What are the odds? - Stat geeks help!

[DanMcS]

What's the equation for at least "r" number of successes, given the probability "p" for success and "N" number of tries? I don't have a good text book on hand: I was using my memory (from 6 years ago in 11th grade) and Google for reference.
-George

Buh, um, the probability for at least 1 success is 1 - (the probability of no successes), which is (1-p)^N, so in total it's 1 - [(1-p)^N]. The probability for at least r successes, um, dunno that one, but I can probably figure it out.

Hmm.

Google had it.

It's (the probability of r successes) + (the probability of r+1 sux) plus (the probability of r+2 sux) + ... + the probability of N successes, basically.

The probability of any number r successes from N trials is:

Choose(N,r) * p^r * (1-p)^(N-r).

For instance, if you're flipping coins, the odds of getting 5 heads out of 7 tries is:

7 choose 5 = 7!/(5!*2!) = 21
p = .5, r=5
p^r is .5^5
(1-p)^(N-r) is .5^2.

Total probability is 21 * .5^7 = .1640625.

The choose(N,r) accounts for the fact that you don't care what order the successes come; 5 successes then two failures is equivalent to 2 failures then 5 successes.

 

[Zoatebix]

Hooray!

Your google-fu is indeed strong. Thank you very much. I'm subscribing to this thread so I can use it as a reference until I find an old textbook or two (or start taking math classes again...)

 

[Majere]

Here are my answers (with working !)

1) What are the odds of me rolling a single natural 20 out of the four dice I roll for a normal round?
P(Not rolling a 20 if I roll 1d20)= 19/20
P(Not 4d20 not rolling a 20) = (19/20)^4 = 0.81450625
P(At least 1 success) = 1- P(0 sucessess)= 0.18549

2) What are the odds of rolling a single natural 20 out of the six dice I roll for a full-round action?
P(0 sucessess) = (19/20)^6= 0.73509
P(At leas 1 success) = 1-P(n=0(6))= 0.2648908

3) What are the odds of rolling 4 natural 20's out of the 4 dice I roll for a normal round?
P(n=4(6))= (1/20)^5 = 1 in 160000

4) What are the odds of rolling 6 natural 20's out of the 6 dice I roll for a full-round action?
P(n=6(6)) = (1/20)^6= 1 in 64x10^6

Crit range now 18-20
P(0(1)) =17/20

1) P(n>0(4))= 1-P(n=0(6))= 0.47799
2) P(n>0(6))= 1-P(n=0(6))= 0.622850
3) P(n=4(4))=(3/20)^4 = 81 in 160000 = 0.00050625
4) P(n=6(6))=(3/20)^6 = 729 in 64million= 0.00001139

 

[Ferret]

Buh, um, the probability for at least 1 success is 1 - (the probability of no successes), which is (1-p)^N, so in total it's 1 - [(1-p)^N]. The probability for at least r successes, um, dunno that one, but I can probably figure it out.

Hmm.

Google had it.

It's (the probability of r successes) + (the probability of r+1 sux) plus (the probability of r+2 sux) + ... + the probability of N successes, basically.

The probability of any number r successes from N trials is:

Choose(N,r) * p^r * (1-p)^(N-r).

For instance, if you're flipping coins, the odds of getting 5 heads out of 7 tries is:

7 choose 5 = 7!/(5!*2!) = 21
p = .5, r=5
p^r is .5^5
(1-p)^(N-r) is .5^2.

Total probability is 21 * .5^7 = .1640625.

The choose(N,r) accounts for the fact that you don't care what order the successes come; 5 successes then two failures is equivalent to 2 failures then 5 successes.

How does that work (Just curious)? I would have PMed or emailed but I couldn't. How would you add the extra requisite of not getting one number?

 

[DanMcS]

How does that work (Just curious)? I would have PMed or emailed but I couldn't. How would you add the extra requisite of not getting one number?

The probability of getting at least X successes on a given number of trials, Y, is the same as the probability of getting X successes, X+1, ...., up to Y successes.

For instance, the probability of rolling at least 2 20s on 4d20 is the probability of rolling 2 20s plus the probability of rolling 3 20s plus the probability of rolling 4 20s; 4 and 3 both satisfy "at least" 2.

The probability of rolling 4 20s on 4d20 is easy, it's 1 in 20^4.

The probability of rolling 2 20s on 4d20 is like so:
You have 4 individual trials, A, B, C, and D.
You need two of them to be 20, and 2 to be not(20), or 1-19.

Prob of 2 being 20 is (.5)^2. Prob of 2 being 1-19 is (.95)^2

There are a number of ways you can get this result. A and B might be 20, A and C, A and D, B and C, B and D, or C and D.

Turns out, the number of ways of doing this is the number of ways you can choose 2 individual objects from 4 choices, that is, 4 choose 2.

Choosing is a well-defined formula, with J choose K being
J!/ (K!*(J-K)!). [! is the factorial function]

That is, the number of ways you can choose 2 things from 4 things is
4! / 2!*2!, or 6.

So your probability of 2 successes on 4 trials is .5^2 * .95^2 * 6.

The formula is general. The probability of exactly X successes on Y trials is always:

P(success)^X * P(failure)^(Y-X) * Y!/ (X! * (Y-X)!)

So the probability of 3 20s on 4d20 is
.5^3 * .95^1 * 4!/(3!*1!)

Then you add them all up.

I'm not sure I understand the last question; what do you mean 'the extra requisite of not getting one number?' A specific example would help; do you mean something like, the odds of rolling 2 20s on 4d20 and not rolling any 1s?