Odds to beat a crowd

[Lottery Larry]

Am I calculating this correctly? If there are 4 other players in, and I estimate my chances of beating them individually as:
70%
90%
80%
95%

then my chance to beat ALL of them is about 48%

If I'm 80% against each player, I'm just under 41% to beat them all.

Correct?

Thanks
LL

[WhiteWolf]

[ QUOTE ]
Am I calculating this correctly? If there are 4 other players in, and I estimate my chances of beating them individually as:
70%
90%
80%
95%

then my chance to beat ALL of them is about 48%

If I'm 80% against each player, I'm just under 41% to beat them all.

Correct?

Thanks
LL

[/ QUOTE ]

Your calculations are correct, *IF* your matchup against each player is independent (ie your chance that you beat the first player does not affect the chance that you beat the second). However, you cannot always assume the matchups will be independent.

To use poker to illustrate this, hands in 7-card stud are mostly independent, so the above assumptions (mostly) apply. However, hold'em hands, which all share the same 5-card board, tend to run closer together. As a result, the independence assumption does not apply. For example, AA all-in preflop against 1 random hand is usually around 80% to win. Using an independent model, you would expect the chance of winning against 9 random hands would be close to 13% (barely +EV). However, since the final hand values of your opponents are somewhat dependent on eachother (ie if there are not 3 to a flush on the board, no one beats you with a flush), your true chances in practice are around 30% (hugely +EV).

HTH,

The Wolf

[SunOfBeach]

if these 4 events are independent (and thats doubtful in a 5person pot), then this would be correct. there are likely some correlations here, however.

[AaronBrown]

I agree with SunofBeach and WhiteWolf, except that I would say more strongly that I can think of no Poker situation in which this computation is relevant. Even if the hands themselves are close to independent, as in games without community cards, your chance of beating other players will almost always be highly dependent.

In any actual situation, I would expect your probability of beating everyone to be different (usually higher) than your calculation.

[Lottery Larry]

[ QUOTE ]
Even if the hands themselves are close to independent, as in games without community cards, your chance of beating other players will almost always be highly dependent.

In any actual situation, I would expect your probability of beating everyone to be different (usually higher) than your calculation.

[/ QUOTE ]

And WW wrote "However, hold'em hands, which all share the same 5-card board, tend to run closer together. As a result, the independence assumption does not apply"

I'm not quite getting what you all mean by the "independance assumption." Obviously if there is a 3-flush on board, then the hand ranges I'm assigning them have to change, and therefore my "change to beat individually" also changes. I'm not associating that idea with interdependence correctly, I believe??
Are you referring to hands counterfeiting one another with shared cards?

I was intending to use this in two ways:
1) as a rough estimate of the chances of success, if I can successfully determine a range of hands each opponent would hold (that's a risky assumption on my part! :P )

If I ignore betting tendencies and the like, knowing that I'm cutting out 70% of the real world influences that would be in effect, why can't the simple percentages be used?

I thought a rough percentage would be fairly accurate, but if the 13% vs. 30% for Aces is correct, then I may be assuming too much.


BTW- In my mind, I was thinking something like holding a big pair such as AA against loose players and doing a quick calculation of my chances against them (my original examples).

2) as a teaching tool for some players I'm working with; the lessons to be taught are ones on facing multiple opponents and the effects they have, as a group, on your hand(s).



Also, is there a limit on how I can use these calculations in a rough manner? For example, if I think I'm behind everyone and give myself a 30% chance against four others individually, my chance to beat everyone CAN'T be less than 1%, can it?

Is there some rule-of-thumb similar to counting outs on the flop, turn and river, where your addition can't make your hand more than around a 50% chance of winning? At least, that's how I understand the out-counting rule-of-thumb.

Thanks again for the help.

[ohnonotthat]

The best example of there being [virtually] total independence would be if your hand was made and could not be improved (an A-high straight) while all of your opponents were behind (fl. draws, two-prs, trips) but not drawing dead.

That said, it will never be necessary - and will probably never be possible - to calculate these numbers in anything close to an accurate manner.

It's unnecessary because the "close" decisions will amount to nothing at the end of the year.

- It's impossible because calling time while you "count the stars in the sky" is not possible.

If you are constantly getting your stack in the pot with a 4-1 edge, a 3-1 edge, a 2-1, edge or even a 3-2 edge does it really matter whether your 2-2 is facing an opponent's AK/suited (making you ~ a 1% 'dog) or unsuited (making you ~ a 2% fav) ?

If you answered yes I'm afraid you don't quite get it.

Don't sweat the petty things (and don't pet the sweaty things).

- I saw it on a tee shirt . . . funny AND true

[OrangeKing]

The "independence assumption" everyone is talking about works something like this.

Your calculation - figuring out what your odds are against each player individually, and then multiplying those winning percentages to see how often you beat all those hands - would be absolutely 100% correct if you were facing each of those hands individually in different hands, but not when you're facing all of them as a group in the same hand. Once you're in a hand against 5 players, they will almost certainly share a lot of outs with each other (i.e. - you might be up against KQ and KJ, each of which counts the two remaining kings as cards that improve them), which works in your advantage. So while you might be 80% against KQ and 80% against KJ (approximate numbers for purposes of this example, don't kill me :P), you won't be 64% against the two of them as a group - you'll more likely be around 70%.

[AaronBrown]

If you're holding a drawing hand, you'll hit it or you won't. If you do, you're likely to beat everyone, if not, you're likely to beat no one. In that case your probability of beating everyone is higher than you get by multiplication.

If you're playing against drawing hands, you could be in a situation where the cards that make one person's hands are different than the cards that make someone else's. In that case your chance of winning is less than you get by multiplication.

Suppose each of the four players has 9 outs on the river. Your chance of beating any one is 1 - 9/38 = 29/38 = 76%. If you raise that to the fourth power, you get 34%.

But if all their outs are different, there are only 2 cards that let you beat everyone, that's 2/38 = 5%. If they're all waiting for the same 9 cards, your chance of beating them all is your chance of beating any one of them, 76%.

The 34% would only be correct if there are 24 cards that beat you. That could be true if each player has five cards that help only him, plus four cards that help everyone. But there's no reason to assume this would be the case.

[Lottery Larry]

Thanks to you and others (including my PM) for the help.