I can't remember my combinatorics but I was hoping someone could answer
the following question:

Given card X is on the board, what is the probability that another player
holds card X (assuming that I do not hold card X)? Assume also that 5
players are in the hand.

If someone could give me the formula for this I would appreciate it. It
seems to me it should be:

p = ((number of players)*2 CHOOSE 3) / ((number of players) * 2)!

But this formula gives a probability that is way too low.

Thanks for the help.

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> Given card X is on the board, what is the probability that another player
> holds card X (assuming that I do not hold card X)? Assume also that 5
> players are in the hand.

It sounds like you're talking at the river. So I work from that
assumption.

Let's create some relationships (I'm a programmer...so that's how I
think)

NOOP=NumberOfOtherPlayers
CIOH=(Number of cards in other players hands)NOOP*2
UCEH=(Unseen cards at the end of the hand) 52-2-5-CIOH

So what we want to calculate is the probability 3 of that rank are
still in the UCEH:

We calculate the number of combinations of cards in other people's
hands given the cards we want are in the UCEH, then we calulate the
total number of combinations of cards in other people's hands to get
the probability no one holds at least one of a certain rank that's on
the board.

(UCEH - 3) choose CIOH
------------------------
UCEH choose CIOH

So for 5 (other) players I get:

NOOP=5
CIOH=10
UCEH=35

(32!)/((22!)(10!))
P(No one paired a board card of a certain rank) = --------------------
(35!)/((25!)(10!))

P= 35.14%

(My arithmatic is the most suspect thing here...I used the windows
calculator...and it doesn't have parens)

The end equation ends up being:

(52-2-5-(NumberOfOtherPlayers*2)- 3) choose (NumberOfOtherPlayers*2)
----------------------------------------------------------------------
52-2-5-(NumberOfOtherPlayers*2) choose (NumberOfOtherPlayers*2)


I apologize if my formatting didn't translate well.

--Michael

"HDouble" <anonymous@hotmail.com> wrote:
> Given card X is on the board, what is the probability that another player
> holds card X (assuming that I do not hold card X)? Assume also that 5
> players are in the hand.
>

Assuming there is exactly one of X on the flop, and not taking into
account these other players' pre-flop strategy (i.e. every 2-card
combination is equally likely for them to hold):

# other players p(none hold any Xs) p(at least one has at least
one X)
1 44c2/47c2 = 87.5% 12.5%
2 44c4/47c4 = 76.1% 23.9%
3 44c6/47c6 = 65.7% 34.3%
4 44c8/47c8 = 56.4% 43.6%
5 44c10/47c10 = 47.9% 52.1%

So if there are five other players in the game with you who are
willing to hold anything, the odds are better than even that one of
them will have paired a specific flop card. If you meant X on the
board at the river vs. the flop the odds are slightly higher, still
assuming that there's only one of X on the board.

Stacy

this is the only one that is accurate - the formula is simply solve for
all situations where that card does not exist = in yr scenario, 10 cards
are dealt to the players and three are up. This leaves 39 cards. What
are the odds that none of the players have any particular card then is
expressed by 36/39 times 35/38 times 34/37 and so on till you reach
29/32(8 cards). What you will get is a result around 49% so it is 51%
that any particular card is paired by one of the remaining four players.
On Aug 4 2003 3:01AM, stacy_friedman wrote:

> "HDouble" <anonymous@hotmail.com> wrote:
> > Given card X is on the board, what is the probability that another player
> > holds card X (assuming that I do not hold card X)? Assume also that 5
> > players are in the hand.
> >
>
> Assuming there is exactly one of X on the flop, and not taking into
> account these other players' pre-flop strategy (i.e. every 2-card
> combination is equally likely for them to hold):
>
> # other players p(none hold any Xs) p(at least one has at least
> one X)
> 1 44c2/47c2 = 87.5% 12.5%
> 2 44c4/47c4 = 76.1% 23.9%
> 3 44c6/47c6 = 65.7% 34.3%
> 4 44c8/47c8 = 56.4% 43.6%
> 5 44c10/47c10 = 47.9% 52.1%
>
> So if there are five other players in the game with you who are
> willing to hold anything, the odds are better than even that one of
> them will have paired a specific flop card. If you meant X on the
> board at the river vs. the flop the odds are slightly higher, still
> assuming that there's only one of X on the board.
>
> Stacy

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Thanks all for the replies.

So it looks like the general rule is, given that we have second best pair
and there are less than 5 players who see the flop, we should assume we
have the best hand (if our kicker is good). Of course if there is an Ace
or a King on the board, second best pair isn't as strong because people
are less willing to fold them.

This excludes the chances that someone has hit bottom 2 pair or trips with
the bottom or middle pair, so probably our second best pair-good kicker is
only a favorite when there are only 2 other players seeing the flop with
us.

Is this right or did I miss something?

Thanks.




On Aug 3 2003 10:45PM, HDouble wrote:

> I can't remember my combinatorics but I was hoping someone could answer
> the following question:
>
> Given card X is on the board, what is the probability that another player
> holds card X (assuming that I do not hold card X)? Assume also that 5
> players are in the hand.
>
> If someone could give me the formula for this I would appreciate it. It
> seems to me it should be:
>
> p = ((number of players)*2 CHOOSE 3) / ((number of players) * 2)!
>
> But this formula gives a probability that is way too low.
>
> Thanks for the help.

__________________________________________________ _______________
Posted using RecPoker.com - http://www.recpoker.com

HDouble wrote:
> I can't remember my combinatorics but I was hoping someone could answer
> the following question:
>
> Given card X is on the board, what is the probability that another player
> holds card X (assuming that I do not hold card X)? Assume also that 5
> players are in the hand.

Everyone who answered seems to have a slightly different take on what
you're asking. But from your follow-up, it sounds like what you really
want to know is, "If I flop middle pair on an unpaire flop, what are the
chances that no one has a higher pair?" This is a bit different than
how you originally phrased it, because it includes the possibility that
someone has a pocket pair larger than your split pair.

First, the probability that someone flopped top pair or top set is 1
minus the probability that none of your remaining live opponents has one
of the pairing cards. You can see five cards so there are 47 unknown
cards. Your opponents hold eight of these. There are C(47,8) =
(47x46x45x44x43x42x41x40)/(8x7x6x5x4x3x2x1) ways to choose eight cards
from 47 (we don't need to do all the multiplication and division just yet).

Of those 47 cards, three are of the rank of interest, leaving 44 that
don't pair the top card on the flop. So there are C(44,8) =
(44x43x42x41x40x39x38x37)/(8x7x6x5x4x3x2x1) ways to choose eight cards
from 44 which don't contain the three pairing cards.

The probability that no one has one or two of the pairing cards is
therefore C(44/8) / C(47/8). Both of these numbers are fractions, and
dividing by a fraction is the same as multiplying by its reciprocal,
that is X/(1/Y) = X x Y. So C(44/8) / C(47/8) =

(44x43x42x41x40x39x38x37)/(8x7x6x5x4x3x2x1) x
(8x7x6x5x4x3x2x1)/(47x46x45x44x43x42x41x40) =

(44x43x42x41x40x39x38x37)/
(47x46x45x44x43x42x41x40) =

39x38x37/
47x46x45 = 54834/97290 = 0.5636 = 56.36%

That's how often no one will have top pair or top set on the flop,
assuming your four opponents play random cards. You can perform similar
calculations for the turn and river.

However, there's still a possibility that someone has a pocket pair
higher than your split pair. There's also the possibility that a higher
card will come on the turn or river and make someone a bigger pair. The
calculations for these scenarios get tedious and depend on the rank of
your pair.

Lin

HDouble wrote:
> Thanks all for the replies.
>
> So it looks like the general rule is, given that we have second best pair
> and there are less than 5 players who see the flop, we should assume we
> have the best hand (if our kicker is good). Of course if there is an Ace
> or a King on the board, second best pair isn't as strong because people
> are less willing to fold them.
>
> This excludes the chances that someone has hit bottom 2 pair or trips with
> the bottom or middle pair, so probably our second best pair-good kicker is
> only a favorite when there are only 2 other players seeing the flop with
> us.
>
> Is this right or did I miss something?


See my other post.

You don't have to be better than even money to make money on a bet. If
you'll win the pot 1 out of X times, then you make money on a bet if you
have more than X callers. You also make money if a bet will in the pot
outright 1 out of Y times and the pot is paying better than Y:1 on your
bet. Sometimes also a bet will improve your chances by knocking out
enough players to move you into the overall profit column.

That doesn't mean you should always play 2nd pair like it's a winner,
but sometimes it can be right to do so, even against multiple opponents,
simply because it's profitable in some circumstances (typically when you
have draws to bigger hands), not because you're going to win a lot of
the pots with second pair vs a large field.

Lin

of course you have stated the problem in your answer- your opponents do
not play random cards(at least mine don't). So all the probability
matters very little relative to the way specific opponents play - it only
matters if you are playing showdown and there are no cards turned after
the flop.....
On Aug 4 2003 2:32PM, Linda K Sherman wrote:

> HDouble wrote:
> > I can't remember my combinatorics but I was hoping someone could answer
> > the following question:
> >
> > Given card X is on the board, what is the probability that another player
> > holds card X (assuming that I do not hold card X)? Assume also that 5
> > players are in the hand.
>
> Everyone who answered seems to have a slightly different take on what
> you're asking. But from your follow-up, it sounds like what you really
> want to know is, "If I flop middle pair on an unpaire flop, what are the
> chances that no one has a higher pair?" This is a bit different than
> how you originally phrased it, because it includes the possibility that
> someone has a pocket pair larger than your split pair.
>
> First, the probability that someone flopped top pair or top set is 1
> minus the probability that none of your remaining live opponents has one
> of the pairing cards. You can see five cards so there are 47 unknown
> cards. Your opponents hold eight of these. There are C(47,8) =
> (47x46x45x44x43x42x41x40)/(8x7x6x5x4x3x2x1) ways to choose eight cards
> from 47 (we don't need to do all the multiplication and division just yet).
>
> Of those 47 cards, three are of the rank of interest, leaving 44 that
> don't pair the top card on the flop. So there are C(44,8) =
> (44x43x42x41x40x39x38x37)/(8x7x6x5x4x3x2x1) ways to choose eight cards
> from 44 which don't contain the three pairing cards.
>
> The probability that no one has one or two of the pairing cards is
> therefore C(44/8) / C(47/8). Both of these numbers are fractions, and
> dividing by a fraction is the same as multiplying by its reciprocal,
> that is X/(1/Y) = X x Y. So C(44/8) / C(47/8) =
>
> (44x43x42x41x40x39x38x37)/(8x7x6x5x4x3x2x1) x
> (8x7x6x5x4x3x2x1)/(47x46x45x44x43x42x41x40) =
>
> (44x43x42x41x40x39x38x37)/
> (47x46x45x44x43x42x41x40) =
>
> 39x38x37/
> 47x46x45 = 54834/97290 = 0.5636 = 56.36%
>
> That's how often no one will have top pair or top set on the flop,
> assuming your four opponents play random cards. You can perform similar
> calculations for the turn and river.
>
> However, there's still a possibility that someone has a pocket pair
> higher than your split pair. There's also the possibility that a higher
> card will come on the turn or river and make someone a bigger pair. The
> calculations for these scenarios get tedious and depend on the rank of
> your pair.
>
> Lin

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