Probability: Flushes in Hold 'Em

[DBMiller]

I guess I'll make one more probability post before the US money is withdrawn.

I was trying to find another good question, and decided to look at flushes in Hold-Em. We'll make this a two-parter. And I'll just give you my computations up front instead of doing a poll like last time. I will at least put my answers as separate replies so you can think about it before reading the replies. If you really want to do the math yourself, don't scroll down to later replies until you have done so.


Question #1:

You are spectating at the final 10-Player table of a tournament where the top 9 places get paid. The Big Blind is All-In and everybody calls, and everybody checks it down to the river, hoping that someone will knock out the BB and everybody else makes the money. The board does not pair, and contains three hearts that cannot make a straight flush. What is the probability that someone at the table has a flush?


Question #2:

You were the BB at the above table who is All-In, and you hold 2h3h. What is the probability that you will win the hand, i.e., it is the only flush?

[DBMiller]

Question #1:

You are spectating at the final 10-Player table of a tournament where the top 9 places get paid. The Big Blind is All-In and everybody calls, and everybody checks it down to the river, hoping that someone will knock out the BB and everybody else makes the money. The board does not pair, and contains three hearts that cannot make a straight flush. What is the probability that someone at the table has a flush?

This question deals with the issue on an "a priori" basis - Knowing nothing about any hand at the table, the only thing you see is the 5 known cards on the board.

So what do we know?


There are 47 unknown cards between the deck and all the players' hands
There are 10 hearts distributed between the deck and all the players' hands

To have a flush, a player needs to have their hole cards be two of the ten remaining hearts. This can happen in C(10,2) = 45 different ways. Or if you want to look at it another way, there are 10 possible hearts for the first card, 9 for the second, and order does not matter. So 10 * 9 / 2 = 45.

The number of possible hands a player could have had is C(47,2) = 1081. So the probability of one specific player having a flush is 45/1081, or 4.16%.

And finally since there are 10 players at the table, the probability of someone having a flush is 450/1081, or 41.63%.

Edit: I made a mistake here, to be explained in a followup

[DBMiller]

Question #1:

You are spectating at the final 10-Player table of a tournament where the top 9 places get paid. The Big Blind is All-In and everybody calls, and everybody checks it down to the river, hoping that someone will knock out the BB and everybody else makes the money. The board does not pair, and contains three hearts that cannot make a straight flush. What is the probability that someone at the table has a flush?


Question #2:

You were the BB at the above table who is All-In, and you hold 2h3h. What is the probability that you will win the hand, i.e., it is the only flush?

The second question deals with conditional probability. We know more about the cards - Namely we have the worst possible flush, and we can be beat by any other flush at the table.

So what do we know?


There are only 45 unknown cards
There are only 8 hearts left to choose from
There are 9 other players to consider
There can be up to 4 more flushes :eek:

For another player to have a flush can now happen in C(8,2) = 28 ways out of C(45,2) = 990 possible hands. Any one of 9 players could have the hand, so the probability of at least one opponent holding a flush is 9 * 28 / 990 = 14/55 (About one in four)

For two more opponents to have a flush, means a second player has C(6,2) = 15 ways out of C(43,2) = 903 hands. Since we don't care which two players they are, it can be any two of the nine, or C(9,2) = 36 combinations. Putting these numbers together with two of the previous values, (28 * 15 * 36) / (990 * 903), means two opponents will have flushes 8 out of every 473 deals. (One in fifty-nine)

For three more opponents to have a flush, we are adding C(4,2) = 6 flushes and C(41,2) = 820 hands, with C(9,3) = 36 ways to pick the three opponents. Reducing fractions gets us to 28/96965

And lastly for four opponents, there's only one flush remaining out of C(39,2) = 741 hands, C(9,4) = 126 ways to pick which 4 players have flushes. Reducing fractions gets us to 14/23950355

Now we can finally determine if the BB is the only flush. That probability is computed by the equation...

P = 1 - P(1) + P(2) - P(3) + P(4)

I won't go into too much here, other than to say it uses the "inclusion-exclusion" method of combining probabilities. When determining if one opponent had a flush, I didn't care what the other players cards were, but they could be any of the remaining cards and might also contain a flush.

So if you think about it, all the times I counted where two more opponents had a flush were also counted when we were looking at when only one opponent had a flush. Similarly, the twos were counted in the threes, and the threes in the fours. This equation just removes the multiple occurances.

Plugging all the numbers from earlier into this equation says the probability that the BB will win the hand with the flush is 76.2%.

[Ladyluk3]

omg db you certainly put a lot of thought into this one, im very impressed. but you can be sure if i am at that table and think i have the winning hand, someone will definitely have the flush. :roflmao: :roflmao: :roflmao:

[AINTITDUCKY]

:dance: I'm in awe. I would not even know where to start figuring this out mathematically. :oops:

I would think that if I am the BB sitting with 2H,3H, and even with 3 other hearts showing I am going to get beat by a higher flush. I would not have guessed the odds at 76% would have been that high for this hand to win.

Your totally awesome DB. :worship:

[DBMiller]

And finally since there are 10 players at the table, the probability of someone having a flush is 450/1081, or 41.63%.


:oops: I made a mistake (Call Ripley's :silly:).

I had it in my notes that I needed to do inclusion/exclusion on the above numbers, and was going to steal from my work on question #2. But the number of unknown players/cards changed and I forgot to go back and complete it.

I fell into the classic mistake of "overcounting".

To explain what overcounting is, let's take a look at flipping a coin; either three times in succession, or three coins at once. Since we all agree that the probability of a "true" (non-weighted) coin coming up heads is .5, you can't just say that the probability that there will be at least one head in that collection of three flips is 3 x .5 = 1.5 ... A probability of greater than 1 is impossible!

So what went wrong? In this case it is easy just to list the possible outcomes...

Coin# # of Heads
----- ----------
1 2 3 0 1 2 3
----- ----------
H H H + + + Y
H H T + + Y N
H T H + + Y N
H T T + Y N N
T H H + + Y N
T H T + Y N N
T T H + Y N N
T T T Y N N N


In the above "chart" each of the possible 8 outcomes for the coins is listed on the left. Whether or not that flip contained a certain number of heads is listed on the right. Note that not only is there just a Yes or No that the case had that many heads, in some boxes there is a "+" indicating that there were MORE than the number we were expecting.

So for the above chart we get zero heads exactly once, but zero+ heads all eight times. Likewise just one head comes up three times, yet 1+ heads seven times. Two heads comes up three times as well, with 2+ heads four times. And finally three heads comes up just once.

So depending on our counting method, we get 8 + 7 + 4 + 1 = 20 items counted, yet there were only eight possible cases.

But you may note that 8-7=1, 7-4=3, and 4-1=3, and the correct individual answers are obtained by subtracting the extra counted items from each other.

In coin flips, this is easy to map out and see. But depending on which piece of information you are looking for, it's often easier to do the inverse operation, which we did in an earlier problem posted as a poll. The probability of there being one OR MORE heads is easier determined by computing the probability that the coins will ALL be tails. And .5 x .5 x .5 = .125 or 1/8, and so there will be one or more heads 7/8 or the time, which we also counted in the above chart.

Applying that methodology to the flush question, there are 45/1081 ways for a person to make the flush, so 1036/1081 ways to not make it. And since there are ten players at the table, (1036 / 1081) ^ 10 = .6536 probability of no flush at the table, or 34.64% chance there is at least one flush. Compare this to the original 41.63% and you'll see that I over-counted just a bit :oops:

This number is not always as accurate as the counting method, where we can specifically see and count each and every possibility. Also the cards already dealt influence what hands can still be made (A single heart dealt to a player reduces the number of possible flushes other players can make - While coin flips are completely independant). But when dealing with large numbers of poker hands, it's close enough for government work.

[101stAB714]

:oops: I made a mistake (Call Ripley's :silly:).

So what went wrong? In this case it is easy just to list the possible outcomes...it's close enough for government work.
Hmmmm Mistake??? Let me guess you used the local RNG to do your calculations.