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In the card game Pitch using 1 deck with 52 cards and 4 people playing and six cards are dealt to each person and I have a partner and I have a King and wish to bid in the suit the King is in using the King as high what is the percentage my opponents have the Ace in the same suit as my King out of their 12 cards?

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2 Answers 2

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Assuming you have only 1 king to chose from, and you don't have the matching ace in your hand: you have in your hand 6 cards that aren't the ace. So there are 46 cards left in the game, amongst which only one is the ace of interest.

Your opponents each have 6 cards, so in total they have 12 of those 46 cards.

One way to handle the real question (especially when you're unsure whether or not the draws are independent: does it matter or not that the second card dealt to one of your opponents comes from a smaller stack?) is to look at the negation of it: it will be easier to assess what the probability is that none o f the 12 cards of your opponents are the ace in question. We consider here probability as: the number of valid situations among the possible situations.

Now, the possible situations are all possible draws of 12 cards from 46 cards. There are C(12,46) of those. Which ones of those are 'valid'? Exactly those draws that look as if they came from a stack of cards that didn't hold the ace! That is: all the draws we can make from a stack of 45 (non-ace) cards! Similarly as before, there are C(12, 45) of those.

Since $C(a,b)=\frac{b!}{a!(b-a)!}$, you can work the probability out to: $$\frac{C(12,45)}{C(12,46}=\frac{17}{23}$$ so, finally, the probability of not having this happen is $1-\frac{17}{23}=6/23$.

This is exactly what we obtain if we do assume that the draws were independent. To demonstrate, wo we look at the problem in another light: The 12 cards are drawn and handed face down to your opponent (this is what makes it easier to see the 'independence' of the draws). After that, the cards (still all face down) are all numbered: the cards your opponents were dealt get numbers 1 to 12, the rest of the 46 cards get numbers 13 up to 46.

Now, we look again at probability as number of valid situations among the possible situations: the ace is going to be one of the 46 cards, so you can look at it as if you were to assign a random number from 1 to 46 to the ace. Now, there are 46 possible situations. Of these, only 12 are going to give the ace to your opponents (namely if the random number was between 1 and 12).

Obviously, this gives, again: $\frac{12}{46}=\frac{6}{23}$.

I love the smell of probability in the morning. Here's hoping I haven't just made your homework.

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They have 12 cards, there are 46 unseen cards... I like 12/46 = 26%.

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