# In this particular game of probability, what tactic increases the chances of payoff the most?

Note: This is a simulation that me and my friends are fiddling with on a virtual game.

So essentially, please assume that there is a game which has Gambling in it.

Gambling works by using /dice(/ symbolizes a command, /dice is assumed to be absolutely fair).

The games always involve a player and a dealer, at the start of the game the player places a bet with the dealer(Say 10 units), if the player wins then he receives a double payoff(20 units in this case) and if he loses he gets nothing back.

For this simulation please only consider the games of 1x, 2x and 3x. 1x is when the player rolls once and the dealer then rolls, and the one with the larger total wins. Same total is a draw. In 2x, the dice is rolled twice by both parties and in 3x it is rolled thrice.

Originally the probabilities of winning are .5 for both sides, but how can the chances of getting the most profitable payoff be increased?

Currently a popular one is to use a doubling technique, If you bet 10 units and lose you bet 20 units, lose and 40 units and so on. There is a limit on the maximum bet however(Say 100 units, so if the player begins at 10 units, he would only be able to double upto 4 times).

Now, is the doubling technique beneficial to the player or the dealer? Also, are there any other techniques that might be able to beat this one?

Thanks alot for your answers, and again, this is a virtual simulation in a virtual game and hence /dice is considered absolutely fair.

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If the game is fair, no betting strategy can change that. You can make various probability distributions of winnings and losings, but the expectation will always be zero. Let's ignore draws, so each player has $\frac 12$ chance of winning. Let the initial bet be $1$. If you can double $n$ times, your chance of losing all of them is $\frac 1{2^{n+1}}$ and you lose $2^{n+1}-1$. The other $1-\frac 1{2^{n+1}}$ of the time you win $1$. The expectation is then $1(1-\frac 1{2^{n+1}})-\frac {2^{n+1}-1}{2^{n+1}}=0$