odds of getting a pocket pair 3x in a row

[mulebennett]

sitting at a 10-handed hold'em game, what are the odds of being dealt three straight pocket pairs? i can't remember the last time i was dealt three in a row.. thanks for the numbers as always

[TomCollins]

3/51^3. = .02%, or 4912-1

I saw someone get a pocket pair 3 times in a row, and flop a set all 3 times. What are the odds of THAT?

PS- He lost 2 of the hands, and split one.

[cwes]

The probability of getting a pp is 1/17.

Before you get dealt the first hand, the probability, that the next three hands will be pps is just (1/17)^3 = (1/4913).

After you've been dealt the first pp the probability that the next two hands also will be pps is (1/17)^2 = (1/289).

If you already had two pps in a row the probability of getting a third is the probability of getting a pp i.e. 1/17.

So you expect this (three pps in a row) to occur every 4913 hands you play. Shouldn't be too rare. It just happened to me yesterday (KK,QQ,JJ, two of them were good /images/graemlins/smile.gif ).

[Tuben]

I got QQ,QQ,KK,AA,AA,AA in one sesion played like 2 hours.

Won 2 of thoose AA: 2 times rest a lost.

I once drew pocket aces on 2 hands in a row, then skipped a hand, then got aces again. I won one pre-flop, one on the flop, and busted a player on the third AA.

The other day I got five pocket pairs in a row. I ended up not tripping on any of them but still won 3/5.

[sekrah]

Had it happen to me this weekend.

Had Q-Q and K-K cracked on back to back hands at a 1/2 NL game at Seneca Allagany.

I asked the dealer to wash the deck,. next hand, 8-8, and I flop a set of 8's to win a pot.

[sekrah]

In an empire limit game, I got A-A, A-A, A-K on 3 consecutive hands. I lost both times with the A-A to K-3 (flush) and K-4 (flopped boat k-4-4), then I won with the A-K, on Ace-high.

[spaminator101]

1/18cubed=1/5832

[spaminator101]

its 17 to 1 or 1/18 so you cube 18 not17

[TaintedRogue]

[ QUOTE ]
3/51^3. = .02%, or 4912-1

I saw someone get a pocket pair 3 times in a row, and flop a set all 3 times. What are the odds of THAT?

PS- He lost 2 of the hands, and split one.

[/ QUOTE ]

I remember once......2 years ago.....at the Borgata in Atlantic City.......I was dealt JJ in the UTG+1 seat, 3 orbits in a row......
In all three instances, UTG open/raised.
I the first & third hand, I flopped a set.
I won all three.
Guess what........I could give a rat's ass what the odds where of that :-)

No. The odds are 16 to 1.

[Luzion]

You guys are sorta using faulty logic. Yes, odds of getting three pockets pairs in a row is (1/17)^3 but ONLY if you are playing exactly three hands. And who only plays exactly three hands?

The question is, can anyone figure out what are the odds of getting three pocket pairs in a row at least once after playing a stretch of 1000 hands? Because I thought about it and can't solve it.

[DougOzzzz]

[ QUOTE ]
You guys are sorta using faulty logic. Yes, odds of getting three pockets pairs in a row is (1/17)^3 but ONLY if you are playing exactly three hands. And who only plays exactly three hands?

The question is, can anyone figure out what are the odds of getting three pocket pairs in a row at least once after playing a stretch of 1000 hands? Because I thought about it and can't solve it.

[/ QUOTE ]

One quick and nearly correct way to solve this is to simply take 1-(1-1/4913)^998 since there are 998 "opportunities" to get 3 PP's in a row. This obtains an answer of 18.38%. However, it's not correct because every time you get a pocket pair, and then fail to get 3 in a row, you are using up 1 (or 2, if you get 2 in a row but fail to get 3) of your extra opportunities. So you really have more like 938 chances at getting 3 PP's in a row. I'm not sure how to solve this exactly because it depends the number of chances you have depends on how many PPs you get (and also, the number of PP's you get increases the chances of getting 3 in a row, obviously).

I have an alternate solution that I will post in a few minutes.

[DougOzzzz]

My alternate solution will have to wait. I typed it up and my browser closed. Essentially I was going to compute the probability of N PP's from 0 to 1000 in 1000 hands, and then compute the probability of having 3 consecutive PP's for each N. The answer would be the sum of the products for each N.

Might be a better way to do it.

[Luzion]

[ QUOTE ]
My alternate solution will have to wait. I typed it up and my browser closed. Essentially I was going to compute the probability of N PP's from 0 to 1000 in 1000 hands, and then compute the probability of having 3 consecutive PP's for each N. The answer would be the sum of the products for each N.

Might be a better way to do it.

[/ QUOTE ]

That sounds like a brute-force method for doing it. It would be correct, but awfully time consuming.

I also dont know how you would compute the probability of N pocket pairs being next to each other. You would have to draw a diagram and keep track of all combinations manually.

How about making it easier for now? What is the probability of getting 3 pocket pairs in a row, if you play a stretch of 10 hands?

[DougOzzzz]

[ QUOTE ]
[ QUOTE ]
My alternate solution will have to wait. I typed it up and my browser closed. Essentially I was going to compute the probability of N PP's from 0 to 1000 in 1000 hands, and then compute the probability of having 3 consecutive PP's for each N. The answer would be the sum of the products for each N.

Might be a better way to do it.

[/ QUOTE ]

That sounds like a brute-force method for doing it. It would be correct, but awfully time consuming.

I also dont know how you would compute the probability of N pocket pairs being next to each other.

[/ QUOTE ]

Yeah I was doing it in Excel, but my method was incorrect anyways. I got an answer CLOSE to correct. Essentially this is the formula that I used:
probability of 3 straight PP's given an N = (998!/(N-3)!(1000-N)!)/1000 Choose N.

However this is not correct because it double counts several cases.

[DougOzzzz]

I'm not sure there's an easy method for solving it BTW. I remember on Baseball Prospectus someone did a study on the probability of Ichiro having a 55 game hitting streak in a 162 game season. Very similar problem and I don't think he came up with anything better than an estimate.

For your example though, the quick and dirty calculation is far closer to the actual answer than the hitting streak example.

[BruceZ]

[ QUOTE ]
The question is, can anyone figure out what are the odds of getting three pocket pairs in a row at least once after playing a stretch of 1000 hands? Because I thought about it and can't solve it.

[/ QUOTE ]

I've shown how to do this type of problem a few times on here. Here is the simplest method know, though it can also be done with matrices as a Markov process which may be less calculation.

Let P(n) be the probability of a sequence of 3 pocket pairs in a row by the nth hand. Then:

P(1) = 0
P(2) = 0
P(3) = (1/17)^3
P(4) = P(3) + (16/17)*(1/17)^3
P(n) = P(n-1) + [1 - P(n-4)]*(16/17)*(1/17)^3 for n > 4

P(1000) =~ 17.4%

That is, the probability that it happens in n hands is the probability that it happens in n-1 hands plus the probability that it happens first on the nth hand. To happen first on the nth hand, it must not happen by hand n-4, then we must not get a pair on hand n-3 with probability 16/17, followed by 3 pairs in a row with probability (1/17)^3. This can be computed in a single column in Excel.

[DougOzzzz]

[ QUOTE ]
[ QUOTE ]
The question is, can anyone figure out what are the odds of getting three pocket pairs in a row at least once after playing a stretch of 1000 hands? Because I thought about it and can't solve it.

[/ QUOTE ]

I've shown how to do this type of problem a few times on here. Here is the simplest method know, though it can also be done with matrices as a Markov process which may be less calculation.

Let P(n) be the probability of a sequence of 3 pocket pairs in a row by the nth hand. Then:

P(1) = 0
P(2) = 0
P(3) = (1/17)^3
P(4) = P(3) + (16/17)*(1/17)^3
P(n) = P(n-1) + [1 - P(n-4)]*(16/17)*(1/17)^3 for n > 4

P(1000) =~ 17.4%

That is, the probability that it happens in n hands is the probability that it happens in n-1 hands plus the probability that it happens first on the nth hand. To happen first on the nth hand, it must not happen by hand n-4, then we must not get a pair on hand n-3 with probability 16/17, followed by 3 pairs in a row with probability (1/17)^3. This can be computed in a single column in Excel.

[/ QUOTE ]

Nice solution. I feel dumb for not having thought of it myself. Then again, you and phzon and a couple others here often make me feel that way.

So, for the Ichiro hitting streak problem, assuming:

last years statistics for Ichiro:
.372 BA
4.372671 AB's per game
Distributed as: 4 ABs in 62.73% of games
5 ABs in 37.27% of games
Chances of getting a hit in a single game:
86.6636%

Now, using your formula, the chances of Ichiro having a 55 game hitting streak in a 162 game season (not missing any games) would then be 0.001106131, or, a little less than 1 in 900. Is that correct?

[BruceZ]

[ QUOTE ]
Now, using your formula, the chances of Ichiro having a 55 game hitting streak in a 162 game season (not missing any games) would then be 0.001106131, or, a little less than 1 in 900. Is that correct?

[/ QUOTE ]

No, I get 0.005815 or about 1 in 172.

P(1) to P(54) = 0
P(55) = (0.866636)^55
P(56) = P(55) + (1 - 0.866636)*(0.866636)^55
P(n) = P(n-1) + [1 - P(n-56)]*(1 - 0.866636)*(0.866636)^55, for n > 56
...
P(162) = 0.5815%

[mtgordon]

Here's one that'll blow your mind. Earlier today I got KK 3 times in a row on bodog. And after the first hand I got KK on Pacific as well. I don't feel like doing the math again, but it was something like 0.000093% chance of getting KK 3 times in a row.

[DougOzzzz]

[ QUOTE ]
[ QUOTE ]
Now, using your formula, the chances of Ichiro having a 55 game hitting streak in a 162 game season (not missing any games) would then be 0.001106131, or, a little less than 1 in 900. Is that correct?

[/ QUOTE ]

No, I get 0.005815 or about 1 in 172.

P(1) to P(54) = 0
P(55) = (0.866636)^55
P(56) = P(55) + (1 - 0.866636)*(0.866636)^55
P(n) = P(n-1) + [1 - P(n-56)]*(1 - 0.866636)*(0.866636)^55, for n > 56
...
P(162) = 0.5815%

[/ QUOTE ]

Okay, I had an error in my formula. For some reason I was multiplying by (1-.86636)^2.
In other words, my formula was: P(n) = P(n-1) + [1 - P(n-56)]*(1 - 0.866636)^2*(0.866636)^55, for n > 56
I get the same answer now.

[DougOzzzz]

Here's another question for BruceZ.

Given the same information about Ichiro, what is the median longest hitting streak during a 162 game season?

I think I could figure this out using the following method:

For each n, 0 to 162, find the probability that Ichiro will have a streak of this length using your formula. The probability that this is the longest streak of the season is P(n)-P(n+1). The median longest streak would be the first value where the sum from 0 to n of P(n)-P(n+1) >= .5.