threemonksX Joined: Nov 17, 2015
• Posts: 4
November 17th, 2015 at 11:25:46 PM permalink
at a nine player table what are the odds of two players being dealt two cards of the same suit, for example, I am dealt two clubs and another player is also dealt two clubs. Thank you.
Dodsferd Joined: Jun 10, 2015
• Posts: 375
November 17th, 2015 at 11:28:04 PM permalink
Quote: threemonksX

at a nine player table what are the odds of two players being dealt two cards of the same suit, for example, I am dealt two clubs and another player is also dealt two clubs. Thank you.

What style of poker are you talking about? I can't think of any poker games that have more than one deck used at a time, so you shouldn't be dealt the same card as someone else.

Edit:

Welcome to the forum!
This feeling is heavy, makes my body ache and I'm ready; To fall into the sky and I see now, the reason why. My heart is heavy, takes me to a place I can't breathe. Only then I know why I see the warning sign.
BlueEagle Joined: Jun 7, 2015
• Posts: 249
November 17th, 2015 at 11:44:23 PM permalink
Quote: threemonksX

at a nine player table what are the odds of two players being dealt two cards of the same suit, for example, I am dealt two clubs and another player is also dealt two clubs. Thank you.

I'm not one of the math geeks so I can't answer your question about two players being dealt the same suit. However, according to Google, the odds of a player being dealt any two suited cards is 3.3 to 1 (or 23.3% chance).
threemonksX Joined: Nov 17, 2015
• Posts: 4
November 17th, 2015 at 11:46:57 PM permalink
Dodferds thanks fot the reply. Maybe I wasn't clear. Its a nine player hold'em table. what are the odds I wil be dealt two suited cards and another player will be dealt two cards of the same suit. not the same suit and rank just the same suit.
charliepatrick Joined: Jun 17, 2011
• Posts: 2577
November 18th, 2015 at 12:52:59 AM permalink
^ The easiest way to think about it is that the first card dealt to you is any one of 52 cards and has its suit - for the rest of this thread let's assume it is a Spade. This leaves 12 Spades [cards in the same suit] from 51 cards in the deck. Thus the chances of you getting two suited cards is 12/51 (23.53%).

As to other people it gets complicated. If it was head-to-head then you'd be looking at 11/50 * 10/49 (4.49%). This also caters for the next person at the table. As you work round the table the chances change depending on how many Spades are already held by other players. A spreadsheet, with all the possible permutations, could work it out.

Another way is to look at [nearly] 2^16 ways that 11 Spades and 39 other cards could be distributed in the next 16 cards dealt and see how many result in two in a player's hand.
beachbumbabs Joined: May 21, 2013
• Posts: 14234
November 18th, 2015 at 2:02:59 AM permalink
Quote: charliepatrick

^ The easiest way to think about it is that the first card dealt to you is any one of 52 cards and has its suit - for the rest of this thread let's assume it is a Spade. This leaves 12 Spades [cards in the same suit] from 51 cards in the deck. Thus the chances of you getting two suited cards is 12/51 (23.53%).

As to other people it gets complicated. If it was head-to-head then you'd be looking at 11/50 * 10/49 (4.49%). This also caters for the next person at the table. As you work round the table the chances change depending on how many Spades are already held by other players. A spreadsheet, with all the possible permutations, could work it out.

Another way is to look at [nearly] 2^16 ways that 11 Spades and 39 other cards could be distributed in the next 16 cards dealt and see how many result in two in a player's hand.

Yeah, that makes a lot more sense. Thanks, CharliePatrick. I'm going to delete mine so it doesn't distract from your answer.
If the House lost every hand, they wouldn't deal the game.
Wizard Joined: Oct 14, 2009
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November 18th, 2015 at 6:22:41 AM permalink
Quote: threemonksX

at a nine player table what are the odds of two players being dealt two cards of the same suit, for example, I am dealt two clubs and another player is also dealt two clubs. Thank you.

The math on this is pretty messy. However, I can give an estimate pretty easily, which should be close to the exact answer.

1. The probability that any specific player has suited cards are 12/51, which is the probability the second card matches the first in suit.

2. The probability that any specific other player has two cards of the same suit is (11/50)*(10/49) = 110/2450.

3. There are combin(10,2)=10*9/2=45 ways to choose 2 players out of 10.

4. That probability there are no pairs of players where both have both cards of the same suit is (1-(12/51)*(110/2450))^45 = 62.01%.

5. The probability of at least one such pair is 1-62.01% = 37.99%.

Let's just call it 38%.
It's not whether you win or lose; it's whether or not you had a good bet.
threemonksX Joined: Nov 17, 2015
• Posts: 4
November 18th, 2015 at 7:59:50 PM permalink
Thanks to everyone who took the time to reply to my question.
Threemonks Joined: Jul 3, 2015
• Posts: 30
November 19th, 2015 at 8:59:03 AM permalink
Once, as a learning experience I calculated the exact solution in for 6max. Below is the math for that in case anyone is curious (the Wiz is right that it's messy). I used inclusion-exclusion. For 9max it's the same thing just a few more lines to add and subtract (if full precision is desired). Part of the trickiness is that there can be more than one group of 2+ players with the same suit (e.g. two players can have Clubs and another two can have Hearts).

First we look at the cases of just 1 group of 2+ players:
4 * [
C(6,2)*C(13,4) / C(52,4) - . . . (Line 1)
2* C(6,3)*C(13,6) / C(52,6) + . . . (Line 2)
3* C(6,2)*C(13,5) / C(52,8) - . . . (Line 3)
4* 6*C(13,3) / C(52,10) + . . . (Line 4)
5* 13 / C(52,12) ] . . . (Line 5)

That first line, however, double-counts the case of 2 groups of 2 players, so next we subtract it:
- C(4,2)*C(6,4)*C(13,4)^2 *(3!!)^2 / C(52,8) / 7!! . . . (Line 6)

After subtracting that, the case of 3 groups of 2 players isn't counted, so we add it back in:
+ 4*C(13,4)^3 *(3!!)^3 / C(52,12) / 11!! . . . (Line 7)

Next we consider the case of one group of 3 + one group of 2. It was counted 4 times by Line 1, -2x by Line 2 (after applying the coef of -2), and -3x by Line 6. Therefore its current tally is 4-2-3 = -1, so it must be added back twice.
+2* 4*3*6*C(13,4)*C(13,6)*(3!!)(5!!) / C(52,10) / 9!! . . . (Line 8)

Next we look at the case of 2 groups of 3 players. It was counted:
6x by Line 1
-4x by Line 2 after applying the coef
-9x by Line 6
12x by Line 8 w/coef
So its tally is +5, therefore we now subtract it 4x.

-4* C(4,2)*C(13,6)^2 *(5!!)^2 / C(52,12) / 11!! . . . (Line 9)

Finally, there's the case of one group of 2 & one group of 4. It was counted:
7x by Line 1
-8x by Line 2 w/coef
3x by Line 3 w/coef
-6x by Line 6 w/coef
8x by Line 8 w/coef
for a total of +4 so we subtract it 3x.

-3* 4*3*C(13,4)*C(13,5)*(3!!)(7!!) / C(52,12) / 11!! . . . (Line 10)

After all that, the end result is 14.058060 %

Again, that's for 6max. I myself don't feel like calculating 9max, but it would be good practice for someone wanting to test whether they understand the math shown above.
Ayecarumba Joined: Nov 17, 2009
• Posts: 6763
November 19th, 2015 at 12:06:42 PM permalink
Quote: Wizard

The math on this is pretty messy. However, I can give an estimate pretty easily, which should be close to the exact answer.

1. The probability that any specific player has suited cards are 12/51, which is the probability the second card matches the first in suit.

2. The probability that any specific other player has two cards of the same suit is (11/50)*(10/49) = 110/2450.

3. There are combin(10,2)=10*9/2=45 ways to choose 2 players out of 10.

4. That probability there are no pairs of players where both have both cards of the same suit is (1-(12/51)*(110/2450))^45 = 62.01%.

5. The probability of at least one such pair is 1-62.01% = 37.99%.

Let's just call it 38%.

The OP requested the number for a nine person table, which I think comes to ~32%. Based on my experience, this seems high, but assuming that small cards are often folded pre-flop, it is certainly possible.
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