Saw the One Drop tournament tonight on ESPN. The hand of the night was AA vs AA and Drinan lost an all-in bet to a flush. For the curious here is the probability that would happen.

For one player, there are 12 cards remaining for each of his suits. The board has to have either 4 or 5 of one of the two suits. Therefore,

Pr(flush for Player A) =2*[ C(12,4)*C(36,1) + C(12,5)*C(36,0)]/C(48,5) = 2.17%.

Since both players can’t flush, the probability one of the two hits is double the above or 4.34%, so while not high, it may be higher than a lot of people would guess.

most times if allowed in cash games the players will split the pot bfore seeing the cards.
but it is good for new players to see that these things happen and it isnt the longest shot in the world or a tremendous bad beat.

So a few questions...

What's "C(x,y)" and how do you solve it?
Why include the 5 card flush? With AA vs AA it takes a 4 card flush - any 4 cards. The 5th card is irrelevant after #4 hits.
What are the odds of getting AA v AA in the first place?

It's pretty damn rare to see AA v AA - then to see a 4 card flush on top of that - seems extremely long shot to me...

Had this happen live. 2 5 nl 1k buyin I was full stacked I hit 4 flusher too bad other guy was short at around 400. I will never forget it im sure the other guy won't either, he almost had a coronary.

Video of the hand:


Drinan's reaction was remarkably philosophical. Indeed, this "worst beat in history" actually seems rather dull when the reactions are subdued like this.

Statmanhal's equation was interesting. Personally, I use Equilab to get the same answer.

Player - Equity - Win - Tie
Hero 50.00% 2.17% 47.83% { AA }
Vill 50.00% 2.17% 47.83% { AA }

You have to substract the times when there's a (non KQJT9) straight flush on the board.

Really inaccurate calculations.

It's right around a 2% hand to win so it's like 45-50 to 1.

Statistically, this hand is worse (the AAAA vs. Royal Flush hand at the 08 main) but the one drop is 100x the buy in so i'd say the Drinan hand is the worst beat of all time.

Drinan satellited in to the one drop so it's not like he lost a million of his own or his backers $.

You're right, so when you do this carrying all calcs to 4 decimal places, the 2.17% and 4.34% results become 2.17% and 4.34%.

1. C(N,r) is the number of ways (combinations, combos) to select r elements out of a total of N distinct elements. It is defined as:

C(N,r) = N!/{r!*(N-r)!}, where X! = X * (X-1)*(X-2) * …. *3*2*1 [C(N,0) is defined to be equal to 1]

Example: A basketball team has 8 players. How many possible 5-man starting teams can be formed?

N = 8, r = 5

C(8,5) = 8!/(5!*3!) = 8*7*6 / (3*2*1) = 56

2. Yes 4 flush cards will do it but so will 5, so you need to count all success possibilities (and avoid double counts). The first term in the brackets, C(12,4)*C(36,1) is the number of ways of selecting 4 flush cards out of the 12 remaining multiplied by the number of ways of selecting 1 non-flush card out of the 36 non-flush cards in the deck. The second term is the number of ways of selecting 5 of the 12 flush cards. The factor C(36,0) is 0 out of 36 and by definition equals 1. The 2 multiplier is for the two suits held by the player.

As was pointed out, I should have subtracted out the 7*4=28 combos of straight flushes that lead to a chop from the 18,000+ combos that give a win to a specific player.

Note that as in all probability problems like this there are other ways it can be solved. For me, this was the easiest. Check out the Mathenoobics thread in this forum.

3. AA vs AA. Assume a heads-up game. The probability the first player gets AA is 4/52 * 3/51. There are now two aces left in the deck of 50 cards, so the probability the second player gets AA given the first has AA is 2/50*1/49. Therefore the probability for AA vs AA is

Pr(AA vs AA) = 4*3*2*1/(52*51*50*49) = 0.00000369.

For n players, the answer is C(n,2)*0.00000369; e.g., for 6 players, Pr = 15* 0.00000369 = 0.0000554 or about once every 18,000 hands. For the One Drop, I believe it started with 42 players or say 7 tables of 6 players each. Assuming 30 hands per hour for 8 hours of play, that’s 7*30*8 = 1680 hands. So, seeing it over several days with the number of hands decreasing each day is quite unlikely but unlikely doesn’t mean impossible.

Which would result in an 8.8885% chance of it happening at least once that day.

Of course, that's just AA vs AA. The chance of AA vs AA and a flush is 4.34% / 18048 = 1 in 415860

So that will happen about once every 248 days assuming 1680 hands a day.

Yes, I don't know why I used the term "quite unlikely"; perhaps I focused more on just a specific 2-player result (1 in 270,000) rather than any 2 of 6 players (1 in 18,000).

because this hand is defined "bad beat" , would not be more just "cooler" or "worst beat"

I'm not impressed at all.