jlim
• Posts: 7
Joined: Jan 2, 2016
January 2nd, 2016 at 1:31:15 PM permalink
Hi guys,

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
PeeMcGee
• Posts: 115
Joined: Jul 23, 2014
January 2nd, 2016 at 8:22:36 PM permalink
So I’m assuming the following:
(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
jlim
• Posts: 7
Joined: Jan 2, 2016
January 2nd, 2016 at 9:50:19 PM permalink

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).
beachbumbabs
• Posts: 14268
Joined: May 21, 2013
January 2nd, 2016 at 9:59:23 PM permalink
Quote: jlim

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.
If the House lost every hand, they wouldn't deal the game.
jlim
• Posts: 7
Joined: Jan 2, 2016
January 2nd, 2016 at 10:24:44 PM permalink

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?
beachbumbabs
• Posts: 14268
Joined: May 21, 2013
January 2nd, 2016 at 11:18:11 PM permalink
Quote: jlim

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
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.
If the House lost every hand, they wouldn't deal the game.
jlim
• Posts: 7
Joined: Jan 2, 2016
January 3rd, 2016 at 8:18:27 AM permalink

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
Last edited by: jlim on Jan 3, 2016
PeeMcGee
• Posts: 115
Joined: Jul 23, 2014
January 3rd, 2016 at 10:04:04 AM permalink
Yes, my post allowed for the community to pair up. If you want no pairs on the community then the four cards (five community cards minus the one card that matches one of your hole cards) must all be different rank. So you want to choose 4 ranks from 11 (13 ranks minus the 2 ranks of your hole cards) and with each rank you are selecting 1 card from a pool of 4. So math’d out…

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
jlim
• Posts: 7
Joined: Jan 2, 2016
January 3rd, 2016 at 12:51:29 PM permalink

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 ?
Last edited by: jlim on Jan 3, 2016
PeeMcGee
• Posts: 115
Joined: Jul 23, 2014
January 3rd, 2016 at 3:24:32 PM permalink
50!/45! or P(50,5) can work for the denominator if you are assuming order matters for the numerator as well (which I didn’t). So, in your OP when you have the numerator as (6*44*40*36*32) you have to consider that the matching card can be in any of the five “slots”. Therefore, you would have to do…

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.
• Posts: 30
Joined: Jul 3, 2015
January 8th, 2016 at 1:40:00 PM permalink
I'll attempt excluding straights, flushes, and the board pairing. It depends on which preflop hand Hero has, so let's say our hand is AKo.

I'll be using card combinations, so my denominator is C(50,5).

I'll start by calculating the valid suit permutations for the board. But first, I should explain the phrase "suit permutations", which I'm using in the interest of precise language (at the expense of immediate clarity). A card is defined by a rank and a suit. A hand is a combination of cards. A combo of cards can be seen either as suit perms assigned to rank combos, or rank perms assigned to suit combos. In other words, either ranks must be permuted or suits must, but not both. If you used rank perms and suit perms, the result would be card permutations. If you used rank combos and suit combos, the result would be meaningless for poker. You'd be saying there's no difference between {Ah, Kc} vs {Ac, Kh} on a board with four hearts and the two hands would split the pot because both would have equal claim to having an Ace High flush. That's because the cards would be not fully defined and so the hands wouldn't be card combinations. {Ah, Kc} would really just be two separate sets {A, K} and {h, c}, and elements from one set would not be assigned to the other, so you wouldn't actually have what we know as "cards". By permuting one of the sets, you assign the suits to the ranks (or vice versa) and then you have cards.

With that abstraction out of the way, here are the numbers.

For the number of suit perms not resulting in a flush, we have to exclude 5 of a suit and 4 of one of our suits. There are 4 monotone perms and 2*3*5=30 perms with four of one of our suits (2 suits that can be the 4-flush on board, 3 possibilities for the other suit on board, and 5 possibilities for which rank gets the oddball suit). There are 4^5 total suit perms, so the number of non-flush suit perms is 4^5 - 34 = 990

Next I'll focus on the ranks. Since I counted suit perms, I have to count rank combos (and this, btw, is easier than doing it the other way around). We need 5 different ranks (board must be unpaired), one of which must be an A or K (but not both), and we can't have TJQ or 2345. There are 2 choices between A and K. There are C(11,4) choices for the other 4 ranks, but 1 of those combos is 2345 and 8 of them involve TJQ. So the number of valid rank combos is 2*[C(11,4) - 9] = 642

Combining the suit perms with the rank combos gives us card combos, so our numerator is 990 * 642 = 635580. Dividing that by C(50,5), we obtain a probability of almost exactly 30%.

Edit -- whoops, goofy mistake, I failed to account for the fact that 2 of the ranks only have 3 remaining, and 2 of the suits only have 12 remaining. I'll fix this when I get a chance.

Edit #2 -- I think all that needs to be fixed are my suit permutations. It's not because 2 only have 12 remaining, it's because we already know an A or K is required on the board and that it can't be the same suit as the hole card of that rank. The rank combo calculation is fine because fixing the suits will account for there only being 3 A's and 3 K's. When we pick, say, one of the 3 aces to be on the board, we're really just picking from the set of 3 suits.

Now, it's true that 4^5 suit permutations are possible, but they can't all be combined with the set of valid rank combos. For instance, if our hand is As Kd, and the board contains an Ace, then already the suit perm {d,d,d,d,d} is impossible. So what we have to count are the suit perms that can be combined with the rank combos.

There are 3*4^4 total suit perms for the rank combos, and of those, 3 + 1+4 result in flushes (3 possible suits for a monotone board, 2 possible suits for a 4-flush board, 1 or 4 possibilities for which rank is the oddball suit depending on which suit the flush is). So the number of valid suit perms is 3(4^4) - 8 = 760

760 * 642 / C(50,5) = 23.03%

Edit #3 -- arghh, above I said 3 + 1+4 but it should be 3 + 2+3*4. There are 3 possible suits for the monotone board, 2 possible suits for the 4-flush board, 2 possibilities for the oddball suit if it's the hero's rank, 3 possibilities for the oddball suit if it's not hero's rank, and 4 possibilities for which rank is the oddball suit if not hero's rank. So the 760 should instead be 751.

751 * 642 / C(50,5) = 22.76%

At least for now...
Last edited by: AlmondBread on Jan 8, 2016
jlim
• Posts: 7
Joined: Jan 2, 2016
January 9th, 2016 at 12:46:33 AM permalink

I'm not quite sure if I follow on why you did C(2,1) for the 3oak calculation. Does the C(2,1) represent matching one of the two cards dealt in the beginning? Otherwise if i'm calculating the probability for only the community card which already takes account of the situation, I'm not sure if that C(2,1) is necessary.

Let me know,
Thanks.
jlim
• Posts: 7
Joined: Jan 2, 2016
January 9th, 2016 at 12:50:31 AM permalink
Quote: PeeMcGee

50!/45! or P(50,5) can work for the denominator if you are assuming order matters for the numerator as well (which I didn’t). So, in your OP when you have the numerator as (6*44*40*36*32) you have to consider that the matching card can be in any of the five “slots”. Therefore, you would have to do…

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.

I just replied without quoting you, let me know when you see it.