equity - ez for you to answer

[DrVanNostrin]

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Lets say I am in a hand against 2 opponents.

I have the first guy beat 54% of the time.

I have the 2nd guy beat 38% of the time.

What is my equity against both players? (please explain)

ty in advance

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It depends on player how often player 1 will beat player 2. If player 1 will never beat player 2 your equity is 38%. This could happen if players 1 and 2 had the same ranking cards but player 2 had a flush draw. You could also be in a situation where player 1 will beat you almost everytime player 2 doesn't. You have AQ, player 1 has AK, and player 2 has QQ. If you end up beating player 2 you'll almost surely lose to player 1. This isn't a perfect example because you can win, but it's possible to have no equity in this spot.

let A = beating player 1
let B = beating player 2
P(A & B) = P(A)*P(B|A) = P(B)*P(A|B)

If A and B are independant events P(B|A) = P(B) and P(A|B) = P(A).

The problem is neither P(B|A) or P(A|B) are specified or can be determined from the given info.

[T50_Omaha8]

Not enough information.

I'll use an example to try to show you why.

Let's say you have JsTs.

Opponent 1 has 33.

Thus equity against player 1 is 53.25%

Hand Pot equity Wins Ties
33 46.75% 4,743,270 120,354
JsTs 53.25% 5,410,200 120,354

Now let's assume player 2 has AQo. Our equity against player 2 is 39.56%

Hand Pot equity Wins Ties
JsTs 39.56% 673,184 8,523
AsQd 60.44% 1,030,597 8,523

Our equity against both players is 33.72%.

Hand Pot equity Wins Ties
JsTs 33.72% 2,761,722 35,007
AsQd 35.46% 2,904,492 35,007
33 30.82% 2,523,303 35,007

Now let's change player 2's hand. We change it to TT, which DECREASES our equity against him to 38.69%

Hand Pot equity Wins Ties
JsTs 38.69% 1,940,247 94,662
TT 61.31% 3,102,003 94,662

But what happens to our equity against both of them? Our equity in the 3-way pot INCREASES to 34.69%

Hand Pot equity Wins Ties
JsTs 34.69% 8,321,166 505,188
33 18.50% 4,535,046 89,082
TT 46.81% 11,312,172 505,188


I hope this serves as a counterexample for calculating your equity against two opponents given individual equity vs each.

[AaronBrown]

T50_Omaha8 is correct.

The key is whether the main uncertainty is whether you will improve or whether the other hands will improve. At one extreme, you have a draw the comes through 38% of the time and gives you the nuts. If it fails, you always lose to hand 2, but you still sometimes beat hand 1. Now you beat both hands 38% of the time, the highest possible amount.

Or, suppose your hand never improves. You only win if both hand 1 and hand 2 fail to improve. If the events are independent then your chance of beating both is 0.54*0.38 = 21%.

At the other extreme, assume you're down to the river card. 19 of the 42 unseen cards win for player 1 and 27 win for player 2. If that includes all 42 unseen cards (that is, if there are only 4 cards that make both player 1 and player 2 beat you), then you have zero chance of winning.

[UtzChips]

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can anybody tell me if he is correct ?

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I have to warn you, Gary Carson says I'm insane. I'd hate to be both insane and wrong, but I have to admit it's possible. [img]/images/graemlins/tongue.gif[/img]

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Is Gary posting anywhere? I haven't heard from him in a few years. I use to get a lot of feedback from on RGP, however, that has become........well.......you know.

[uDevil]

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can anybody tell me if he is correct ?

[/ QUOTE ]

I have to warn you, Gary Carson says I'm insane. I'd hate to be both insane and wrong, but I have to admit it's possible. [img]/images/graemlins/tongue.gif[/img]

[/ QUOTE ]

Is Gary posting anywhere? I haven't heard from him in a few years. I use to get a lot of feedback from on RGP, however, that has become........well.......you know.

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He has started a blog:

http://www.mathandpoker.com

He has some interesting material on the site.

RGP may have changed, but Gary has not. He still knows how to hold a grudge, and he still has a penchant for attacking people (as demonstrated by his posts on Ed Miller's blog, http://www.notedpokerauthority.com ).