Rather complex question

[WuTank]

First i want to apologize for poor English, but here we go:
Me and a friend are talking about a problem.We came to it when i obsed him playing a sng on PP.
Here was the situation:
He had 7k otb and blinds were 300,600.Both Blinds had 4k,lets say 4,2k so 7BB.

I said its +EV to shove any 2 here, and since he is a rather analytical player he said i should prove it with maths.
So i thought of it like this:
x is the equity of a random hand against villians range.
y is the calling FQ of them.

7x*y + 1,5(1-y) = 7(1-x)*y

I think this is correct.When we set y=1(he calls with every hand) the equity is 50%.
We stoved a little bit and tried to find the Nash-Equilibrium.
In the end we failed because we weren`t able to find a way to get x as a function of y.
So we basicly need a way to express how the calling range influences the equity against a random Hand.
Has anyone a function for this or another way to approach this problem?

[PantsOnFire]

I won't address your math directly.

Let's say that SB will always fold and BB has 4000. You are risking 4000 to win 4900. With 0% fold equity you have a pot equity of 4000/8900 or 45%. So instead of any two cards, this play is mathematically correct if you had a hand that had 45% equity against a random hand.

Even if BB is all-in for his last 600 after posting, you would be risking 600 to win 900 for an equity of 600/1500 or 40%. This means that you still can't go with any two since there are some poor hands that are less than 40% against random. Now if you were SB here, you could go with any two because you would have 25% pot equity which covers all hole cards.

[jogsxyz]

You can't easily prove it with math. You need to write a computer program solving this with monte carlo. Think there's charts somewhere in the internet with the proper jam or fold parameters.

[RichGambler]

[ QUOTE ]
First i want to apologize for poor English, but here we go:
Me and a friend are talking about a problem.We came to it when i obsed him playing a sng on PP.
Here was the situation:
He had 7k otb and blinds were 300,600.Both Blinds had 4k,lets say 4,2k so 7BB.

I said its +EV to shove any 2 here, and since he is a rather analytical player he said i should prove it with maths.
So i thought of it like this:
x is the equity of a random hand against villians range.
y is the calling FQ of them.

7x*y + 1,5(1-y) = 7(1-x)*y

I think this is correct.When we set y=1(he calls with every hand) the equity is 50%.
We stoved a little bit and tried to find the Nash-Equilibrium.
In the end we failed because we weren`t able to find a way to get x as a function of y.
So we basicly need a way to express how the calling range influences the equity against a random Hand.
Has anyone a function for this or another way to approach this problem?

[/ QUOTE ]

Realistically they will call with 15% of the hands.
That means you have 70% folding equity vs 2 players.

That means you will win 2070 chips before they call. Your random hand have 35%/65% equity and you will be losing 30% of your invested $$$ in the long run.

30% from 4000 chips = 1200 since 2070>1200 then it means pushing with any 2 is correct play.

[jukofyork]

[ QUOTE ]
First i want to apologize for poor English, but here we go:
Me and a friend are talking about a problem.We came to it when i obsed him playing a sng on PP.
Here was the situation:
He had 7k otb and blinds were 300,600.Both Blinds had 4k,lets say 4,2k so 7BB.

I said its +EV to shove any 2 here, and since he is a rather analytical player he said i should prove it with maths.
So i thought of it like this:
x is the equity of a random hand against villians range.
y is the calling FQ of them.

7x*y + 1,5(1-y) = 7(1-x)*y

I think this is correct.When we set y=1(he calls with every hand) the equity is 50%.
We stoved a little bit and tried to find the Nash-Equilibrium.
In the end we failed because we weren`t able to find a way to get x as a function of y.
So we basicly need a way to express how the calling range influences the equity against a random Hand.
Has anyone a function for this or another way to approach this problem?

[/ QUOTE ]
Not really sure what you are trying to do here, but it looks interesting. [img]/images/graemlins/smile.gif[/img]

As far as expressing x as a function of y, then using Mathematica:

Reduce[14*x*y + 3*(1-y) == 14*(1-x)*y,{x}]

x = (-3 + 17y) / 28y {where Y!=0}

Hope this helps - Juk [img]/images/graemlins/smile.gif[/img]

[AaronBrown]

As PantsOnFire said, it's certainly not positive EV to shove with any hand.

You are correct that if there are only two players, and the other player always calls, you have positive EV by going all-in with any hand. You have a 50% chance of winning and because of the blinds, you win more when you win than you lose when you lose.

However even in this case, going all-in with any two cards is not your maximum EV. You win 5,100 when you win and lose 4,200 when you lose, so you need 4,200/(5,100 + 4,200) = 45% chance of winning to be positive EV. That means you should push with J3 or better, and fold with anything less. You will play 65% of your hands, and win 56% of the time you play, for an average profit of 966. If you push all the time, your average profit is 450 (half of the 900 blinds).

You could do better still by betting different amounts with different hands.

The problem is more complicated with two players, and players who do not always call. But it's still not correct to always push.