If you assume the other players have any 2 cards with equal probability given the betting patters (probably not a great assumption!) then the easiest way of finding the probability that someone has a higher spade is by finding the probability that nobody has a higher spade first.

This is (41/45)*(40/44)*(39/43)*(38/42)*(37/41)*(36/40) which is roughly 0.55,

so the probability someone DOES have a higher spade is 1 - 0.55 = 0.45.

There are 6 cards in your opponents' hands, each of which is one of the 45 remaining cards in the deck. The probability the first is not one of the 4 higher spades is (45-4)/45 = 41/45. The probability that neither the first or the second are higher spades is P(first is not a high spade)*P(second is not a high spade GIVEN that the first is not a high spade). If we know that the first is not a high spade, then the chance that the second is not a high spade is 40/44 (since one card is now accounted for by the first card). If you continue this process you get the formula above.

You should never be adding probabilities in situations like this...if you have a 1% chance of winning something and you try it twice, the chance that you win at least once is NOT 2%!!! It is 1-(1-.01)*(1-.01) = .0199. The difference here is negligible but if you have two coinflips, the probability you win one is not .5+.5 = 1 but .75 (even though it may seem like .35 sometimes

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