Odds of 2 players with trips or better?

[trader01]

I posted this in the probability forum but nobody answered, so maybe someone here knows. What are the odds that on a 6 player table, where every player goes through to the river without folding, that at least 2 players are going to end up with trips or better as a hand? On what percentage of hands should we expect to see something like this? I have no idea how to calculate it, so if somebody here knows it would be appreciated. Also, if it's easy to do, the same calculation with 2 pair instead of trips would be useful...

[Fluorescent Pig]

Oink.

[whipsaw]

50%

[MasterLJ]

That's a pretty ridiculous problem to solve. There are a lot of cases. None of them are hard, but you'd have to count them all.

[trader01]

[ QUOTE ]
That's a pretty ridiculous problem to solve. There are a lot of cases. None of them are hard, but you'd have to count them all.

[/ QUOTE ]

I could be wrong, but I figured there would be standard percentages that people knew about that would tell you what the chances were of getting ending up with trips, a straight, a flush, or whatever at the river before any cards were even dealt to you. (So we might say that for any given hand, before you are dealt any cards, the chances of ending up with trips at the river is a standard 8% or whatever (I don't really know what the real chances are I'm just making that up), a straight is 5%, etc. etc...)

If you knew these percentages it should be fairly simple to figure out what the chances were that 2 or more players ended up with any kind of trips, straight, flush, or higher, no? Of course, I don't know what those percentages are (but I assume that someone might)... also my combinatorial math isn't the greatest, so maybe someone else might know how to calculate it with confidence...

[Son_Of_Arathorn]

[ QUOTE ]
I posted this in the probability forum but nobody answered, so maybe someone here knows. What are the odds that on a 6 player table, where every player goes through to the river without folding, that at least 2 players are going to end up with trips or better as a hand? On what percentage of hands should we expect to see something like this? I have no idea how to calculate it, so if somebody here knows it would be appreciated. Also, if it's easy to do, the same calculation with 2 pair instead of trips would be useful...

[/ QUOTE ]

Ok well it takes 3 cards to make trips right? And two people need to make them out of 6, using 7 cards.

So we take the three cards and times them by two, this makes six. 6 x 6 is 36, 36 x 7 is something like 220.


So we arrive at 220, the magic number. What do we do with this? Well there are 52 cards in a deck so we end up with odds of 52/220 or pretty much 1 in 4. Of course online there is no dealer but a machine deals it so we add 1 to the number of cards to represent this. We thus end up at 2-4.

So the odds of 2 people getting trips or better in any given hand is approx. 1-2, or 50%.

Been nice to help. [img]/images/graemlins/smile.gif[/img]

[SkinnyLittleTwig]

"but there's only a 4% chance that he has the flush, so i can safely call all in with my pair of threes on a four-flush board!!!"

[igetjokes]

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50%

[/ QUOTE ]

either two people have trips, or they don't, so 50% is the correct answer.

[Son_Of_Arathorn]

[ QUOTE ]
[ QUOTE ]
50%

[/ QUOTE ]

either two people have trips, or they don't, so 50% is the correct answer.

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Yes I came much to the same conclusion but using hardcore maths.

[durkahdurkah]

32.34855321%





repeating obv.