiles with the pips visible and two tiles face down.

Highest double from either row is placed first. If no such double is available, the heaviest domino is placed. However, the first domino must be playable by the visible dominoes in the other row. Skip all dominoes without a visible tile in the other row that can be played off of it. If two playable dominoes are tied as the heaviest, the player chooses which is placed first.

During each turn, if both rows have the same number of total tiles (visible and unseen), any visible tile can be played. Otherwise, only visible tiles from the row with the larger number of tiles can be played.

The number of visible tiles in each row must always be greater than the number of unseen tiles. That is, when a row has two visible and two unseen tiles, reveal one of the unseen tiles, and when a row has one visible and one unseen tile, reveal the unseen tile.

If a playable tile is available, it must be played. If no tiles can be played, flip one of the unseen tiles from the row with the most tiles. If both rows have the same number of tiles, flip any unseen tile. If the flipped tile is not playable, continue flipping eligible unseen tiles until a playable tile is revealed, which must be played.

If the row with the most tiles has no unseen tiles (or if both rows with the same number of tiles have no unseen tiles) and no playable tile is available, the game is blocked and ends.

The player wins if the last tile is played. Otherwise, the player’s score is the number of pips on the unplayed tiles, including any unseen tiles. Lower scores are better.

After shuffling the tiles face down, 14 tiles are drawn and arranged in two rows, with the four tiles on the right remaining face down.

The highest double is 5-5, on the top row, and since there is a five on the bottom row, it is the first tile that is placed.

The next tile must come from the bottom row, since it has more tiles. The tile must have a five.

Now, any of the eight visible tiles can be played, since both rows have the same number of tiles; however, the only tile that can be played is 5-3.

The next tile must come from the top row, but none of the visible tiles can be played. Therefore, the an unseen tile on this row is revealed.

The newly revealed tile can be played.

Note that a tile from either row can now be played, even though the top row has more visible tiles. The *total* number of tiles (visible and non-visible) in each row determines which tiles can be played.

At this point, the bottom row has two visible and two unseen tiles. Since the number of visible tiles must be greater than the number of unseen tiles, one of the unseen tiles is revealed.

The game continues until either all tiles are played or the game is blocked because all tiles have been revealed and no eligible tile can be played.

]]>

This is an idea that I first encountered over a decade ago on the blog of a Czech theoretical physicist (which is now closed to the public). He proposed the idea of an “exponential percent” that could be used as a more rational way to express changes in absolute quantities. I agree completely that it makes more sense to “count the logarithms” than to count absolute changes in the quantities themselves.

The problem that Dr. Motl brings up is not new. Plenty of people have noticed the shortcomings of using percentages to express changes in a quantity. For example, a 10% increase followed by a 10% decrease does not get one back to the original value. In addition, it is not straightforward to combine percent differences and the result of such an attempt can be deceptive as Darrell Huff pointed out in his classic book, *How to Lie with Statistics*. He cleverly demonstrated that a 100% increase in the price of one commodity (bread) and a 50% decrease in another (milk) can be made to look like a 20% increase or decrease in their combined price, depending on the price chosen as a reference point (before or after the change). Mr. Huff’s solution for this dilemma was to switch to using the geometric mean when combining two or more percent differences.

Dr. Motl’s solution is not new either, and it already has a name. What he calls an “exponential percent” is already known as a *centineper* (cNp). This unit, based on the neper, is an established unit—similar to the more familiar decibel (dB), which is an essential unit for anyone working in acoustics or with audio equipment. While the neper is not part of the International System of Units (SI), it is accepted for use alongside the SI. A change expressed in centinepers is also sometimes called the “log change.”

Although the centineper is quite useful, here I propose an alternative notation and way of thinking about this general concept. The new notation that I propose here accomplishes the same thing as Huff’s geometric mean when combining two or more percent differences—mathematically it’s identical—but hopefully this notation will be more intuitive.

I call the measure of change that I propose “percent doubling,” and it is defined as follows. If a quantity \(q\) changes from an initial value of \(q_{\text{i}}\) to a final value of \(q_{\text{f}}\), the percent doubling rate \(r_{{\times}2}\) is the value that satisfies\[

q_{\text{f}} = q_{\text{i}} 2^{r_{{\times}2}/100}

\]Thus, a value of \(r_{{\times}2} = 100\) results in\[

q_{\text{f}} = 2 q_{\text{i}}

\]or a doubling of the original amount. In terms of \(r_{{\times}2}\), this can be written\[

r_{{\times}2} = 100 \times \log_2 \frac{q_{\text{f}}}{q_{\text{i}}}

\]To denote quantities expressed this way, either “\(\%_{{\times}2}\)” or “\(\%{\times}2\)” can be used. The former is more compact, the latter is more straightforward and doesn’t require a change in font size. Another possibility is to use the abbreviation “pcd.”

Because of the choice of scale, percent doubling is somewhat similar to the percent difference, which is defined as\[

\Delta\%q = \frac{q_{\text{f}} – q_{\text{i}}}{q_{\text{i}}} \times 100\%

\]In fact, percent doubling and percent difference are very similar for differences that either are very small (near 0%) or are very large (near 100%), as shown below.

Percent Difference | Percent Doubling |
---|---|

0.0 | 0.0 |

5.0 | 7.0 |

10.0 | 13.8 |

20.0 | 26.3 |

30.0 | 37.9 |

40.0 | 48.5 |

50.0 | 58.5 |

60.0 | 67.8 |

70.0 | 76.6 |

80.0 | 84.8 |

90.0 | 92.6 |

95.0 | 96.3 |

100.0 | 100.0 |

The difference between the two values is largest in the middle region, near 50%. Outside of the 0% to 100% range, the similarity ends and the logarithmic nature of percent doubling dominates.

For example, a percent difference of \(-100\%\) indicates that nothing is left in the original quantity, but a difference of \(-100\%_{{\times}2}\) means that only *half* of the original quantity is gone. In this case, it might make sense to adopt a separate notation for negative percent doubling. This new quantity—which we can call “percent halving” (or perhaps “half percent”) and express as \(\%_{{\div}2}\) or \(\%{\div}2\) to be consistent with the notation introduced above—is defined as\[

r_{{\div}2} = 100\times\log_2\frac{q_{\text{i}}}{q_{\text{f}}}

\]since \(q_{\text{f}} = \frac{1}{2} q_{\text{i}} \Rightarrow r_{{\div}2} = 100\%_{{\div}2}\). It is immediately clear that this new quantity is simply the negative of the percent doubling:\[

r_{{\div}2} = -100 \times\log_2\frac{q_{\text{f}}}{q_{\text{i}}} = -r_{{\times}2}

\]This is the first example of a generalization of this concept. While a factor of two was chosen because \(+100\%\) means twice the original amount, we can also define a “percent \(n\)-times,” denoted “\(\%_{{\times}n}\)” or “\(\%{\times}n\),” which is given by the formula\[

r_{{\times}n} = 100\times\log_n\frac{q_{\text{f}}}{q_{\text{i}}}

\]Thus, \(r_{{\div}2}\) is equivalent to \(r_{{\times}1/2}\).

With this more general notation, Motl’s “exponential percent” (or “E-percent”), which he denoted as “E%,” can be written in my notation as “\(\%{\times}e\).” In fact, the only difference between our ideas is a choice of notation. A short table below shows equivalent notation in the two systems.

E-Percent | Percent Doubling |
---|---|

E% | \(\%{\times}e\) |

E\(_2\)% | \(\%{\times}2\) |

E\(_n\)% | \(\%{\times}n\) |

As Motl points out, one advantage of the E-percent (and the centineper) is that it approximates the percent difference for very small differences. Expressed in terms of the percent difference, the E-percent is\[

\mathrm{E\%} = 100 \times \ln \left(\frac{\Delta\%q}{100} + 1\right)

\] Since\[

\ln (x+1) = x – \frac{1}{2} x^2 + \frac{1}{3} x^3 – \cdots

\]this quantity can be written\[

\mathrm{E\%} = \Delta\%q + O\left((\Delta\%q)^2\right)

\]or if \(\Delta\%q\) is small,\[

\mathrm{E\%} \approx \Delta\%q

\]

Percent doubling does not have this property. It is the choice of scaling that is *exactly* equal to the percent difference at values of 0 and 100 so as to preserve the meanings of 0% and 100%.

As mentioned above, Motl’s E-percent is actually a centineper, a unit that is already defined:\[

1\,\mathrm{E\%} = 1\,\mathrm{cNp}

\] Furthermore, the neper is similar to the decibel in that they both measure the same thing: logarithmic-scale increases and decreases. The difference between the two units is that the decibel uses the decadic (base-10) logarithm, whereas the neper uses the natural logarithm. Therefore, the two units are related by a scaling factor:\[

1\,\mathrm{Np} = 20 \log_{10} ( e )\,\mathrm{dB}

\approx 8.69\,\mathrm{dB}

\]

The percent doubling also measures logarithmic-scale increases decreases, but the measurement is according to factors of two. That is, it uses base-2 logarithms. Therefore, it is related to these other two units by scaling factors:\[

1\%_{{\times}2} = \frac{1}{\ln 2}\,\mathrm{cNp}

\approx 1.44\,\mathrm{cNp}

\]and\[

1\%_{{\times}2} = \frac{0.2 \log_{10} ( e )}{\ln 2}\,\mathrm{dB}

\approx 0.125\,\mathrm{dB}

\]

It is important to emphasize that there is nothing conceptually new with the introduction of this unit of measure. It’s much like replacing miles with kilometers or inches with centimeters. (And in fact, the modern inch is defined as exactly 2.54 cm.) The scale is chosen for convenience. For example, feet and inches lend themselves well to factions, since 12 can be divided by 2, 3, 4, and 6 (whereas 10 is divisible only by 2 and 5). Centimeters and their associated metric units, on the other hand, are more amenable to decimal arithmetic.

In this case, a scale is proposed that lends itself well to a two-digit representation of a change that is similar to the familiar percentage convention that is often used for changes in quantities. The end points of this scale are the same—i.e., 0 means no change and 100 means twice the amount. Therefore, this new scale should be intuitive to anyone accustomed to working with percentages.

Outside of the range from 0 to 100, percent doubling loses its intuitive similarity with percent differences. For example, a percent difference of 200% means an addition that is twice the original amount. That is, the original amount has increased by a factor of three. A percent doubling of 200%\(_{{\times}2}\), however, means that the original amount has been doubled twice. The original amount has increased by a factor of *four*, and the new addition is *three* times the original amount.

A key feature of percent doubling is that it allows successive relative changes to be combined into a total relative change that can be determined by combining the parts by addition. This lends itself well to time-dependent processes, such as growth or decay, and in fact percent doubling is closely related to these processes.

Constant exponential growth and decay are governed by the following equation:\[

q ( t ) = q_0\, e^{\lambda t}

\]where \(q_0\) is the initial quantity, \(q(t)\) is the quantity at time \(t\), and \(\lambda\) is a constant that determines the rate of growth or decay (\(\lambda > 0\) for growth, \(\lambda < 0\) for decay). Such dynamical systems (especially when \(\lambda > 0\)) are sometimes described by the e-folding time\[

t_{\text{e}} = \frac{1}{\lambda}

\]which is the time required for the quantity to increase by a factor of \(e \approx 2.1416\). Closely related to this quantity is the doubling time, which is the time required for the quantity to increase by a factor of two. The converse of the doubling time for decreasing quantities is the half-life, which is commonly used to quantify the rate of radioactive decay in nuclear physics. In terms of the relevant constant \(\lambda\) (growth constant if \(\lambda > 0\), decay constant if \(\lambda < 0\)), these two times are\[

\text{doubling time}\qquad

t_{\text{d}} = \frac{\ln 2}{\lambda}

\] \[

\text{half-life}\qquad

t_{1/2} = – \frac{\ln 2}{\lambda}

\]

For systems of constant exponential growth, percent doubling is the percentage of elapsed time \(t\) to the doubling time\[

\%_{{\times}2} = 100 \times \frac{t}{t_{\text{d}}}

\]Similarly, percent halving is the percentage of elapsed time to the half-life\[

\%_{{\div}2} = 100 \times \frac{t}{t_{1/2}}

\]

A common term encountered when considering growth over a time period is the Compound Annual Growth Rate (CAGR), which can be useful for comparing growth rates of various quantities in a common domain and is often employed to compare financial growth. This term is defined as\[

\text{CAGR}(t_0, t_n) = \left [ \frac{q(t_n)}{q(t_0)} \right ]^{\frac{1}{t_n – t_0}} – 1

\] where \(q(t_0)\) is the initial value, \(q(t_n)\) is the final value, and \(t\) is measured in years. CAGR expresses an increase over several years as if it were the result of exponential growth over the time period with a constant relative increase each year, which allows comparisons of the rate of growth for increases over different time periods (e.g., the growth of one quantity over five years versus the growth of another quantity over ten years).

Although not exactly equivalent to CAGR, percent doubling can be used for the same purpose as CAGR for comparing growth over differing time periods. In fact, percent doubling is superior for comparison purposes because CAGR does not provide a completely reliable measure of the rate of growth. For example, a 10% CAGR is not twice the rate of increase of a 5% CAGR. There is more after two years at 5% (10.25% total) than one year at 10%. On the other hand, two years of 5%\(_{{\times}2}\) growth is *exactly* equivalent to one year of 10%\(_{{\times}2}\) growth. Percent doubling also simplifies the math involved since the total value over the entire period can be divided by the number of time intervals (e.g., years) to arrive at a per-interval value.

To see how the two measures of growth are related, it is useful to note that the definition of CAGR implies\[

\ln (\text{CAGR} + 1) = \frac{\ln q_n – \ln q_0}{n}

\]where \(q_n \equiv q(t_n)\) is the quantity after \(n\) intervals, \(q_0 \equiv q(t_0)\) is the initial quantity, and \(t_n – t_0 = n\). Meanwhile, the same increase expressed as percent doubling is\[

r_{{\times}2} = 100 \times \log_2 \frac{q_n}{q_0}

= 100 \times \frac{\ln q_n – \ln q_0}{\ln 2}

\]so that the average percent doubling over each interval is\[

\bar{r}_{{\times}2} = \frac{100}{\ln 2} \times

\frac{\ln q_n – \ln q_0}{n}

\]Both the expression for CAGR and the expression for percent doubling include the following term on the right hand side of the equations:\[

\frac{\ln q_n – \ln q_0}{n}

\]

This similarity can be explored further. Since\[

\ln (x+1) = x – \frac{1}{2} x^2 + \frac{1}{3} x^3 – \cdots

\]small values of CAGR can be approximated with\[

\text{CAGR} \approx \frac{\ln q_n – \ln q_0}{n}

\]Therefore, if CAGR is expressed as a percentage, it is related to the average percent doubling per year by\[

\text{CAGR(%)} \approx \ln 2 \times \bar{r}_{{\times}2} \approx 0.693\, \bar{r}_{{\times}2}

\]in the limit of small growth.

Here I have introduced a new terminology and notation for expressing relative changes. It is qualitatively similar to percent change, but it has the advantage that multiple successive changes can be combined through addition, which is not possible when using percent differences.

This new notation uses a logarithmic scale, which is nothing new. Both the neper and the decibel provide the same functionality for comparing quantities. The only novel feature is the choice of scale.

The new terminology has the following features and limitations:

- It is similar to the more familiar percent change: 0% and 100% mean exactly the same thing in both systems.
- Two or more changes can be combined through addition to reach a total change through this system.
- Doubling the pcd means twice the exponential growth.
- It is best applied to fractional changes: changes outside of \(0\%_{{\times}2}\) to \(100\%_{{\times}2}\) quickly loose their intuitive association with percent change.
- Its converse, the percent halving (\(\%_{{\div}2}\)), performs a similar feature for decreasing quantities.
- It is useful for describing exponential growth and decay.
- It can serve a function similar to Compound Annual Growth Rate (CAGR) for growth rates, while simplifying the mathematics required and providing a more genuine measure of geometric growth.

In Blackjack, as played in casinos, only the player has freedom of action. The dealer’s actions are controlled by the rules of the game. Therefore, if betting is ignored and the player also performs the dealer’s actions according to the rules, Blackjack can be considered to be a game of solitaire. In fact, the popular solitaire variant Canfield was originally devised as a gambling game. The player purchased a deck of cards and played the solitaire game—under the watchful eye of a house employee—until completion. The player was then paid according to the number of cards that were successfully moved to the foundations.

Playing one hand of Blackjack, win or lose, is fairly straightforward. To genuinely enjoy the game, however, it is necessary to play a series of deals and keep a running score, and in fact, this is essential to gain the benefits of splitting and doubling down. This requires some way to keep score.

In the solitaire game described here, the cards themselves are used to track the score. Each game consists of one time through one or more shuffled decks. The player plays three hands in each deal. When the end of the deck is reached, the score determines whether the player wins (beats the house), loses, or draws. This is a fast-playing solitaire game, which can be used to practice and sharpen one’s Blackjack skills, including card counting. Any Blackjack variant can be used as the basis of this game.

Blackjack Solitaire uses the following layout:

Spaces 1, 2, and 3 are used for the player’s three hands. The dealer’s cards go in *D*. *B* and *W* are used for scoring. Blackjack uses all 52 cards of the deck. More than one deck can be used. After the deck or decks have been shuffled, the top card is removed (burned), revealed, and placed face up (reversed) near the bottom of the deck, about a dozen cards from the bottom.

Starting on the right, deal one card face up to each of the player’s spaces: 1, 2, and 3. Deal one card face up to *D*. Deal a second card face up to the three player spaces, again going from right to left. If the variant of Blackjack calls for a hole card, place it face down on top of the dealer’s face-up card. I usually play the European variant that doesn’t use a hole card.

The cards in each hand should be stacked and offset vertically so that the value of the underlying cards are visible, the same way that cards are stacked in the tableau in most solitaire games. The player’s hands should stack by adding cards downward; the dealer’s hand should stack by adding cards upward. An example of this stacking is shown below:

With the cards having been dealt, the player plays each of the hands in turn, starting with the hand on the right (space 1). Splitting and doubling down are allowed according to the rules of the Blackjack variant. Splits are formed by adding a new hand beside the existing stacks. Doubling down is indicated by placing one card face down on the two initial cards.

If any of the hands doesn’t bust, turn over the dealer’s hole card (or deal a card from the deck if there is no hole card) and play the dealer’s hand according to the rules of the variant. When the dealer is finished, each hand is evaluated and scored by placing cards from the the hand face down in *W*. The number of cards for each outcome is given below:

Dealer wins | 0 cards |

Player wins (no double down) | 2 cards |

Push | 1 card |

Player wins after doubling down | 3 cards |

Blackjack | 1 card straight and |

1 card turned sideways |

The remaining cards that are not placed in *W* are added to the bottom of the deck. Finally, one card from the dealer’s hand is placed face down in *B*. In addition, a card is place sideways in *B* for every hand (resulting from splits) in excess of the original three and for every double-down hand that lost.

Continue to deal and play more hands. When the reversed card reaches the top of the deck, place it face down, sideways across the dealer’s hand to indicate that the deck has been completed and that the game ends after the current deal. Each deck of cards should result in four or five deals. The game is scored from the cards in *W* and *B* as follows:

Each straight card in W |
+10 |

Each sideways card in W |
+15 |

Each straight card in B |
−30 |

Each sideways card in B |
−10 |

If Blackjack pays 2:1 in the variant used, then sideways cards in *W* are worth +20. Positive scores win; negative scores lose.

You know a couple who has two children. At least one of the children is a girl. What is the probability that they have two girls?

This is an ambiguous problem, which leads to different answers depending on the assumptions that are used. Not enough information has been provided to produce a definite answer, and the unstated assumptions fill in the space needed to complete the logic.

Here I investigate this problem and explain the ambiguity.

This problem was initially proposed by Martin Gardner in 1959, who phrased it as the following two questions:

- Mr. Jones has two children. The older child is a girl. What is the probability that both children are girls?
- Mr. Smith has two children. At least one of them is a boy. What is the probability that both children are boys?

The original purpose of these two questions was to demonstrate that they have different answers, because of a subtle difference in the information provided. The problem is that perhaps they do or perhaps they do not.

Nobody doubts the answer to Question 1. We are provided the sex of the older child. The sex of the other child is equally likely one way as the other (ignoring that, in the real world, the frequency of having sons and daughters is not exactly 50/50). Therefore, the probability is one half.

Question 2 has been worded to be very similar to Question 1 in an attempt to lead the reader to assume that the reasoning works the same way and that the answer is the same. A different answer is presented, however, usually by the following reasoning.

Each child could be a boy or a girl. Therefore, denoting the sex of the two children by two letters (“B” for boy and “G” for girl), with the first letter indicating the sex of child one and the second letter indicating the sex of child two, there are four possibilities:

BB | GB ----+---- BG | GG

Each of these possibilities is equally likely, occurring with a probability of one fourth. The information that “at least one of them is a boy” allows us to discard “GG” as impossible. Therefore, three possibilities remain, each equally likely: BB, BG, and GB. Only one of these possibilities is that there are two boys, and thus, the probability of Mr. Smith having two boys is one third—not one half as in Question 1.

There is nothing wrong with this reasoning, and indeed, as Gardner later pointed out, if one were to collect a list of all of the families with two children, say from the phone book, and eliminate all of the families with no boys, then roughly one third of the remaining families will have two boys. Nevertheless, this reasoning relies on an unstated assumption, as I shall demonstrate below.

If the odds of having a boy and having a girl are equal, then the same situation can be examined by the toss of two fair coins. Consider a game in which I use a cup to toss two coins, which are hidden from view under the cup. Then I peek under the cup and announce the value, heads or tails, of one of the coins, which is determined at random—say the first coin I see or the closest coin to me or whatever. Is there a way for someone who has not seen the coins to guess the result for *both* coins with this information and be correct more than half of the time?

Well, if I announce that one of the coins is heads, then one knows that there are two coins with a 50/50 outcome of heads and tails, and at least one of which is heads. Since this seems similar to the situation of two children with at least one boy, it is tempting to conclude that one should guess there is one coin heads up and one coin tails up, since the reasoning above indicated that the probability of two boys (or two heads) is half of the probability of only one boy (or a heads and a tails).

This conclusion is incorrect, however, and if you don’t believe me, feel free to try this experiment yourself. You will find that you are correct only about half of the time. Obviously, this situation requires a different analysis than the one provided above.

Since the person trying to guess the result of the coin toss does not know what value (heads or tails) will be called out when the coins have been tossed and are examined, he will need to reason without that information. He could choose to decide that the other coin is always heads or that it is always tails, but that’s just the standard way of betting on a coin toss. The resulting probability will always be 50%, and nothing can change that.

The other possibility is to use the information given about one of the coins to guess the value about the other. In this case, he can choose to guess that the two coins are alike or that they are different. (To be comprehensive, I should mention that the person can also employ a “mixed strategy,” in which he assigns a probability to the two choices and picks at random, but this generalization, commonly used in game theory, doesn’t affect the results discussed here.) This is the strategy mentioned above, when it is reasoned (incorrectly) that being different is twice as likely as being the same.

The problem is that knowing the result of one of the coins tells nothing about whether the two coins are alike or different. Therefore, there is no way to use this information to formulate a strategy to improve the odds of success. Nevertheless, it is natural to ask whether there is a way somehow to rig the system and set up a game, like the card version of the Bertrand’s Box Paradox, in which the odds end up in one’s favor.

One way to do this is to have a third person handle the coins and report the result of the toss, someone who is honest and supposedly “neutral,” but who behaves in a predictable way. Suppose that I have arranged with this person to *always* call out that one of the coins is heads if either of the coins shows a head. This additional information is quite useful, for if he calls tails, then I know for certain that the result of the coin toss is two tails.

Even if he calls out heads, I have information that can increase my odds of guessing correctly. Of the four possibilities that are evenly likely—two with the coins alike and two with coins different—observing that tails wasn’t called eliminates one of the possibilities—that the two coins are alike (two tails). Therefore, it is advantageous to guess that the coins are different, since that result is twice as likely as the result that both coins are heads.

So what is going on here? We have two situations that, on the surface, seem like the same problem—there are two coins, at least one of them is tails—but yield different probabilities for the result of the coin toss. To understand what is going on, it’s important to note that the key difference that changed the probabilities is the knowledge that the information to be provided was *biased towards heads.*

It is this knowledge that is the implicit assumption that leads to the usual answer to the Two Child Problem. This is similar to the reasoning that is often given in the Monty Hall Problem, where the unstated assumption is that Monty will always reveal a door with a goat. Unless that assumption is used, then there is no advantage in switching. Most people, however, apply this assumption without even thinking about it, because having the game show host reveal the prize during the middle of the game for no reason goes against the spirit of the game.

Returning to the Two Child Problem, the implicit assumption that leads one to determine that two girls is less likely is that the information provided was restricted to girls. That is, there never was a possibility to learn about the couple’s potential sons. If, on the other hand, the information had been provided about one of the couple’s children, regardless of sex, then the information could have been “at least one child is a boy” or “at least one child is a girl,” depending on the sex of the children, but either statement would have been equally likely to have been provided. In that case, the odds of the sibling being a boy or a girl are even, just as in the experiment with the two coins.

As deeper analysis of these “paradoxes” demonstrates, implicit assumptions can have a substantial impact on how additional information affects estimates of probability. Therefore, extra care should be taken to ensure that *all* of the assumptions have been identified and their applicability has been assessed. If the assumption does not fit within the information that has been provided, then it should be discarded, at least until such time as it can be justified by additional information.

So if you know a couple who has two children, the sex of whom you do not know, and you meet them on the street with their daughter, then you can assume that the odds are even that their other child is a girl. If, on the other hand, you see them taking their daughter to a Girl Scout meeting, then you can safely reason that the odds are two to one that their other child is a boy. Why? Because they would not be taking a boy to a Girl Scout meeting, and that is enough information to change the odds.

]]>The problem is described in a report published earlier this year by the American Council of Trustees and Alumni. An excerpt of the really telling part is provided below (emphasis mine):

In late summer of 2015, ACTA commissioned the research firm GfK to survey recent American college graduates and the public at large about their understanding of our free institutions of government. Our questions were drawn from standard high school civics curricula. They emphasized the content of the U.S. Constitution and the basic workings of our government. A smaller number of questions also asked about prominent figures currently serving in the federal government.

The results were abysmal. For example:

- Only 20.6% of respondents could identify James Madison as the Father of the Constitution. More than 60% thought the answer was Thomas Jefferson—despite the fact that Jefferson, as U.S. ambassador to France, was not present during the Constitutional Convention.
- College graduates performed little better: Only 28.4% named Madison, and 59.2% chose Jefferson.
- How do Americans amend the Constitution? More than half of college graduates didn’t know. Almost 60% of college graduates failed to identify correctly a requirement for ratifying a constitutional amendment.
- We live in a dangerous world—but almost 40% of college graduates didn’t know that Congress has the power to declare war.
- College graduates were even confused about the term lengths of members of Congress. Almost half could not recognize that senators are elected to six-year terms and representatives are elected to two-year terms.
- Less than half of college graduates knew that presidential impeachments are tried before the U.S. Senate.
- And 9.6% of college graduates marked that Judith Sheindlin—“Judge Judy”—was on the Supreme Court!
Many of the figures may actually understate how poorly our colleges are doing

because older respondents performed significantly better than younger ones. For example, 98.2% of college graduates over the age of 65 knew that the president cannot establish taxes—but only 73.8% of college graduates aged 25-34 answered correctly.Most college graduates over age 65 knew how to amend the Constitution—76.7% answered correctly. But among college graduates aged 25-34, less than a third chose the right answer, and over half answered that the president must ratify an amendment, failing to comprehend how the division of powers among coequal branches protects citizens’ rights.

Notice that the questions in the survey were drawn from standard *high school* civics curricula. Kids should be *entering* college knowing this stuff, not graduating without knowing it.

Think about this the next time someone (particularly a millennial) tries to tell you that “educated people” think this or think that. This report indicates that they don’t have the proper analytical skills to have an informed opinion (compared to most of their international peers or to previous generations), even the ones with the highest degrees.

Pay particular attention to the part that addresses “years of schooling” and “conferring of credentials and certificates.” Anyone who thinks that Sanders’s plan for “free college” would to do anything to help the country or the economy is fooling himself.

Keep in mind that this is a report produced by ETS — the folks who do the SAT/GRE/etc. tests. I’ve included a few excerpts from their summary below:

]]>Millennials may be on track to be our most educated generation ever, but they consistently score below many of their international peers in literacy, numeracy, and problem solving in technology-rich environments (PS-TRE).

… despite having the highest levels of educational attainment of any previous American generation, these young adults [those born after 1980 and between 16-34 years of age] on average demonstrate relatively weak skills in literacy, numeracy, and problem solving in technology-rich environments compared to their international peers. These findings hold true when looking at millennials overall, our best performing and most educated, those who are native born, and those from the highest socioeconomic background. Equally troubling is that these findings represent a decrease in literacy and numeracy skills for U.S. adults when compared with results from previous adult surveys.

The findings also offer a clear caution to anyone who believes that our policies around education should focus primarily on years of schooling or trusts that the conferring of credentials and certificates alone is enough. While it is true that, on average, the more years of schooling one completes, the more skills one acquires, this report suggests that far too many are graduating high school and completing postsecondary educational programs without receiving adequate skills.

- In literacy, U.S. millennials scored lower than 15 of the 22 participating countries. Only millennials in Spain and Italy had lower scores.
- In numeracy, U.S. millennials ranked last, along with Italy and Spain.
- Our best-educated millennials those with a master’s or research degree only scored higher than their peers in Ireland, Poland, and Spain.

Upon learning about Bendford’s Law, my colleague decided to put it to a test. So he grabbed an ENDF file, `endf66a`

, and pulled as many values as he could from it. (ENDF stands for “Evaluated Nuclear Data File.” It is a large database of nuclear data, such as cross-sections of various nuclides. For our purposes here, it is a large collection of real-world numerical data.) He collected the leading digit in the mantissa of each floating-point number in the file and examined the frequency of occurrence of each digit. The plot of the results that he produced is shown below.

After seeing his empirical verification of this law, I tried to explain why this law works in a couple of ways.

The first way is to consider the population of data, and here we’re talking about values that span multiple orders of magnitude. For example, if we were talking about lengths, the list could include values measured in centimeters, meters, and kilometers. Now for all of these units to be represented in our population we’d need the number of values expressed in centimeters to be roughly equivalent to the number of values expressed in kilometers. Naturally, such a population wouldn’t even come close to being distributed uniformly over the range of values—there are simply too many centimeters in a kilometer to sample from them in the same way that one would sample over centimeters in one meter. Mathematically, this means that, if \(f(x)\) is the probability density function (p.d.f.) of our population, then\[

\int_{10\,\text{cm}}^{1\,\text{m}} f(x)\,dx \approx

\int_{100\,\text{m}}^{1\,\text{km}} f(x)\,dx

\]which means that\[

\bar f_{(1\,\text{m})} \approx 1000\times\bar f_{(1\,\text{km})}

\]where \(\bar f_{(a)}\) is the average value of the p.d.f. over the neighborhood where \(x \approx a\).

Therefore, it makes sense to look at the value of \(f(x) \cdot x\) plotted versus \(x\) on a semi-log scale, keeping in mind that the probability of \(X\) (a random value drawn from the population) being between \(x_1\) and \(x_2\) is\[

P(x_1 < X < x_2) = \ln(10) \int_{\log_{10}(x_1)}^{\log_{10}(x_2)}
f(10^\xi)\cdot 10^\xi\,d\xi
\]That is, when \(f(x) \cdot x\) is plotted versus \(x\), with a logarithmic scale for \(x\), the area under the curve represents the relative probability of \(X\) being in a particular region. (\(\xi\) gives the linear distance along the logarithmic scale: \(\xi = \log_{10}x\).)

A (made-up) example of such a distribution is shown below.

Here the regions where the leading digit would be 1 are marked off. Compare this to the same figure with the regions where the leading digit would be 5 are highlighted.

As we can see, the areas marked off in in the first figure are significantly larger than the areas marked off in the second, which means that 1 is more likely than 5 to show up as the first digit of a number that is randomly drawn from this distribution, and this is the case for many collections of numbers that span multiple scales.

But perhaps someone objects to the p.d.f. that I used as an example. OK. Then consider this. To simplify the situation let’s consider only integers and let’s look at the range of integers from 0 to 9. If I have a population of these integers that is uniformly distributed, then any random variable that I produce is as likely to be one digit as another. But what happens if I double my range of possibilities by expanding to the right?

Now, I’m looking at the range of numbers from 0 to 19. Once again, if the distribution is uniform, I’m equally as likely to draw any of the numbers, but now the probability of drawing a number with 1 as the leading digit has changed from 10% to 55%! Over half of the time I’m going to get a number that starts with 1.

This example is contrived, of course, but it readily generalizes. For example, consider what happens if I expand the range to extend from 0 to 49. Although the probability of getting a number that begins with a 1 is no longer 55%, it’s still larger than getting a number that begins with a 5, 6, 7, 8, or 9.

This phenomenon is an artifact of the way we *represent* real values, not the real values themselves. To see this, it is instructive to consider just one number—for example, the fine-structure constant:\[

\alpha \approx 7.297\times 10^{-3}

\]This is a physical constant. It’s dimensionless. There is no mathematical basis for this constant. As Richard Feynmann once wrote,

It’s one of the greatest damn mysteries of physics: a magic number that comes to us with no understanding by man. You might say the “hand of God” wrote that number, and “we don’t know how He pushed his pencil.”

So this is about an arbitrary a number as one can find. It happens to begin with a 7, not a 1, but that’s just because we use a number system with ten digits. (Ignore, for the moment, that its reciprocal, \(\alpha^{-1} \approx 137.036\), an equally arbitrary number, *does* begin with a 1.) The number is what it is, regardless of how we write it. We can think of it as a single point on a line that represents all real numbers (the \(x\)-axis of the complex plane). What digits we use to write that number depend on where we lay down the grid lines that denote 1, 2, 3, and so on. If the real values are scaled logarithmically, this line looks like the following: The point that is \(\alpha\) also is shown above and falls between the ticks for 7 and 8, resulting in a number that begins with 7. But what happens when a different base (number of digits) is used to represent the same number. In octal (base 8, which is often used in computer science, because \(8 = 2^3\)), the real line looks like (Keep in mind that now the octal “10” is really an “8” in decimal notation.) In this system of numbers, the numerical representation of \(\alpha\) begins with a 3. If we double the number of digits to 16 (hexadecimal), we find the following: The leading digit is now 1.

It is important to note that the real line itself and all of the real values it represents, including \(\alpha\), haven’t changed. They are the same in all three diagrams. By changing the number of digits in our number system, we change only where the *grid lines* (tick marks) are located. All of these number systems agree on the location of 1 (and zero and infinity), but everything else changes. Below, I have tabulated the value of \(\alpha\) for number systems with 3 to 16 digits (base 3 to base 16):

Base | \(\alpha\) |
---|---|

3 | 1.210e-12 |

4 | 1.313e-10 |

5 | 4.240e-4 |

6 | 1.324e-3 |

7 | 2.334e-3 |

8 | 3.571e-3 |

9 | 5.278e-3 |

10 | 7.297e-3 |

11 | 9.793e-3 |

12 | 1.074e-3 |

13 | 1.305e-2 |

14 | 1.605e-2 |

15 | 1.996e-2 |

16 | 1.DE4e-2 |

In 8 out of the 14 number schemes (57%), the representation of \(\alpha\) begins with a 1. This is not surprising when you think about it, because although the locations of the grid lines change, the structure of the grid lines remains similar. The space along the (logarithmically scaled) real line where a 1 is the leading digit is always the largest space. Therefore, 1 is the most likely leading digit, regardless of the base.

For a truly arbitrary number, chosen without any restrictions, it is not difficult to predict that the likelihood of the digit \(n\) being the first digit is\[

P(n) = \log_b(n+1) – \log_b(n) \] where \(b\) is the base of the number system being used. This can be deduced geometrically from the figures above.

As long as one uses a notation system for numbers consisting of a series of symbols in which each successive symbol in the series constitutes a value that is \(b\) times smaller than the symbol before it (where \(b\) is the number of symbols used), the symbol denoting the smallest (non-zero) value (in our case, 1) will be the most likely symbol to appear first in the series.

]]>Since many “thirders” seem to be fond of relying on betting analogies to reason through the problem and explain their arguments, a useful substitute for the coin toss is a lottery. A typical lottery provides a very small chance of winning accompanied by a very large payoff (which is why lotteries are so popular). So here we shall examine what happens when Sleeping Beauty plays the lottery.

The events of the original Sleeping Beauty Problem occur as described before. That is, Beauty is put to sleep on Sunday and woken on Monday. She is then put to sleep with her memory erased and woken again on Tuesday *only* if a certain random event happens.

The difference between the two problems is that the coin toss in the original problem is replaced with a lottery. Before she goes to sleep on Sunday night, Beauty chooses a “lucky” set of lottery numbers. (Note: Each set of numbers constitutes one lottery ticket and one chance to win the lottery.) Then, sometime after she is asleep, the lottery’s winning numbers are drawn at random. If the numbers that Beauty chose match the winning numbers, she will be woken on Monday, have her memory erased, and woken again on Tuesday. If her numbers do not match the winning numbers, she will be woken only on Monday.

As in the original problem, the circumstances of her wakings are identical, so she cannot tell what day it is or the result of the lottery. The new question is the following: What should Beauty think about her probability of winning the lottery upon awakening?

Note that winning the lottery in this problem corresponds with the coin toss coming up tails in the original problem, and losing the lottery corresponds to the toss coming up heads.

To quantify the problem, assume that there are \(n\) possible sets of lottery numbers and each set has an equal chance of being drawn. Therefore, the probability of winning the lottery is 1 in \(n\). Typically, \(n\) is a large number—e.g., one in a million.

The following notation is used below to indicate the conditions of the experiment:\[

\begin{aligned}

W &{}= \text{Beauty picked the winning set of numbers}\\

L &{}= \text{Beauty did not pick the winning set of numbers}\\

D_1 &{}= \text{It is the first day (Monday)}\\

D_2 &{}= \text{It is the second day (Tuesday)}

\end{aligned}

\]

With the problem defined, it is useful to consider the typical approaches that are used by the “thirders” to tackle a problem such as this. To this purpose, I have divided the majority of the thirders into three categories:

**The Waking Thirders** – This group fixates on the datum that Sleeping Beauty wakes up

**The Vegas Thirders** – This group evaluates probabilities by turning them into some sort of gambling proposition

**The Monday-Morning Thirders** – This group fixates on what happens on Monday morning

Note: The fact that I have divided the “thirders” into three categories is not intended to imply that they are evenly divided into these categories. In fact, many “thirders” use, or at least explore, more than one of these approaches to this problem, so the process of categorization is somewhat nuanced.

Little changes in the analysis provided by the “waking thirders.” Once again, they observe three states that Beauty can find herself when she wakes:

(1) Beauty lost the lottery and it’s Monday

(2) Beauty won the lottery and it’s Monday

(3) Beauty won the lottery and it’s Tuesday

Since it is impossible for her to distinguish between the three, they are all assumed to be equally as likely:\[

P(L \cap D_1) = P(W \cap D_1) = P(W \cap D_2)

\]Because these are the only possibilities, their probabilities must sum to one. Therefore, each state has a probability of 1/3, and since two out of the three states is a win for Beauty, they conclude that Beauty should believe upon waking that she has a 2/3 chance of having won the lottery.

For the “Vegas” crowd to be able to analyze this situation, additional assumptions are necessary, because they are interested in the payoff. Their analysis requires that we know how much she won to determine how likely she was to win.

Therefore, let’s stipulate that the winner is paid according to the odds of winning. That is, the winner of a lottery with a one-in-a-million chance of winning is paid $1 million on a $1 lottery ticket. (The winnings will be taxed, of course, which is how the state will pull in its take, but that’s not relevant to this analysis.) Furthermore, we must assume that the possibility of two or more people selecting the same set of numbers is either so unlikely as to be negligible or that it is simply not allowed by the rules.

Beauty chooses her “lucky” numbers on Sunday night. Each time she wakes, she is given an opportunity to purchase a lottery ticket. To avoid the possibility of having to split the winnings, we can posit that the lotteries on Monday and Tuesday are separate lotteries, which use the same numbers that were drawn on Sunday. Therefore, it is possible for Beauty to win the entire lottery prize twice.

Note that if \(n = 2\), this problem reduces to the problem of the flipped coin, with Beauty betting on tails. Therefore, this lottery problem can be considered to be a generalization of the original Sleeping Beauty Problem, and the reasoning considered here is a generalization of the reasoning used by “thirders” for the original problem.

To make this problem concrete, a “Vegas thirder” would suggest something like a $1 lottery ticket for a $1 million prize with a 1-in-a-million chance of picking the winning number. Then, the “thirder” would consider the situation in which the “experiment” (lottery drawing) is repeated many, many times—say, one million times so that we can expect that Beauty picks the winning number once.

With the conditions in place, the totals can be compiled. If Beauty buys a lottery ticket every time she wakes, then she will have spent $1,000,001 in lottery tickets (because she would have bought an additional ticket on Tuesday the time that she picked the winning number), and she will have won $2,000,000. The “thirder” then calculates the odds of winning the lottery as\[

\begin{aligned}

\text{odds} &{}= (\text{amount won}):(\text{amount lost})\\

&{}= 2,000,000:999,999 \approx 2:1

\end{aligned}

\]This is then interpreted as 2:1 odds of picking the winning number or a probability of \(P(W) = 2/3\).

The answer given by the “Monday-morning thirders” is different than the answer given by the other two groups, because this group actually considers the probabilities of the random number generator (i.e., coin toss in the original problem) in their arguments. They reason that, since Beauty must wake on Monday regardless of the result of the lottery, her chance of having the winning number on Monday are\[

\begin{aligned}

P(W|D_1) &{} = \frac{1}{n}\\

P(L|D_1) &{} = \frac{n-1}{n}

\end{aligned}

\]They note that the probability of Beauty picking the wrong number is\[

P(L) = P(L|D_1)\cdot P(D_1) + P(L|D_2)\cdot P(D_2)

\]Since Beauty will not be woken on Tuesday (\(D_2\)) if she did not pick the winning number, \(P(L|D_2) = 0\). Therefore,\[

P(L) = P(L|D_1)\cdot P(D_1) = \frac{n-1}{n} P(D_1)

\]So the answer depends on the probability of the day being Monday. Since there are three indiscernible states in which Beauty can wake and two of these states occur on Monday, the “thirder” concludes that \(P(D_1) = 2/3\). Thus, the probabilities associated with Beauty picking the winning number are\[

\begin{aligned}

P(L) &{} = \frac{2n – 2}{3n}\\

P(W) &{} = 1 – P(L) = \frac{n + 2}{3n}

\end{aligned}

\]

In the limit that \(n\) becomes large,\[

\begin{aligned}

P(L) &{} \sim 2/3\\

P(W) &{} \sim 1/3

\end{aligned}

\]This is closer to the correct answer than the other two arguments, but it is still incorrect.

The correct set of probabilities is given below.\[

\begin{aligned}

P(W) &{}= \frac{1}{n}\\

P(L) &{}= \frac{n – 1}{n}\\

P(D_1|L) &{}= 1\\

P(D_2|L) &{}= 0\\

P(D_1|W) &{}= 1/2\\

P(D_2|W) &{}= 1/2\\

P(W \cap D_1) &{}= P(W) \cdot P(D_1|W) = \frac{1}{2n}\\

P(W \cap D_2) &{}= P(W) \cdot P(D_2|W) = \frac{1}{2n}\\

P(L \cap D_1) &{}= P(L) \cdot P(D_1|L) = \frac{n-1}{n}\\

P(L \cap D_2) &{}= P(L) \cdot P(D_2|L) = 0\\

P(D_1) &{}= P(W) \cdot P(D_1|W) + P(L) \cdot P(D_1|L) =

\frac{2n-1}{2n}\\

P(D_2) &{}= P(W) \cdot P(D_2|W) + P(L) \cdot P(D_2|L) =

\frac{1}{2n}\\

P(W|D_1) &{}= \frac{P(D_1|W) \cdot P(W)}{P(D_1)} =

\frac{1}{2n – 1}\\

P(L|D_1) &{}= \frac{P(D_1|L) \cdot P(L)}{P(D_1)} =

\frac{2n – 2}{2n – 1}\\

P(W|D_2) &{}= \frac{P(D_2|W) \cdot P(W)}{P(D_2)} = 1\\

P(L|D_2) &{}= \frac{P(D_2|L) \cdot P(L)}{P(D_2)} = 0

\end{aligned}

\]These results are consistent with the original problem, which corresponds to \(n = 2\), \(W = T\), and \(L = H\).

By changing the probabilities of the outcomes of the event that determines the number of times that Beauty wakes, the flaws in the reasoning commonly used by the “thirders” become obvious.

My advice to the “thirders” is the following: Try to find some drug that produces limited memory loss, then go buy a lottery ticket—the larger the payout the better. With the help of a friend, you can go to sleep on the night of the drawing confident that when you awake, you’ll have between a 33% and a 67% chance (depending on your reasoning) of being a lottery winner and becoming a new multimillionaire. Good luck!

Note: This entry uses some free clip art from openclipart.org and cliparts.co.

]]>

The Sleeping Beauty Problem is usually described as follows:

Beauty is going to be the subject of an experiment that will take place over three days. On Sunday, Beauty will be told the plan for the experiment, and then given a drug and put to sleep. The drug will cause her to sleep until Wednesday, but the experimenters plan to wake her up some number of times. To decide how many times to wake her up, they toss a coin. If the coin lands heads, they will wake her on Monday briefly, and then return her to sleep. If the coin lands tails, they will instead wake her briefly on both Monday and Tuesday. After they put her back to sleep on Monday, they will erase her memory, so that upon waking on Tuesday, her last memories will be having gone to sleep on Sunday. The circumstances of her wakings are identical, so she cannot tell from her environment what day it is or which way the coin landed. What should Beauty think about the probability of heads and tails upon awakening?

Alternatively, the question that is often asked is, “What is the probability that the coin came up heads?”

A popular, incorrect answer to this problem is that Beauty should believe that, because she woke up, the fair coin is twice as likely to have come up tails as to come up heads. Since this results in a probability of heads of 1/3, the people who argue this point of view are often called “thirders.”

In the most naive argument of this type that is put forward, the proponent points out that Beauty can wake in one of three states:

(1) The coin toss was heads and it’s Monday

(2) The coin toss was tails and it’s Monday

(3) The coin toss was tails and it’s Tuesday

The proponent then (implicitly or explicitly) assumes that all three states are equally likely to occur—i.e., each occurs with the same probability as the others. Since the coin came up heads in only one of the three states, he or she concludes that the probability of heads is 1/3.

More complicated, but equally wrong, arguments are addressed below.

Here we will not only provide the correct answer to the problem (which is trivial) but also rigorously explore the estimates of probabilities that Beauty should assign to other parts of the experiment based on what information she has available.

In the interest of conserving space, let’s introduce the following notation for conditions of the experiment:\[

\begin{aligned}

H &{}= \text{The coin came up heads}\\

T &{}= \text{The coin came up tails}\\

D_1 &{}= \text{It is the first day (Monday)}\\

D_2 &{}= \text{It is the second day (Tuesday)}

\end{aligned}

\]The notation used above for Monday and Tuesday readily generalizes to the scenario proposed by Nick Bostrom whereby Beauty is wakes for many, many (e.g., a million) days if the coin comes up tails (more below).

The assumption that the coin is fair implies\[

\begin{aligned}

P(H) &{}= 1/2\\

P(T) &{}= 1/2

\end{aligned}

\]If the coin comes up heads, Beauty will be woken on Monday, but not on Tuesday. This means\[

\begin{aligned}

P(D_1|H) &{}= 1\\

P(D_2|H) &{}= 0

\end{aligned}

\]If the coin comes up tails, Beauty will be woken on both Monday *and* Tuesday. She does not know which day it is when she wakes, but she knows that if tails shows up, she will be woken on *both* days, so in this situation the frequency of Monday and Tuesday awakenings will be the same. Therefore, they have an equal probability:\[

\begin{aligned}

P(D_1|T) &{}= 1/2\\

P(D_2|T) &{}= 1/2

\end{aligned}

\]

From that, the grid of all four possibilities falls out:\[

\begin{alignedat}{2}

P(H \cap D_1) &{}= P(H) \cdot P(D_1|H) & & {}= 1/2\\

P(H \cap D_2) &{}= P(H) \cdot P(D_2|H) & & {}= 0\\

P(T \cap D_1) &{}= P(T) \cdot P(D_1|T) & & {}= 1/4\\

P(T \cap D_2) &{}= P(T) \cdot P(D_2|T) & & {}= 1/4

\end{alignedat}\]

Let’s get one thing out of the way, Beauty is *always* going to wake up:\[

P(\text{wake}) = P(\text{wake}|H) = P(\text{wake}|T) =

P(\text{wake}|D_1) = P(\text{wake}|D_2) = 1

\]So the fact that she wakes and is “in the moment” doesn’t give her or us any more information than she started with. Since Beauty has not been provided with any new information, the probability of the coin coming up heads remains \(P(H) = 1/2\), which is the *right* answer to the problem.

With that out of the way, we can use the information above to determine the probability of Beauty waking on a Monday,\[

P(D_1) = P(H) \cdot P(D_1|H) + P(T) \cdot P(D_1|T) = 3/4

\]and the probability of it being a Tuesday,\[

P(D_2) = P(H) \cdot P(D_2|H) + P(T) \cdot P(D_2|T) = 1/4

\]If Beauty is informed that she has risen on a Monday, then she can reevaluate the probability of the coin landing on heads or tails by applying Bayes’s Theorem:\[

\begin{alignedat}{2}

P(H|D_1) &{}= \frac{P(D_1|H) \cdot P(H)}{P(D_1)} & &{}= 2/3\\

P(T|D_1) &{}= \frac{P(D_1|T) \cdot P(T)}{P(D_1)} & &{}= 1/3

\end{alignedat}

\]Similarly, if she is told that it is a Tuesday, then she knows for certain that the coin toss was tails:\[

\begin{alignedat}{2}

P(H|D_2) &{}= \frac{P(D_2|H) \cdot P(H)}{P(D_2)} & &{}= 0\\

P(T|D_2) &{}= \frac{P(D_2|T) \cdot P(T)}{P(D_2)} & &{}= 1

\end{alignedat}

\]

Although the flaws in logic employed by the “thirders” encompass a wide range of fallacies, a couple of mistakes appear quite frequently. Here we’ll examine why they are wrong.

The mistake that most “thirders,” including the original guy who published the first article on this, make is to reason that, since she *must* wake up on Monday—regardless of the coin flip—we must consider her situation at *that* time. The usual explanation that is given is that the experimenters might not have flipped the coin until *after* Beauty had been woken on Monday, because the coin doesn’t affect that part of the experiment. The coin toss determines only whether Beauty has her memory erased and is put back to bed. Nothing in the experiment requires that the coin is flipped before Beauty is woken and asked about the result of the coin toss.

They then argue that it is ridiculous, in the situation where Beauty is asked on Monday about the probability of heads on a coin toss that has not yet happened, for her to say anything but 1/2. In one sense, they have a valid point. If Beauty is woken and told that it is Monday *and* that the coin toss hasn’t yet occurred, then she should reasonably conclude that the probability of the (future) coin toss coming up heads is 1/2. However, this is *not* the question that is being asked in the problem.

By focusing on the result of a coin toss on Monday—regardless of whether the coin has actually been tossed or not—the “thirders” have eliminated the possibility that the coin toss could have resulted in Beauty waking on Tuesday. In their line of reasoning, it’s simply not possible. Therefore, the coin toss that they are considering is *not* the coin toss that Beauty is asked about in the problem.

Following up on this mistake by applying it to probability calculations, they then (erroneously) reason that\[

\begin{aligned}

P(H|D_1) &{}= 1/2 \\

P(T|D_1) &{}= 1/2

\end{aligned}\rlap{\qquad\text{(wrong)}}

\]and work backward through Bayes’s Theorem to conclude the wrong answer:\[

P(H) = 1/3 \rlap{\qquad\text{(wrong)}}

\]

The source of their error is that they have changed the definition of the problem to a different problem in which the day is always Monday:\[

\begin{aligned}

P(D_1) &{}= 1 \\

P(D_2) &{}= 0

\end{aligned}

\]Naturally, the coin toss can have no effect on this:\[

\begin{alignedat}{2}

P(D_1|H) &{}= P(D_1|T) & &{}= 1 \\

P(D_2|H) &{}= P(D_2|T) & &{}= 0

\end{alignedat}

\]It is important to note that the claim made above about the probability of heads and tails on Monday, which is wrong in the original problem, is *correct* in this new, different problem,\[

\begin{alignedat}{2}

P(H|D_1) &{}= \frac{P(D_1|H) \cdot P(H)}{P(D_1)} & &{}= 1/2\\

P(T|D_1) &{}= \frac{P(D_1|T) \cdot P(T)}{P(D_1)} & &{}= 1/2

\end{alignedat}

\]but *only* if \(P(H) = P(T) = 1/2\), so even their modified problem doesn’t demonstrate their claim that the probability of heads is 1/3. When the assumptions are explicitly stated and the math is done correctly, it merely reinforces the proof that the probability is 1/2.

The other mistake that is commonly encountered in “thirder” reasoning is to treat the Sleeping Beauty problem as a betting proposition. There are several variants as to how the wager is presented—whether only one bet or multiple bets are considered, whether odds are given on the bet, the amount of the bet, etc.—but they all suffer from the same critical error. There is even confusion over what constitutes a “bet.” While it is generally agreed that Beauty bets on the result of the toss of the coin *every* time she wakes up, some people count the number of bets as the number of times the coin is tossed \(n_{\text{t}}\), which can be controlled, since it’s the number of times Beauty goes to sleep on Sunday. Meanwhile, other people consider every waking to be a separate bet, even though this number is randomly determined by the coin tosses. The best that can be said is that the number of times Beauty wakes \(N_{\text{w}}\) is expected to be\[

E(N_{\text{w}}) = n_{\text{t}} P(H) + 2 n_{\text{t}} P(T)

= 3 n_{\text{t}} / 2

\]That is, Beauty can be expected to wake about 50% more than the number of times that she was put to sleep on Sunday, but the exact number in each series of experiments depends on the outcome of the coin tosses.

To avoid this ambiguity, let’s define “toss” to be every time the coin is tossed (and Beauty is put to sleep on Sunday) and “bet” as every time she has a chance to win or lose money (which is every time that Beauty wakes and is interviewed). Beauty has two options. She either can decide what she will bet when she wakes up each morning, or she can decide what to bet when she goes to sleep on Sunday and consistently place the same bet each time she wakes. Either way, her expected winnings will be the same. She can decide always to bet heads, always to bet tails, or to bet heads randomly at a specified fraction of the time \(f_{\text{H}}\).

Her expected winnings are\[

E(W) = f_{\text{H}}\cdot\bigl(P(H) – 2P(T)\bigr) +

(1 – f_{\text{H}})\cdot\bigl(2P(T) – P(H)\bigr)

\]which becomes\[

E(W) = (2f_{\text{H}} – 1)\cdot\bigl(P(H) – 2P(T)\bigr)

\]So if Beauty always bets heads (\(f_{\text{H}} = 1\)),\[

E(W) = -1/2

\] and if Beauty always bets tails (\(f_{\text{H}} = 0\)),\[

E(W) = 1/2

\]If she bets heads only part of the time (\(0 < f_{\text{H}} < 1\)) then the expected winnings vary linearly with \(f_{\text{H}}\) between these two extremes.

So obviously, Beauty’s best strategy that maximizes her winnings is to always bet tails. For every, two dollars she offers to wager, she is expected to win one, which leads the “thirders” to mistakenly conclude that the odds of the coin coming up heads is 1 to 2 or that heads has a probability of \(P(H) = 1/3\).

However, this is not the case. If she seals in her bet (or bets, in case of tails) on Sunday by always betting either heads or tails, she is making a wager with equally likely results on the toss, but an *uneven* payoff on the result of the toss. The payoff table looks like the following:

Bet Heads | Bet Tails | |

Result Heads | Win 1 | Lose 1 |

Result Tails | Lose 2 | Win 2 |

The “thirders” have confused betting odds with statistical odds. They are not the same thing. Therefore, a wise Beauty can make easy money by always betting tails and taking the two-to-one payoff, but she should not delude herself that the probabilities of the coin toss are anything but even.

Philosopher Nick Bostrom has proposed a scenario—possibly inspired by the movie *Groundhog Day*—in which Beauty, instead of being woken only on Monday and Tuesday when the coin comes up tails, is woken many, many times—perhaps as many as a *million* times. (Never mind that a person who lives to be 100 years old will have less than 37,000 days on which to wake up. This is a philosopher talking.)

The idea behind this argument is that the set of possible days on which Beauty can wake up is overwhelmed by all of these additional days making it “more absurd” for someone to argue that the *one* day with heads out of all of these possible days should be so frequent. This is nonsense, of course.

The coin will come up heads 1/2 of the time, and the experiment will end on Monday every time this happens. The other half of the time, Beauty will be woken on one of the days, each of which is no more likely than the rest without additional information. Thus, if there are \(n_{\text{T}}\) such days, then\[

P(D_i|T) = 1/n_{\text{T}}, \quad i = 1, 2, \ldots, n_{\text{T}}

\] So the problem generalizes as follows\[

\begin{alignedat}{2}

P(D_1|H) &{}= 1\\

P(D_i|H) &{}= 0, \quad & i &{}= 2, 3, \dots, n_{\text{T}} \\

P(D_i|T) &{}= 1/n_{\text{T}}, \quad & i &{}= 1, 2, \dots, n_{\text{T}} \\

\end{alignedat}

\]and\[

\begin{alignedat}{3}

P(H \cap D_1) &{}= P(H) \cdot P(D_1|H) & & {}= 1/2 & & \\

P(H \cap D_i) &{}= P(H) \cdot P(D_i|H) & & {}= 0, & \quad

i &{}= 2, 3, \dots, n_{\text{T}} \\

P(T \cap D_i) &{}= P(T) \cdot P(D_i|T) & & {}= \frac{1}{2n_{\text{T}}}, & \quad

i &{}= 1, 2, \dots, n_{\text{T}}

\end{alignedat}\]

There is nothing in this more general problem to imply anything other than \(P(H) = 1/2\).

Finally, there are a few people who claim that the framing of the problem is too ambiguous to evaluate the probability of the coin landing on heads. I don’t quite understand how they have confused themselves enough to reach this conclusion, but they are clearly wrong, as has been demonstrated above.

The correct answer to the Sleeping Beauty Problem is that Beauty has received no additional information upon waking. Therefore the probability that the coin has or will come up heads is 1/2, which is exactly the same as it was when she went to sleep on Sunday. A fair coin is, after all, a fair coin.

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In the spirit of Thanksgiving, and wanting to take a break from reading, thinking, and writing about nuclear energy, I’m offering my tried and true cooking instructions for something completely different.

By Sunday night you will be stuffed, fed up, literally, and figuratively, with turkey. Instead of food fit for pilgrims, try food invented in the wide open west—chili. Cook this dish on Saturday. Eat it on Sunday.

These instructions take about an hour to complete. This chili has more vegetables and beans than some people might like, but we’re all trying to eat healthy. Although the name of this dish has the word “nuclear” in it, it isn’t that hot on the Scoville scale. If you want some other choices for nuclear chili there are lots of recipes on Google

The beer adds sweetness to the vegetables, as does the brandy, and is a good for cooking generally. In terms of the beer, which is an essential ingredient, you’ll still have five cans or bottles left to share with friends so there’s always that.

However, I recommend Negra Modelo for drinking with this dish and Budweiser or any American pilsner for cooking it. Alternatives for drinking include local western favorites, Moose Drool or Black Butte Porter, and regional amber ales Alaskan Amber, Fat Tire, or Anchor Steam. Do not cook with “light” beer. It’s a very bad idea.

Scoville, Idaho, is the destination for Union Pacific rail freight for the Idaho National Laboratory (INL) way out on the Arco desert.

There is no town by that name, but legend has it that way back in the 50s & 60s, when the place was called the National Reactor Testing Station, back shift workers on cold winter nights relished the lure of hot chili hence the use of the use of the name “Scoville” for shipping information.

Overnight temperatures on the Arco desert can plunge to −20 °F or more. Unfortunately, the guys running the reactors couldn’t drink beer, but they did have coffee. It’s still that way today.

This is “second-day chili.” That means after you make it, put it in the unheated garage to cool, then refrigerate it, and reheat the next day. The flavors will have had time to mix with the ingredients, and on a cold Idaho night what you need that warms the body and the soul is a bowl of hot chili with fresh, warm cornbread on the side.

If you make a double portion, you can serve it for dinner over a hot Idaho baked potato with salad. Enjoy.

1 lb chopped or ground beef (15% fat)

1 large onion

1 sweet red pepper

1 sweet green pepper

10–12 medium size mushrooms

1 can pinto beans (plain, no “sauce”)

1 can black beans

1 can chopped tomatoes

1 can small, white “shoepeg” corn

1 12 oz can beer

1 cup hot beef broth

1 tablespoon cooking brandy

2 tablespoons finely chopped jalapeno peppers

2–4 tablespoons red chili powder

1/2 teaspoon black pepper

1/2 teaspoon salt

1/2 teaspoon coarse powdered garlic

1/2 teaspoon cumin

- Chop the vegetables into small pieces and brown them in cooking

oil. Add 1 tablespoon of cooking brandy near the end. Drain thoroughly. - Brown the meat separately and drain the fat.
- Combine all the ingredients in a large pot. Be sure to drain the

beans, and tomatoes before adding. Simmer slowly for at least

60–120 min. Stir occasionally. - Set aside and refrigerate when cool.
- Reheat the next day. Serve with cornbread. Garnish with shredded

sharp cheddar cheese.

Feeds 2–4 adults.

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