Flopping FH and quads

[RedSoxFan]

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

[jay_shark]

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]

[RedSoxFan]

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

[jay_shark]

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 .

[BruceZ]

[ 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.

[jay_shark]

[ 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 .

[jay_shark]

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 .

[RedSoxFan]

[ 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?

[RedSoxFan]

Om- can ignore questions 2 and 3- got answers from your reply to my set over set post from couple of days ago.

[jay_shark]

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} .