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| In [[probability theory]], the '''craps principle''' is a theorem about [[Event (probability theory)|event]] [[probabilities]] under repeated [[Independent and identically-distributed random variables|iid]] trials. Let <math>E_1</math> and <math>E_2</math> denote two [[mutually exclusive]] events which might occur on a given trial. Then for each trial, the [[conditional probability]] that <math>E_1</math> occurs given that <math>E_1</math> or <math>E_2</math> occur is
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| :<math>\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]}</math>
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| The events <math>E_1</math> and <math>E_2</math> need not be [[collectively exhaustive]].
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| ==Proof==
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| Since <math>E_1</math> and <math>E_2</math> are mutually exclusive,
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| :<math> \operatorname{P}[E_1\cup E_2]=\operatorname{P}[E_1]+\operatorname{P}[E_2]</math>
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| Also due to mutual exclusion,
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| :<math> E_1\cap(E_1\cup E_2)=E_1</math>
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| By [[conditional probability]],
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| :<math> \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]</math>
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| Combining these three yields the desired result.
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| ==Application==
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| If the trials are repetitions of a game between two players, and the events are
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| :<math>E_1:\mathrm{ player\ 1\ wins}</math>
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| :<math>E_2:\mathrm{ player\ 2\ wins}</math>
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| 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 (tie)|draw]] does not occur). In fact, the result is only affected by the relative marginal probabilities of winning <math>\operatorname{P}[E_1]</math> and <math>\operatorname{P}[E_2]</math> ; in particular, the probability of a draw is irrelevant.
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| ===Stopping===
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| 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.
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| ==Etymology==
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| 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:
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| :<math>E_1:\textrm{ the\ original\ roll\ (called\ 'the\ point')\ is\ rolled\ (a\ win) }</math>
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| :<math>E_2:\textrm{ a\ 7\ is\ rolled\ (a\ loss) }</math>
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| Since <math>E_1</math> and <math>E_2</math> are mutually exclusive, the craps principle applies. For example, if the original roll was a 4, then the probability of winning is
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| :<math>\frac{3/36}{3/36 + 6/36}=\frac{1}{3}</math>
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| This avoids having to sum the [[infinite series]] corresponding to all the possible outcomes:
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| :<math>\sum_{i=0}^{\infty}\operatorname{P}[\textrm{first\ }i\textrm{\ rolls\ are\ ties,\ }(i+1)^\textrm{th}\textrm{\ roll\ is\ 'the\ point'}]</math> | |
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| Mathematically, we can express the probability of rolling <math>i</math> ties followed by rolling the point:
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| :<math>\operatorname{P}[\textrm{first\ }i\textrm{\ rolls\ are\ ties,\ }(i+1)^\textrm{th}\textrm{\ roll\ is\ 'the\ point'}]
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| = (1-\operatorname{P}[E_1]-\operatorname{P}[E_2])^i\operatorname{P}[E_1]
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| </math>
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| The summation becomes an infinite [[geometric series]]:
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| :<math>\sum_{i=0}^{\infty} (1-\operatorname{P}[E_1]-\operatorname{P}[E_2])^i\operatorname{P}[E_1]
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| = \operatorname{P}[E_1] \sum_{i=0}^{\infty} (1-\operatorname{P}[E_1]-\operatorname{P}[E_2])^i
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| </math>
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| ::<math> = \frac{\operatorname{P}[E_1]}{1-(1-\operatorname{P}[E_1]-\operatorname{P}[E_2])}
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| = \frac{\operatorname{P}[E_1]}{\operatorname{P}[E_1]+\operatorname{P}[E_2]}
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| </math>
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| which agrees with the earlier result.
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| ==References==
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| {{cite book |author=Pitman, Jim |title=Probability |publisher=Springer-Verlag |location=Berlin |year=1993 |pages= |isbn=0-387-97974-3 |oclc= |doi=}}
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| [[Category:Statistical theorems]]
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| [[Category:Probability theory]]
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| [[Category:Statistical principles]]
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