I’ve thought of a gambling problem (read: probability problem) that I can’t seem to solve using the methods I know of. The problem is to find an optimal playing strategy for the following game:

Pay me $100. I give you one chip (worth $1). You can decide to keep your $1 and walk away (down $99) or flip a coin with me. If it comes up heads, then you have three chips. If it comes up tails, you walk away with nothing ($100 down). If you have three chips, then you can walk away with $3 ($97 down), or try for nine chips. And so on.

The thing that makes this game hard to analyze is its infinite expectation. Every time you go for the coin, your payback is greater than your odds. For instance (ignore the $100 for now): I have one chip. I can walk away with $1. Or I can flip and get 0 1/2 of the time, and 3 1/2 of the time, for an expectation of 1.5 chips. Clearly I should flip. The exact same reasoning happens at every stage of the game. You hit every time, and you end up losing it all, because the coin is going to come up tails eventually.

It has something to do with that $100 for one chip (Keep in mind that hat’s the *only* way you can play this game). Because without it, the game is pretty easy to analyze. Give me some money, and I’ll triple it with 1/2 probability, or take it all. To play that in real life, you just play with amounts of money such that if you lose, you can play again. Play with $10. Get $30. Play with $10 again. Lose it. Keep playing like that and you can get as rich as you want.

For the simpler game, (without the $100) you can calculate the stake you need to win $x with probability y. Then, in the more complex game, you need that stake plus $100.

Oops, that’s not quite right – you need to calculate the stake for the simpler game for winning $x+100

Oh, hang on – it’s not that simple, because you need to pay the $100 every time you enter the game. If you always keep going (because you haven’t made your designated profit), the game terminates on average every M=sum_x(x*2^-x) moves, so you have to win at a rate of

at least $100/M in order to make money. Which I don’t think you can do, but the proof sounds a bit tedious.

I think it is possible to make money on the game. After some playing around with Mathematica, I found it to be a bankroll problem. If you always try to get to the twelvth stage in the game (probability 0.02% (1/2^12), payback $531,441 (3^12)) and then quit if you get there, then you

canmake it to a profit, since your expectation is $130. If you play 2^12 = 4096 games, you’ll have a 50% chance of having gotten to the twelvth stage. In order to play that many games, you need a bankroll of $409,600. But notice how the payback is about 30% more than that.If you go to 14 rolls, the payback is $4,782,969 with probability 1/16,384. In other words, you need $1,638,400 to have a 50% shot at getting anything back. But notice, $4,000,000 * 50% > $1,600,000. So now, finally, you’re really making money in the long run. You just have to have an investor’s market to do it.

Well, there really isn’t a good “strategy” except not to play. It’s a little weird, because there are discrete steps with scales that don’t really align nicely, but here’s my reasoning:

You’re out $100 to start with. Our coin flips are independent, so calculating the probability of n heads in a row is quite simple. In order to make more than $100 in this game, we need 5 heads in a row, which gives us a probability of 1/32, assuming a fair coin. The winnings of this is $243. In order for the player to break even across an infinite number of games with an initial stake of $100, the 5th head would need to yield $3200, the 6th $6400, and so on.

Of course, I could be completely off, given how much attention I gave my probability course ;)

Ha! This would be a horrible game to play in Colorado, where the bet limit is $100. However it is interesting in it’s implications for taxes and other times when money is demanded up front. Then I always think of real world applications first. I could imagine a number of situations. 1. $99 is taken from you and $1 is taken from me or any distribution in between. 2. We go with scenario 1 and then also vary the payout between the two of us. Ultimate question being who achieves a higher expectation.