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:

Here's my general math way to do the bold gambling puzzle. Say we have $1 and want to reach $5.

— Christopher D. Long (@octonion) April 24, 2017

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, , satisfies the equation:

We can solve this pretty easily to see that .

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:

You have $1 and want to reach $pi, and can bet up to your bankroll each time, winning with probability 1/2, losing with probability 1/2.

— Christopher D. Long (@octonion) April 24, 2017

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.

Studying shuffling and Shannon entropy part 2

We did a fun project about Shannon Entropy and Shuffling yesterday:

Chard Shuffling and Shannon Entropy

That project was based largely on an old Stackexchange post (well, comment) here:

See the first comment on this Stackexchange post

Today I wanted to extend that project a little bit and thought it would be fun to look at a different kind of shuffle to see if there was a difference in entropy.

Here’s the shuffle for today as well as what the boys think will happen with this shuffle:

Next we took some time off camera to enter the card numbers in our spreadsheet. Here’s what we found for the new entropy after one of these new shuffles:

Finally, we took even more time off camera to do 4 more shuffles and write down the order of all the cards. After that we did 5 successive shuffles and wrote down the numbers after the 10th shuffle.

The kids didn’t think the cards were as mixed as they were in yesterday’s project, and here’s what the entropy calculation said:

I really enjoyed these two projects. It was especially fun to see how kids could get their arms around the idea of entropy even though the math itself is pretty advanced.