Probability of at least one player out of five getting flush in Texas Holdem

Question

I have been working on a problem which is the following: A game of Texas Holdem poker is played. There are three cards on the table and these cards are all the same color (e.g. hearts). In total a deck with 52 cards is used and 49 cards left after the initial 3 cards are placed on the table.

There are five players around the table which are dealth two cards each.

What is the probability that at least one of the players gets flush (e.g. 2 hearts on hand combined with 3 hearts on table).

My initial idea was to use the total number of combinations (without repetition) where the sample space is C(49, 2) and the number of possible combinations to get the flush is C(10, 2).

C(10, 2) / C(49, 2) = 3,8%

The probability of the first player getting a card of the same color as on the table is (10/49) and the second time (90/2352). So in total the probability of the first player getting two cards of the same color which is also on the table should be (10/49) * (90/2352).

Another alternative seems to be to use the binomial formula which is:

Pr(x=k) = c(n, k) * p^k * (1-p)^n-k

Where n = number of trials and k = number of successes

Which with our data translates to:

Pr(x = k) c(5, 1) * [(10/49) * (90/2352)]^1 * {1-[(10/49)*(90/2352)]}^4 = 3,8%

Is this a reasonable approach and would you say that it gives the correct probabilities?

Answer 1

Easier to say, what is the chance that no one hit the flush when the flop comes up with 3 of one suit. And then, what is the compliment of that.

$1-\left(1-\frac {10\choose 2}{49\choose 2}\right)^5$

Answer 2

You are correct that the probability that the first player has a probability of about $0.038$ of getting two hearts. However, it can be a little tricky to extend this approach to find the probability that at least one player of the five has two hearts. One way to do it is to use the Principle of Inclusion and Exclusion (PIE).

Let's say the game has "Property $i$" if player $i$ has two hearts, for $i=1,2,3,4,5$. Let $S_j$ be the total of the probabilities of the games with $j$ of the properties, for $j=1,2,3,4,5$. Then $$\begin{align} S_1 &= \binom{5}{1} \binom{10}{2} / \binom{49}{2} \\ S_2 &= \binom{5}{2} \binom{10}{4} / \binom{49}{4} \\ S_3 &= \binom{5}{3} \binom{10}{6} / \binom{49}{6} \\ S_4 &= \binom{5}{4} \binom{10}{8} / \binom{49}{8} \\ S_5 &= \binom{5}{5} \binom{10}{10} / \binom{49}{10} \\ \end{align}$$ By PIE, the probability that a game has at least one of the properties, i.e. at least one player has two hearts, is $$S_1-S_2+S_3-S_4+S_5 = \boxed{0.181565}$$