How is the side bet ride the line calculated?

I Sort of understand what to do, the probrability of losing is 2×((3/32) × (6/9))+ 2× ((4/32) × (6/10)) + 2×((5/32) × (6/11))=49/110

The probrabilty of winning using the Ride the line Rules is 6/32+2/32+(3/32*1/3)+(4/32*2/5)+(5/32*5/11)+(5/32*5/11)+(4/32*2/5)+(3/32*1/3)=61/110

Im just confused how to plug all that into the return tables. Why is 1 win exactly a 24.7% chance? Does that mean out of all the possible winning combination that is 24 % of them, for instance the probability of exactly 1 win is 61/110, the probrability of exactly 2 wins is (61/110)^2 and after adding up everything 1 win is 24.7% of everything but 0 wins? Or is that wrong?


0 0.445454545
1 0.247024793
2 0.136986476
3 0.075965228
4 0.042126172
5 0.023360877
6 0.012954668
7 0.007183952
8 0.003983828
9 0.002209214
10 0.001225109
11 0.001525136

It means that, from any particular starting point, the probability of getting 0 pass line wins before sevening out is 0.445454545, the probability of getting exactly 1 pass line win before sevening out is 0.247024793, the probability of getting exactly 2 is 0.136986476, and so on.

Multiply the probability of N wins by what you get if you get N wins, and add up all of the values for all possible Ns to get the expected return.

So say if i got only 2 wins and the pay was -1 i would get a return of -0.136986. If it hit 5 wins and i got paid 4 i would get a return of 0.093443508. What i don't understand is why exactly is getting 1 pass win 0.247024793? I do know what all from 1-11 wins equals 61/110, but that also is the same answer as just doing 6/32+2/32+(3/32*1/3)+(4/32*2/5)+(5/32*5/11)+(5/32*5/11)+(4/32*2/5)+(3/32*1/3)=61/110, 61/110^2 is 0.307520 which would be the odds of winning 2 passlines in a row, but obviously it isn't for this and im stumped.

I had some fun creating a version of craps that uses X amount of sides, say d10 d8 ect. I need to know the formula to plug it in from d6 craps.

Quote: RaoulDukem

So say if i got only 2 wins and the pay was -1 i would get a return of -0.136986. If it hit 5 wins and i got paid 4 i would get a return of 0.093443508. What i don't understand is why exactly is getting 1 pass win 0.247024793?
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The probability of getting exactly one pass win = the probability that the first comeout will result in a pass win multiplied by the probability that the second comeout will result in a seven-out, ending the bet. (61/110)^2 is the probability of getting two or more, but in order to calculate the return properly, you need to know how often you will get each particular number of wins before a loss.

i tried 0.445454545454*0.554545454545454=0.24702479 For 1 point
2 Points 0.44545454545454545454545454*0.30752066=0.136986475 so i think i got it. Yay :)

I believe your formula for x total wins is (61/110)^x * 49/110

Quote: RaoulDukem

...Why is 1 win exactly a 24.7% chance? Does that mean out of all the possible winning combination that is 24 % of them...

Assume the chances of winning is p and losing is q (q=1-p). In this case we ignore come outs of 2,3 or 12. A win is either a natural (7 or 11) or making the point, a loss is failing to make the point. So p = 488/880 and q = 392/880 (normally it's 488 and 502 from 990, but 110 of those are Craps).

To get no wins the shooter just has to lose the first time, so probability = q.
To get exactly one win the shooter has to win the first and then lose the second, so probability = p q.
To get exactly n wins the shoort has to win the first n and then lose the one after that, so probability = pⁿ q.
The get 11 or more wins the shooter has to win the first 11 and it now doesn't matter what happens, so probability = p¹¹.

Plug in appropriate p and q to get the figures quoted. Typically one win followed by one loss, for an even money bet, will be slightly less than 1/4.