1. ## Poker Problem

Here is the problem:

There are 52 cards total in a deck of cards.

There are 4 suits of cards in the deck, spades, hearts, clubs, and diamonds.

They are divided equally into 13 cards of each suit.

You have been dealt 3 cards that are spades.

There are 5 other players also playing, that were also dealt three cards each.

Leaving 34 cards left in the deck, 3 cards each in their hands, and 3 cards in your hand.

It is unknown to you what cards are left in the deck, and what cards are in the other players hands.

There are 5 cards that will be dealt onto the table, that all the players may use to improve their hand.

What percentage chance is there, that out of those 5 cards to be dealt, that two will be spades?

2. ## Re: Poker Problem

Hello, GTMotion!

There are 52 cards total in a deck of cards.
There are 4 suits of cards in the deck, spades, hearts, clubs, and diamonds.
They are divided equally into 13 cards of each suit.

You have been dealt 3 cards that are Spades.
There are 5 other players also playing, that were also dealt three cards each.
Leaving 34 cards left in the deck, 3 cards each in their hands, and 3 cards in your hand.
It is unknown to you what cards are left in the deck, and what cards are in the other players hands.

There are 5 cards that will be dealt onto the table, that all the players may use to improve their hand.

What percentage chance is there, that out of those 5 cards to be dealt, that two will be Spades?

I will assume you are hoping for a Flush.
And you hope that among the 5 cards dealt, at least two are Spades.

Believe it or not, we can ignore the cards dealt to the other five players.
The problem is between you and the remaining deck of cards.

You have 3 of the Spades.
The remaining 49 cards contain: 10 Spades and 39 Others.

There are: . ${49\choose5} \,=\,1,\!906,\!884$ possible deals.

There are four ways to get your Flush.

$\begin{array}{cccc}\text{2 Spades, 3 Others:} & {10\choose2}{39\choose3} & =& 411,\!255 \\ \\[-3mm] \text{3 Spades, 2 Others:} & {10\choose3}{39\choose2} &=& \;88,\!920 \\ \\[-3mm] \text{4 Spades, \;1 Other:} & {10\choose4}{39\choose1} &=& \;\;\; 8,\!190 \\ \\[-3mm] \text{5 Spades, 0 Others:} & {10\choose5}{39\choose0} &=& \;\;\;\;\; 252 \\ \\[-4mm] \hline & \text{Total:} && 508,\!617\end{array}$

Therefore: . $P(\text{Flush}) \;=\;\frac{508,\!617}{1,\!906,\!884} \;=\;0.266726765 \;\approx\;26.7\%$

3. ## Re: Poker Problem

Thank you so much! There is a game called super holdem where people get 3 cards instead of 2, and people would always say 3 suited cards is a 10% chance of a flush, and I always thought that was incorrect but never could prove it, now I can!

4. ## Re: Poker Problem

In case this makes you feel "safer": ran simulation, a million games twice,
and got 266,741 and 266,796. So Soroban bang on!