# Sharing a simple and surprising mathematical idea from “How to gamble if you must” with kids

Yesterday I got a neat book in the mail – I’d seen Nassim Taleb recommend it on Twitter:

One warning – this is not a “popular math” book, it is pretty math heavy. Flipping through the first 1/4 of the book, I really enjoyed the presentation and was once again reminded of the surprising fact that foundational research on basic gambling problems was being done in the late 1950s and early 1960s. An accessible and incredibly interesting account of some of this work can be found in Ed Thorp’s autobiography “A man for all Markets.”

One nice example from the beginning of the book relates to gambling in a 50/50 game called “red and black.” Think of the game as trying to guess the color of a card pulled from a randomly shuffled deck, or just betting on a coin flip. If you want to turn, say, \$100 into \$1,000 by betting on this game, what is your best strategy?

IF you are interested, a shorter account of this problem (with accompanying practice problems) can be found in this nice summary paper by Kyle Seigrist published by the Mathematical Association of America.

Summary of the ideas from “How to gamble if you must” by Kyle Seigrist and the Mathematical Association of America

Because this particular gambling problem is accessible to kids, for today’s project I wanted to introduce the idea of 50/50 gambles and ask them what they thought the optimal gambling strategy would be. The specific question is what is the best strategy to follow if you want to try to turn \$100 into \$1,000?

They had some absolutely terrific ideas. My 6th grade son practically suggested the betting strategy from the Kelly criterion!

Next we turned to the computer to study this game in Mathematica. We looked at some simple betting ideas first. So, if we want to turn \$100 into \$1,000 in this game, what happens if we bet \$100 on each bet? What happens if we bet \$50 on each bet?

After seeing the surprising results from the fist set of trials, we looked at the gambling strategies that the boys proposed. First we looked at a version of the strategy that my old son suggested -> basically bet the maximum amount every time (except when you don’t need to bet the max amount to reach \$1,000).

Are you more or less likely to turn \$100 into \$1,000 with this strategy?

Now we checked the betting strategy that my younger son suggested -> bet 1/2 your money each time (except when you don’t need to bet that much to reach \$1,000).

The boys had some pretty interesting ideas about what would happen here.

So, definitely a fun project and the result is pretty surprising (at least to me!) -> in 50/50 games your betting strategy doesn’t matter.

# Christopher Long’s fun generalization of an Expii problem

Twitter is really great place to see fun math. Before showing the fun generalization, though, just to avoid spoilers I want to show the original problem. Here’s the direct link:

https://www.expii.com/solve/69/5

and here’s the problem itself:

So, I’ll pause here to not ruin the problem for anyone who wants to work on it.

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Ok . . . here’s the really cool set of tweets I saw from Christopher Long this morning:

which continued as follows:

To see how delightful Long’s general solution is, maybe walking through my chicken scratch solution which happened to be still sitting on my desk will be helpful:

Here’s a sketch of my approach.

In order to maximize your chance of winning a bet like the one in the problem (one that you expect to lose) you should bet as much as you can at each step (subject to a maximum of the amount you need to win) at each stage.

So,

(i) at step 1 the probability of getting the tree to grow to 40 feet is 1/5 and probability of losing is 4/5.

(ii) Assuming you win, you now have a 1/5 probability of getting the tree to grow to 80 feet and a 4/5 probability of losing.

(iii) Assuming you win, you now have a 1/5 probability of getting the tree to grow to 100 feet and a 4/5 probability of having it shrink to 60 feet.

(iv) If you win on stage (iii) you win (1 out of 125 times). If you lose, you now have a 1/5 probability of having the tree now grow to 100 feet and a 4/5 probability of having the tree shrink to 20 feet.

So, after the 4th branch in my picture you’ve either won (probability 1/125 + 4/625), returned to 20 feet (probability 16/625) or lost (the only other case).

Thus, your probability of winning the game from the start, $x$, satisfies the equation:

$x = 1/125 + 4/625 + (16/625)*x$

We can solve this pretty easily to see that $x = 3/203$.

The really fun – and honestly, amazing – thing about Long’s solution is that he notices that the pattern in the branches of the binary tree corresponds exactly to the pattern in the digits of the binary expansion of 1/5. For clarity, the 1/5 here comes from the growth multiple – 20 feet growing to 100 feet – and not from the probability which, by coincidence, also has a 1/5 in it.

Anyway, Long’s solution also allows you to immediately see how to solve any problem like the Expii one, and, for extra fun, problems where the growth multiple is irrational:

The answer is in Long’s timeline, but it is a good challenge to see if you can work out the answer just from the tweets I’ve included here. Since he skips a bit of algebra in his tweets, working through his tweets is also an important way to make sure that you really understand his work.

I think the sequence of tweets from Long are a great thing to show kids who are learning math – especially kids learning probability and stats. Those tweets really show how a mathematician thinks about a problem.

# A probability problem from James Tanton’s “Solve This”

We were unpacking a bunch of books today and ran across James Tanton’s Solve This and Theoni Pappaas’s The Adventures of Penrose The Mathematical Cat. I asked the boys to find a problem they’d like to talk about and they found a gem from Tanton’s book:

16.1 A Fair Game?

Peter has ten coins, Pennelope has nine. Peter and Pennelope agree to toss all their coins simultaneously. Whoever receives the largest number of heads will win. In case of a tie Pennelope will be declared the winner, so as to offset the advantage Peter has to begin with. Given this agreement, who is most likely to win?

We began by talking about the problem and the boys had some ideas right away. Their initial guess is that Peter will win, but they want to explore the problem by looking at an easier case first.

After discussing the problem we started looking at the easier case in which Peter has one coin and Pennelope has 1 coin. One of the nice things about this simple version of the problem is that you can actually talk through all of the cases. Talking through these cases is a really nice introduction to counting and probability. The other nice thing about the simple case, of course, is the surprising answer!

Given the surprising answer from the last video, we decided to check out the next easiest case – Peter has 3 coins and Pennelope has 2. Fortunately we’ve already understood the cases that Pennelope will encounter here, so we just have to understand the situations that can come up for Peter.

Again, this exercise is a nice exercise in counting and also this time the snap cubes we are using for heads and tails help show the symmetry involved in the various cases. The boys do a nice job here working through this slightly more complicated example:

Finally we decided to try to tackle the original problem. My younger son has a great symmetry idea what we talk about for a bit, but it seemed that approaching the problem this way was still a bit out of reach for them.

They wanted to calculate, but I didn’t 🙂 As sort of an on-the-fly compromise, I thought working a little bit with Pascal’s triangle would still be a good exercise and also show them that the calculation they were looking to do would be pretty ugly!

So, maybe not the most satisfying end to the project since we didn’t completely solve the problem. Nonetheless, I like all of the great math practice that we got in this project – lots of good ideas from introductory probability and counting, and lots of good arithmetic practice, too.

Also, it was nice to see the boys starting to think about using symmetry to help solve problems. Even if they aren’t yet quite able to put those ideas to work in their solution yet, I’m glad those ideas are starting to come to the surface.

Tomorrow we are starting to work through Art of Problem Solving’s Introduction to Counting and Probability. Today’s project was a great way to motivate that summer project!

# The triangle inequality and geometric probability

I finished up a section on the triangle inequality today with my older son and wanted to show him a surprising application. This example – taken from geometric probability – is a little over his head, but was still fun to talk through.

The problem is this:

You have a stick with length 1. If you chop the stick into 3 pieces, what is the probability that those three pieces can form a triangle?

Fun question – we started out with him giving his thoughts on the problem, and the problem of thinking through probability when you have an infinite number of choices:

With the initial thoughts out of the way, we started thinking through a few specific ways to cut the stick into pieces. The important observation from this part of the talk is that if any piece has length greater than 1/2, we won’t be able to form a triangle:

Finally we get to the geometric probability. We have to start with a special property of an equilateral triangle. We’d talked about this property of an equilateral triangle yesterday. My son remembered the property, but not the proof. We spent a minute or two reviewing the proof, but the proof wasn’t the main point.

We can use an equilateral triangle with height of 1 to model our probability problem in a surprising way. With our new model, we see that the amazing connection to geometry tells us that the probability of our broken up stick forming a triangle is 1/4.

So, a fun little project that is a little more advanced than what we’ve covered so far. It touches on geometric probability which is an incredible subject on its own, it touches a little on properties of infinity, and finally the relationship between algebra and geometry.

All from the simple ideas behind the triangle inequality. Amazing!