However, what makes it cumbersome is a slightly complicated sample space and conditional probabilities! In this lesson, students learn about the game of craps as the avenue for some sophisticated theoretical probability calculations, and they learn why, in the long run, the house always wins. The probability of this happening is 1 out of 10 lakh. There is a probability of getting a desired card when we randomly pick one out of 52. For example, the probability of picking up an ace in a 52 deck of cards is 4/52; since there are 4 aces in the deck. The odds of picking up any other card is therefore 52/52 – 4/52 = 48/52.

Hey Wizard,
I'm a CJ/CIS major working in a Math department, and I've been asked to convert an open ended question test to multiple choice test. I found your site 'The Wizard of Odds', and was directed here to ask my question. What is the probability of rolling a 2 given your roll only being even, when rolling a 20 sided die? Thank you for your help.
-R

I know what you probably meant, but you actually need to specify what's on the 20-sided die. If it's 10 through 200 in increments of 10, the probability is zero.
Under the assumption that the d20 contains integers 1..20, here are the answers I'd list on a four-choice pick:
a) 1/2
b) 1/5
c) 1/10
d) 1/20
Edit: What's a 'CJ' major?

I know what you probably meant, but you actually need to specify what's on the 20-sided die. If it's 10 through 200 in increments of 10, the probability is zero.
Under the assumption that the d20 contains integers 1..20, here are the answers I'd list on a four-choice pick:
a) 1/2
b) 1/5
c) 1/10
d) 1/20
Edit: What's a 'CJ' major?

Funny, I read it differently for the same reason. I 'decided' the question stipulated the 20 sides were numbered 2 thru 20, evens only, with 2 identical sides per number, before I read ME's answer. Be interested to see the clarification.
I'm guessing 'Criminal Justice/Crime Scene Investigation' for the major.

If you're saying 'I've got a d20. The next roll will be an even number. What are the odds of that number being a 2?' Then I'd have to say the odds are 1 in 10. But how does one guarantee that no odd number will be rolled?

It's a conditional probability problem. It's guaranteed because the question says it's guaranteed. It's a math question, not a physics question.Designing multiple-choice tests is one of my favorite things to do. I see this question, and all I think is 'gosh, i wonder how they're doing answer placement. Is it randomized? Rotated? both?'
My HS world geography teacher was a sadistic guy. He put patterns into all his tests, but he also rotated the correct answers. So my 100% score could go a,b,c,d,a,b,c,d... while my friend could have the same answers but need to respond d,c,b,a,d,c,b,a... Made colluding and/or selling answers a real pain.

CJ - Criminal Justice
The numbers on the 20 sided dice are 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20. If this reply sounds smartalically, I appologize. I don't really understand what you mean needing to 'specify what's on the 20-sided die.' Anyhow, which is the answer; a, b, c, or d?

There are 10 even numbers. If you know for sure that it is even, then the answer is 1/10.
You can also solve this by going through the formula for conditional probability. To be honest I always have to look it up (check wikipedia for conditional probability), but you will end up dividing 1/20 by 10/20 and get 1/10.

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

${\displaystyle \mathrm{P}[E_1\text{before}{E}_2]=\mathrm{P}\left[E_1\mid E_1\cup E_2\right]=\frac{\mathrm{P}[E_1]}{\mathrm{P}[E_1]+\mathrm{P}[E_2]}}$

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]

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

${\displaystyle \mathrm{P}(A)=\mathrm{P}(E_1)\mathrm{P}(A\mid E_1)+\mathrm{P}(E_2)\mathrm{P}(A\mid E_2)+\mathrm{P}(B)\mathrm{P}(A\mid B)=\mathrm{P}(E_1)+\mathrm{P}(B)\mathrm{P}(A\mid B)}$

and ${\displaystyle \mathrm{P}(B)=1-\mathrm{P}(E_1)-\mathrm{P}(E_2)}$. Since the trials are i.i.d., we have ${\displaystyle \mathrm{P}(A\mid B)=\mathrm{P}(A)}$. Using ${\displaystyle \mathrm{P}(A{E}_1)=1,\mathrm{P}(A{E}_2)=0}$ and solving the displayed equation for ${\displaystyle \mathrm{P}(A)}$ gives the formula

${\displaystyle \mathrm{P}(A)=\frac{\mathrm{P}(E_1)}{\mathrm{P}(E_1)+\mathrm{P}(E_2)}}$.

Application[edit]

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

${\displaystyle E_1:\mathrm{p}\mathrm{l}\mathrm{a}\mathrm{y}\mathrm{e}\mathrm{r}1\mathrm{w}\mathrm{i}\mathrm{n}\mathrm{s}}$

${\displaystyle E_2:\mathrm{p}\mathrm{l}\mathrm{a}\mathrm{y}\mathrm{e}\mathrm{r}2\mathrm{w}\mathrm{i}\mathrm{n}\mathrm{s}}$

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 ${\displaystyle \mathrm{P}[E_1]}$ and ${\displaystyle \mathrm{P}[E_2]}$ ; 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:

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

${\displaystyle E_2:\text{a 7 is rolled (a loss)}}$

Conditional Probability Formula

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:

Conditional Probability Worksheet

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

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

${\displaystyle \mathrm{P}[\text{first i rolls are ties,}(i+1{)}^{\text{th}}\text{roll is ‘the point’}]=(1-\mathrm{P}[E_1]-\mathrm{P}[E_2]{)}^i\mathrm{P}[E_1]}$

The summation becomes an infinite geometric series:

${\displaystyle \sum _{i=0}^{\mathrm{\infty }}(1-\mathrm{P}[E_1]-\mathrm{P}[E_2]{)}^i\mathrm{P}[E_1]=\mathrm{P}[E_1]\sum _{i=0}^{\mathrm{\infty }}(1-\mathrm{P}[E_1]-\mathrm{P}[E_2]{)}^i}$

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

which agrees with the earlier result.

References[edit]

1. ^Susan Holmes (1998-12-07). 'The Craps principle 10/16'. statweb.stanford.edu. Retrieved 2016-03-17.
2. ^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]

Craps Conditional Probability Definition

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

Craps Conditional Probability Rule