# % of identical flops

• Bronze
Joined: 07.05.2008
last night at one of my gigs there was 2 hands exactly after each other with the each same flop. but to make it even cooler. both flops contained the same tens. TTT. Anyone know how to work out how often that would happen?
• 7 replies
• Bronze
Joined: 24.03.2008
5,68738923809458810789659871374x10^(-11)
or approximately
0,000 000 005 7%
• Bronze
Joined: 07.05.2008
holy shit, how did u get that number?
• Bronze
Joined: 19.04.2008
I don't think it's a difficult calculation..

so were are looking at two flops, first one is completely random

you only have to calculate the possibility for those exact cards to fall on the second flop. You have to count the two cards you were dealt, so the odds are:

6* (1/52) * (1/51) * (1/50) = 6/132600 = 4.52* 10^-5
• Bronze
Joined: 07.05.2008
oh ok, got ya, thanks.
I'm not good at that kind of maths.
• Bronze
Joined: 04.07.2008
I get one in 22,100

given the three cards can be in any order:

3/52 x 2/51 x 1/50
• Bronze
Joined: 24.03.2008
Originally posted by justkyle88
holy shit, how did u get that number?
I assume you meant you want to know the odds of two absolutely identical flops coming in direct sequence. Let's say that you want to know the odds of T T T falling twice in a row. The odds of T being the first card is 1/52. There are 51 cards left, so the odds of T falling equals 1/51, and consequently T 's odds are 1/50. So the odds of that flop are 1/(52*51*50).

The odds of the same flop falling twice are then [1/(52*51*50)] * [1/(52*51*50)] or simplified 1/(52*51*50)^2. Worked out, this equals 1/17,582,760,000 or the number I posted earlier.

Of course, the above calculation assumes that the first of the two flops hasn't been dealt yet. If the flop you're looking at is T T T , the odds of the next one being identical is only 1/(52*51*50) or 0,000754%.
• Bronze
Joined: 07.05.2008
Thanks Berzerger.