# Winning game of craps? [closed]

This is the suggested solution, I could understand how all the P(E)s are obtained, but I don't understand the coefficients of P(I)s...(0,0,1/3,4/9,5/9...etc)

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## closed as not a real question by Henning Makholm, Ayman Hourieh, William, Ｊ. Ｍ., NorbertOct 3 '12 at 20:11

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sorry, im using images b/c im not sure how to use the proper notation in the editor, i looked through advanced help, but still nothing...it takes me to mathjax and provided a download link of mathjax, but there's no usage on the page... –  user133466 Sep 19 '12 at 19:55
For MathJax help, consult the following threads on math.meta: Do we have an equation editing HOWTO?; where is the latex reference please; MathJax basic tutorial and quick reference. There is a bit of a learning curve, but others will be respectful for just for making the attempt. –  Arthur Fischer Sep 19 '12 at 20:10
@ArthurFischer Thank you! The MathJax basic tutorial and quick reference was particularly useful! –  user133466 Sep 21 '12 at 21:58

I agree with you. The calculation of $P(E_5)$, the chance that given you start with a $5$ you finally win is $\frac 25$ because on each roll you have $\frac 4{36}$ chance of rolling $5$, $\frac 6{36}$ of rolling a $7$ and $\frac {26}{36}$ of something else. $P(E_5)$ then equals $\frac 4{4+6}=\frac 25$. $P(E_7)=1$ because you have already won. The definition of the $P(I)$'s is confusing and you should just have $P(W)=\frac 1{36}P(E_2)+\frac 2{36}P(E_3)+\ldots \frac 1{36}P(E_{12}) \\ =\frac 1{36}0+\frac2{36}0+\frac 3{36}\frac 13+\frac 4{36}\frac 25+\frac 5{36}\frac 5{11}+\frac 6{36}1+\frac 5{36}\frac 5{11}+\frac 4{36}\frac 25+\frac 3{36}\frac 13+\frac 2{36}1+\frac 1{36}0$, which duly gives $\frac {244}{495}$

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made effort.... –  user133466 Sep 19 '12 at 20:06
why do we have $\frac 1{36}$0, $\frac 2{36}$0, and $\frac 1{36}$0 (the last term). Those are the instances where we Win, shouldn't we add those fractions to the total probability of Winning? –  user133466 Sep 19 '12 at 20:37
@user133466: the first term is throwing a $2$ on the first roll, the second a $3$ and the last a $12$. They are immediate losses. The ones with $1$ multiplying the fraction are $7$ and $11$, immediate wins. –  Ross Millikan Sep 19 '12 at 20:39
you are right, my mistake –  user133466 Sep 19 '12 at 20:40
$P(W)$ is the probability of winning based on your pass-line bet.
Now, let's say that the come-out roll was a 4. You're now looking to compute the probability that another 4 comes up before a 7. That probability is $\frac{1}{3}$. That's where those coefficients come from.