## Meta Solving a Probability Puzzle

I recently posted my new favorite probability puzzle from the fall 2019 issue of Emissary, submitted by Peter Winkler.

Puzzle.Alice and Bob each have a biased coin that flips heads with probability 51% and 100 dollars to play with. The buzzer rings, and Alice and Bob begin flipping their coins once per minute and betting one dollar on each flip against the house. The bet is at even odds: each round, each of them either loses or wins a dollar. Alice always bets on heads, and poor Bob, always on tails. As it happens, however, both eventually go broke. Who is more likely to have gone broke first?

Follow-up question: They play the same game as above, but this time Alice and Bob are flipping the same coin (biased 51% toward heads). Again, assuming both eventually go broke, who is more likely to have lost their money first?

One might assume that Bob obviously got a bad deal. He will lose his money very fast. So the answer to both questions must be that Bob loses his money first. If you know me, you should realize that I wouldn’t have given you this puzzle if the answer was that intuitive. I love counter-intuitive puzzles.

Bob has a disadvantage, so he is guaranteed to lose eventually. Alice has an advantage, so she might lose, or she might not. If she goes broke, that means there should be more tails in the flips than if she doesn’t go broke. How does this influence the second scenario, in which they use the same coin? As we are expecting a lot of tails in the flips, the situation should be better for Bob as compared to the first scenario. Given that I posed this puzzle, one might decide that in the follow-up question Bob doesn’t go broke first. Is it a tie?

But, if Bob is the first to go broke in the first scenario, why would I include the first part of the puzzle at all? If you know me, you should wonder what makes the first question interesting. Or maybe, you know Peter Winkler, in which case you should also wonder the same thing.

I gave you enough meta information to guess the amazing counter-intuitive answers to these two questions: in the first scenario, they tie; and in the second scenario, Alice goes broke first with higher probability.

To prove that they tie in the first scenario, let me first define a **switch** on a sequence of coin flips. A switch changes each head into tails and vice versa. The switch creates a one-to-one correspondence between the sequences of flips where Bob goes broke with the sequences where Alice goes broke. Suppose *p* is the probability that Alice goes broke with a particular sequence *a*, and *q* is the probability that Bob goes broke with a `switched’ sequence *b*. Then *p = rq*, where *r* = (49/51)^{200}. The reason is that sequence *a* has exactly 200 more tails than sequence *b*. The same ratio *r* works for any pair of switched sequences. Bob is guaranteed to go broke eventually. But to calculate the conditional probability of Alice going broke with sequence *a*, given that she does go broke, we need to divide the probability of the sequence *a* occurring by the probability that Alice goes broke. The resulting probability distribution on sequences of length *n* has to be the same as for Bob because it has to sum to one. If Alice goes broke, then the probability that she goes broke after *n* flips is the same as for Bob. That means they tie.

To answer the second question, we use the switch again. The switch creates a one-to-one correspondence between sequences of flips where Alice goes broke first and Bob second, and vice versa. The sequence for which Alice loses second has 200 more tails than the corresponding sequence where Bob loses second. Thus, in each pair of two switched sequences, the probability of the sequence where Alice loses second is equal to *r* times the probability of the sequence where Bob loses second. Thus, the probability of Alice winning is *r* times the probability of Bob winning, and *r* is a very small number.

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## Yuzhou Gu:

I am wondering the following: Suppose Alice goes broke with a particular sequence a. If there are k heads, then there are k+100 tails. In the switched sequence b, there are k+100 heads and k tails. So sequence b has 100 (rather than 200) more heads than sequence a, and the ratio between probabilities should be (49/51)^100.

9 February 2022, 9:08 pm## 17 Best Mathematics Blogs in 2022 – Mass Blog:

[…] Brimming with uncommon Math information, Dr. Tanya Khovanova manages this exceptional blog. She is a freelance mathematician, a Mathletics coach at the Advanced Math and Sciences Academy in Massachusetts, and a math research coordinator at MIT. Although the website seems outdated, it is still highly active. From algorithms, sequences, statistics, and even math humor, there are plenty of categories to choose from. For starters, take a look at this blog post on how to solve a probability puzzle. […]

27 March 2022, 10:49 am