Having a hard time getting this.
If i have 2 non paired cards and the board doesn't pair,
how often in 100 tries will i have 2 pair on the river?

There are 52 – 2 = 50 remaining cards. Of the five-card board, two have to match your hole cards and three don’t match nor pair the board. The probability is

C(3,1)*C(3,1)*C(11,3)*C(4,1)^3/C(50,5)

= 4.5%

.

The above represents the probability of getting an unpaired board and two pair, which I took to be consistent with the thread title. However, the posting states that the unpaired board is a given. For that case the probability of two pair given an unpaired board is

0.045/0.5074 = 8.84%.

The denominator is the probability of an unpaired board given hero does not have a pair.

Disregard, statmanhal is fine.

It's not 0.045 in the numerator then since it's more likely to get 2 pair when the board isn't paired.

3*3*C(11,3)*4^3 /
[3*3*C(11,3)*4^3 + 2*3*C(11,4)*4^4 + C(11,5)*4^5]

≈ 11.57%

≈ 8.84%

Shouldn't the second term in the denominator be 2*3*C(11,4)*4^4?

Yes, and that gives the same answer you had. You were taking the 4.5% of total boards and dividing that by the percentage of boards that have all 5 cards different which is fine. Sorry about that.

I get the same answer in a different way.

Pick five ranks at random with equal probability.

The chance that both your pocket cards are among the five is 5/13 x 1/3 = 5/39
The chance that exactly one of your pocket cards is among the five is 5/13 x 2/3 + 8/13 x 5/12 = 20/39
The chance that neither of your pocket cards is among the five is 8/13 x 7/12 = 14/39

All we need is the ratio of 5:20:14.

Now compute the ratio of the number of ways we can arrange the suits in each of the three cases, it's 9:12:16.

Multiply term by term and get 45:240:224, which sum to 509. So the probability is 45/509 of two pair, 240/509 of one pair and 224/509 of no pair.

Hey, I found an error in a BruceZ probability calculation. This may be my greatest accomplishment in a long technical career!

That's the downside of always being right, everyone wants to pick you off.

That's why I wisely make many mistakes.

I make a lot more mistakes than you, but, unfortunately, it's not because I'm wiser.

I make more than everyone combined. Thanks everyone. I'm still a little confused. Let me ask it this way ...
I have 3, 5, i'm at a full ring table, if i took that hand to the river 100 times on average how many times would i have two pair? Meaning pairing the 3 and pairing the 5.
Is the answer approx 8 times out of 100? Thanks again everyone!

Do you mean _exactly_ or _at least_ two pair? And if you pair 3 and 5, but the board contains another (larger) pair too, does it count (e.g. board: 3588A, where you have paired both 3 and 5, but your hand is 8855A)?

With you holding a 3, 5:

A: No board stipulation: 4.5 times out of 100 hands, a dealt board will have one 3, one 5 and the other three cards will all be different.

B: Given that the board does not have a pair: 8.8 times out of 100 one card will be a 3 and another a 5.

C. No board stipulation: The chance of just pairing the 3 and 5 and no conditions on the other three cards is 9*C(44,3)/C(50,5) = 5.6%. This would include, for example, a board of 3,5,7,7,7

Here is an answer to a more general question, that I think is more useful.

If you have unpaired pocket cards, there are 6 of the remaining 50 cards that improve your hand, ignoring straights and flushes. The board is 5 of those 50 cards, selected at random. The probability that it will include k of the 6 cards that help you is C(6,k)*C(44,5-k)/C(50,5) for k=0 to 5.

The rows below show number of helping cards and probabilities:

0 51% nothing
1 38% one pair
2 9.4% two pair or trips
3 0.89% full house or quads
4 0.031% double trips or quads and a pair
5 0.0003% quads and trips

In general, you should either memorize from a book or (better in my opinion) work out for yourself and memorize, the probabilities of getting various numbers of good cards in common situations. There's less to memorize if you think about it that way than if you ask more specific questions like what are the chances of getting exactly two pair.

Great points! and i thought i had thought of everything.
In this case i am only interested in pairing my two cards.
As there is quite a few ways of doing better.

Yes no board stipulation. Thanks alot guys.
i'm getting good at the math after the flop its preflop to river when i still have major problemo

Its worth distinguishing cases in 2 and 3, and relatively easy to figure at least approximately.

Having hit the first helper card, 2 of the remaining 5 helper cards make us trips and 3 of the remaining 5 helper cards make us quads. So we should have trips 40% (3.76) of the 9.4% of the time and 2p 60% (5.64)

When we have 2pr and and hit a 3rd helper card, that third helper card always gives us a fh, and never quads. When we have trips and hit a 3rd helper card, of the four helper cards available one makes us quads and three fill us up.

So with three helper cards.

60% -- we cannot make quads with exactly 3 helpers and must make a fh
30% -- we hit one of the three out of four helper cards that fills up our trips
10% -- we hit the case card of our rank that gives us quads.

So 90%of the time we have exactly 3 helper cards, we make a FH and the other 10% we make quads.

0 51% nothing
1 38% one pair
2 9.4%{5.64% two pair + 3.76% trips}
3 0.9213%{.801% full house + .089% quads}
4 0.031% {.8194 double trips +.012586% quads and a pair}
5 0.0003% quads and trips

We can simplify a little bit and make it more technically correct by combining the cases where we have 4 and 5 helper cards to get us a FH or quads into the appropriate case in the 3 helper cards scenario (since in cases 4 and 5, we only really have 3 helper cards)


0 51% nothing
1 38% one pair
2 9.4%{5.64% two pair + 3.76% trips}
3 0.9213%{.8194% full house + .101886% quads}

It should be easy enough to calculate.

You have 50 cards left in the deck, and you are going to pick 5 of them. Of those 50 cards, 3 are threes and 3 are fives. So, you will _not_ pair both of your cards if
1) the board contains no threes OR
2) the board contains no fives
However, since there are boards that contain neither threes nor fives, these will be counted twice, so you have to subtract
3) the board contains neither threes nor fives

the probability of 1) is easy: 47/50 * 46/49 * 45/48 * 44/47 * 43/46.
The probability of 2) is the same as 1)
the probability of 3) is 44/50 * 43/49 * 42/48 * 41/47 * 40/46

So, the probability of _not_ pairing both 3 and 5 is 1) + 2) - 3) =
2 * 47/50 * 46/49 * 45/48 * 44/47 * 43/46 - 44/50 * 43/49 * 42/48 * 41/47 * 40/46

This can be written as

2* 47!/42! * 45!/50! - 44!/39! * 45!/50! =
45!/50! * (2* 47!/42! - 44!/39!)
= 0,93539145537956

Therefore, the probability of pairing both 3 and 5 (the three remaining board cards can be anything; 3's or 5's or other cards, paired or not etc) is
1 - 0,93539145537956 = 0,06460854462044 (~6,5%)

For problems of this sort, there are always several ways to solve them. The above analysis is for the probability of getting at least a pair or better given an unpaired hand.

Using combinations for not meeiting this criterion we have the following:

Given a hand of 3,5, the probability of

No 3 and no 5 on the board= C(44,5)/C(50,5)=51.3%

Pair either the 3 or 5 and other card not on the board = 2*C(3,1)C(44,4)/C(50,5) = 38.4%

Trips of either 3 or 5 and other card not on the board = 2*C(3,2)C(44,3)/C(50,5) = 3.8%

Quads of either 3 or 5 and other card not on the board= 2*C(3,3)C(44,2)/C(50,5) =0.1%

Adding up these probabilities and subtracting from 1 gives the above answer of 6.5% after rounding.

This is most certainly true.

I tried (perhaps unsuccessfully) to give a method that would be easy for someone with no knowledge of probability calculations to understand (and also understand the reasoning behind it).

I can't stand it anymore.

Say the board is A2345.

There are ten different two pair.

Each two pair has nine holdem hand two card combos. So there are 90 qualifying combos.

That 90 is out of the number of two card hands that can be made from the remaining 47 non board cards. 47x46/2 which is 1081.

Except that the OP specified non pair hands. There are six combos each of sixes through kings and three each aces through fives. That is 63. Leaving 1018.

90/1018=.084

I guess you noticed that OP also changed his mind about what he was asking mid-thread?

Whether he did or not the original answerers were needlessly complicated.

The OP asked for the probability of pairing both his different ranked hole cards under the condition that the board doesn't pair. That condition was the complication because the board can have 0, 1, or 2 of his ranks, and there are a different number of those ranks left than the others. What's simpler than saying that there are 3*3*C(11,3)*4^3 boards that match both, 6*C(11,4)*4^4 boards that match 1, and C(11,5)*4^5 boards that match neither? That's all I did with

3*3*C(11,3)*4^3 /
[3*3*C(11,3)*4^3 + 2*3*C(11,4)*4^4 + C(11,5)*4^5]

≈ 8.84%

It's long, but it's not complicated. You can do the same thing with fractions if you want, but you still have to multiply one by C(5,2). I assume statmanhal did something similar before I did in the 2nd response of the thread except initially he ignored the condition and just had the numerator/C(50,5). Then he applied the condition by dividing by the denominator /C(50,5). I don't find this any more complicated than Aaron Brown's method. He used fractions and considered matching the hand to the board, but he still had to consider matching 0, 1, or 2 separately, and then consider the suits.