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 the probability that E_1 occurs before E_2 equals the conditional probability that E_1 occurs given that E_1 or E_2 occur on the next trial, which is

:\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 E_1 and E_2 need not be collectively exhaustive (if they are, the result is trivial).{{Cite web| title = The Craps principle 10/16| author = Susan Holmes| work = statweb.stanford.edu| date = 1998-12-07| access-date = 2016-03-17| url = http://statweb.stanford.edu/~susan/courses/s116/node63.html}}{{cite book|author=Jennifer Ouellette|title=The Calculus Diaries: How Math Can Help You Lose Weight, Win in Vegas, and Survive a Zombie Apocalypse|url=https://books.google.com/books?id=fQzJoqBwtT4C&pg=PT50|date=31 August 2010|publisher=Penguin Publishing Group|isbn=978-1-101-45903-4|pages=50–}}

Proof

Let A be the event that E_1 occurs before E_2. Let B be the event that neither E_1 nor E_2 occurs on a given trial. Since B, E_1 and E_2 are mutually exclusive and collectively exhaustive for the first trial, we have

: \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 \operatorname{P}(B) = 1 - \operatorname{P}(E_1) - \operatorname{P}(E_2).

Since the trials are i.i.d., we have \operatorname{P}(A \mid B) = \operatorname{P}(A). Using \operatorname{P}(A|E_1)=1,\quad \operatorname{P}(A|E_2)=0 and solving the displayed equation for \operatorname{P}(A) gives the formula

:\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

: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=

If the game is played repeatedly until someone wins, then the conditional probability above is the probability that the player wins the game. This is illustrated below for the original game of craps, using an alternative proof.

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:

:E_1:\text{ the original roll (called ‘the point’) is rolled (a win) }

:E_2:\text{ 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}[\text{first i rolls are ties,}(i+1)^{\text{th}}\text{roll is ‘the point’}]

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

:\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:

:\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

{{reflist}}

=Notes=

  • {{cite book |author=Pitman, Jim |title=Probability |publisher=Springer-Verlag |location=Berlin |year=1993 |page=210 |isbn=0-387-97974-3 |url=https://books.google.com/books?id=L6IWgaCuilwC&pg=PA210 }}

{{Craps}}

Category:Theorems in statistics

Category:Theorems in probability theory

Category:Statistical principles