Odds of this?

[franknagaijr]

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The problem with the argument that the approximation is exact, is shouldn't you be using conditional probability?


[/ QUOTE ]

RNU - I took that into account and did the calcs with the line of reasoning you suggest, and the nubers came out the same. Message me your e-mail if you want the spreadsheet.

[RobNottsUk]

OK, I believe you. I dunno how on earth you manage to make that calc rigorously.

Still in the general case, it is true that this multiplication is not exact, and will be worse the more likely the single opponent condition is.

So even if this exact this time, what Buzz has said, stands in my view.

[franknagaijr]

I'm not a math guy, but I play one on TV. I'm making up terminology as I go along.

As far as I can tell, the difference between the challenge Buzz posed with the two potential cards for a higher flush vs the original problem, involving two or more aces in one hand, is that the original problem can only be solved in discrete packets, hand by hand for 8 handsif you will. Buzz' problem can be solved as if all 8 hands were really one big hand with 32 cards. If the original problem had been, 'odds that one or more hand included an ace', this would have been a collective probability issue, and we would be better off solving for the whole group, to discount cases where more than one hand met the condition.

So to summarize:
If only one hand can meet the condition - Multiply times number of hands should be okay.
If multiple hands can simultaneously meet the condition - Need to ratchet down the number, or calc as a collective set instead of multiplying

[RobNottsUk]

I was discussing the original problem, not Buzz's example.

If you don't know what conditional probability is, then I'm afraid your calc probably on the AAnn v Annn issue is unlikely to be right.

Conditional probability is hairy and best avoided, it's a 1st year Uni Maths topic, and something I found too complicated to make practical use of for poker.

The core issue, is that only 1 hand can meet the condition, if you decide that it has AAnn, and it's the first dealt.

But the 2nd dealt, and n+1th, depend on the previous cards, which may contain 1 Ace. If 2 Aces have been dealt then it's a bust...

See the probability of later hands, is conditional on other events occuring with a certain probability.

[franknagaijr]

No, I think I do understand what you're talking about. I ran it out to three hands in conditional probability, and was satisfied that the various probability trees still came up with the same answer.

Again, PM me your e-mail, and I'll ship you an XL spreadsheet.

[RobNottsUk]

Weird, I'm coming round now though. I'll choose a simpler example. This has definitely been worth thinking about, though I don't think the posts have been that easy to understand.

In the K-hi flush Hold'em example, deciding the liklihood of it being best on the river, when a 3-flush is on board.

You look for Axs, where As is paired by 1 of 7 sooted partner cards S. 45 unseens cards, 1 As, 7 S and 37 offsuit cards O.

A hand can be, OO, OS, AsO, or SS, which all significantly affect the later chances of finding AsS in another hand; once you know what the cards are.

There the simple multiplication is a definite reasonable approximation, even if it's not rigourous it's close enough for practical play.

I'm coming round though to this idea that it's exact if there can be only 1 hit.

With AxS, as you don't know the other hands, and generating all possible dealt hands to 8 opponents, and seeing how often AxS would occur, would be valid. Perhaps because the other hands are unknown, there are simply more chances of As being dealt, and it's partner card being an S or an O is the same as with the single hand situation.

The only problem is, in an actual game, if on the river, I've not got a better idea of whether opponents are on a flush draw, or have a non-flushing hand, then I'm not doing a good job. This type of reasoning can lead to annoying situations where I've folded a pair + likely best K-hi flush draw on the turn, to an opponent with at least top pair and raises a player who probably had a Q-hi or worse draw only, but just might have the nut flush draw for once. Just had to revise the outs down, and without being sure of extra river bets with better flush on river, it made it a marginal fold, so the over-player of his draw won the pot on the river thanks to the made hand folding me out.