I thought about this outta no where. I want to make a casino game where the odds are against the house, but the catch is that the game rules force the player to play until the odds change against him.(adsbygoogle = window.adsbygoogle || []).push({});

For example, make the odds 80% against the house (player has 80% chance of winning). If the player puts down a dollar at first and each time he wins, the casino pays him another dollar (doesn't double his money). He is forced to play a certain number of times, but if he loses he loses his initial dollar.

Here's how I thought about it (from the player's POV):

amount of money won = number of times the game was played = X

probability of winning each time = P

amount lost = $1

Number of times the game has to be played in order to break even:

(amount of money won)*(probability of winning) - (amount lost)*(probability of losing) = 0

X*P^X - ($1)*(1- P^X) = 0

which simplifies to: (P^X)*(X+1) - 1 = 0

All the casino has to do to increase the odds in its favor is to force the player to play at least 1 more time than X (which depends on P). Anyone see any problems with my logic or math?

PS. I initially thought this problem would be a lot simpler than this .

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# Simple (but interesting) gambling problem

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