math on this

someone posted these numbers when solving an all bet and other scenarios, how did they arrive at these answers?
if i was trying to solve for rolling a 5 and 9 before a 7, i would do (4/10)*(4/10). 4 is how many ways to make a 5, out of 10 because there are 6 ways to make a 7, same thing for rolling a 9, but i get 0.16 as an answer.

numbers before a P Q true Odds Pays p*pays -Q edge
2,3,4,5,6 7 0.02635391 0.97364609 36.94503358 34 0.89603294 -0.97364609 -0.07761315
all 7 0.005257704 0.994742296 189.1970898 174 0.914840496 -0.994742296 -0.0799018
2,3,4,5 7 0.03019481 0.96980519 32.1182743 30 0.9058443 -0.96980519 -0.06396089
3,4,5,6 7 0.07370718 0.92629282 12.56719929 12 0.88448616 -0.92629282 -0.04180666
2,3,4,5 6 0.04395604 0.95604396 21.75000205 20 0.8791208 -0.95604396 -0.07692316
3,4,5,6 2 0.5361361 0.4638639 0.865198035 0.75 0.402102075 -0.4638639 -0.061761825
2,3,12 7 0.0190476 0.98095238 51.49999738 49 0.93333338 -0.98095238 -0.047619
2,3,11,12 7 0.01147186 0.98852814 86.1698225 80 0.9177488 -0.98852814 -0.07077934
2,4,6,8,10,12 7 0.0135697 0.9864303 72.69359676 69 0.9363093 -0.9864303 -0.050121
5,9 7 0.2285714 0.7714286 3.375000547 3 0.6857142 -0.7714286 -0.0857144
4,6 7 0.2164502 0.7835498 3.620000351 3 0.6493506 -0.7835498 -0.1341992
4,5,6,8,9,10 7 0.06216816 0.93783184 15.08540449 14 0.87035424 -0.93783184 -0.0674776

These can be calculated using the inclusion exclusion method, which is very easy for 5 and 9 before 7. You calculate p(7 before 5) + p(7 before 9) - p(7 before 5 or 9) = 6/10 + 6/10 - 6/14 = 27/35. This is the probability that the 7 will appear before one or the other. The probability that both will appear before the 7 is 1 - 27/35 = 8/35 = 0.228571428.

If you google "fire bet inclusion exclusion," you can find some more info.

appreciate it, helped a lot

Quote: CrystalMath

These can be calculated using the inclusion exclusion method, which is very easy for 5 and 9 before 7. You calculate p(7 before 5) + p(7 before 9) - p(7 before 5 or 9) = 6/10 + 6/10 - 6/14 = 27/35. This is the probability that the 7 will appear before one or the other. The probability that both will appear before the 7 is 1 - 27/35 = 8/35 = 0.228571428.

If you google "fire bet inclusion exclusion," you can find some more info.

How can that be right? 7 has 6 ways to roll, 5 and 9, 4 each. So the 7 appearing "before one or the other" is 8 to 6 AGAINST, not 27/35.

Quote: Joeshlabotnik

How can that be right? 7 has 6 ways to roll, 5 and 9, 4 each. So the 7 appearing "before one or the other" is 8 to 6 AGAINST, not 27/35.

Maybe it's just the wording. To win, you must roll a 5 and 9 before the 7. To lose, you just need the 7 to appear before one or the other, including the following sequences:

7 5 9 - lose
5 7 9 - lose
9 7 5 - lose
In all of these cases, the 7 appears "before one or the other."

Your interpretation only covers the first one (and any similar variation where the 7 comes first).

Quote: CrystalMath

Maybe it's just the wording. To win, you must roll a 5 and 9 before the 7. To lose, you just need the 7 to appear before one or the other, including the following sequences:

7 5 9 - lose
5 7 9 - lose
9 7 5 - lose
In all of these cases, the 7 appears "before one or the other."

Your interpretation only covers the first one (and any similar variation where the 7 comes first).

In the last two cases, "one or the other" DOES, in fact, appear before the 7. What you meant to say, I infer, is that a 7 appears before BOTH a 5 and 9 appear. That would jive with the 25/32 figure.

What I calculated was the probability that a 7 appears before one or both of the 5 and 9.

I think this wording is solid.

ME, how would you do this if you wanted to calculate the same thing but with two 7's, not just one?

ie: the odds of rolling two 7's before two 6's?

Quote: RS

ME, how would you do this if you wanted to calculate the same thing but with two 7's, not just one?

ie: the odds of rolling two 7's before two 6's?

Here's how I would do this specific problem:

Calculate the probability of two 7s before any 6 and the probability of a 6&7 followed by a 7.

p(77) = (6/11)^2
p(76 or 67 followed by 7) = (6/11*5/11*2)*6/11

Total probability = 756/1331

Ah, I didn't even think of that.

And if you wanted to do three 7's before two 6's....you'd do a similar thing?

prob(three 7's before one 6)
prob(7 and 6 followed by two 7's)


So on and so on....?



While I got you here...I think you made a post about calculating ROR recently for single outcomes (win/loss). How would you go about that but with multiple outcomes (ie: roll one die, a 1-3 lose 1 unit, 4 wins 0.5 units, 5 wins 1 unit, 6 wins 2 units)?

Quote: RS

Ah, I didn't even think of that.

And if you wanted to do three 7's before two 6's....you'd do a similar thing?

prob(three 7's before one 6)
prob(7 and 6 followed by two 7's)


So on and so on....?



While I got you here...I think you made a post about calculating ROR recently for single outcomes (win/loss). How would you go about that but with multiple outcomes (ie: roll one die, a 1-3 lose 1 unit, 4 wins 0.5 units, 5 wins 1 unit, 6 wins 2 units)?

For three 7s before two 6s, it's close to what you showed:

p(777) + p(776 in any order, followed by 7)

I didn't post any ROR info for that, but I did post ROR info for DIAD. For that, I just used a simulation.