# Odds of this happening are?

• Bronze
Joined: 01.05.2012
So here's one for somebody who likes Poker Maths to work out.

This happened to me twice yesterday within 80 hands.

I got it all in on the flop both times where I was winning set over set, both times on the turn the miracle card came to give them Quads. Talk about being coolered huh?

So what's the odds of this happening in such a small timescale - or does that not even matter?
• 5 replies
• Bronze
Joined: 21.08.2012
well I once read in NLHE, odds of set over set are 1 in 100 approximately (somone on 2+2 did the math). And the odds of someone hitting a one outer after this occurance alll in on the flop is like 4%. That's all the math I know : ) so odds of this happening twice (in 80 hands) is .... very unlikely : )
• Bronze
Joined: 25.01.2009
Let's see, considering you hold a pair (let us use 77 as an example) and your opponent has another pair (let us assume 44), we still have 48 cards left. Considering that we have 17296 possible flops. To a set- over-set to occur we need one of the remaining 7s and one of the remaning 4s, so we have about 176 ways of a flop like this ocurring (2*2*44), considering that our probability of success is ~1.01^(-2) or ~1.02%, let's call it event A.

After that the probability of having a quad on the turn is around ~2.22%, lets call It event B.

Considering that the probability of event A and B happening is around 0.0226% (probability of A*probability of B)

That was a quick calculation I made here by hand and calculator, I might be wrong.

Now about the probability of that happening twice in 80 hands, I won't make the calculation, but the probability of that happening is approximately zero.

By the way, that's the calculation on a heads-up situation, if we consider that more pairs see the flop the odds can change.

EDIT

When I say "after that the probability of having a quad on the turn is around 2.22%" I'm talking about the probability of YOUR OPPONENT turning a quad, the probability of YOU or YOUR OPPONENT turning quads is around 4.44% and, considering that, the probability of EITHER YOU OR YOUR OPPONENT having set-over-set and then quads is about 0.0452%.

Cheers
• Bronze
Joined: 01.05.2012
Originally posted by tieppofer
Let's see, considering you hold a pair (let us use 77 as an example) and your opponent has another pair (let us assume 44), we still have 48 cards left. Considering that we have 17296 possible flops. To a set- over-set to occur we need one of the remaining 7s and one of the remaning 4s, so we have about 176 ways of a flop like this ocurring (2*2*44), considering that our probability of success is ~1.01^(-2) or ~1.02%, let's call it event A.

After that the probability of having a quad on the turn is around ~2.22%, lets call It event B.

Considering that the probability of event A and B happening is around 0.0226% (probability of A*probability of B)

That was a quick calculation I made here by hand and calculator, I might be wrong.

Now about the probability of that happening twice in 80 hands, I won't make the calculation, but the probability of that happening is approximately zero.

By the way, that's the calculation on a heads-up situation, if we consider that more pairs see the flop the odds can change.

EDIT

When I say "after that the probability of having a quad on the turn is around 2.22%" I'm talking about the probability of YOUR OPPONENT turning a quad, the probability of YOU or YOUR OPPONENT turning quads is around 4.44% and, considering that, the probability of EITHER YOU OR YOUR OPPONENT having set-over-set and then quads is about 0.0452%.

Cheers
Thanks for taking the time to do that, I appreciate it.

Obviously I knew I had been extremely unlucky not once but twice, but just wanted to put a number on it.

Cheers.
• Bronze
Joined: 25.01.2009
Originally posted by andiofwbafc
Originally posted by tieppofer
Let's see, considering you hold a pair (let us use 77 as an example) and your opponent has another pair (let us assume 44), we still have 48 cards left. Considering that we have 17296 possible flops. To a set- over-set to occur we need one of the remaining 7s and one of the remaning 4s, so we have about 176 ways of a flop like this ocurring (2*2*44), considering that our probability of success is ~1.01^(-2) or ~1.02%, let's call it event A.

After that the probability of having a quad on the turn is around ~2.22%, lets call It event B.

Considering that the probability of event A and B happening is around 0.0226% (probability of A*probability of B)

That was a quick calculation I made here by hand and calculator, I might be wrong.

Now about the probability of that happening twice in 80 hands, I won't make the calculation, but the probability of that happening is approximately zero.

By the way, that's the calculation on a heads-up situation, if we consider that more pairs see the flop the odds can change.

EDIT

When I say "after that the probability of having a quad on the turn is around 2.22%" I'm talking about the probability of YOUR OPPONENT turning a quad, the probability of YOU or YOUR OPPONENT turning quads is around 4.44% and, considering that, the probability of EITHER YOU OR YOUR OPPONENT having set-over-set and then quads is about 0.0452%.

Cheers
Thanks for taking the time to do that, I appreciate it.

Obviously I knew I had been extremely unlucky not once but twice, but just wanted to put a number on it.

Cheers.
Twice in a range of 80 hands is not only unlikely, but its almost impossible... so yeah, u get the prize of Unlucky of The Year :p

Cheers
• Bronze
Joined: 01.05.2012
Originally posted by tieppofer
Twice in a range of 80 hands is not only unlikely, but its almost impossible... so yeah, u get the prize of Unlucky of The Year :p

Cheers
Story of my poker life that.