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#1
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Details on Two Statements by Sklansky
1. In TOP he says odds of flopping a set are "about 8/1". What is the math for this?
2. In HEFAP he provides a table of probabilites for completing a hand by the river. But he doesn't provide the math. Example: For 9 outs the probability of completing the hand is 35%. Again, what is the math for this? |
#2
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Re: Details on Two Statements by Sklansky
For making a set (or quads) on the flop the easiest way is to calculate the odds of not making it and subtracting from 1. Of the 50 cards in the deck other than your pair, 48 will not make you a set so the probability of not making a set or quads are (48/50)*(47/49)*(46/48) = .882 so the odds of making it are 1 - .882 = .118. To convert to odds .882/.118 = 7.5:1 against.
For 9 outs we can solve the same way. With 2 cards to come there are 47 possible cards, 9 of which makes your hand, so the probability of not making your hand is (38/47)*(37/46) = .65 so the probability of making it is 1-.65 = .35 which is 1.86:1 against. |
#3
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Re: Details on Two Statements by Sklansky
For making a set (or quads)...
You calculation also includes full houses. Lost Wages |
#4
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Re: Details on Two Statements by Sklansky
You calculation also includes full houses.
So it does. I guess I should have said flopping trips or better (was ignoring impact of cards which don't match your hole pair in rank). |
#5
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Re: Details on Two Statements by Sklansky
Simple. Elegant. Thank you.
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#6
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Re: Details on Two Statements by Sklansky
For 9 outs: There are 47 possible cards, 9 of which makes your hand, so the probability of not making your hand is (38/47)*(37/46) = .65 so the probability of making it is 1-.65 = .35"
This can be generalized to 1-((47-A17)/47)*((46-A17)/46) for N outs. |
#7
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Re: Details on Two Statements by Sklansky
1) Odds of flopping a set (or quads):
C(m,n) = the number of combinations of m objects chosen n at a time. For almost all purposes, m>=n and m and n are nonnegative integers. E.g., C(52,5) = number of draw poker hands and C(52,2) = number of holdem hands. It is calculated by the formula C(m,n) = (m!)/[((m-n)!)(n!)] where k! denotes the product 1 x 2 x...x k. If you have a pocket pair, there are now only 50 cards to choose from and three are chosen at random to make up the flop; therefore, there are C(50,3) = 19600 possible flops. Quads: 48 possible combinations (choose the two case cards that match the rank and any other) Exactly trips: 2 (matching rank) x C(48,2) (any two of the other) = 2 x 1128 = 2256 Thus, the probability of making exactly trips is 2256/19600 or about 0.1151020 or 7 97/141 to 1 against or about 7.687943 to 1 against. Including quads, the proability increases to 2304/19600 or about 0.1175510 or 7 73/144 to 1 against or about 7.506944 to 1 against which is about 7.5 to 1 against. 2) Probability of completing a hand with k outs where k>=1. Actually, the table could have been extended from k=1 to 23 to be complete. Since there are 47 unknown cards, and if k were bigger than 23, just take j=47-k and look the number for j on the table and subtract the percentage from 100%. The simple reason is you are just looking at outs for the competition. And of course, for k=0, the answer is trivial! And in LHE, you only need a very small portion of this table! One easy way to compute the table is consider the chances of missing instead. There are 47 unseen cards and therefore C(47,2)=1081 possible combinations of turn and river cards. If there are k "key cards" to make the hand, there are then 47-k cards that miss. There are simply C(47-k,2) such combinations of two "misses". Thus, the probability of a miss is C(47-k,2)/C(47,2) and obviously if this is subtracted from 1, the probability of hitting is found. Strangely, in my edition of HPFAP, the entry for 1 out is 4.4% but a calculation shows C(46,2)/C(47,2) = 0.9574468 so that the last entry should read 4.3% instead. All the entries for the common situations are correct. |
#8
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Includes full houses!
Of course, as previously mentioned, the numbers for "exactly
trips" also include full houses! |
#9
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Re: Details on Two Statements by Sklansky
Thank you for the tutorial. I now am armed to do the work I need.
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#10
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Re: Details on Two Statements by Sklansky
Both of these questions are discussed in detail in the archives.
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