Question about how Number of Players effects Odds of Hands Winning?

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lawd

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I've been playing around with Poker Stove to figure out the odds of playing certain hands in particular situations (ex. all in the low pair pre-flop v. 1 opponent; flop with 2 set cards, and me having 2 set cards (i.e. 4 set cards v. someone with a High Pair). I noticed that these odds tend to change when the amount of players at the table changes.

I can't seem to figure out why, from a stats perspective, it makes no sense.

Why would a 77 All-In Pre-flop have different odds of winning versus a single player on a table composed of (a) 10 players as opposed to (b) 2 players?


With 2 players the odds are 66% with a 33, with 10 players the odds are 59% versus any random player.


Basically, at the theoretical level I'm trying to understand how the amount of players at the table affects the odds of certain hands being the statistical favorite versus one other player. If we assume that the other cards are in the form of a random distribution (which poker stove is supposed to do), this makes no sense.

is this a bug in the program or am i missing something?
 
SavagePenguin

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Why would a 77 All-In Pre-flop have different odds of winning versus a single player on a table composed of (a) 10 players as opposed to (b) 2 players?

I'm not a Poker Stove user, but I *thought* that you only enter your hand, and then assign a hand to the players who are are going to play their hands.

That is, if you give yourself A/A and tell it that 8 other people have "random" hands, Poker Stove assumes that nine people are playing their cards that hand.
I mean, I don't think you can say "At a 9 handed table I have Js/9s in the SB, what's my equity against the one possible caller's random hand." It would just be "what's Js/9s equity against one random hand." IE, the number of players who fold isn't considered.
Again, I could be wrong.
 
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slycbnew

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No bug - when you have a hand like 77 against a single random hand, there's a certain percentage chance that other hand will have two overcards (example). Against two random hands, there's now a percentage chance that both random hands will be overcards that don't match each other - i.e., now you have to deal w four overcards. Etc.
 
thepokerkid123

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I'm not sure where you're getting your numbers from...

10 handed (33 vs 9 random hands) you win 11% of the time.
Against two players (33 vs 2 random hands) you win 34% of the time.
Heads up (33 vs 1 random hand) you win 53% of the time.

These numbers are all from pokerstove.

With 2 players the odds are 66% with a 33, with 10 players the odds are 59% versus any random player.
This doesn't make sense. I'm not sure how you're getting the numbers, but as far as I know your odds vs a given player wont change based on the number of all ins as long as the hands are random. If the hands are not random... I'm not really sure how 33 will do, but TT+ do a lot better because of card removal on A's.

I don't know how you're getting the odds that you are out of pokerstove, I suspect either your version is glitchy or you're misinterpreting something. Possibly that pokerstove assumes all hands go to showdown, so increasing the number of players decreases your equity.
 
SavagePenguin

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lawd said:
Why would a 77 All-In Pre-flop have different odds of winning versus a single player on a table composed of (a) 10 players as opposed to (b) 2 players?

No bug - when you have a hand like 77 against a single random hand, there's a certain percentage chance that other hand will have two overcards (example). Against two random hands, there's now a percentage chance that both random hands will be overcards that don't match each other - i.e., now you have to deal w four overcards. Etc.

Huh?
Of course the odds change if you are playing against more players, as there are more cards in the deck that can help them.

But from what I got, he's saying that if he plays at a HU table his all-in-pre with 7/7 shows a different equity percentage than if he's all-in in a SB vs BB situation at a full table.
His question confused me because I didn't think you could specify table size.
 
slycbnew

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My bad, misread it. I'm staring at pokerstove now and not sure what the question is actually about, as you say there's no table size option.
 
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lawd

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I'm not sure where you're getting your numbers from...

10 handed (33 vs 9 random hands) you win 11% of the time.
Against two players (33 vs 2 random hands) you win 34% of the time.
Heads up (33 vs 1 random hand) you win 53% of the time.

These numbers are all from pokerstove.

This doesn't make sense. I'm not sure how you're getting the numbers, but as far as I know your odds vs a given player wont change based on the number of all ins as long as the hands are random. If the hands are not random... I'm not really sure how 33 will do, but TT+ do a lot better because of card removal on A's.

I don't know how you're getting the odds that you are out of pokerstove, I suspect either your version is glitchy or you're misinterpreting something. Possibly that pokerstove assumes all hands go to showdown, so increasing the number of players decreases your equity.


for 2 players I use the odds that are given on pokerstove.

for 10 players, I do the following Calculation:
(1 Random Player chance to win) + (33 pair holder chance to win) = percentage that these 2 win the pot [i.e. these are the 2 isolated players]

(33 pair holder chance to win)/(percentage that these 2 win the pot) = odds if the 33 pair goes up against 1 random player.
 
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lawd

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I'm not a Poker Stove user, but I *thought* that you only enter your hand, and then assign a hand to the players who are are going to play their hands.

That is, if you give yourself A/A and tell it that 8 other people have "random" hands, Poker Stove assumes that nine people are playing their cards that hand.
I mean, I don't think you can say "At a 9 handed table I have Js/9s in the SB, what's my equity against the one possible caller's random hand." It would just be "what's Js/9s equity against one random hand." IE, the number of players who fold isn't considered.
Again, I could be wrong.

I don't think pokerstove makes an assumpition that they will play their hands, it just tells you the odds they will win the pot given the hand ranges you specify for them.
In this case i specified:
(1) 33 v. 1 random hand (i.e. any potential hand)
(2) 33 v. 9 random hands, with 1 player isolated for an all-in (assuming that this players hand quality is completely random)
 
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lawd

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Yes, i meant the raw equity changes. Sorry if that was confusing, I'm new so I'm still learning the terms used in poker analysis.
 
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I think the software considers less chances of the a draw when there are less cards in the deck. I couldn't find any other mathematical answer.
 
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lawd

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swingpro: the amount of cards in the deck is exactly the same though versus known cards (i.e. the pair of 33s, i.e. 4% of the deck whether you have 2 or 10 players)

i'm starting to think there may be a bug in how pokerstove calculates odds...
 
Stu_Ungar

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Its blindingly obvious why the odds of winning decrease with the number of players: Blockers

77 does well when against 66, but if one player holds 66 then the next player is less likely to hold 66! And the ratio of (88-AA) vs (66-22) shifts. because you are taking hands out of the range you beat, its more and more likely that one of the remaining players holds a hand in the range that beats you. Its not limited to PPs I just used that as an example to illustrate what's going on.

Again with ranges, a single player can hold his entire range, but multiple players mean that in any simulation, when one of your opponents holds a hand you beat, the rest of your opponents ranges shift to hands that you don't beat. The more players the more of a shift.
 
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lawd

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^ can you elaborate? perhaps give an example? I'm having trouble understanding the concept...


I agree in a way in that I noticed the same phenomenon with simulations of other hands (about 4-8% shifts), but i don't get the why.
 
Stu_Ungar

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There is an example.

if you hold 77 and deal ONLY pocket pairs (for the purposes of an example) you will find that as the number of players increases it becomes increasingly difficult to deal them al PP that you beat because when you give one player a PP you beat, the next has fewer to chiise from and as the first didnt beat you, the "bag of pocket pairs" now contains a higher ratio of PP that beat you.

The next player gets a PP that you beat, the bag of PPs now contains an even higher ratio before of PP that beat you, and so on and so on.

THis happens for entire ranges, but just focus on PPs as its easier to grasp.
 
Stu_Ungar

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Second easier example.

There is a bag with marbles in.

1 is red and 9 are black.

You loose if your opponent pulls ouut the red marble but win if he pulls out a black one. MARBLES ARE NOT REPLACES (this is what you are not seeing with the cards they are not put back!)

As you increase the number of opponents, it becomes more and more likely you will lose, with 10 opponents it becomes an absolute certainty you will lose.
 
SavagePenguin

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With that bag of marble analogy, if you reach in and grab a marble your odds are 1 in 10 of getting the red. 10%

If 5 strangers each grab a marble but don't show you, 50% of the time you won't have a chance at getting the red because one of them will have it.
But 50% of the time you'll be 20% to get the marble.

Half the time you are 20% to win, and half the time you are 0%, which averages to 10%. Because their holdings of previous drawers are unknown and random, the decreased odds counteract the increased odds and it all works out the same in the long run. So it doesn't matter if one person draws for the marble of if 9 draw before you, as you're going to get that marble 10% of the time.

But if you assign specific cards or a specific range of cards to the villains, that effects what is available in the deck and does change the odds. So for every black marble we know a villain has, we know our chances of winning go up.

I don't think pokerstove makes an assumpition that they will play their hands, it just tells you the odds they will win the pot given the hand ranges you specify for them.
Those are the odds of them winning if they play the hand to showdown, which assumes that they play the hand because you can't win the showdown any other way. :)
 
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Stu_Ungar

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With that bag of marble analogy, if you reach in and grab a marble your odds are 1 in 10 of getting the red. 10%

If 5 strangers each grab a marble but don't show you, 50% of the time you won't have a chance at getting the red because one of them will have it.
But 50% of the time you'll be 20% to get the marble.

Half the time you are 20% to win, and half the time you are 0%, which averages to 10%. Because their holdings of previous drawers are unknown and random, the decreased odds counteract the increased odds and it all works out the same in the long run. So it doesn't matter if one person draws for the marble of if 9 draw before you, as you're going to get that marble 10% of the time.

But if you assign specific cards or a specific range of cards to the villains, that effects what is available in the deck and does change the odds. So for every black marble we know a villain has, we know our chances of winning go up.


Those are the odds of them winning if they play the hand to showdown, which assumes that they play the hand because you can't win the showdown any other way. :)

In the example I gave, you never draw a marble. Only your opponents.

I was trying to keep it simple! And show how card removal aka taking a marble out of the bag works.

The point was the more players the more likely the marble that beats you is drawn. Because each time you remove a black marble, the odds of drawing the red increase.

When there are 10 players, it is guaranteed that the marble that beats you is drawn.
 
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lawd

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thanks stu, that really made it clear for me what the reason behind my observations was. very useful concept for understanding 5-15% shifts in equity of certain hands v. amount of players
 
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