equity - ez for you to answer

Propping Fool · #1 05-22-2007, 02:10 AM

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 · #2 05-22-2007, 02:46 AM

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 · #3 05-22-2007, 03:01 AM

can anybody tell me if he is correct ?

uDevil · #4 05-22-2007, 03:14 AM

[ 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]

ncray · #5 05-22-2007, 03:37 AM

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 · #6 05-22-2007, 03:42 AM

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.

uDevil · #7 05-22-2007, 04:47 AM

[ 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]

Propping Fool · #8 05-22-2007, 08:37 AM

28% is right.

I recreated this exact situation in pokerstove and 28% is correct.

Deal yourself KQs and hu preflop vs one player with a range of 40%. (54%)

KQs vs a guy with 9.8% range (38%)

then deal KQs vs both of them and your equity is 28%

nice work, we can put the str8 jacket back in the closet until tomorrow.

ty

DWarrior · #9 05-22-2007, 10:02 AM

That could be a good estimate vs ranges. However, vs specific hands, the times you come ahead of player 1, you're inevitably more likely to be ahead of player 2 (because usually you have improved)

For example, say you have AKo and are up against T8o and 99
Equity vs T8o: 64%
Equity vs 99: 44.7%

Your equity vs both is 41.4%, about the same as that vs 99. This is because you need to at least pair up to beat 99, and when you pair, T8o is far behind.

Now let's replace 99 with J2o:
Equity vs T8o: 64%
Equity vs J2o: 68.5%

However, your equity vs both is 48.5%. This is because a lot of your equity vs each hand comes from times when neither of you pair and you win with a high card. However, your equity basically goes down to that vs a pair because now you have to avoid 4 cards pairing, or pair your own.

Hope this helps.

Carlson411 · #10 05-22-2007, 11:15 AM

[ 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.