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LennySkutnik
Joined: 11 Feb 2008
Posts: 4
Location: B&M, PS
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 Posted: Mon Feb 11, 2008 3:59 am Post subject: A9 vs K9, AK9 rainbow flop, probabilities question |
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The chance that K9 will catch a draw and beat A9 is 8.38%. How do I make this calculation by hand.
I can come up with the prob. that K9 misses his draw,
(45/47)*(44/46)= 91.58%
therefore the complementary probability that he hits his draw is 1-91.58 = 8.42%, which is close but not quite the answer (due to the joint prob. that A9 also makes A when K9 makes K).
Please show me exactly how to make this calculation by hand. Thanks. |
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ciaran
ITH Support
Joined: 10 Sep 2004
Posts: 4781
Location: Alpharetta, GA
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| Posted: Mon Feb 11, 2008 5:28 am Post subject: |
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You need the chance that he hits the K on one of the two remaining cards, but doesn't hit the A on the other one. BTW, you know 7 cards, so you're working from 45 remaining on the turn and 44 on the river.
2/45*42/44 is the chance he gets there on the turn and doesn't get rivered. 4.24%
41/45*2/44 is the chance that the turn is a blank and that he hits the river. 4.14%
Summed, that's 8.38%. |
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ciaran
ITH Support
Joined: 10 Sep 2004
Posts: 4781
Location: Alpharetta, GA
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| Posted: Mon Feb 11, 2008 5:38 am Post subject: |
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And, for thoroughness:
43/45*42/44 is 91.21% chance he misses altogether, leaving 8.79% that he hits something.
2/45*2/44*2 is the chance that he hits a K on one of the two remaining cards and an A on the other (it's the same probability twice because it doesn't matter which comes on the turn and which on the river for our purposes), and that's the .4% remainder between the 8.79% chance that he hits something and the 8.38% that he outdraws (ignoring the .01% that I'm off due to rounding errors). |
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LennySkutnik
Joined: 11 Feb 2008
Posts: 4
Location: B&M, PS
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| Posted: Mon Feb 11, 2008 6:27 pm Post subject: |
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| ciaran wrote: | You need the chance that he hits the K on one of the two remaining cards, but doesn't hit the A on the other one. BTW, you know 7 cards, so you're working from 45 remaining on the turn and 44 on the river.
Event A 2/45*42/44 is the chance he gets there on the turn and doesn't get rivered. 4.24%
Event B 41/45*2/44 is the chance that the turn is a blank and that he hits the river. 4.14%
Summed, that's 8.38%. |
Major brain fart. I forgot that since his cards are also known you can remove them from the denominator.
Why aren't you factoring in K9 hitting on both the turn and the river for quads?
Event C 2/45*1/44 = 0.101% Do you then have to figure in the joint prob of Event A and Event C? |
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LennySkutnik
Joined: 11 Feb 2008
Posts: 4
Location: B&M, PS
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| Posted: Mon Feb 11, 2008 6:27 pm Post subject: |
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| ciaran wrote: | You need the chance that he hits the K on one of the two remaining cards, but doesn't hit the A on the other one. BTW, you know 7 cards, so you're working from 45 remaining on the turn and 44 on the river.
Event A 2/45*42/44 is the chance he gets there on the turn and doesn't get rivered. 4.24%
Event B 41/45*2/44 is the chance that the turn is a blank and that he hits the river. 4.14%
Summed, that's 8.38%. |
Major brain fart. I forgot that since his cards are also known you can remove them from the denominator.
Why aren't you factoring in K9 hitting on both the turn and the river for quads?
Event C 2/45*1/44 = 0.101% Do you then have to figure in the joint prob of Event A and Event C? |
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ciaran
ITH Support
Joined: 10 Sep 2004
Posts: 4781
Location: Alpharetta, GA
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| Posted: Mon Feb 11, 2008 6:54 pm Post subject: |
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| LennySkutnik wrote: | Why aren't you factoring in K9 hitting on both the turn and the river for quads?
Event C 2/45*1/44 = 0.101% Do you then have to figure in the joint prob of Event A and Event C? |
I am, we're checking for the K on the turn and "not the A" on the river (which includes the case you note where the case K hits). It's just that it doesn't matter if he catches the K on both streets, so it doesn't need to be explicitly separated out. |
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LennySkutnik
Joined: 11 Feb 2008
Posts: 4
Location: B&M, PS
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| Posted: Mon Feb 11, 2008 7:21 pm Post subject: |
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| ciaran wrote: | | LennySkutnik wrote: | Why aren't you factoring in K9 hitting on both the turn and the river for quads?
Event C 2/45*1/44 = 0.101% Do you then have to figure in the joint prob of Event A and Event C? |
I am, we're checking for the K on the turn and "not the A" on the river (which includes the case you note where the case K hits). . |
Double-doh! brain fart. Thanks! |
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