[Rio DiNero]

I have a limited math backround, and don't even know where to start. I have read that it is about 3%, but IDK how that number was calculated. Like to hear what the math guys at SBR think, seems like it would be a very difficult math problem.

[u21c3f6]

Three passes or specifically 3 points which is quite different and a more involved calculation?

Joe.

[u21c3f6]

I have a limited math backround, and don't even know where to start. I have read that it is about 3%, but IDK how that number was calculated. Like to hear what the math guys at SBR think, seems like it would be a very difficult math problem.

Assuming specifically that you throw a point and make that point three times in a row is as follows:

The chance of throwing a point is 24/36 and the chance of making that point is approx 9.75/24 or approx .6667 * .406 = .2707 to throw and make one point. Three times in a row is approx: .2707 ^ 3 = .0198

Joe.

[Rio DiNero]

Three passes or specifically 3 points which is quite different and a more involved calculation?

Joe.

Yes, this is what I should have specified. I want to include 7 and 11 opening roll.
Thanks

[MonkeyF0cker]

Yes, this is what I should have specified. I want to include 7 and 11 opening roll.
Thanks

That would be a combinatorial problem.

Let's call A the probability of winning on the come out roll. B is the probability of winning the point.

A = .2222
B = .2707

What we're looking for are all of the permutations with repetition where the event will take place. To find the number, it would simply be the number of choices (2) to the number of events (3)... Or 2^3 = 8.

Here are the different permutations:

AAA
AAB
ABB
ABA
BAA
BAB
BBA
BBB

To find the probability of each occurring, we simply multiply their respective probabilities together. And to find the whole, we add them up.

AAA = 0.010970645048
AAB = 0.013365227788
ABB = 0.016282480478
ABA = 0.013365227788
BAA = 0.013365227788
BAB = 0.016282480478
BBA = 0.016282480478
BBB = 0.019836487243

Adding them all together, we get 0.119750257089 or 11.975% of winning 3 pass line wagers in a row.

[MadTiger]

Great work, MonkeyF0cker.

[MadTiger]

Assuming specifically that you throw a point and make that point three times in a row is as follows: The chance of throwing a point is 24/36 and the chance of making that point is approx 9.75/24 or approx .6667 * .406 = .2707 to throw and make one point. Three times in a row is approx: .2707 ^ 3 = .0198 Joe.

Thanks for your efforts.

[MonkeyF0cker]

LOL. I actually overthought the problem. Although, it's the same result. You can simply add the two probabilities together and cube them. (.2222 + .2707)^3 = 0.119750257089.

Duh.

[Rio DiNero]

LOL. I actually overthought the problem. Although, it's the same result. You can simply add the two probabilities together and cube them. (.2222 + .2707)^3 = 0.119750257089.

Duh.

Thanks for your insight.