Calculating probablity of being dealt particular Omaha hand

[I dunno]

What is the easiest way of calculating the probability of being dealt A3QQ, suits irrelevant?

[maddis]

step 1: the chance of getting an ace as your first card is 4/52. 4/51 that you get a 3 next. 4/50 that you get a Q as your third card and 3/49 that you get a Q as your forth card. multiplicating this gives you the probability to get A3QQ. BUT it gives you only the probability to get it in this order.

step 2: think of the possible ways that this hand can be dealt to you: A3QQ, 3AQQ, AQ3Q, 3QAQ, QA3Q, Q3AQ, QAQ3, Q3QA, QQA3, QQ3A, AQQ3, 3QQA. = 12 possibilies
puh i hope this is all. and i hope that you dont have to differantiate both Queens, but i think not. (feedback plz!)
you have to multiplicate
12 x (4/52 * 4/51 * 4/50 * 3/49) and you got it! i hope [img]/images/graemlins/smile.gif[/img] again: feedback please if this is correct

[bachfan]

There are four ways to get dealt an ace, four ways to get dealt a three, and six ways to get dealt two queens.

So 4 * 4 * 6 = 96 A3QQ hands in existence.

There are 52C4 total omaha hands (look up combinatorics for more) = 270725. (<--- save this number for future such computations).

so 96 hands out of 270725 = 0.035%

Cheers,
bachfan