[Closed] Someone tell me WHAT THE CHANCES of this happening is? Any math smart ppl?

Originally posted by BrokesoBroke
I was dealt KK in the CO, BU 3bets me, I 4bet, he shoves 100bb, opens AA. I lost the hand. THE NEXT HAND, I'm in the HJ dealt KK again, I open, the next BU 3bets me, I 4bet, he shoves, opens AA... And wins...

Now I thought to myself, what are the chances of this happening on the same table?

The following day which is today... obviously on another table... I'm dealt AA... I open UTG, I get 3b by a 50bb stack, I 4b, he shoves... He opens up KK... I won that hand... THE NEXT HAND, I'm dealt AA again, UTG opens, I 3bet to 10bb, BU shoves for 87bb, and UTG shoves on top of that for 100bb... I snap call, and UTG had KK... Rivered a K...

But what are the chances of this happening to the same player two hands in a row having either KK twice or AA twice and at the same time another player has AA or KK????

odds are slim!

6 combos for any specific pocket pair total possible starting hand combos is 1326

Ratio 6:1326 = 1:221

Any two random hands both having a specifc pocket pair:

1*1:221*221 = 1:48841

Happened to me today as well AA twice in BU then CO made a grand total of $0.15 at NL2. ♠_cool:

Did your opponents have KK? What are the odds of that happening two times in a row when you hold AA? Or the other way around, you having KK two hands in a row running into AA?

Originally posted by ExMagnusxL

1*1:221*221 = 1:48841

Op has kk so its 1 to 221 rest of the table has AA so its 1 to 221*5 and then all ^2 (2 times) and its 8,70*10^7 just guess dont think its right

I lost AA to KK on 2 different tables vs the same opp at the same time. He floped K on both tables almost at once, I was going to have heart attack. Happened a third time on a different table, all yesterday.

Hey lucky man

Probability of being dealt a particular pocket pair (KK):

0.45%

(6 ways you can make a pair with 2 cards/all possible combinations of 2 cards (52x51/2 -> divided by 2 so you don't count a same hand twice: 3s4c is the same as 4c3s)

At a table with N players, the probability that other player has another particular PP (AA):

(N-1)x0.48% (= 3.92% for N=9)

(6 / (50x49/2) <-you are discounting the cards you have been dealt )

Then, for FR (9 players) the probability that other player is holding AA while you are holding KK (or viceversa):

0.45%x3.92% = 0.017%

Now, if you look at the event as a succession you could say:

probability of this happening in hand 1: 0.017%
probability of this happening in hand 1 and 2: 0.017%x0.017% = 2.9^(-6)%, or around 3 in 100,000,000

(People out there, please correct me if I am wrong)

The odds of winning the euromillion are 1 in 76,275,360. I think you just had it even better! However, I don't envy you...

S.

Edit: (by the way, how many hands have you played so far?)

@topic creator

your answer would be 1:48841*1:48841 = 1 : 23 854 823 538
But the thing is there are more than 1000s of billion hand played in the world.
So it is possible. ♠_cool:

oops i did the maths wrong it is 1 : 2 385 443 281
1 in 2 billion

I have just noticed that you are looking at the same player every time. In that case,

Player A holding AA while you hold KK: 0.45%x0.48% = 0.002%

Chances of this happening twice in a row: (0.002%)^2 = 4^-8 %, i.e.,

4 in 10,000,000,000 (4 in 10 billions)

(r4t4n123, I think the small difference comes from my calculation of the probability of the other player having AA once you were dealt KK -> 2 cards are not in the deck anymore)

S.

edit: to wrap it up. If my calculations are correct, then

You being dealt KK and someone else AA twice in a row:
3 in 100,000,000

You beind dealt KK and the same player having AA twice in a row:
4 in 10,000,000,000

-

Originally posted by argentina71
I have just noticed that you are looking at the same player every time.

You noticed wrong BU cannot but same in both hands.

i still think its 50:50

Once I get AA at the same table tree times in a row.. ♦ What are chances for that??

Unfortunatelly, I just collected blinds in all tree hands.. ♠_confused: