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# Probability in games of chance

posted in: Games of chance One of the player’s goal is to find an optimal method to win the game. We say that the players play a martingale. It should not be confused with the martingale probability which is a stochastic process such that its expected value knowing the information available on a certain date s, denoted Fs, is the value of:

E (Xt | Fs) = Xs (with s ≤ t).

However, there is a link between these two terms. If the gain of a player in a game of chance is a martingale (stochastic process), then the hope of gain is constant. If the gain of a player in a game of chance is a sub-martingale (stochastic process), then the hope of gain is growing, then it is a martingale (in the technical sense of the game). Note also that the mathematical term is derived historically from the term of the game technique.

### Example of game where each game is a game of pure chance

Let’s take an example of a simple game in which, each game, the player wins his bet with probability p and loses his bet with probability 1-p, the achievements of the game are independent: bet on black at roulette game for 10 achievements. For each ball run, the probability of winning his initial bet is p = (18 black squares) / (37 possible numbers) ≈ 0.486. The expected gain is p- (1-p) ≈ -0.027.

• Using the technique to bet each time the same amount (\$1, for example), the expected gain is ≈ -0.27. That is to say, on average over 10 games, the player will have lost  €0.27.
• Using the classic martingale which is to build twice the amount wagered in the previous section (\$1, then \$2, then \$4, etc.) until the player wins then stops playing. If the player wins, the gain will be \$1; eg for a victory in the fifth game, he will earn \$16 and will spend 1 + 2 + 4 + 8 = 15 \$. Otherwise, it will have lost 1 + 2 + 4 + 8 + 16 + 32 + 64 + 128 + 256 + 512 = 1023 \$. The average gain is then (1 – (1-p)10) – 1023 (1 – p)10 ≈ -0.318 \$ It should be noted that if we increase the number of parties, the average gain tends to 1\$, the original bet. The expected number of games is, on the other side, infinity.

### Example game where each game is a reasoned game of chance

Take the example of draw poker, that is to say, a game in which five cards are randomly dealt to each player and the rest of the game does not depend on randomness. In this case, it is useful to know the probability of each hand to fully appreciate the value of the game in hand and thus be able to adopt a vetting and bluff adapted technique. The used probabilities are then from combinatorics.

When several successive parts, the technique used depends also on the way of playing the other players.

### 0 Responses

1. ##### Bandar Domino Online Indonesia
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i’m really aware that gambling can not be apart from math formula, but i think it’s not easy to understand it, and using that formula on betting.