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A friend and I were playing Cribbage the other day and, at one point, my hand had a 6 count in the form of 15-2, 15-4 and a pair for 6. My friend then counted their hand, and had exactly the same count. Then I counted my crib (I was the dealer), and found that the crib also had exactly the same count. We were wondering what the odds of that happening are or is there a formula that we could use to predict other possible hand odds.

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    $\begingroup$ These kinds of weird personal coincidence problems aren't really amenable to a rigorous answer. Yes, one can work out the probability of some hand or combination of hands, but the post hoc decision that some or other hand is remarkable is not controlled by that analysis. $\endgroup$ – Sycorax Sep 22 '17 at 22:44
  • $\begingroup$ Thanks for the reply, and I was not surprised at the reply. Just thought that someone with more knowledge regarding calculating odds might have a better idea, $\endgroup$ – Russel Bairstow Sep 24 '17 at 15:12
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    $\begingroup$ Possible duplicate of Determining statistical probability related to a string event of dates $\endgroup$ – kjetil b halvorsen Sep 9 '18 at 7:24
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    $\begingroup$ @Sycorax Do you want to turn that comment into an answer? $\endgroup$ – Peter Flom Sep 9 '18 at 11:25
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    $\begingroup$ I do not see why people think this is a duplicate as the OP explicitly asks for a general formula for predicting hand odds. I think it is similar to asking what is the probability of being dealt a straight flush in poker or all 13 of the same suit at whist. $\endgroup$ – mdewey Sep 20 '18 at 16:14
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These kinds of weird personal coincidence problems aren't really amenable to a rigorous answer. Yes, one can work out the probability of some hand or combination of hands, but the post hoc decision that some or other hand is remarkable is not controlled by that analysis.

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  • $\begingroup$ I am not sure I agree. There is only a finite number of ways of distributing 12 cards from 52 and then only a finite number of ways each player can put two into the crib so in theory it must be possible to calculate the probability of both players and the crib scoring six (or both getting 15-2, 15-4 and a pair if that is what is meant). The OP does ask for a general formula implying he is not just interested in this coincidence. $\endgroup$ – mdewey Sep 19 '18 at 16:13

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