equity - ez for you to answer

[Propping Fool]

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

[uDevil]

I'm not sure about this, but I'd say 48%.

Assume your equity is equal to the percentage of the total number of chips in play that you hold.

Opponent 1 has 46% as many chips as you do.
Opponent 2 has 62% as many chips as you do.

If you hold x chips, the total number of chips in play is

.46x+.62x+x=2.08x

so you hold x= 1/2.08(total number of chips), i.e. 48% of the total.

[Propping Fool]

can anybody tell me if he is correct ?

[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]

[ncray]

Given the information, it's impossible to tell. Let A be the event that you beat the first guy. Then P(A) = .54. Let B be the event that you beat the second guy. P(B) = .38. You are looking for the probability of the event where you beat both players P(A and B). Well P(A and B) = P(A) + P(B) - P(A or B) , and you don't know P(A or B), but it's between .54 and .92, so P(A and B) is between 0 and .38.

[ncray]

I guess just to make things a bit more clear. P(A and B) = 0 in the case where the 54% of the time you are beating the first guy, the second guy is beating you, and in the 38% (different from the 54%) time you are beating the second guy, the first guy is beating you.

P(A and B) = .38 in the case where the 54% of the time you are beating the first guy, you are beating the second guy 38% of the total time (or 70.3% of the 54% of the time). The remaining 16% of the time you are beating the first guy, the second guy is beating you.

[Carlson411]

[ QUOTE ]
[ QUOTE ]
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 ]

He's right. I just went on the advance calculator on pokerlistings.com. You can check out your question there.

[UtzChips]

[ QUOTE ]
[ QUOTE ]
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.

[uDevil]

[ QUOTE ]
[ QUOTE ]
[ QUOTE ]
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.

[/ QUOTE ]

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 ).

[uDevil]

[ QUOTE ]
I'm not sure about this, but I'd say 48%.

Assume your equity is equal to the percentage of the total number of chips in play that you hold.

Opponent 1 has 46% as many chips as you do.
Opponent 2 has 62% as many chips as you do.

If you hold x chips, the total number of chips in play is

.46x+.62x+x=2.08x

so you hold x= 1/2.08(total number of chips), i.e. 48% of the total.

[/ QUOTE ]

Very fuzzy thinking on my part (I'm really worried about my sanity now). Sorry.

Still no guarantee, but trying again:

Let

T1= total number of chips between you and opponent 1.
T2= total number of chips between you and opponent 2.

You have .54*T1=.38*T2 chips, so T2=1.42*T1
Opponent 1 has .46*T1 chips
Opponent 2 has .62*T2 chips

The total number of chips in play is

T= .46*T1+.62*T2+.54*T1= .46*T1+.62*1.42*T1+.54*T1= 1.88*T1

so T1= .532*T

But you have .54*T1, which is .54*.532*T= .287*T

I.E. you have 28.7% of the total number of chips, so 28.7% equity against both opponents.

Should have at least checked my previous answer for reasonableness (thanks ncray). Maybe Gary is right. [img]/images/graemlins/frown.gif[/img]