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What is the probability of a straight with no face cards?
My answer:
6C1 x (4C1)^5 - 6C1 x 4C1 and then whole thing divided by 52C5
TA's answer:
6C1 x (4C1)^5 - 5C1 x 4C1 and then whole thing divided by 52C5
Did I do my cases wrong? I'll explain my analysis in case it is confusing.
6C1 : low card of straight from A to 6 (6 possible cards)
4C1^5 : picking suit of each card in straight
-6C1 x 4C1: Subtract straight flush possibilities.
Please explain to me which one is right. FYI, a normal straight probability is:
10C1 x (4C1)^5 - 10C1 x 4C1 and then whole thing divided by 52C5
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It looks to me that you did it right. The 5 in the TA's answer may be from the five cards in a hand, which is irrelevant. For each of the 6 no-face-card-straights, there are 4 ways it can be a flush, as you said.
(although if I was doing the problem, given your wording, I wouldn't subtract anything at all. Isn't a straight flush still a straight?)
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You seem right. Have you asked the TA whether they may have made a mistake/gotten a reason why you are wrong from them?
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Did you ask if the Ace was a "face card" in that definition?
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Even if it is, the first 6 and the second 5 should both be 5 then, because they both mean "the number of straights without a face card".
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Your answer seems correct. When subtracting, you are subtracting a straight flush possibility from each of the straight possibilities; thus you should be able to factor 6C1 from your answer.
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Thank you all. The face card does not include Ace otherwise the first part would be wrong too.
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probably that you will get help on TL: 1/1
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EPT
