Omaha Question

[KungFuSandwich]

In a 10man game and no one folds. If you flop middle set what are the odds of one of the other 9 having top set?

Formulas would be helpful: I would like to know how to apply this to less people or bottom set. Also would like to know how to adjust the odds after the turn.

Thanks alot! [img]/images/graemlins/spade.gif[/img]

[KungFuSandwich]

If one of you will just give me the Formula, Ill do the math.

[fnord_too]

Well, let's assume a specific scanario to make things easier:

You have TT64 and the flop is KT2r. Also assuming no one raising preflop reveals nothing about their hand.

So, there are 9 4 card hands out there, and there are 45 unknown cards distributed between those hands and the stub. Here is where is gets tricky, because the hands are not independant. That is, if Hand 1 has one king, the chance of anyone has gone way down, and if Hands 1 and 2 each have a king, they have gone away completely. I would have to think a bit to figure out an exact answer, (an I believe it would be a little messy, I'll give you an approach at the end.)

If you were up against one hand, it is easy:

There are 6 pairs of kings (not counting trip kings) and 42 non king cards, so there are 6*42*41 pair of king hands and C(45,4) total hands. Also, there are 1*42 ways to hace three kings. So we have a total of 10,374 ways of having top set and 148,995 possible hands. This is about a 7% chance your first opponent has top set. If the hands were independant, I think that works out to about a 48% chance that one of your opponents has top set, but they are not independant. (Though, for a quick and dirty treating them as independant probably gets you pretty close).

To get an exact answer, you could built a probability tree. To do this, caluclate the ways hand 1 can have 0, 1, 2, or 3 Kings. Then, on each branch (except 2 and 3, since then you have satisfied the condition), calculate the ways hand 2 can have 0, 1, 2, or 3 (if possible) kings. Rinse and repeat, keeping track of the total kings out on each branch. At the end, sum all the 2 and 3 kings probabilities and there you go.

Of course, when you have middle set, even without top set being out, you could be losing at the time or an underdog to a draw. Also, we are creating a pretty unrealistic situation with 10 people seeing the flop with no revealing preflop action.

[KungFuSandwich]

Ill get to work, Thanks

[BruceZ]

[ QUOTE ]
You have TT64 and the flop is KT2r. Also assuming no one raising preflop reveals nothing about their hand.

So, there are 9 4 card hands out there, and there are 45 unknown cards distributed between those hands and the stub. Here is where is gets tricky, because the hands are not independant. That is, if Hand 1 has one king, the chance of anyone has gone way down, and if Hands 1 and 2 each have a king, they have gone away completely. I would have to think a bit to figure out an exact answer, (an I believe it would be a little messy, I'll give you an approach at the end.)

If you were up against one hand, it is easy:

There are 6 pairs of kings (not counting trip kings) and 42 non king cards, so there are 6*42*41 pair of king hands and C(45,4) total hands. Also, there are 1*42 ways to hace three kings. So we have a total of 10,374 ways of having top set and 148,995 possible hands. This is about a 7% chance your first opponent has top set. If the hands were independant, I think that works out to about a 48% chance that one of your opponents has top set, but they are not independant. (Though, for a quick and dirty treating them as independant probably gets you pretty close).

To get an exact answer, you could built a probability tree. To do this, caluclate the ways hand 1 can have 0, 1, 2, or 3 Kings. Then, on each branch (except 2 and 3, since then you have satisfied the condition), calculate the ways hand 2 can have 0, 1, 2, or 3 (if possible) kings. Rinse and repeat, keeping track of the total kings out on each branch. At the end, sum all the 2 and 3 kings probabilities and there you go.

[/ QUOTE ]

This post represents a misunderstanding of epidemic proportions. It is NOT any more difficult to do this computation for 9 hands as it is for 1 hand in this case, and the fact that the hands are not independent has nothing to do with it. If you know the probability of 1 particular opponent having top set, then we can simply multiply this probability by 9 to obtain the EXACT probability that one of 9 opponents has top set. The reason this is exact is because only 1 opponent can have top set, i.e., the event of the players having top set are mutually exclusive so their probabilities can be added.

Additionally, there are only 3 pairs of kings available not 6, since one is on the flop, and the number of hands with a pair of kings is 3*C(42,2) = 3*42*41/2, not 3*42*41 since this would double count every hand.

Here is the probability that at least one of N players has top set when you hold middle set.

N*[ 3*C(42,2) + 42 ] / C(45,4)

For N=9, this is 15.9%. In general, this is approximately N*1.76%.

On the turn, the probability is:

N*[ 3*C(41,2) + 41 ] / C(44,4)

For N=9 this is 16.6%, and in general this is approximately N*1.84%.

That's all - you don't need any probability trees. These probabilities are the same for your opponents holding bottom pair. Now if you hold bottom pair, the computation of the exact probability of your opponents holding top pair OR middle pair becomes a little more tricky since in this case it is possible for 2 opponents to hold these hands, so simply multiplying by 9 would double count these cases. This double counting is handled by the inclusion-exclusion principle. I hope you see the difference in these two cases.

It is critical to understand the difference between mutual exclusivity and independence, and I have continually stressed this issue. See this post for an explanation of the difference. Too often I see "independence" given as a reason that a certain method is not exact, when in fact the issue is mutual exclusivity. In this case, failing to appreciate mutual exclusivity leads to not recognizing a simple solution when one exists, and doing a great deal of unnecessary extra work.

[fnord_too]

Doh! Right, 3 pairs of kings, monday morning.

Also, I'm pretty sure you are right about the 9*P(1 opponent). Again, I blame a monday morning with too little sleep.

Your turn solution assumes that the top card did not pair, (which would make quads for anyone who flopped top set).

I do understand probability, but my liklihood of screwing something up when I answer on the fly appears directly proportional to the observability of my calculations. (Of course this could be for the same reason that the number of bugs found in software rises with the number of users [img]/images/graemlins/smile.gif[/img] )