Two Plus Two Older Archives Am I making this AK calculation correctly
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#11
11-24-2002, 02:00 PM
 marbles Senior Member Join Date: Oct 2002 Location: Wauwatosa, WI Posts: 568
your intuition has gotten the best of you!

Dynasty,

Another angle to consider:

Consider this as a pass/fail test with 18 trials (for the 18 cards dealt to everyone but you). A success is an ace or king, a failure is anything else. Here's how the probabilities would break out:

P(zero): .057
P(one): .228
P(two): .346
P(three): .255
P(four): .096
P(five): .017
P(six): .001

So yes, the most likely result is two ace/kings being dealt out when you hold AK against 9 opponents. But notice that when you take the weighted average of the 7 possible results, the expected value is actually 2.16 successes. If exactly two opponents hold ace and/or king, that's actually LESS than you should expect. There are actually an extra 0.16 ace/kings left in the deck for the flop.

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