Calculating 2 AhAd in a row in the next 3 hands?
 

I know how to calculate the chances of getting AhAd in the next 2 hands: 1 / (1/1326^2). Can someone tell how to calculate the chances of getting at least 2 AhAd in the next 3 hands? How about 4, 5 and 10?


Thanks

Re: Calculating 2 AhAd in a row in the next 3 hands?
 

There are three ways of getting AhAd in exactly two out your next three hands: 1st+2nd, 1st+3rd, and 2nd+3rd. Plus you may also get AhAd in all three of your next three hands.

So the probability of getting AhAd in at least two of your next three hands is:

(1/1326)^3 + 3*(1/1326)^2*(1325/1326)=0.0000017, or 587,000-to-1 against.

For a quick approximation, since 1/1326 is so small, you can safely neglect all terms but the lowest-degree, and pretend that 1325/1326 is equal to 1.

For 2-out-of-4, it's:

(1/1326)^4 + 4*(1/1326)^3*(1325/1326) + 6*(1/1326)^2*(1325/1326)^2 = 0.0000034, or about 293,000-to-1 against.

In general, if p is the probability of something happening (in this case 1/1326), and you want to know the chances of that "something" happening at least r times in the next n tries, the "quick approximation" is:

C(n,r)*p^r

(this only works when p is small)

2-out-of-5: 10*(1/1326)^2 = 0.0000057; 176,000-to-1 against.
2-out-of-10: 45*(1/1326)^2 = 0.000026; 39,000-to-1 against.

where C(n,r) is the number of ways of choosing r items out n total.

Of course, if you were interested in any pocket Aces, not just red pocket Aces, the probabilities would be about 36 times larger.

The full answer in the general case is:

C(n,n)*p^n + C(n,n-1)*p^(n-1)*(1-p) + C(n,n-2)*p^(n-2)*(1-p)^2 + ... + C(n,r)*p^r*(1-p)^(n-r).