[jba]

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What are the odds of being dealt AA three out of ten hands?
Sorry for being a moron, and thanks in advance for the replies.

PS Please show your work!

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I assume you mean exactly three times, no more no less?

there's a couple steps to this.

1 - we need to know the odds of getting AA in one hand:

we find this by finding all combos of AA, and divide by all possible hands (all combos of 52 cards):

C(4,2)/C(52,2)

which is the same as :

(4*3/2!)/(52*51/2!)

6/1326, or 1/221


2 - we need to know the odds of not getting AA:

ok easy enough 220/221

3 - to get AA three times, then non-AA seven times, we say:

1/221*1/221*1/221*220/221*220/221*220/221*220/221*220/221*220/221*220/221
=8.97503248e-8 = 1/11,142,021

4 - but it's easier than that, cause we can also get non-AA the first seven times then AA the last three times, or AA/non-AA/AA/non-AA/AA/non-AA/......etc. I think the number of possible combos here can be written as 10!/(7!*3!) = 120. (If someone reading this can confirm, I'm a little shaky on this point.)


now we know there are 120 orders in which we can achieve the goal, so the final answer is:

120/11,142,021 = 1/92,850