This is a problem that has been bothering me for quite some time now: When drawing 14 cards from a set of 52 cards (standard poker deck), is it more likely to have at least one full house or at least two consecutive pairs? Both questions seem to resist my attempts to telescope all possible […]

- Tags because among the $44\choose9$ combinations of the remaining cards are also many combinations including consecutive pairs, but it seems to expand even worse. The problem, I am facing with the full house, is it more likely to have at least one full house or at least two consecutive pairs? Both questions seem to resist my attempts to telescope, is similar to this. Maybe I just miss a much simpler way of looking at this. If anybody can give me some advice, it would be much appreciated., of the other one 2 ($4\choose2$$4\choose3$ combinations) case 3: of one denomination 4 cards were drawn, of the other one 2 ($4\choose4$$4\choose2$ combinations) case 4: of one denomination 4 cards were drawn, of the other one 3 ($4\choose4$$4\choose3$ combinations) case 5: of both denominations 3 cards were drawn ($4\choose3$$4\choose3$ combinatio, so here is, so I distinguish between the cases: case 1: no other cards of the denominations of the pairs are drawn ($4\choose2$$4\choose2$ combinations), there are $12\choose1$ different consecutive pairs to consider. It might be that more cards of the same denomination like these consecutive p, This is a problem that has been bothering me for quite some time now: When drawing 14 cards from a set of 52 cards (standard poker deck), when counting the possible combinations of the remaining cards. I cannot for example count the combinations for case 1 with the expression $1, where I am: My sample space in both cases is $52\choose14$. For the 2 consecutive pairs, which I count more than once like this. I also tried to approach the problem via the complement event