Tough question for the odds gurus here

I've seen some pretty complex math here, people figuring odds. I posted a question last week that nobody answered, so I'm trying again.

In a 9 player game, what are the odds against 3 players flopping sets?

It seems it would be the odds of 3 players of 9 having pocket pairs, times the chances of each of them flopping their set.

Since it happened to me, I figure it's probably the rarest thing I've ever seen, and I'd like to know just how rare.

Well, to hold a pair before the flop is 16 to 1; so, in a nine handed game, 1 player would have a pocket pair (16/9) to 1 which is 1.78 to 1; therefore, 3 would have a pocket pair (1.78*3) to 1, which is 5.33 to 1. The probability of making a set with a pocket pair on the flop is 7.51 to 1, so three people making a set would be (7.51^3) to 1, which is 423.7 to 1, which is then multiplied by the 5.33 to 1 odds we got earlier, giving us 2257.6 to 1.

So by my calculations, and I hope someone corrects me if I'm wrong: In a 9 handed game 3 players with pocket pairs will make their set on the flop 1 time out of 2257.6 hands.

*Odds for holding a pair preflop and making a set on the flop were taken from Doyle Brunson's Super System.

dam nice calculations war eagle

looks about right to me cuz i've only ever seen it happen once

WarEagle1266 said:

Well, to hold a pair before the flop is 16 to 1; so, in a nine handed game, 1 player would have a pocket pair (16/9) to 1 which is 1.78 to 1; therefore, 3 would have a pocket pair (1.78*3) to 1, which is 5.33 to 1. The probability of making a set with a pocket pair on the flop is 7.51 to 1, so three people making a set would be (7.51^3) to 1, which is 423.7 to 1, which is then multiplied by the 5.33 to 1 odds we got earlier, giving us 2257.6 to 1.

So by my calculations, and I hope someone corrects me if I'm wrong: In a 9 handed game 3 players with pocket pairs will make their set on the flop 1 time out of 2257.6 hands.

*Odds for holding a pair preflop and making a set on the flop were taken from Doyle Brunson's Super System.

Click to expand...

It's a good try, but not quite correct. You forget to take into account the fact that one player flopping a set affects the odds of the next player flopping his (because there are only 2 cards that can make his set). You also made a similar error preflop.

It is going to become pretty complex, so I'm going to do this in two parts.

Right now we'll figure out the odds of 3 players hitting sets when there are 3 players going to the flop, all with pocket pairs.

Obviously, we must assume they have different pocket pairs, or this would be impossible. Basically, we need to have one of each card flop. For example, say the players have 44 55 and 66. We need the flop to be 4 5 6 or some rearrangement of that.

First the chances that the first card is a 4 5 or 6 is 6/46. Then the second card must be one of the other two numbers. There will be 4 outs out of 45 unknown cards so the chances is 4/45. Similarly the chances that the third flop card will be the remaining pair is 2/44.

Multiplying these gives 6/46 * 4/45 * 2/44 = .000527, or 1 in 1897.5.

So one of every 1897.5 times three players all have pocket pairs and see the flop, all three will hit their set.

More to come.

Part 2

After doing the calculations, it turns out that preflop, one player having a pocket pair has almost no effect on the chances that another player has a different pocket pair. It produces less than a 0.5% error to simply assume that the odds of each of the three players getting a pocket pair is 3/51 in each case.

So assume (3/51)^3 is the chance that three players have a pocket pair when there are three players. Multiplying by 9 choose 3 should give us the correct answer. So (3/51)^3 * 84 = 0.01710.

I'm about 95% sure that this is correct to a negligible margin of error, but if someone feels that this is wrong please say something.

Anyway.... chances that 3 people have a pair (0.01710) * chances that all three people hit their set (.000527) gives us the chances that, in any random hand, three people will hit a set on the flop is .000009010. This is 1 in 110,983.

As I said, I'm only about 95% sure this is correct, so feel free to ask questions or question my assumptions.

Had to delete my previous post to post this, because apparently I can only post 7 messages a day as a new member; wish I knew that before I wrote my long reply and then lost it all. Here's the short version: Not sure about your 2nd calculations, but I found a different way of looking at it.

If one player deals out 2 face down cards and 3 for the flop, the odds that he will have a pocket pair and make a set in the same hand is (16*11.76) to 1, which is 188.16 to 1. So, if 9 players are playing, it is 3 times as likely that this situation will happen for just one player, or (188.16/9) to 1, which is 20.9 to 1. So, the chances of 3 players having a pocket pair and making a set on the flop would increase the odds by the third order, being (20.9^3) to 1, which is 9138.1 to 1 odds. This seems like the most simple solution to me, which usually means it's right. What do you think James?

Don't ask me. In the carbon freeroll I was trying to argue that A,J suited is a coin flip vs, J,J. Not!....

I still think that you are missing the fact that one player hitting a set decreases the chance that the other player hits their set (greatly).

To demonstrate this, take your logic one step further. What are the chances of 4 players having pocket pairs and all hitting their sets on the flop? By the math above, approximately 190802:1. But of course, the true answer is zero - four players can't all hit their set. That's because the first three players hitting sets decreases the chances that the fourth person does (in this case to zero). Extending my method above, you would get zero as one of the terms in the numerator will become zero, so my method checks out at least in that respect.

Unfortunately, the simplest answer is not usually the right one in probability, which is why I find it so interesting.

So, we have answers of about 2000-1, and about 100,000-1, and they both look right. 111,000-1 feels more right, because at 2000-1, it would be half as likely as making quads from a pocket pair. I've seen quads from at least a dozen times, and this is the first time I ever saw this.

jamesdadeliverer said:

After doing the calculations, it turns out that preflop, one player having a pocket pair has almost no effect on the chances that another player has a different pocket pair. It produces less than a 0.5% error to simply assume that the odds of each of the three players getting a pocket pair is 3/51 in each case.

So assume (3/51)^3 is the chance that three players have a pocket pair when there are three players. Multiplying by 9 choose 3 should give us the correct answer. So (3/51)^3 * 84 = 0.01710.

I'm about 95% sure that this is correct to a negligible margin of error, but if someone feels that this is wrong please say something.

Anyway.... chances that 3 people have a pair (0.01710) * chances that all three people hit their set (.000527) gives us the chances that, in any random hand, three people will hit a set on the flop is .000009010. This is 1 in 110,983.

As I said, I'm only about 95% sure this is correct, so feel free to ask questions or question my assumptions.

Click to expand...

My first question is where did the 84 come from? I did the math but came to a different conclusion. Here is what I did:

I agree with you post flop. (6/46)*(4/45)*(2/44)=0.000527

Preflop however I see a problem. If you have 9 people, the chance that any one will get a pocket pair is (3/51)*9= 27/51 (which is fairly common, 1.8888 to 1). Once it is established one player has a pocket pair, the chances of a second player is (3/49)*8. Note a few things, because it doesn't matter which order the cards are dealt nothing changes in the math. The 49 represents the deck minus the first pocket pair and the first hole card of the second player. (3/49)*8=24/49. Repeat for a 3rd player where 47 is the deck minus the first two pocket pairs and player three's first hole card. (3/47)*7=21/47. Multiply them all together. (27/51)*(24/49)*(21/47)=0.11586 or 8.63-to-1. It seems correct because the likelyhood of 3 players getting pocket pairs preflop is not uncommon.

So if we assume 3 players are dealt pocket pairs and all see the flop then (0.11586)*(0.000527)=0.0000611 or a 16,377-to-1 chance of all seeing the flop and each hitting a set.

I think that my method is closest. Your shorthand method makes sense but I think you screwed something up. (3/51)^3 makes perfect sense, but that 84 is what is messing up your results. Again I'm not sure where it came from. If you take (3/51)^3 and multiply it by 9*8*7 (which represents the possibility of remaining players being dealt a pocket pair), then you get [(3/51)^3]*9*8*7=0.1023 which is significantly closer to the 0.11586 that I got.

The shorthand method is fine to approximate as you did, but that 84 still doesn't make sense. Could you please explain where you got it from if it wasn't an error? Using my shorthand method, 9*8*7=504 where your 84 would be.

From my observation, I have it as 12+ times as rare as a pocket pair hitting quads, which is 1000-1. So the answer of 16000-1 seems intuitively right to me.

You're preflop stuff seems intuitively way off to me. 3 players definitely don't have a pocket pair that much of the time.

9 choose 3 = 84. Basically, that means that when there are 9 people, there are 84 ways for 3 of them to be chosen. If I am thinking about it right, there should then be 84 times as many ways for 9 players to have 3 pocket pairs as for 3 players to have pocket pairs, which is what we estimated before.

This doesn't fully explain the discrepancy between your numbers and mine, and I'm not sure exactly why it is.

I'm going to go gather some statistical evidence to approximate a probability.

Man look

Swear to god dude if it happened online it really dosent matter, they can do whatever they want in the name of variance. I have seen so much straight bull, that can not be explained by any form of logic, so you just play at your own risk "its a jungle" out there

Hi, my name is James, and this is an odds thread. Can I help you? Would you like some water?