Craps Binomial Distribution probability question

Suppose 3 of my last 9 established points have won (33.3%). How would I calculate the probability of my cumulative established point bet winning percentage reaching 50% at some point in the near future (next 20 outcomes).

Obviously 3 of my next 3 would get me there but so would 4 of the next 5, 5 of the next 7, 6 of the next 9, etc.

Cumulative probability of multiple events from a binomial distribution that are related but, unless I’m mistaken, neither truly independent or dependent.

Put $1000 on Big Red and $100 on Horn-Hi-Yo. That will give you a 70% chance of profitability

I did some brute-force counting, and got this:

Max # of points	% wins	Exact fractions
6	11.5396	4,704,234,083 / 40,766,034,375
8	15.0445	55,657,525,830,853 / 369,951,761,953,125
10	17.6744	16,021,369,024,356,108,869 / 90,647,430,472,564,453,125
12	19.7130	3,603,656,580,691,310,069,627 / 18,280,565,145,300,498,046,875
14	21.3356	48,265,780,372,572,206,210,149,613 / 226,221,993,673,093,663,330,078,125
16	22.6545	37,672,132,316,993,686,932,253,321,948,903 / 166,290,131,999,249,324,572,357,177,734,375
18	23.7450	358,332,102,004,145,002,992,465,170,863,946,437 / 1,509,082,947,893,187,620,494,141,387,939,453,125
20	24.6596	1,375,861,721,632,210,814,398,551,119,395,808,615,599 / 5,579,415,010,127,313,119,104,728,298,187,255,859,375
22	25.4357	1,877,744,672,433,658,315,086,479,004,967,370,856,629,675,729 / 7,382,319,279,424,802,443,846,634,676,821,994,781,494,140,625
24	26.1009	7,948,261,466,740,790,660,955,176,635,737,518,475,568,579,014,061 / 30,452,067,027,627,310,080,867,368,041,890,728,473,663,330,078,125
26	26.6759	810,915,723,660,380,914,527,989,412,742,862,704,985,977,195,748,188,381 / 3,039,877,591,032,896,228,822,585,014,781,741,969,883,441,925,048,828,125

The first column is the maximum number of points after the first 9; the second is the percentage chance of reaching 50% overall (including your first 9) within that many additional points; the third column is the exact fraction for the second value

This is really helpful and I appreciate it. I ran it for the first few scenarios I mentioned above and got similar results. Is there a calculation that aggregates the results of the next 20 outcomes into 1 percentage, the probability of getting back to 50% at any point in the next 20 outcomes?

It's an odd question

If you've won a third of your points, you probably are ahead with your bankroll, since chances are that you have come out ahead during the rolls that resolved in the come-out. Even if you only rolled points of 4 or 10, you’d come out even, in the points-resolving phase, making a third of them, as they pay 2:1 . You’d have had to have been unlucky in the come-outs that resolved to be losing overall. Edit: Let me change that to "possibly are ahead", see next post

If you reach 50% you’d be killing it just about no matter what, since each free odds bet pays better than even.

of course I assume you would be betting rightside if you’re trying to make the points

After posting that, I realized you need to do better than a third, to expect to be ahead, as the ones that pay 6:5 or 3:2 will get you behind.

Definitely, though, you don't need to hit 50%

what % do you need to win on average , to be ahead making your points? I'll work on that

Quote: Crapsjunkie

This is really helpful and I appreciate it. I ran it for the first few scenarios I mentioned above and got similar results. Is there a calculation that aggregates the results of the next 20 outcomes into 1 percentage, the probability of getting back to 50% at any point in the next 20 outcomes?
link to original post

Those are already aggregates. The 20 row, for example, is the probability of getting back to 50% points won at any point in the next 20 points. It could be anywhere from the next six to the next 20.

I assume that you specifically meant points, and that comeout rolls of 2, 3, 7, 11, and 12 were to be ignored - correct?

My apologies. That’s exactly what I needed then. Thanks for your work and help. I really appreciate it.

for the bets that pay 6:5, it’s fair odds, so if you only win 1/3, you win 6 units on a win 33% of the time vs losing 5 units 67% of the time , which is an EV of -1.33 units

for the bets that pay 3:2, winning 1/3 of the time and getting 3 while 2/3 of the time losing 2 is an EV of -0.333

we can say this shows 1/3 doesn't cut it

You need to win 5/11 for the 6 or 8 point and the prob of rolling those is 10/36 [to break even on making points]

you need to win 2/5 for the 5 or 9 , prob of rolling 8/36

and 1/3 for 4 or 10, prob of rolling 6/36

so it seems to be weighted 10+8+6 = 24 total weight

10/24 win 5/11

8/24 win 2/5

6/24 win 1/3

10/24*5/11+8/24*2/5+6/24*1/3=

0.4060606060606061

or

1 out of 2.463 , roughly 2 out of 5

this shows you have to do better than 1/3, kind of looks right, but I would have to guess is probably not right

but that’s my 2 cents. I used to run this past Mission. I hope you don't feel I hijacked but this became a more interesting question to me

Interesting results as well, OdiousGambit.

ThatDonGuy, would you mind sharing your methodology so that I can use it with other scenarios and thresholds?

Quote: Crapsjunkie

Interesting results as well, OdiousGambit.

ThatDonGuy, would you mind sharing your methodology so that I can use it with other scenarios and thresholds?
link to original post

The "methodology" is simply this:

Let MaxPoints be the maximum number of upcoming points - say, 20
Let p be the probability of making an established point - this is (1/4 x 1/3) + (1/3 x 2/5) + (5/12 x 5/11) = 67/165

Results[] is an array of MaxPoints numbers, all starting at -1; each 1 represents a win, and each -1 represents a loss; note that the first term in the array is Results[0], the second Results[1], and so on

Let WinningProb = 0; this will be the probability of you reaching 50% Wins

Repeat the following until Results[0] > 1:
Let Current = -6, since your previous 9 passes were 3 wins and 6 losses, and you trying to reach the point where they are equal.
Starting with Results[0], add Results[N] to Current until either Current = 0, which means you have reached the 50% Wins point, or you have added all the values through Results[MaxPoints - 1] and Current is still < 0, which means you never got to 50%
Now, if Current = 0, calculate the probability of this particular result:
Let CurrentProb = 1
Check Results[0] through Results[N], where N is the last number you checked; for each 1, multiply CurrentProb by p, and for each -1, multiply CurrentProb by (1 - p)
When this is done, add CurrentProb to Winning Prob
Regardless of whether Current = 0 or not, go to the next possible result:
If N < MaxPoints - 1, then set all Results[N + 1] through Results[MaxPoints - 1] to -1
Add 2 to Results[N]
While N > 0 and Results[N-1] > 1, set Results [N-1] to 1, subtract 1 from N, and add 2 to Results[N]

Since 1/3 * 2/3 = 2/9 of the passline return comes from winning during comeout rolls and 2/3 comes from winning points, then (1/2 - 2/9) * 3/2 = 5/12 of all points must be won to break even.

Quote: odiousgambit

for the bets that pay 6:5, it’s fair odds, so if you only win 1/3, you win 6 units on a win 33% of the time vs losing 5 units 67% of the time , which is an EV of -1.33 units

for the bets that pay 3:2, winning 1/3 of the time and getting 3 while 2/3 of the time losing 2 is an EV of -0.333

we can say this shows 1/3 doesn't cut it

You need to win 5/11 for the 6 or 8 point and the prob of rolling those is 10/36 [to break even on making points]

you need to win 2/5 for the 5 or 9 , prob of rolling 8/36

and 1/3 for 4 or 10, prob of rolling 6/36

so it seems to be weighted 10+8+6 = 24 total weight

10/24 win 5/11

8/24 win 2/5

6/24 win 1/3

10/24*5/11+8/24*2/5+6/24*1/3=

0.4060606060606061

or

1 out of 2.463 , roughly 2 out of 5

this shows you have to do better than 1/3, kind of looks right, but I would have to guess is probably not right

but that’s my 2 cents. I used to run this past Mission. I hope you don't feel I hijacked but this became a more interesting question to me
link to original post

That number (67/165 =~ .406) is the chance of winning a point (aka not sevening out). So as long as you win the expected number of points then you will obviously break even on the odds bet since it pays fair. However you will still lose the EV on the passline bet. You’d need to win more than 67/165 points to overcome that…5/12 with no odds

This is fantastic! Thanks again!