At the end of the Slate Money Podcast this week there’s a discussion of a dice game that Felix Salmon asked about in last week’s podcast. Here’s the podcast from this week:
The game is easy to explain. You roll a fair 6-side die N times, where N is any number you pick. You also choose an amount of money to bet – say X. If you never roll a 6 in your N rolls, you win 2^N times your money back. If any of your rolls are a 6, you get $0 back. Salmon’s questions are -> (1) How much money would you bet if you could play this game once, and (2) how many rolls would you select?
I thought this game would be fun to talk through with my older son. Here I explain the game and he talks about a few of the ideas he thinks will be important for answering Salmon’s questions. He has some interesting ideas about “high risk” and “low risk” strategies. We also talk through a few simple cases:
In the last video my son was calculating the probability of winning the game in N rolls by calculating the probability of not losing. That’s, unfortunately, a fairly complicated way to approach the problem so I wanted to talk a little more so he could see that a direct calculation of the probability of winning wasn’t actually too hard. We talked through that calculation here. We also find that if you roll 4 times you have roughly a 50/50 chance of winning the game.
Before we played the game he wanted to calculate the expected value for your winning in this game. Here we do that calculation and find the surprising answer. We then play the game. He decided to bet $100 and roll three times, and . . .
This was a fun problem to talk through with my son, and I’m excited to talk through it with my younger son tomorrow to see if he reaches a different conclusion. It had never occurred to me to talk through this or any version of the St. Petersburg Paradox with the boys before, so thanks to Felix Salmon for sharing this problem.