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The Craps principle 10/16

Suppose a game that has to be continued as long as neither player has won a game, where $P(A\; wins)=p_a$, $P(B\; wins)=p_b$ and P(Draw)=pd, with pa+pb+pd=1, I showed that the probability of A winning is:

\begin{displaymath}P(A \;wins)=p_a(p_d+p_d^2+p_d^3 +\ldots)=\frac{p_a}{1-p_d}=\frac{p_a}{p_a+p_b}\end{displaymath}

Intuitively, we know this, the only probabilities that matter are the relative chances of A and B of winning, pd does not matter.

I went on to define craps:
The game is played with 2 dice, a first throw can finish in either a loss (if I throw 12,3 or 2), a win (if I throw 7 or 11) or a draw-replay, (if I throw 4, 5, 6, 8, 9, or 10, from hereon I have to remember what I threw in this case, it will be called my ``point''), if I have a replay number, I have to go on throwing the two dice, until either I get the same number I had at first, then I win, or I get a 7, (I lose).

Here are two craps websites, I found useful:
http://www.charm.net/~shack/game/begcraps.html
http://www.cris.com/~Atiara/DiceOdds.html


next up previous index
Next: Example of computation for Up: Independence 10/14 and 10/16 Previous: Independence of more than
Susan Holmes
1998-12-07