Patwalshh is correct. The simplest way to look at it mathematicly is to work out the chances that the the flop won't be paired; The second card must not match the first, which is any 48 out of the 51 remaining cards, then the third card must not match either the first or the second, which is any 44 of the 50 remaining cards. (48/51)x(44/50)=0.828 So there's an 82.8% chance the flop won't be paired, therefore a 17.2% chance it will be paired. Thats the same as 1 in every 5.8 flops being paired. There are no variables. With each new flop there is exactly a 17.2% chance it will be paired. It makes no difference how many players are dealt in; the flop is always 3 random cards from 52. Over small samples you can easily see big variations, ie over 100 hands either 27 paired flops, or only 7 paired flops would be nothing to get excited about. As the samples get bigger the proportion should get much closer to 17.2%. Over 1,000 hands it'll probably be within 1% (16.2% to 18.2%). Over say 100,000 hands it'll probably be within 0.1% (17.1% to 17.3%). Posted by GaryQQQ
Exactly right with the bold. You really need to look at HUGE SAMPLE SIZES, to get a realistic figure. 100k hands minimum, more like 1m+ to be within 0.5-0.05% of the SD. I really like the debate within this thread.
@Talon, it doesn't matter at all about other players in the game because their cards are RANDOM, if we knew their cards then the calculation would be different, but you wouldn't care because you'd know their cards!
Patwalshh is correct. The simplest way to look at it mathematicly is to work out the chances that the the flop won't be paired; The second card must not match the first, which is any 48 out of the 51 remaining cards, then the third card must not match either the first or the second, which is any 44 of the 50 remaining cards. (48/51)x(44/50)=0.828 So there's an 82.8% chance the flop won't be paired, therefore a 17.2% chance it will be paired. Thats the same as 1 in every 5.8 flops being paired. There are no variables. With each new flop there is exactly a 17.2% chance it will be paired. It makes no difference how many players are dealt in; the flop is always 3 random cards from 52. Over small samples you can easily see big variations, ie over 100 hands either 27 paired flops, or only 7 paired flops would be nothing to get excited about. As the samples get bigger the proportion should get much closer to 17.2%. Over 1,000 hands it'll probably be within 1% (16.2% to 18.2%). Over say 100,000 hands it'll probably be within 0.1% (17.1% to 17.3%). Posted by GaryQQQ
What about your own hole cards though?
I agree that we can and should ignore the issue about how many other people are playing.
But the chances of the flop having exactly 2 the same rank must be different depending on whether or not we have a pocket pair.
All the times we have a pocket pair, the 50 unknown cards contain 12 ranks with 4 cards, and 1 rank with 2 cards.
Whenever we do not have a pocket pair, it is 11 ranks with 4 cards, plus 2 with 3 cards.
I am too lazy (and probably too incompetent!) to do the maths, but surely the outcome is different?
Of course, if we're ignoring our own hand, and just averaging it out over all the millions of hands we are dealt in our lives, then I agree with your method.
I have just looked at last 200 flops in games I was playing.
In the first 100 12 were paired
In the 2nd 100 15 were paired
I must admit while playing I would have said it was more than that, they appear like buses none for a while then you get a few together which makes you think there were more than there was!
In Response to Re: Odds of paired flop? : What about your own hole cards though? I agree that we can and should ignore the issue about how many other people are playing. But the chances of the flop having exactly 2 the same rank must be different depending on whether or not we have a pocket pair. All the times we have a pocket pair, the 50 unknown cards contain 12 ranks with 4 cards, and 1 rank with 2 cards. Whenever we do not have a pocket pair, it is 11 ranks with 4 cards, plus 2 with 3 cards. I am too lazy (and probably too incompetent!) to do the maths, but surely the outcome is different? Of course, if we're ignoring our own hand, and just averaging it out over all the millions of hands we are dealt in our lives, then I agree with your method. Posted by Padzz77
OK, I have tried to work this out, and actually the difference seems negligible.
If we start off with a pocket pair, then I believe we can expect a paired board 17.38% of the time.
Whereas if we start off with 2 different hole cards, then I think it is 17.16% of the time.
This is a trivial difference and one which would not show up, I imagine, unless the sample sizes where in the millions.
(Of course, as I said before, my maths may not be up to scratch.)
I have done further looking into the maths of it and realised that it is impossible to give an answer to this question, only a range is truly possible.
I then looked at a six handed table and then worked out the percentages of the best and worse case secnarios for the hole cards. The 2 answers were 23.5% and 16.2%. So this is the range of percentages for this particular scenario. Each individual hand will have its own odds due to the distribution of the hole cards but will always fall within this range, provided we dont change the number of people in the hand.
Any attempt to narrow this down would require the use of probablilities to determine the most likely distribution of hole cards, which would naturally have its own built in error factor.
So we have the range of a paired flop in a six handed game as being between 1 in 4 and 1 in 6. Probably not as accurate an answer as most would think possible but is as accurate as it can be. The odds would vary of course depending on how many players were playing. But as a site that operates predominantly 6 handed i thought that this would be the best way to go.
The odds only depend on our own hole cards (assuming we have looked at them). It doesnt matter how many other people are playing, or what their hole cards are. We dont even have to know how many other people are playing. All we need to know is that it is a standard 52 card deck, and that 3 will be dealt to the flop. Whether there will be burn cards or not also does not matter.
Everything other than our own hole cards is unknown. Thus there are 50 unknown cards (ie we dont know if they are in the deck, have been dealt to others, or have been burned). Each one of these 50 cards is exactly as likely as each other to be dealt first to the flop, second to the flop or third to the flop. So each card has a 2% chance of being dealt in each position. Thus each of those 50 cards has a 6% chance of appearing on the flop and a 94% chance of not appearing. If it does not appear, then that might be because it is in someone else's hand, or it might be burned, or it might still be in the deck. We do not know and, for this issue, it does not matter.
Scenario 1
We hold a pocket pair, with (P,P). Ie 2 cards each of rank P. And x, y, z are 3 different ranks which are different to P.
In each deck, when we know our hole cards, there are 50 cards left.
2 of these are rank P.
Any of 48 cards can be rank, x. (ie any card in the deck apart from one of the 2 which is rank P)
Once x is chosen, any of 44 cards can be rank, y.
Once both x and y are chosen, any of 40 cards can be rank, z.
The first card on the flop is chosen 1 out of 50 unkowns. The second is 1/49 and the third is 1/48.
So a flop which is (P,x,y) or (x,P,y) or (x,y,P) all give us a pair on an unpaired board, where the order listed in the brackets is the order in which the flop is dealt. (The order of "x" and "y" does not matter as "x" is the first blank to be dealt and "y" is the second).
The percentage chance of each of these 3 is [(2 x 48 x 44)/(50 x 49 x 48)] x 100% = 3.592%.
So, in total, the chance of us having a set on an unpaired board is 3 x 3.592% = 10.78%.
Whereas a flop of (x, y, z) fails to improve our pocket pair, on an unpaired board.
The percentage chance is [48/50 x 44/49 x 40/48] x 100% = 71.84%
So overall, the chance of an unpaired board when we start off with a pocket pair is 82.62%. This means, of course, the chance of a paired board is 17.38%.
Scenario 2
We do not have a pocket pair. If we have 2 different cards, (H,L), then there are 3 of each of these left in the deck.
Any of 44 cards can be rank, x. (ie any card in the deck apart from one of the 6 which are rank H or L)
Once x is chosen, any of 40 cards can be rank, y.
Once both x and y are chosen, any of 36 cards can be rank, z
Any of these outcomes is an unpaired flop that gives us one pair:
(H, x, y); (x, H, y); (x, y, H); (L, x, y); (x, L, y); (x, y, L).
So there are 6 possibilities each of which is [(3 x 44 x 40)/(50 x 49 x 48)] x 100% = 4.490%
So, in total, the chance of us having a pair on an unpaired board is 6 x 4.490% = 26.94%.
Any of these outcomes is an unpaired flop that gives us two pair:
Comments
I agree that we can and should ignore the issue about how many other people are playing.
But the chances of the flop having exactly 2 the same rank must be different depending on whether or not we have a pocket pair.
All the times we have a pocket pair, the 50 unknown cards contain 12 ranks with 4 cards, and 1 rank with 2 cards.
Whenever we do not have a pocket pair, it is 11 ranks with 4 cards, plus 2 with 3 cards.
I am too lazy (and probably too incompetent!) to do the maths, but surely the outcome is different?
Of course, if we're ignoring our own hand, and just averaging it out over all the millions of hands we are dealt in our lives, then I agree with your method.
I then looked at a six handed table and then worked out the percentages of the best and worse case secnarios for the hole cards. The 2 answers were 23.5% and 16.2%. So this is the range of percentages for this particular scenario. Each individual hand will have its own odds due to the distribution of the hole cards but will always fall within this range, provided we dont change the number of people in the hand.
Any attempt to narrow this down would require the use of probablilities to determine the most likely distribution of hole cards, which would naturally have its own built in error factor.
So we have the range of a paired flop in a six handed game as being between 1 in 4 and 1 in 6. Probably not as accurate an answer as most would think possible but is as accurate as it can be. The odds would vary of course depending on how many players were playing. But as a site that operates predominantly 6 handed i thought that this would be the best way to go.
We hold a pocket pair, with (P,P). Ie 2 cards each of rank P. And x, y, z are 3 different ranks which are different to P.
In each deck, when we know our hole cards, there are 50 cards left.
2 of these are rank P.
Any of 48 cards can be rank, x. (ie any card in the deck apart from one of the 2 which is rank P)
Once x is chosen, any of 44 cards can be rank, y.
Once both x and y are chosen, any of 40 cards can be rank, z.
The first card on the flop is chosen 1 out of 50 unkowns. The second is 1/49 and the third is 1/48.
So a flop which is (P,x,y) or (x,P,y) or (x,y,P) all give us a pair on an unpaired board, where the order listed in the brackets is the order in which the flop is dealt. (The order of "x" and "y" does not matter as "x" is the first blank to be dealt and "y" is the second).
The percentage chance of each of these 3 is [(2 x 48 x 44)/(50 x 49 x 48)] x 100% = 3.592%.
So, in total, the chance of us having a set on an unpaired board is 3 x 3.592% = 10.78%.
Whereas a flop of (x, y, z) fails to improve our pocket pair, on an unpaired board.
The percentage chance is [48/50 x 44/49 x 40/48] x 100% = 71.84%
So overall, the chance of an unpaired board when we start off with a pocket pair is 82.62%. This means, of course, the chance of a paired board is 17.38%.
Scenario 2
We do not have a pocket pair. If we have 2 different cards, (H,L), then there are 3 of each of these left in the deck.
Any of 44 cards can be rank, x. (ie any card in the deck apart from one of the 6 which are rank H or L)
Once x is chosen, any of 40 cards can be rank, y.
Once both x and y are chosen, any of 36 cards can be rank, z
Any of these outcomes is an unpaired flop that gives us one pair:
(H, x, y); (x, H, y); (x, y, H); (L, x, y); (x, L, y); (x, y, L).
So there are 6 possibilities each of which is [(3 x 44 x 40)/(50 x 49 x 48)] x 100% = 4.490%
So, in total, the chance of us having a pair on an unpaired board is 6 x 4.490% = 26.94%.
Any of these outcomes is an unpaired flop that gives us two pair:
(H, L, x); (H, x, L); (x, H, L); (L, H, x); (L, x, H); (x, L, H).
So there are 6 possibilities each of which is [(3 x 3 x 44)/(50 x 49 x 48)] x 100% = 0.3367%
So, in total, the chance of us having two pair on an unpaired board is 6 x 0.3367% = 2.02%.
Whereas a flop of (x, y, z) fails to give us any pair, on an unpaired board.
The percentage chance is [44/50 x 40/49 x 36/48] x 100% = 53.88%.
So altogether, the chance of an unpaired flop when we start without a pocket pair is [26.94 + 2.02 + 53.88] is 82.84%. Thus the chance of a 17.16%.
(IMHO)