#1
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Odds of hitting Broadway...
After the fourth time I was a victim to this happening in the space of about nine SNGs, I tried to figure out what the odds were, and I'm not sure how.
If I have two non-paired cards 10 or higher in my hand, what are the odds of my making the ace-high straight if I see all five cards? |
#2
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Re: Odds of hitting Broadway...
Ok, I think I figured out how to do this, can someone tell me if this looks right?
Assume, without loss of generality (since any two non-paired 10+ cards should be as good as any other), I have JT, and my opponent has no 10+ cards (say, a lower pair). I need an ace, a king, and a queen. There are 4 ways to get the ace, out of 50 cards. After that, there are 4 ways to get the king, out of 49, and 4 ways to get the queen, out of 48. And then two cards that can be anything. Then multiply by 5! (= 120) since I don't care about the order. So (4/50) * (4/49) * (4/48) * 120 = 64/117600 * 120 = 0.065. If instead of a lower pair, my opponent had, say, AK, it would be: (3/50) * (3/49) * (4/48) * 120 = 36/117600 * 120 = 0.037. These seem like the right magnitude to me (unlike the numbers I was getting before, that were in the 35% range), am I making any major mistakes here? |
#3
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Re: Odds of hitting Broadway...
I believe you're double counting:
[C(4,1) * C(4,1) * C(4,1) * C(47,2)]/C(50,5) = 3.265% If your opponent held A-K, I think it should be: [C(4,1) * C(3,1) * C(3,1) * C(45,2)]/C(48,5) = 2.08% Although that would include the times you lose to a boat/quads. |
#4
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Re: Odds of hitting Broadway...
Can someone explain why it is C(4,1) rather than C(16,1)on the first card.
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#5
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Re: Odds of hitting Broadway...
I'm disregarding order. You need the board to contain an A, K and Q. There's 4 Aces, 4 King and 4 Queens which, by themselves can come a total of 64 ways (different suit combinations).
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#6
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Re: Odds of hitting Broadway...
Ohh yea i forgot about that.
Thanks |
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