Probability Of Winning Craps

In probability theory, the craps principle is a theorem about eventprobabilities 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

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{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[edit]

Since and are mutually exclusive, the craps principle applies. For example, if the original roll was a 4, then the probability of winning is / / + / = This avoids having to sum the infinite series corresponding to all the possible outcomes.
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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} (Amid E_{1})+operatorname {P} (E_{2})operatorname {P} (Amid E_{2})+operatorname {P} (B)operatorname {P} (Amid B)=operatorname {P} (E_{1})+operatorname {P} (B)operatorname {P} (Amid B)}

and ${displaystyle operatorname {P} (B)=1-operatorname {P} (E_{1})-operatorname {P} (E_{2})}$ . Since the trials are i.i.d., we have ${displaystyle operatorname {P} (Amid B)=operatorname {P} (A)}$ . 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[edit]

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.

Stopping[edit]

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[edit]

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

Calculating The Probability Of Winning Craps

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[edit]

Probability Of Winning Craps

^Susan Holmes (1998-12-07). 'The Craps principle 10/16'. statweb.stanford.edu. Retrieved 2016-03-17.
^Jennifer Ouellette (31 August 2010). The Calculus Diaries: How Math Can Help You Lose Weight, Win in Vegas, and Survive a Zombie Apocalypse. Penguin Publishing Group. pp. 50–. ISBN978-1-101-45903-4.

Notes[edit]

Probability Of Winning Craps

Pitman, Jim (1993). Probability. Berlin: Springer-Verlag. p. 210. ISBN0-387-97974-3.

Probability Of Winning A Field Bet In Craps

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