Strange occurance yesterday... Odds?

[Hotrod0823]

Well I was playing a MTT sng and got dealt KK back to back

Now as if the odds of getting KK back to back aren't big enough.....

What are the odds of getting quad K's back to back?

Thanks

[toms2866]

Assuming Texas hold 'em,
Odds KK consecutively = 1:48841
Odds KKKK consecutively = 1:59,830,225

Note: once you have KK, odds of it again are usual 1:221; and similarly, once you hit KKKK, odds of it again are usual 1:7735. Also, when figuring KKKK, I didn't require first two cards to be the KKKK. It's significantly more remote if you require the KKKK to be formed from KK pocket.

[KJL]

The odds of getting dealt any pocket pair is equal to: (52*3)/C(52,2) or 1/8.5
The odds of the board containing both of the cards to match that pocket pair is:
2*1*C(48,3)/C(50,5)=~.016
So the odds of being dealt a pocket pair and making quads with it is eual to:
1/8.5*.016=~.002
The odds of being dealt the exact same pocket pair is equal to:
C(4,20)/C(52,2)=1/221
Once again the odds of the board containing the remaining 2 of these cards is ~.016
So the odds of being dealt a specific pocket pair and making quads with it is:
.016*1/221=~.00007
Therefore the odds of, on any two given hands, being dealt a pocket pair, making quads with it and then have the same thing happen the next hand is equal to:
.00007*.002=.0000001 or .00001%

[BruceZ]

You made some instructive errors on these.

[ QUOTE ]
The odds of getting dealt any pocket pair is equal to: (52*3)/C(52,2) or 1/8.5

[/ QUOTE ]

1/17. You must divide the numerator by 2 if you use combinations in the denominator, or else you will count every combination twice. For example, you are counting AsKc and KcAs as 2 separate hands in the numerator, but only once in the denominator. I see that you are thinking "52 ways to pick the first card, and 3 ways to pick the second card", but that double counts the combinations because it counts each possible ordering of the 2 cards. However, you could do 52/52 * 3/51 = 1/17. Notice the subtle difference to the first method. When you use combinations you don't care about order, but when you use fractions, then you do consider the probability for the first card times the probability for the second card.

Also, for all of these, the number represents the probability, not the odds.


[ QUOTE ]
The odds of the board containing both of the cards to match that pocket pair is:
2*1*C(48,3)/C(50,5)=~.016

[/ QUOTE ]

Just 1*C(48,3)/C(50,5) =~ 0.0082. There is only 1 KK, and then C(48,3) combinations for the remaining 3 cards. We are using combinations, so we don't say "2 kings for the first card times 1 king for the second card". This counts each KK twice, but obviously there is only 1, not 2.


[ QUOTE ]
So the odds of being dealt a pocket pair and making quads with it is equal to:
1/8.5*.016=~.002

[/ QUOTE ]

1/17 * 0.0082 =~ 0.00048.


[ QUOTE ]
The odds of being dealt the exact same pocket pair is equal to:
C(4,20)/C(52,2)=1/221

[/ QUOTE ]

The 1/221 is correct, and you mean C(4,2) = 6, not C(4,20).


[ QUOTE ]
Once again the odds of the board containing the remaining 2 of these cards is ~.016

[/ QUOTE ]

~0.0082.


[ QUOTE ]
So the odds of being dealt a specific pocket pair and making quads with it is:
.016*1/221=~.00007

[/ QUOTE ]

0.0082 * 1/221 =~ 0.000037.


[ QUOTE ]
Therefore the odds of, on any two given hands, being dealt a pocket pair, making quads with it and then have the same thing happen the next hand is equal to:
.00007*.002=.0000001 or .00001%

[/ QUOTE ]

0.00048 * 0.000037 = 0.000000018 or without rounding until the end 1 in 56,378,481. Odds are 56,378,480-to-1.

[Hotrod0823]

Thank you.

[AaronBrown]

I assume this was online, but if it was bricks and mortar, you should pay attention to the suits of the King. On the first hand, your two Kings were together when the cards were gathered in, and the two Kings on the board were also near each other. On the second hand, the two Kings on the board ended up near each other. If they were the two Kings from your hand, or the two from the board last time, it's likely that the shuffle is imperfect, which is something you can exploit.

Only one time in 133 will both cards from a pocket holding show up on the next board with a perfect shuffle, and only 1 time in 16 will two cards from the previous board be on the next board. If you notice it happening significantly more often than that (and it will with many dealers) you'd better pay attention because other players are.

[J.A.K.]

[ QUOTE ]


Only one time in 133 will both cards from a pocket holding show up on the next board with a perfect shuffle, and only 1 time in 16 will two cards from the previous board be on the next board. If you notice it happening significantly more often than that (and it will with many dealers) you'd better pay attention because other players are.

[/ QUOTE ]

How do you process the information? Are you saying I should play Kx or 33 in unconventional ways after seeing x# of boards w/ repeating K's or 3's? Or shall I give another player extra credit for pairing a board card?

[mostsmooth]

why isn’t the answer 100% or like 99.99999999999%? surely there will be another time in the future that somebody gets quad kings back to back?

[why4duck]

If the casino uses an auto shuffler, though, typically they will be dealing from 2 decks, so there's no evidence of an imperfect shuffle in a case like this (back to back).

[BruceZ]

[ QUOTE ]
and only 1 time in 16 will two cards from the previous board be on the next board.

[/ QUOTE ]

Just to clarify, 1 time in 16 you will get exactly 2 cards from the previous board. You will get at least 2 cards 1 time in 15.

P(exactly 2) = C(5,2)*C(47,3)/C(52,5) =~ 1 in 16.

P(at least 2) = 1 - P(0 or 1) = 1 - [C(47,5) + 5*C(47,4)]/C(52,5) =~ 1 in 15.

Here is how often you should expect to see various numbers of cards from the previous board:

1 or more: 1 in 2.4
2 or more: 1 in 15.0
3 or more: 1 in 235
4 or more: 1 in 11,013
5: 1 in 2,598,960