Thanks for reading - I have 3 math questions I'm not sure about...

1) How to calculate the odds of flopping a full house vs. quads.

2) This might be easy but I can't find it where people agree - How to calculate the odds of set over set.

3) How to calculate the odds of that happening in back to back hands.

Full 10 player table, flopping a full house to quads, and on the very next hand flopping bottom set to top set.

Thanks for any help with this, it's mainly for curiosity as I couldn't believe it happened. I laughed my ass off when the V rolled over top set on the very next hand!

50/50
either it happens or it doesn't

Solid, thanks.

By no means should any of this be taken as correct, until verified by a better skilled poster. Though, let's see if I can get on the right direction.

The full house vs. quads is probably the tougher of the two to calculate, so I'll start with the set over set.

I'd think of it this way, though my entire line of thought could be wrong... The odds of getting a pair are 1/17. So the odds of two players getting pockets pairs are 1/17 x 1/17.

The odds to complete the three of a kind by the river are 4.2 to 1

Based upon that, I'd calculate the odds of set over set are 1/17 x 1/17 x 1/4.2 x 1/4.2, or 1 in 5098 hands.

That's assuming that you were asking the result by the river. If both players flopped the set it'd be less likely. It'd be more on the order of 16256 to 1.

...

For the second part, check this thread http://forumserver.twoplustwo.com/25...quads-1302867/

There's too many variables for me to determine the answer, but if you read through it you may be able to determine the answer.

Then just multiply the result by either 5098 or 16256.

That's my thinking anyway. Good luck.

Thanks GoFLOPyourself - much appreciated. In both scenarios it was on the flop. I'm curious about it since I've never seen it happen so I'm wondering how unlikely it actually is. Thanks again for your help.

Depends On how many people are sitting at the table if I understand the question.

It was 10 handed for both hands.

Thanks! Very much appreciated.

No I was wrong hold on

For two players, it's about 0.0003%.

Thanks again for taking the time. I'll look forward to seeing how you came up with that. I didn't even know where to start.

No problem. I am willing to bet that my answer is wrong, but hopefully in the ballpark. I post here to learn, and usually end up doing just that.

Another post from another forum I found, stated that the odds were 7,800:1 at a full ring table of the set over set scenario.

I'm sure someone will come along and answer it correctly.

One thing to consider though, is that asking what the odds of these two hands back to back are certain to dramatically increase the odds of it happening.

Lets say the odds of both scenarios was 10,000:1. Then the odds would be stated as 100,000,000:1, for both scenarios back to back, even though the most significant part of the odds would be derived from the fact that they happened back to back. You could even think of it this way, if the hands were 56 hands apart, and you ask w hat the odds were of it happening exactly 56 hands apart, again it'd 100,000,000:1 ( I think, lol).

With these kind of calculations, the small things matter, a lot.

And I bet the Villian was happy about the result too.

Exactly, that was the thing that made it noteworthy for me. I've had set over set plenty, full house vs quads twice now, but the back to back part is what peaked my curiosity. I figured the odds have to be astronomical. The look on the dealer's face was priceless haha.

I'll start with #2. OP said 10-handed and I'm assuming Hero has to be the one with the lower set. But I did it for Hero and also for any two players.

Let F = Pr(3 different ranks on flop) = C(13,3) * 4^3 / C(52,3)

Any players (hero or not): P(set>set OR set>set>set | 10-handed) =
F *
[ C(10,2) * C(3,2)^3 / C(49,4) / 3!!
- 2 * C(10,3) * C(3,2)^3 / C(49,6) / 5!! ]
=~ 1 in 642

Specifically with hero: P(hero lower set | 10-handed) =
F *
[ 9*C(3,2)^3 / C(49,4) / C(4,2)
- C(9,2) * 2*C(3,2)^3 / C(49,6) / 5!! / 3 ]
=~ 1 in 6420

The part in magenta could have also been: C(6,2) / 3!!

#1. Let F = P(xxy flop) = 13*12 * 6*4 / C(52,3)

Any players: P(quads>set | 10-handed) = F * C(10,2) * C(3,2) / C(49,4) / 3!!
=~ 1 in 27792

Hero: P(set<quads | 10-handed) = F * 9 * C(3,2) / C(49,4) / C(4,2)
=~ 1 in 277924


#3. For back-to-back just multiply the answers to #1 and #2. (But that's for a specific 2 hands dealt, not if you mean any 2 b2b hands in a session of N hands.)

I think #2 has to be framed more precisely. Is it "if I know I have a pair, what are the odds I'll flop a set and someone else will flop a higher set?" or "if I see I've flopped a set, what are the odds someone else has a higher set?" Also, I think it has to depend on what pair you have --the odds are quite different if you're holding KK vs. 22. I would guess the reason you're finding different answers is that people are framing the question in slightly different ways.

I framed it in a third way altogether: "Before the hole cards are dealt, what is the chance I'll get a pair and then an inferior set?"

I interpreted it that way because of OP's context of the hands happening back-to-back.

Btw, I didn't know cranberry tea was a thing and now I must try it! Judging by the experience of cranberry juice (the pure, unsweetened kind), I bet the tea is delightful.