Craps principle

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\cup E_2\right =\frac{\operatorname{P}

^*}{\operatorname{P}^*+\operatorname{P}^*}

The events $E_1$ and $E_2$ need not be collectively exhaustive.

Proof

Since

E_1

and

E_2

are mutually exclusive,

\operatorname{P} E_2 =\operatorname{P}

^*+\operatorname{P}^*

Also due to mutual exclusion,

E_1\cap(E_1\cup E_2)=E_1

By conditional probability,

\operatorname{P} E_2) =\operatorname{P}\left E_1\cup E_2\right \operatorname{P}\left E_2\right

Combining these three yields the desired result.

Application

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 $\operatorname{P} and \operatorname{P}[E_2$ ; in particular, the probability of a draw is irrelevant.

Stopping

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

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}}i\textrm{\ rolls\ are\ ties,\ }(i+1)^\textrm{th}\textrm{\ roll\ is\ 'the\ point'}

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Proof

Application

Stopping

Etymology