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Comment by danbruc | Hacker News Reader

I don't have infinite money and time to play this forever. And there is a certain probability that I will lose $10, $100 or $1000 and have to stop playing because I no longer can or want to afford it. I wouldn't mind losing $10 with near certainty but I would only play if I had a good chance to win some money before losing $1000 and having to stop. And I would not risk losing $1000 if the expected gain is small, say $100, or if I had to play the entire day to get there. Probably not to easy to quantify but certainly doable.

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You might be looking for the Kelly criterion. It provides a quantitative answer to what percent of your bankroll you should bet, given the characteristics of a game. Roughly, it tries to avoid having you run out of money before you start winning.

So an answer to the paradox is that some people, like yourself, have intuition about the Kelly criterion: that they should limit the portion of their bankroll that they bet.

I ran a quick simulation assuming that the banker and the player both start with $100 and the game ends when either party gets to $0. It seems like the break-even entry fee is somewhere around $5.50, anything higher and the player almost always loses first. If the banker has more money than the player, it only makes the possibility of the player winning smaller. And even if the player has more money than the banker, they may have to play hundreds of games to come out ahead. This game doesn't seem very fun.

> I ran a quick simulation assuming that the banker and the player both start with $100 and the game ends when either party gets to $0.

Why are you modeling it that way? Nothing in the game requires that a player spend the bank to $0.

If I pay 2 dollars to play, I have a 50% chance each game of at least breaking even and being able to continue playing.

If I pay 4 dollars to play, I have a 25% chance each game of at least breaking even and being able to continue playing.

The minimum payout is $2. You have 100% chance each game of at least breaking even if you pay $2 to play.

You only pay a fixed sum at the beginning to play. The question is, what is the upper bound on that fixed sum, and why is it so much smaller than your expected winnings?

You pay the fixed sum before every game. If you go with Ian Hacking's suggested low price of $25, I can easily see losing several thousand dollar before you get lucky and hit one of the rare, high value events.

You have to pay for every game. Half of the time you win $2 which means half of the time you lose $3 if a game costs $5.

(you've misunderstood the description.)

Correct, half the time you win $2, but what happens the other half of the time?

With the other half... half of the time you win at least $4. With the other half of that... half of the time you win at least $8. With the other half of that... half of the time you win at least $16.

So, if you simulate the average winnings in a single round, you get data like this:

[8, 2, 8, 2, 8, 32, 8, 8, 4, 4, 2, 4, 4, 16, 2, 2, 16, 4, 8, 2]

Which even with losses of $3 and $1 for most games still works out to a $44 profit at $5 game.

However, and this is the point of the paradox, if you run the numbers for 100 rounds, you start to see average winnings per round like:

[6.48, 6.6, 15.68, 10.02, 10.26, 17.04, 5.86, 11.96, 8.92, 7.34, 6.56, 17.14, 9.92, 9.64, 11.48, 12.44, 19.64, 171.42, 12.82, 5.9]

So if you had played those 20 times, at $5/game * 100 rounds you would have a $27,000 profit on $10000.

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