Craps principle

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In probability theory, the craps principle is a theorem about event probabilities under repeated iid trials. Let E_1 and E_2 denote two mutually exclusive events which might occur on a given trial. Then for each trial, the conditional probability that E_1 occurs given that E_1 or E_2 occur is

\operatorname{P}\left[E_1\mid E_1\cup E_2\right]=\frac{\operatorname{P}[E_1]}{\operatorname{P}[E_1]+\operatorname{P}[E_2]}

The events E_1 and E_2 need not be collectively exhaustive.

Proof[edit]

Since E_1 and E_2 are mutually exclusive,

 \operatorname{P}[E_1\cup E_2]=\operatorname{P}[E_1]+\operatorname{P}[E_2]

Also due to mutual exclusion,

 E_1\cap(E_1\cup E_2)=E_1

By conditional probability,

 \operatorname{P}[E_1\cap(E_1\cup E_2)]=\operatorname{P}\left[E_1\mid E_1\cup E_2\right]\operatorname{P}\left[E_1\cup E_2\right]

Combining these three yields the desired result.

Application[edit]

If the trials are repetitions of a game between two players, and the events are

E_1:\mathrm{ player\ 1\ wins}
E_2:\mathrm{ player\ 2\ wins}

then the craps principle gives the respective conditional probabilities of each player winning a certain repetition, given that someone wins (i.e., given that a draw does not occur). In fact, the result is only affected by the relative marginal probabilities of winning \operatorname{P}[E_1] and \operatorname{P}[E_2] ; in particular, the probability of a draw is irrelevant.

Stopping[edit]

If the game is played repeatedly until someone wins, then the conditional probability above turns out to be the probability that the player wins the game.

Etymology[edit]

If the game being played is craps, then this principle can greatly simplify the computation of the probability of winning in a certain scenario. Specifically, if the first roll is a 4, 5, 6, 8, 9, or 10, then the dice are repeatedly re-rolled until one of two events occurs:

E_1:\textrm{ the\ original\ roll\ (called\ 'the\ point')\ is\ rolled\ (a\ win) }
E_2:\textrm{ a\ 7\ is\ rolled\ (a\ loss) }

Since E_1 and E_2 are mutually exclusive, the craps principle applies. For example, if the original roll was a 4, then the probability of winning is

\frac{3/36}{3/36 + 6/36}=\frac{1}{3}

This avoids having to sum the infinite series corresponding to all the possible outcomes:

\sum_{i=0}^{\infty}\operatorname{P}[\textrm{first\ }i\textrm{\ rolls\ are\ ties,\ }(i+1)^\textrm{th}\textrm{\ roll\ is\ 'the\ point'}]

Mathematically, we can express the probability of rolling i ties followed by rolling the point:

\operatorname{P}[\textrm{first\ }i\textrm{\ rolls\ are\ ties,\ }(i+1)^\textrm{th}\textrm{\ roll\ is\ 'the\ point'}]
 = (1-\operatorname{P}[E_1]-\operatorname{P}[E_2])^i\operatorname{P}[E_1]

The summation becomes an infinite geometric series:

\sum_{i=0}^{\infty} (1-\operatorname{P}[E_1]-\operatorname{P}[E_2])^i\operatorname{P}[E_1]
= \operatorname{P}[E_1] \sum_{i=0}^{\infty} (1-\operatorname{P}[E_1]-\operatorname{P}[E_2])^i
 = \frac{\operatorname{P}[E_1]}{1-(1-\operatorname{P}[E_1]-\operatorname{P}[E_2])}
= \frac{\operatorname{P}[E_1]}{\operatorname{P}[E_1]+\operatorname{P}[E_2]}

which agrees with the earlier result.

References[edit]

Pitman, Jim (1993). Probability. Berlin: Springer-Verlag. ISBN 0-387-97974-3.