# Probability of hitting quads 3 times in a roll!!!!

• Bronze
Joined: 25.09.2010
hey guys!

i was playing 90 man sit and go 3.30\$ KO on full tilt.

ok this guys plays J9 vs AQ

board A59 turn 9 river 9.....

next hand

22 vs QQ the same guy hit quads with 22

next my turn

i had AJ suited vs his.... A4.... he puts me all in after flop

board was A74 turn 4 river 4

this guys just hit quads 3 times in a row. you can imagine how piss the last hand was.

so i got interested on what probability does someone hit quads 3 times in a row?

is it just only on full tilt poker?
• 20 replies
• Basic
Joined: 17.02.2011
The Poker odds of getting four of a kind is 1 in 4,165.00.

That translates to 72,251,192,125:1 for three in a row. That's 72+ billion to one. You have much better odds of winning the lottery.
• Bronze
Joined: 14.09.2009
Dont forget the odds are reset every time
• Basic
Joined: 17.02.2011
I know, I accounted for that didn't I?
• Bronze
Joined: 18.03.2007
funkyowo do u have Hand history of those QUADS ?
• Bronze
Joined: 14.09.2009
Originally posted by Nevermind22
I know, I accounted for that didn't I?
I might be wrong but from what i understand

each time you have 1 in 4165

So basically one you hit it once , you have 1 in 4165 to hit it again and same for the 3rd time. So even though it is very unlikely I think you have more chance of that happening than winning the lottery. Also givcen the number of hands in online poker it is bound to happen sometimes
• Bronze
Joined: 15.03.2009
The initial calculation is correct, because you are measuring the probability of it happening 3 times in a row for the next three hands.

However, the probability of it happening for the 3rd time after 2 of them have been dealt is 1/4165.

It all depends on how the question is asked..
• Coach
Coach
Joined: 03.08.2009
• Bronze
Joined: 27.01.2010
Come on. Is it really that easy to get quads 3 times in a row? It's just.....I have no comment. Lately it happens to me to see quads 3 times in 5 minutes. Is it really possible for some crazy crap happening all the time?! Not just quads, runner runner straight is even worse. You can say I'm just telling crap and go on with the stories about online poker, but how sure are we actually that "billions \$ company" has no reason to mix it up in their advantage? Couldn't they just make it to be as a real poker?
• Basic
Joined: 17.02.2011
hitting quads once is 1 in 1465 or whatever, then hitting again is once again 1 in 1465.

But to to say you're going to hit quads twice in a row is 1465 times 1465 or something.

It's cerainly more difficult to attain 3 quads than it is to win the lottery.
• Bronze
Joined: 27.01.2008

Chance of hitting quads 3times in a row: 1/4165 x 1/4165 x 1/4165 = 1 / 72251192125

As the first reply says. And it's easier winning on lottery, yes.
• Bronze
Joined: 04.02.2011
i hit quad aces in last 2 days like 6 times so sick played only like 50 9man sng's
• Bronze
Joined: 30.08.2010
ipoker? then it's like 2:1 a day
• Bronze
Joined: 30.08.2010
Originally posted by Jkob

Chance of hitting quads 3times in a row: 1/4165 x 1/4165 x 1/4165 = 1 / 72251192125

As the first reply says. And it's easier winning on lottery, yes.
it's still not close to two person liking each other
• Bronze
Joined: 18.07.2010
It's so unlikely I'd be questioning some sort of cheating.
• Bronze
Joined: 25.08.2010
Some rather obsessive nerdiness follows:

Assuming you see the river, the chance of hitting quads is actually higher than 1/4165. If you don't have a pocket-pair, then you can hit quads with either one of your two different ranks by seeing a five-card board with the three remaining cards of the given rank and two more from the 47 remaining cards in the deck, so the probability is
2*C(47,2)/C(50,5) ~ 0.00102, or one in 980.
If you have a pocket-pair, then you have to see the two remaining cards of that rank and three more from the 48 remaining cards in the deck, so the probability is
C(48,3)/C(50,5) ~ 0.00816, or one in 122.5.
One in seventeen hands is a pocket-pair, so the overall probability of hitting quads on a given hand (call it 'q') is
q = 16/17 x 1/980 + 1/17 x 1/122.5 ~ .001441, or about one in 694.17.

In a session of 3,000 hands, the probability of never hitting quads at all is
(1-q)^3000 ~ (1 - 1/694.17)^3000 ~ 0.0132, or about one in 75.6, which is quite low. (Although note we're still assuming you see the river every hand.)

In any two consecutive hands, the probability that you hit quads on both hands is
q^2 ~ (1/694.17)^2 ~ 0.00000208, or about one in 482,000.

To calculate the probability that in a session of 3,000 hands you never hit quads on two hands in a row, first work out the probability that you don't hit quads in both the 1st and 2nd, or both 3rd and 4th, or ... or both 2999th and 3000th hands (call this "the first condition"). This doesn't mean you couldn't hit quads twice in a row on the 2nd and 3rd, or 4th and 5th, or ... or 2998th and 2999th hands, so now work out the probability that you also don't hit quads in any of these ways, given the first condition.
The probability of the first condition is
(1-q^2)^1500. So now we know there are no "pairs of quads" in the 1st and 2nd, or 3rd and 4th, or ... 2999th and 3000th hands.
What's the probability you don't hit quads in both the 2nd and 3rd hands? We already know that the 1st and 2nd hands are not both quads, so the conditional probability that the 2nd hand is quads (call it q') is, by Bayes' theorem,
q' = (1-q) x q / (1-q^2) ~ 0.001438, or about one in 695.17. (This is slightly lower than q, which makes sense.)
We also know that the 3rd and 4th hands are not both quads, so by the same calcuation the conditional probability that the 3rd hand is quads is also q'. Therefore the probability you don't hit quads in both the 2nd and 3rd hands is 1-q'^2, which means the probability you don't hit quads twice in a row on the 2nd and 3rd, or 4th and 5th, or ... or 2998th and 2999th hands is (1-q'^2)^1499.
Putting this all together, the probability of never hitting quads two hands in a row, in a session of 3,000 hands, is
(1-q^2)^1500 x (1-q'^2)^1499 ~ .9938; or in other words, the probability of
hitting quads on two consecutive hands at some point is roughly 0.00620, or about one in 161.

Extending these calculations for the probability of hitting quads on three consecutive hands at some point in a session of 3,000 hands, it turns out to be
1 - (1-q^3)^1000 x (1-(1-q)(1-q^2)q^2/(1-q^3)^2)^1000 x (1-(1-q^3)q^3/(1+q+q^2-2q^3-q^4))^1000 ~ 0.002074, or about one in 482.

This formula doesn't obviously extend to runs of four consecutive quads, and there's almost certainly an existing solution for these kinds of problems that I don't know about.

It's possible I've made an error here, or used a hidden mistaken assumption, but the formula indicates that in one million hands that go to the river, a given player should have about a 49.98% chance of being dealt quads in three consecutive hands at least once.

I think this illustrates the observed fact that even unlikely events are bound to happen sooner or later.

(Sheesh! That took a lot of effort that could have been used to play more poker.)
• Coach
Coach
Joined: 03.08.2009
<3
• Bronze
Joined: 25.09.2010
Originally posted by fembot26
Some rather obsessive nerdiness follows:

Assuming you see the river, the chance of hitting quads is actually higher than 1/4165. If you don't have a pocket-pair, then you can hit quads with either one of your two different ranks by seeing a five-card board with the three remaining cards of the given rank and two more from the 47 remaining cards in the deck, so the probability is
2*C(47,2)/C(50,5) ~ 0.00102, or one in 980.
If you have a pocket-pair, then you have to see the two remaining cards of that rank and three more from the 48 remaining cards in the deck, so the probability is
C(48,3)/C(50,5) ~ 0.00816, or one in 122.5.
One in seventeen hands is a pocket-pair, so the overall probability of hitting quads on a given hand (call it 'q') is
q = 16/17 x 1/980 + 1/17 x 1/122.5 ~ .001441, or about one in 694.17.

In a session of 3,000 hands, the probability of never hitting quads at all is
(1-q)^3000 ~ (1 - 1/694.17)^3000 ~ 0.0132, or about one in 75.6, which is quite low. (Although note we're still assuming you see the river every hand.)

In any two consecutive hands, the probability that you hit quads on both hands is
q^2 ~ (1/694.17)^2 ~ 0.00000208, or about one in 482,000.

To calculate the probability that in a session of 3,000 hands you never hit quads on two hands in a row, first work out the probability that you don't hit quads in both the 1st and 2nd, or both 3rd and 4th, or ... or both 2999th and 3000th hands (call this "the first condition"). This doesn't mean you couldn't hit quads twice in a row on the 2nd and 3rd, or 4th and 5th, or ... or 2998th and 2999th hands, so now work out the probability that you also don't hit quads in any of these ways, given the first condition.
The probability of the first condition is
(1-q^2)^1500. So now we know there are no "pairs of quads" in the 1st and 2nd, or 3rd and 4th, or ... 2999th and 3000th hands.
What's the probability you don't hit quads in both the 2nd and 3rd hands? We already know that the 1st and 2nd hands are not both quads, so the conditional probability that the 2nd hand is quads (call it q') is, by Bayes' theorem,
q' = (1-q) x q / (1-q^2) ~ 0.001438, or about one in 695.17. (This is slightly lower than q, which makes sense.)
We also know that the 3rd and 4th hands are not both quads, so by the same calcuation the conditional probability that the 3rd hand is quads is also q'. Therefore the probability you don't hit quads in both the 2nd and 3rd hands is 1-q'^2, which means the probability you don't hit quads twice in a row on the 2nd and 3rd, or 4th and 5th, or ... or 2998th and 2999th hands is (1-q'^2)^1499.
Putting this all together, the probability of never hitting quads two hands in a row, in a session of 3,000 hands, is
(1-q^2)^1500 x (1-q'^2)^1499 ~ .9938; or in other words, the probability of
hitting quads on two consecutive hands at some point is roughly 0.00620, or about one in 161.

Extending these calculations for the probability of hitting quads on three consecutive hands at some point in a session of 3,000 hands, it turns out to be
1 - (1-q^3)^1000 x (1-(1-q)(1-q^2)q^2/(1-q^3)^2)^1000 x (1-(1-q^3)q^3/(1+q+q^2-2q^3-q^4))^1000 ~ 0.002074, or about one in 482.

This formula doesn't obviously extend to runs of four consecutive quads, and there's almost certainly an existing solution for these kinds of problems that I don't know about.

It's possible I've made an error here, or used a hidden mistaken assumption, but the formula indicates that in one million hands that go to the river, a given player should have about a 49.98% chance of being dealt quads in three consecutive hands at least once.

I think this illustrates the observed fact that even unlikely events are bound to happen sooner or later.

(Sheesh! That took a lot of effort that could have been used to play more poker.)
very interesting maths but this part

"It's possible I've made an error here, or used a hidden mistaken assumption, but the formula indicates that in one million hands that go to the river, a given player should have about a 49.98% chance of being dealt quads in three consecutive hands at least once. "

seems to make sense without you making any error i assume.

cheers
• Bronze
Joined: 02.07.2008
Originally posted by fembot26
Some rather obsessive nerdiness follows:

Assuming you see the river, the chance of hitting quads is actually higher than 1/4165. If you don't have a pocket-pair, then you can hit quads with either one of your two different ranks by seeing a five-card board with the three remaining cards of the given rank and two more from the 47 remaining cards in the deck, so the probability is
2*C(47,2)/C(50,5) ~ 0.00102, or one in 980.
If you have a pocket-pair, then you have to see the two remaining cards of that rank and three more from the 48 remaining cards in the deck, so the probability is
C(48,3)/C(50,5) ~ 0.00816, or one in 122.5.
One in seventeen hands is a pocket-pair, so the overall probability of hitting quads on a given hand (call it 'q') is
q = 16/17 x 1/980 + 1/17 x 1/122.5 ~ .001441, or about one in 694.17.

In a session of 3,000 hands, the probability of never hitting quads at all is
(1-q)^3000 ~ (1 - 1/694.17)^3000 ~ 0.0132, or about one in 75.6, which is quite low. (Although note we're still assuming you see the river every hand.)

In any two consecutive hands, the probability that you hit quads on both hands is
q^2 ~ (1/694.17)^2 ~ 0.00000208, or about one in 482,000.

To calculate the probability that in a session of 3,000 hands you never hit quads on two hands in a row, first work out the probability that you don't hit quads in both the 1st and 2nd, or both 3rd and 4th, or ... or both 2999th and 3000th hands (call this "the first condition"). This doesn't mean you couldn't hit quads twice in a row on the 2nd and 3rd, or 4th and 5th, or ... or 2998th and 2999th hands, so now work out the probability that you also don't hit quads in any of these ways, given the first condition.
The probability of the first condition is
(1-q^2)^1500. So now we know there are no "pairs of quads" in the 1st and 2nd, or 3rd and 4th, or ... 2999th and 3000th hands.
What's the probability you don't hit quads in both the 2nd and 3rd hands? We already know that the 1st and 2nd hands are not both quads, so the conditional probability that the 2nd hand is quads (call it q') is, by Bayes' theorem,
q' = (1-q) x q / (1-q^2) ~ 0.001438, or about one in 695.17. (This is slightly lower than q, which makes sense.)
We also know that the 3rd and 4th hands are not both quads, so by the same calcuation the conditional probability that the 3rd hand is quads is also q'. Therefore the probability you don't hit quads in both the 2nd and 3rd hands is 1-q'^2, which means the probability you don't hit quads twice in a row on the 2nd and 3rd, or 4th and 5th, or ... or 2998th and 2999th hands is (1-q'^2)^1499.
Putting this all together, the probability of never hitting quads two hands in a row, in a session of 3,000 hands, is
(1-q^2)^1500 x (1-q'^2)^1499 ~ .9938; or in other words, the probability of
hitting quads on two consecutive hands at some point is roughly 0.00620, or about one in 161.

Extending these calculations for the probability of hitting quads on three consecutive hands at some point in a session of 3,000 hands, it turns out to be
1 - (1-q^3)^1000 x (1-(1-q)(1-q^2)q^2/(1-q^3)^2)^1000 x (1-(1-q^3)q^3/(1+q+q^2-2q^3-q^4))^1000 ~ 0.002074, or about one in 482.

This formula doesn't obviously extend to runs of four consecutive quads, and there's almost certainly an existing solution for these kinds of problems that I don't know about.

It's possible I've made an error here, or used a hidden mistaken assumption, but the formula indicates that in one million hands that go to the river, a given player should have about a 49.98% chance of being dealt quads in three consecutive hands at least once.

I think this illustrates the observed fact that even unlikely events are bound to happen sooner or later.

(Sheesh! That took a lot of effort that could have been used to play more poker.)
I came
• Bronze
Joined: 02.12.2010
3 times in a row..oh that's so normal!!!