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.

1 Proof
2 Application
- 2.1 Stopping
3 Etymology
4 References

[edit] Proof

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.

[edit] 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 $\operatorname{P}[E_1]$ and $\operatorname{P}[E_2]$ ; in particular, the probability of a draw is irrelevant.

[edit] 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.

[edit] 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:

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

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:

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

$= \frac{\operatorname{P}[E_1]}{1-(1-\operatorname{P}[E_1]-\operatorname{P}[E_2])} = \frac{\operatorname{P}[E_1]}{\operatorname{P}[E_1]+\operatorname{P}[E_2]}$