Question of expected value on the bubble in Step 5 STT

[Gigabet]

Lets say that I am in the small blind with 2400, bb has 1600, and the button has 2600. It is 5 handed, and utg(who has already folded) has 160. Everyone folds to the button, who pushes all in, the range of hands that I put the button on is any two cards, except JJ, QQ, KK, AA, AQ, and AK.

My confidence level for winning the tourney if I call and win is 70%.

My confidence level for placing at least 2nd if I call and win is 90%.

My confidence level for placing at least 3rd if I call and win is 100%.


My confidence level for winning if I fold is 30%.

My confidence level for getting at least second if I fold is 60%.

My confidence level for getting at least 3rd if I fold is 75%.

My confidence level for getting at least 4th if I fold is 100%.

Can someone give me the expected value of calling vs folding using 1065 as the buy in?

[imcastleman]

I think it would be useful to know what your hand is.

[Gigabet]

Sorry, Kd Qd

[schwza]

just out of curiousity, what are the blinds? i'm guessing 100/200, and you think button would raise smaller with a big hand?

[Gigabet]

150/300

[UMTerp]

Come on Giga, did you ask people to do your math homework for you in high school?? /images/graemlins/wink.gif

I'll oblige, but for simplicity's sake, I'm ignoring the "not AA-JJ, AK, AQ thing", and assigning the button any two cards.

KQs vs. two random cards = 0.634 chance to win.

Since you gave those "confidence statistics", I'll use them, and buy-in, blinds, and stack sizes are irrelevant.

If you call and win:

1st - 70%
2nd - 20%
3rd - 10%
4th - 0%
5th - 0%

If you call and lose:

5th = 100%

If you call: (0.634)[(.7)($4500)+(.2)($2500)+(.1)($1800)] + (0.366)($0) = <font color="red">$2428.22</font>

If you fold:

1st - 30%
2nd - 30%
3rd - 15%
4th - 25%
5th - 0%

If you fold: (.3)($4500)+(.3)($2500)+(.15)($1800)+(.25)($1200) = <font color="red">$2670.00</font>

Damn close, and if you throw out the possibility of the big pairs and big aces, it's even closer to 50-50. Still a slight lean towards a fold though.

[Gigabet]

Thank You, and yes, I had my brother do my math homework, he is a math genius, but too busy to figure this out for me, I wouldn't know where to begin to figure this out.

I know all the numbers by rote, i couldn't conclusively prove them to myself, I just trust that the people who do figure it our are right.

[jcm4ccc]

OK, if we take Kd and Qd out of the deck, there are 1250 possible hands. KdQd is .6340040 against a random hand. We need to take out the possibility of JJ, QQ, KK, AA, AQ, and AK.

Total possibility: .6340040 * 1250 = 792.505

Minus:
JJ: .46 * 6 = 2.76
QQ: .35 * 3 = 1.05
KK .13 * 3 = .39
AA .18 * 6 = 1.08
AKo .30 * 9 = 2.70
AKs .28 * 3 = 0.84
AQo .30 * 9 = 2.70
AQs .28 * 3 = 0.84

792.505 – (2.76 + 1.05 + .39 + 1.08 + 2.70 + 0.84 + 2.70 + 0.84) = 780.145

Total possible hands = 1250 – (6 + 3 + 3 + 6 + 9 + 3 + 9 + 3) = 1208

780.145/1208 = .6458 against the hands left (interesting that it barely improves your odds against a random hand to take those hands out)


If you call:

You will win .6458 * .70 = .452 * 4500 = $2034
You will take 2nd .6458 * .20 = .129 * 2500 = $322.50
You will take 3rd .6458 * .10 = .065 * 1800 = $117
You will lose .354 * 0 = $0

Total Expected Value: $2473.50

If you fold:

You will win .30 * 4500 = $1350
You will take 2nd: .30 * 2500 = $750
You will take 3rd: .15 * 1800 = $270
You will take 4th .25 * 1200 = $300

Total Expected Value: $2670


So you have a higher expected value if you fold.

I have a question for you. How do you pronounce Gigabet. Is it:

A) Giggabet
B) Jiggabet
C) Jijjabet
D) Gidget

[Oluwafemi]

[ QUOTE ]
Come on Giga, did you ask people to do your math homework for you in high school?? /images/graemlins/wink.gif

I'll oblige, but for simplicity's sake, I'm ignoring the "not AA-JJ, AK, AQ thing", and assigning the button any two cards.

KQs vs. two random cards = 0.634 chance to win.

Since you gave those "confidence statistics", I'll use them, and buy-in, blinds, and stack sizes are irrelevant.

If you call and win:

1st - 70%
2nd - 20%
3rd - 10%
4th - 0%
5th - 0%

If you call and lose:

5th = 100%

If you call: (0.634)[(.7)($4500)+(.2)($2500)+(.1)($1800)] + (0.366)($0) = <font color="red">$2428.22</font>

If you fold:

1st - 30%
2nd - 30%
3rd - 15%
4th - 25%
5th - 0%

If you fold: (.3)($4500)+(.3)($2500)+(.15)($1800)+(.25)($1200) = <font color="red">$2670.00</font>

Damn close, and if you throw out the possibility of the big pairs and big aces, it's even closer to 50-50. Still a slight lean towards a fold though.

[/ QUOTE ]

and how exactly does the breakdown of this interesting equation come about and how commonplace is it in big buy-ins? how valuable (if at all) is this type of math equation if it were to be used in lower buy-ins?

[Xhiggy]

you're hardly against "any two cards though." although given the situation, the button has a range probably of the top 70% of all possible hands. I would say, given you've thrown away the big pairs and AK, maybe about 60% of the time you're a slight underdog or a 3:2 underdog. and the remaining 40% you're likely dominating, or a 3:2 favorite. I think 70% confidance for a win is on the high side.
all these factors reduce the EV on calling even more, probably making fold the better option.

[imcastleman]

You are not looking at the situation closely enough. The button has everyone covered and the short stack is going to be all in on the next hand. In addition, Gigabet posted this and he knows his opponents VERY well. There is no doubt in my mind that the button could have ANY 2 cards(including 72o).

[schwza]

so what do you think button would do with, say, AK?

[se2schul]

[ QUOTE ]
and how exactly does the breakdown of this interesting equation come about

[/ QUOTE ]

This is just the basic formula for EV: the sum of the probability of each possible outcome of the hand multiplied by its value.

EV = probability of each outcome * value

So, there is an EV for calling and an EV for folding. If your EV is higher calling, you'd call, otherwise you'd fold.

Look at the EV for calling. There are 2 possible scenarios: call for a win, or call for a loss.

EV(calling) = EV(call and win) + EV(call and lose)

Now, calling for a win:
KQ wins 63.4% of the time against random cards, so the EV is .634 * the payout from calling for a win.
EV(call and win) = 0.634 * [(.7)($4500)+(.2)($2500)+(.1)($1800)]
Notice these probabilities are just estimates from Gigabet. If you're no good at making these estimates, then this EV equation will have little merit.

Similarily, workout EV(call and lose):
EV(call and lose) = probability of a loss with KQ * payout
= .366 * $0
= $0

So, EV(calling) = EV(call and win) + EV(call and lose)
= $2428.22 (according to someone else's calcs, I didn't actually punch out the numbers)

Essentially this is saying that everytime you find yourself in the situation outlined above, if you call, you are making $2428.22.

Is this good? Sure, it's profit, so of course it's good. But maybe folding is more profitable.

So, workout the EV for folding, exactly as I did for calling and see which is higher. If folding is higher, then every time you find yourself in that situation, you want to fold, otherwise, you call.

[ QUOTE ]
and how commonplace is it in big buy-ins?

[/ QUOTE ]
This math goes on for ALL types of gambling, be it poker, blackjack, craps, or anything. Often though, players will know what the move that yeilds the highest EV is in a certain scenaio. It obviously must be pretty important at the higher buy-ins, since Gigabet wants it worked out.

[ QUOTE ]
how valuable (if at all) is this type of math equation if it were to be used in lower buy-ins?

[/ QUOTE ]

It applies to all buy-ins. You even use it, even if you didn't work out the numbers exactly. You know that to get proper odds to call with a small pocket pair, you need to have several callers before you. In this case, you use a general rule that provides you with a positive EV situation.

[se2schul]

[ QUOTE ]
I know all the numbers by rote, i couldn't conclusively prove them to myself, I just trust that the people who do figure it our are right.

[/ QUOTE ]

I'll teach you all the math you want in exchange for you teaching me to play cards the way you do. /images/graemlins/grin.gif

[imcastleman]

How did you figure that KQs was favored by 0.634 against any random hand??

[assron]

[ QUOTE ]
How did you figure that KQs was favored by 0.634 against any random hand??

[/ QUOTE ]

pokerstove

[Ryendal]

Only one question. From where are theses 4500$, 2500$, and 1800$ in your equation ?

[The Yugoslavian]

[ QUOTE ]
Only one question. From where are theses 4500$, 2500$, and 1800$ in your equation ?

[/ QUOTE ]

This is from the payout structure of STEP 5 STTs on party poker.

Yugoslav