#1
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Flopping FH and quads
So total bad beat but interested in knowing odds of this event happening (1/xxxxxx hands).
4 handed game 1) Probability that exactly 2 players are dealt pocket pairs. 2) Assuming both see flop- probability one flops FH and other flops quads. 3) Probability the person with the FH loses his stack (I can answer this one- 100%) I had AA and flop was A88. Thought I was good until the dude turned over 88. Can you show your formulas so I can start to learn this stuff and wont have to keep asking you guys to answer my random questions. Thanks |
#2
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Re: Flopping FH and quads
The approximate probability that exactly two players are dealt distinct pocket pairs at a 4 handed table is :
4c2*(78/1326)^2*(1248/1326)^2 ~ 0.01839 or about twice in 100 deals . 2)Given that exactly two players are dealt distinct pocket pairs , the probability one flops a fh and another flops quads is : 2/48c3 ~ 0.000115633 or about 1 in 10,000 . If we combine 1 and 2 (by multiplying ), then the approximate probability that a fh loses to quads is : 0.0000021265 or about 2 in 1 million deals or equivalently about 1 in 500,000 deals. 3) You should be willing to go broke with it unless maybe you have 500,000 in front of you [img]/images/graemlins/smile.gif[/img] |
#3
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Re: Flopping FH and quads
Thanks- I thought it would be pretty unlikely.
Can you explain your equations 4c2*(78/1326)^2*(1248/1326)^2 -understand 4c2 (where 4=n and n is # players at table) but where do the other two parts come from. 2/48c3 -how did you come up with this? understand 48 unknown cards and combinations of 3. Thanks |
#4
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Re: Flopping FH and quads
I will solve the problem for when it is given that you hold a pocket pair .
1) The approximate probability that exactly one other player holds a distinct pocket pair is: 3c1*(72/1225)*(1153/1225)^2 ~ 0.1562 or about 15 in 100 deals . 2)0.1562*2/48c3 ~ 0.000018 or about 2 in 100 000 deals . |
#5
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Re: Flopping FH and quads
[ QUOTE ]
2)Given that exactly two players are dealt distinct pocket pairs , the probability one flops a fh and another flops quads is : 2/48c3 ~ 0.000115633 or about 1 in 10,000 . [/ QUOTE ] Times 2 since either player can make the quads, and there are 2 ways to make each full house. |
#6
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Re: Flopping FH and quads
[ QUOTE ]
Thanks- I thought it would be pretty unlikely. Can you explain your equations 4c2*(78/1326)^2*(1248/1326)^2 -understand 4c2 (where 4=n and n is # players at table) but where do the other two parts come from. 2/48c3 -how did you come up with this? understand 48 unknown cards and combinations of 3. Thanks [/ QUOTE ] Ok , so for the first part I made an independent assumption that the probability exactly two players are dealt pocket pairs is closely related to the probability that you will be dealt exactly 2 pocket pairs in 4 deals . There are 6 ways each pocket pair can be received . {2c2d,2c2h,2c2s,2d2h,2d2s,2h2s} = 6 ways There are 13 distinct possible pairs and so 13*6=78 There are 52c2=1326 different two-card hand combinations that can be dealt to you . 78/1326 is the probability that you will be dealt a pocket pair . The complement of all this is (1326-78)/1326 = 1248/1326 is the probability that you will NOT be dealt a pocket pair . So , we may be dealt exactly two pocket pair in deals 12,13,14,23,24,34 and so we have : P(12 but not 34)=(78/1326)^2*(1248/1326)^2 P(13 but not 24)= " P(14 but not 23)= " P(23 but not 14)= " P(24 but not 13)= " P(34 but not 12) = " If we add all this we get 6*(78/1326)^2*(1248/1326)^2 -------------------------------------------------- So if we assume that you hold 2c2h and your opponent holds 3c3d , then the possible flops for this event is : {2s3h3s} , {2d3h3s} = 2 The total number of possible flops is 48c3 since 4 cards are removed. Now we simply divide these two numbers . |
#7
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Re: Flopping FH and quads
Correct .
As Bruce pointed out , there is a big difference when we make a generic statement about the probability that exactly two players are dealt pocket pairs and that one has flopped a fh and another , quads . Since , player A may have quads and player b has the full house OR player A has a fh and player B has quads . This is why we multiply by 2 . 2) 2*2/48c3*0.01839 ~ 0.00000425 or about 4.2 times in one million deals . |
#8
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Re: Flopping FH and quads
[ QUOTE ]
Correct . As Bruce pointed out , there is a big difference when we make a generic statement about the probability that exactly two players are dealt pocket pairs and that one has flopped a fh and another , quads . Since , player A may have quads and player b has the full house OR player A has a fh and player B has quads . This is why we multiply by 2 . 2) 2*2/48c3*0.01839 ~ 0.00000425 or about 4.2 times in one million deals . [/ QUOTE ] Ok- I was wondering if that made a difference. I'm following most of your calculations. I'm not clear on the following: So , we may be dealt exactly two pocket pair in deals 12,13,14,23,24,34 and so we have : P(12 but not 34)=(78/1326)^2*(1248/1326)^2 P(13 but not 24)= " P(14 but not 23)= " P(23 but not 14)= " P(24 but not 13)= " P(34 but not 12) = " If we add all this we get 6*(78/1326)^2*(1248/1326)^2 1) where are the deals 12,13,14etc coming from? 2) why do we factor in the NOT part? 3) does the 6 come from 4c2? |
#9
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Re: Flopping FH and quads
Om- can ignore questions 2 and 3- got answers from your reply to my set over set post from couple of days ago.
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#10
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Re: Flopping FH and quads
Basically what I did was change the question and addressed the probability that you will be dealt exactly 2 pocket pairs in FOUR independent deals . This is almost equivalent to exactly 2 players being dealt pocket pairs in ONE deal .
Do you see why ? I've assumed independence here ; that is , the probability that another player at your table will be dealt a pocket has very little to do with your cards . In other words , the dependency is very weak and making the independent assumption starts to make a lot of sense . So you may be dealt a pocket pair in deals 1 and 2 but not 3 and 4 ,deals 1 and 3 but not 2 and 4 , deals 1 and 4 but not 2 and 3 ,deals 2 and 3 but not 1 and 4 , deals 2 and 4 but not 1 and 3 and lastly , deals 3 and 4 but not 1 and 2 . This is how I arrived at 4c2 since there are 4c2 different ways to select 2 numbers from the set {1,2,3,4} . |
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