ok so I admit I have stepped into a headache thinking it wouldn't be too bad. I'm not too bright with probability,

so bear with me and give me some advice. I am currently calculating the probability of drawing strictly one pair in

texas hold em'.

So for instance, I would be dealt two cards and one of the two cards would have to match one of the

five cards proceeding. Lets say I was dealt 10 of hearts and Ace of Spades. I want to calculate the chance

of having one card of the same rank.

I calculated (6x44x40x36x32)/50P5=0.0478...

6=all the different cards that can pair one of the two cards I'm initially dealt.

44=52-2(the dealt)-1(the match)-5(the remainders of 6 possibles)

40=44-1(the dealt)-3(the remainder of same rank)

...

I'm not sure if this made sense, but this is what I've got up to in the last 15 minutes.

And I've done some researching but none of the questions set up the equation so that both

the denominator and numerator would be a calculation based on 50 cards, the number after

excluding two initial cards.

I know there's something wrong in this calculation but I just can't understand any other way to approach it.

I think i'm just stuck because I haven't got the hang of the way to approach it.. Let me know!!

It would be much appreciated

(a) given you have two hole cards of different rank

(b) exactly one of the community cards (5 cards) will match the rank of a hole card

Then, you would want 1 card out of 6 possible matching rank cards. The other 4 cards can be any of the remaining 44 cards. Therefore…

C(6,1) * C(44,4) / C(50,5)

= 58179 / 151340

= 0.3844

C(x,y) is the combination function

I was just a little lost on why you did C(44,4). If I want to have strictly one pair by matching one of my dealt cards(2)

with one of the community cards(5) I would have to cancel out the potential matches. Therefore I did C(6,1)*44*40*36*32.

I know this doesn't make sense because the numerator would exceed the denominator, but I just want to know the

reasoning for C(44,4).

"the other 4 cards can be any of the remaining 44 cards."

I agree but that would only be applicable to the 2nd community card(5).

Quote:jlimThanks for the reply,

I was just a little lost on why you did C(44,4). If I want to have strictly one pair by matching one of my dealt cards(2)

with one of the community cards(5) I would have to cancel out the potential matches. Therefore I did C(6,1)*44*40*36*32.

I know this doesn't make sense because the numerator would exceed the denominator, but I just want to know the

reasoning for C(44,4).

"the other 4 cards can be any of the remaining 44 cards."

I agree but that would only be applicable to the 2nd community card(5).

He's isolating out your request for EXACTLY one pair that includes a card from your hand, from "one pair or better", though not eliminating flushes and straights; there will be no 2 pair, no FH, no 3oak or 4oak, etc THAT INCLUDE YOUR HAND in more than 1 pair of it. Let's say you have a 7 and 8 offsuit. There are 3 more 7's and 3 more 8's in the deck. You need one of those, AND not more than 1. So, you've matched 1, and you're not using the other 5 cards. The other 44 cards don't match, but we don't care what they are, just that 4 more cards have to be in the community cards, so you can have any 4 from those.

Perhaps you were looking for him to answer a different question, but you'd have to ask one.

This clears up much more. No this was the answer I was looking for, I just had another restriction set up in my mind. The reason I mentioned the 44*40*36*32 sequence was to eliminate any chance that there would be a one pair set up within the community card. For instance, I could have a matching 10 from my cards and the communities card but I wouldn't want another match within the 5 community cards. So something like this 10S(the match), 4S, 4D, and etc. If something like this happened wouldn't it be technically a two pair inclusive of the one you matched?

Quote:jlimThanks for the reply,

This clears up much more. No this was the answer I was looking for, I just had another restriction set up in my mind. The reason I mentioned the 44*40*36*32 sequence was to eliminate any chance that there would be a one pair set up within the community card. For instance, I could have a matching 10 from my cards and the communities card but I wouldn't want another match within the 5 community cards. So something like this 10S(the match), 4S, 4D, and etc. If something like this happened wouldn't it be technically a two pair inclusive of the one you matched?

Yes, it would be inclusive. You could have a 4oak on the board, and pair the kicker, as an extreme example, or pair the board and have a 3oak there as well for a FH, according to the way he did the math. Probably part of the reason it's around 38% of the hands that you'll pair the board, even if your pair is counterfeited. So if you need to eliminate all 2 pair hands and above, that's a pretty complicated calculation (at least to me), because you have to determine the subset of hands that rise above 1 pair while still using one of your card, not just subtracting all of those possible hands.

So, I think you have to ignore suits for the moment, and just look at ranks.

13 ranks for your 1st hole card

12 ranks for your 2nd.

6 cards that match those two.

11 ranks left

10 ranks "

9 ranks "

8 ranks " gives you 7 cards, ignoring straights and flushes.

Combin(13,1)*Combin(12,1)*Combin(3,1)^2*Combin(11,1)*Combin(10,1)*Combin(9,1)*Combin(8,1)=11119680

*4 suits = 44478720

divided by total possible 5 card hands out of 7 cards = 44478720/133784560 = .332465

However, that 44478720 has to have all possible straights and flushes subtracted, while retaining the 1 pair that includes your hand, to be correct. I'm not sure how to find that subset accurately.

Ok I thought it had to be inclusive too, 38% was a little too high. I just had a couple questions regarding your calculations. I get the approach that you want to calculate by the each ranks rather than by the whole, it's easier that way. But wouldn't the calculation be inaccurate because not all suits will have 11 ranks left. For instance, If I draw two cards both different in suits or perhaps even the same, I would be left with possibly 12 or 11 respectively to begin with(after drawing the 2 dealt cards). Then it wouldn't be 11C1 for I wouldn't have 11 cards to be even considering. I feel that by looking at ranks of individual suits it has created a new problem on its own. The card which is drawn after the first match in the community card could completely change the outcome.

At this point I wouldn't really care about straights and flushes, although they would be the stronger hand, it's the one pair pattern i'm looking for.

Let me know thanks

C(6,1) * C(11,4) * C(4,1)

^{4}/ C(50,5)

= 6*330*256 / 2118760

= 12672 / 52969

= 0.2392

Again, this does allow for straights and flushes

This makes much more sense now, thank you!

And sorry this is a really dumb question but why wouldn't 50!/45! work for the denominator?

One last thing, I'm trying calculate the occurrence for everything up till Royal Flush, does the calculation get any

easier? I'm just guessing that because there would be 'smaller boundaries' to play with. ie) 3oak= (6C1)^2 * 11C3 * (4C1)^3 ?

5*6*44*40*36*32 / P(50,5)

= 5 * 0.0478

= 0.239

For exactly 3oak (with the three other community cards all different ranks)

= C(2,1)*C(3,2)*C(11,3)*C(4,1)

^{3}/ C(50,5)

= 0.0299

You just have to be careful what assumptions you are making…which has the largest influence on difficultly of the calculations in my opinion.