Dinner! Drinks! Denominators!
Mathematics is only a systematic effort of solving puzzles posed by nature—Shakuntala Devi.
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Peter Winkler is featured in the current issue of the New Yorker magazine. Peter is a famous Dartmouth mathematics professor. He is an expert on math puzzles, both for making them and solving them—see his book, simply titled Mathematical Puzzles.
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A recent evening of mathematical dinner theater hosted by Peter is the subject of the article, which is by Dan Rockmore, also of the Dartmouth math department. This event was co-coordinated by Cindy Lawrence, who is executive director of the National Museum of Mathematics:
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The event was attended by a group of investors, economists. and teachers in a suitable alcove of a restaurant in Manhattan. Quoting Rockmore’s article:
Winkler began the evening’s program. The first course of math, delivered during the first course of dinner (a scattering of salads), was a statistics starter called Simpson’s paradox. The paradox explains how apparent biases in large samples can disappear in smaller ones. A famous example: For the University of California at Berkeley’s graduate programs in 1975, over all, men were admitted at a higher rate than women, but, program by program, women were admitted at a higher rate.
About Berkeley, see this first. We covered Simpson’s paradox here. The article goes on the mention the famous trick of how a biased coin can be made to simulate a fair coin by tossing it twice, and continues:
Winkler let loose with the last official mind bender, a gambling thought experiment involving a fictitious couple named Alice and Bob, who are famous in math circles. Each of them has a biased coin—fifty-one-per-cent chance of heads, forty-nine-per-cent chance of tails. They each start with a hundred dollars, flipping the coin and betting against the bank on the outcome. Alice calls heads every time; Bob calls tails. The puzzle: Given that they both go broke, which one is more likely to have gone broke first?
The article relates that most diners guessed Bob, but the correct answer is Alice. Former Times math and science columnist John Tierney, a serious enough math fan to have remarked once that the term “recreational mathematics” is a self-contradiction, reasoned it out: “The longer Alice plays, the less likely she is to go broke.” We finish Tierney’s intent by contraposing: hence given that she too went broke, the more likely it was she that played less time.
Old Friend
Peter is a long time friend of mine. See him in action here. He is a delight to talk to about many things—just about any part of math.
Peter has a paper with the title, “Seven Puzzles You Think You Must Not Have Heard Correctly.” I had already started writing this post when I looked up this paper to get a few more puzzle examples. I found this one:
#4 Unwanted Expansion:
Suppose you have an algebraic expression involving variables, addition, multiplication, and parentheses. You repeatedly attempt to expand it using the distributive law. How do you know that the expression doesn’t continue to expand forever? Comment: Note that applying the distributive law to, say, the outer product in, yields
which has more parentheses than before.
I thought about this for a little. Then I looked up the answer.
One can analyze the expression in terms of depth of trees, but there’s an easier way: set all the variables equal to 2. The point of the distributive law is that its application doesn’t change the value of the expression. The value of the initial expression limits the size of anything you can get from it by expansion.
Peter then says: This proof of stopping is due to Dick Lipton. I must say that surprised me. I am getting old. Oops.
Open Problems
Let’s end with a fun puzzle about the dictionary that is also mentioned in the New Yorker article.
If each number between one and ten billion was written down in English, which odd number would appear first alphabetically?
Given that spaces do not count, here is the answer. The last part changes because the number must be odd. If spaces do count—as in many dictionaries of phrases—then what is the answer?
I’d like to see the math behind the “last official mind bender” as that result clearly doesn’t hold in the extremes (hence I’m skeptical of the suggested answer and seemingly simplistic reasoning). If Bob wins with probability 0% and Alice wins with probability 50% then it’s clear that Bob never loses any more slowly than Alice, and sometimes he will lose faster.
Zero is a funny number so I would be cautious of that special case. To get the intuition, modify the problem slightly so that there’s just a single coin being flipped, and Alice and Bob are both betting on it. Every time Alice wins, Bob loses, and vice versa. At any time t, let f(t) be the number of heads so far minus the number of tails so far. We are interested in the set S of sequences of coin flips for which f(t) = +100 for some t (Bob goes broke) AND f(t’) = -100 for some t’ (Alice goes broke). We can define an involution on S as follows: For each s in S, find the smallest time t at which both Alice and Bob have gone broke, and reverse all the coin flips before t. This involution pairs off elements in S; in each pair, Alice goes broke first in one case and Bob goes broke first in the other case. The key point is that if Alice goes broke first then s has 100 excess heads (before time t) whereas the partner of s has 100 excess tails (before time t) and hence s is more probable than its partner.
Zero may be special in some sense, but it satisfies all the requirements that the given heuristic solution explicitly relies on (namely, just that Bob’s probability of winning is less than Alice’s), so it casts doubt on that argument as a general solution. I understand the intuition, but I also see that that intuition doesn’t universally hold.
In your modified game I think I believe the result. However, that version relies on properties that either don’t exist in the original version (one player’s win is the other’s loss) or at least don’t play a role in the given argument (the probabilities add up to 1), and it’s not obvious that the result translates to the game where Alice and Bob are flipping independent coins. As I wrote, I’d like to see the math behind that case.
To put my thoughts another way, I’m trying to determine the exact requirements on the winning probabilities to make the intended result hold. It MAY hold for a = .51 and b = .49, but it clearly fails for a = .5 and b = 0.
Actually, now that I’ve worked through it, I agree that Joshua Green was right to be skeptical and that the New Yorker seems to have slightly garbled the story. If Alice and Bob have separate coins, then they are equally likely to go broke first. One way to see this is to adapt my involution argument: for each Alice/Bob outcome, swap their sequences of coin flips, and simultaneously swap heads with tails. The probability doesn’t change, but we’ve switched which of Alice/Bob goes broke first. I suspect Winkler presented a two-part puzzle, with part one having two coins and part two having one coin, and the New Yorker reported the part one puzzle but the part two solution.
Some poking around reveals that Winkler posed this puzzle in the Fall 2019 Emissary, and indeed there are two parts to it. https://www.msri.org/system/cms/files/973/files/original/Emissary-2019-Fall-Web.pdf
I met Peter during a poster session for the Discrete Math Days conference during my senior year of my undergrad. I had no idea what he looked like at the time, and was presenting my research to him. This included going over some of his own previous results. Eventually somebody came up to me, pointed at his name on my poster, and then pointed at Peter and whispered, “That’s him!”.
My guess for the “spaces count” version: eight billion, eight hundred eight million, eight hundred eight thousand, eight hundred eighty-five.