Calculating the chance that all will fold...

[J.A.K.]

I am UTG w/ XX. I estimate that the remaining 8 players at my table will only call my push with top 10% of hands. What is the chance one of them will call?

Is it 1-.9^8?

Thanks!

[SamIAm]

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What is the chance at least one of them will call?

Is it 1-.9^8?

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Yep.
-Sam

[bachfan]

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[ QUOTE ]
What is the chance at least one of them will call?

Is it 1-.9^8?

[/ QUOTE ]
Yep.
-Sam

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Almost. It depends what XX is. [img]/images/graemlins/smile.gif[/img]

I've thought about whipping together a steal calculator to compute this for various numbers of players and ranges for each player... Useful?

[creedofhubris]

[ QUOTE ]
[ QUOTE ]
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What is the chance at least one of them will call?

Is it 1-.9^8?

[/ QUOTE ]
Yep.
-Sam

[/ QUOTE ]

Almost. It depends what XX is. [img]/images/graemlins/smile.gif[/img]

I've thought about whipping together a steal calculator to compute this for various numbers of players and ranges for each player... Useful?

[/ QUOTE ]

yes.

[curious123]

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[ QUOTE ]
I've thought about whipping together a steal calculator to compute this for various numbers of players and ranges for each player... Useful?

[/ QUOTE ]

yes.

[/ QUOTE ]

Make it happen bach. [img]/images/graemlins/smile.gif[/img]

[bachfan]

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[ QUOTE ]
[ QUOTE ]
I've thought about whipping together a steal calculator to compute this for various numbers of players and ranges for each player... Useful?

[/ QUOTE ]

yes.

[/ QUOTE ]

Make it happen bach. [img]/images/graemlins/smile.gif[/img]

[/ QUOTE ]

Sir yes sir! I'm on it! [img]/images/graemlins/cool.gif[/img]

I'll try whipping it up as my next tool when I find the time/motivation.

-bachfan

[AaronBrown]

Not quite, because the hands are not independent. The top 10% of hands contain lots of Aces and no cards 2 through 5 (depending on exactly how you define top 10%).

Let's just think about Aces. 15% of hands have Aces. An Ace hand is in the top 10% if paired with any card Ten or higher, or suited 8 or 9. So given that your first card is an Ace, there are 21 cards out of 51 that put it in the top 10%. Call that 40%. That means an Aceless hand has only about a 5% chance of being in the top 10%.

A random player has a 15% chance of having an Ace. If a player folds, there is only a 10% chance she had an Ace (because fewer Ace hands than Aceless hands are folded). Therefore, the more players fold, the more likely it is the remaining players have Aces, thus the more likely it is that they will call.

My guess is this effect raises your chance of being called from 1 - 0.9^8 = 57% to 58% or 59%. Not a huge difference, but measurable.

[mykey1961]

[ QUOTE ]
I am UTG w/ XX. I estimate that the remaining 8 players at my table will only call my push with top 10% of hands. What is the chance one of them will call?

Is it 1-.9^8?

Thanks!

[/ QUOTE ]


What have they seen you push with that makes them think they can call with the top 10%? 10% seems a little high to me.

[jay_shark]

To find the exact solution you have to use the principle of inclusion/exclusion .

Take 8 events and label them 1,2,,...,7,8.

1= event player one calls with the top 10 % of hands .
{66+,a9s+kjs+a10+} which is about 10.1% of hands .
2= event player 2 calls with the top 10 % of hands .

You are interested in p(1u2u3u4u5u6u7u8) . So to find the exact probability , you have to work out several frivolous calculations which can be tedious .

At a poker table , you don't have time to do these calculations and they really aren't necessary unless you wish to do them on your own spare time . You may simply use Booles inequality in many cases which states that p(1u2u...u8) < p1+p2+...p8 .However , it is important to note that the smaller the value of p , the better the approximation .

[bachfan]

This is done . No docs, lame ui,but you can compute these kinds of questions now. Happy new year.

For instance, if we have J7o utg short handed, and everyone will only call with the top 10% of hands or so, then there is a 59% chance that all will fold (see the first table, first row "vs. 0" column to see number of times nobody called).

Now, if we are in a full game and four people fold first, but we know they WOULD HAVE PLAYED the top 10% of hands, then we only steal successfully about 57% of the time