In probability theory, the craps principle is a theorem about event probabilities under repeated iid trials. Let ${\displaystyle E_{1}}$ and ${\displaystyle E_{2}}$ denote two mutually exclusive events which might occur on a given trial. Then the probability that ${\displaystyle E_{1}}$ occurs before ${\displaystyle E_{2}}$ equals the conditional probability that ${\displaystyle E_{1}}$ occurs given that ${\displaystyle E_{1}}$ or ${\displaystyle E_{2}}$ occur on the next trial, which is

${\displaystyle \operatorname {P} [E_{1}\,\,{\text{before}}\,\,E_{2}]=\operatorname {P} \left[E_{1}\mid E_{1}\cup E_{2}\right]={\frac {\operatorname {P} [E_{1}]}{\operatorname {P} [E_{1}]+\operatorname {P} [E_{2}]}}}$

The events ${\displaystyle E_{1}}$ and ${\displaystyle E_{2}}$ need not be collectively exhaustive (if they are, the result is trivial).^[1][2]

Proof

Let ${\displaystyle A}$ be the event that ${\displaystyle E_{1}}$ occurs before ${\displaystyle E_{2}}$. Let ${\displaystyle B}$ be the event that neither ${\displaystyle E_{1}}$ nor ${\displaystyle E_{2}}$ occurs on a given trial. Since ${\displaystyle B}$, ${\displaystyle E_{1}}$ and ${\displaystyle E_{2}}$ are mutually exclusive and collectively exhaustive for the first trial, we have

${\displaystyle \operatorname {P} (A)=\operatorname {P} (E_{1})\operatorname {P} (A\mid E_{1})+\operatorname {P} (E_{2})\operatorname {P} (A\mid E_{2})+\operatorname {P} (B)\operatorname {P} (A\mid B)=\operatorname {P} (E_{1})+\operatorname {P} (B)\operatorname {P} (A\mid B)}$

and ${\displaystyle \operatorname {P} (B)=1-\operatorname {P} (E_{1})-\operatorname {P} (E_{2})}$. Since the trials are i.i.d., we have ${\displaystyle \operatorname {P} (A\mid B)=\operatorname {P} (A)}$. Using ${\displaystyle \operatorname {P} (A|E_{1})=1,\quad \operatorname {P} (A|E_{2})=0}$ and solving the displayed equation for ${\displaystyle \operatorname {P} (A)}$ gives the formula

${\displaystyle \operatorname {P} (A)={\frac {\operatorname {P} (E_{1})}{\operatorname {P} (E_{1})+\operatorname {P} (E_{2})}}}$.

Application

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

${\displaystyle E_{1}:\mathrm {player\ 1\ wins} }$

${\displaystyle E_{2}:\mathrm {player\ 2\ wins} }$

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 ${\displaystyle \operatorname {P} [E_{1}]}$ and ${\displaystyle \operatorname {P} [E_{2}]}$ ; in particular, the probability of a draw is irrelevant.

Craps example

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:

${\displaystyle E_{1}:{\text{ the original roll (called ‘the point’) is rolled (a win) }}}$

${\displaystyle E_{2}:{\text{ a 7 is rolled (a loss) }}}$

Since ${\displaystyle E_{1}}$ and ${\displaystyle E_{2}}$ are mutually exclusive, the craps principle applies. For example, if the original roll was a 4, then the probability of winning is

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

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

${\displaystyle \sum _{i=0}^{\infty }\operatorname {P} [{\text{first i rolls are ties,}}(i+1)^{\text{th}}{\text{roll is ‘the point’}}]}$

Mathematically, we can express the probability of rolling ${\displaystyle i}$ ties followed by rolling the point:

${\displaystyle \operatorname {P} [{\text{first i rolls are ties, }}(i+1)^{\text{th}}{\text{roll is ‘the point’}}]=(1-\operatorname {P} [E_{1}]-\operatorname {P} [E_{2}])^{i}\operatorname {P} [E_{1}]}$

The summation becomes an infinite geometric series:

${\displaystyle \sum _{i=0}^{\infty }(1-\operatorname {P} [E_{1}]-\operatorname {P} [E_{2}])^{i}\operatorname {P} [E_{1}]=\operatorname {P} [E_{1}]\sum _{i=0}^{\infty }(1-\operatorname {P} [E_{1}]-\operatorname {P} [E_{2}])^{i}}$

${\displaystyle ={\frac {\operatorname {P} [E_{1}]}{1-(1-\operatorname {P} [E_{1}]-\operatorname {P} [E_{2}])}}={\frac {\operatorname {P} [E_{1}]}{\operatorname {P} [E_{1}]+\operatorname {P} [E_{2}]}}}$

which agrees with the earlier result.

References