Aug 30, 2015 - Paradoxes, Probability, Statistics    Comments Off on Sleeping Beauty Plays the Lottery

## Sleeping Beauty Plays the Lottery

I’ve already examined the classic Sleeping Beauty Problem and pointed out some of the pitfalls that many people fail to avoid when trying to solve the problem. I also examined Nick Bostrom’s so-called “Extreme Beauty” modification to the problem, in which Beauty wakes many, many times if the coin toss comes up tails. However, there is another “extreme” variant of this problem, the variant in which the coin toss is replaced with another two-result random process that has extremely uneven odds. That is, in this “extreme” problem, one of the possible results is extremely unlikely. Examining this variant with the methods of reasoning commonly used by the “thirders” can be enlightening and can provide some illustration of why they are wrong.

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 Problem

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}

## The “Thirders”

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.

### The Waking Thirders’ Solution

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.

### The Vegas Thirders’ Solution

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 Monday-Morning Thirders’ Solution

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 $$L = H$$.