3 players dealt the same thing

[Siegmund]

Had a fun hand yesterday: the board showed 54432, and at the showdown, all 3 remaining players faced A-6.

A quick intuition test: is having three players be dealt A-6 (suited or not) on the same hand more or less likely than having two players dealt AA, as in the previous thread?

Not hard to calculate but a good brainteaser to keep y'all awake (and make sure you got the same answer I did).

[gaming_mouse]

Without doing the calc, my gut tells me that its MUCH more unlikely to have 3 A6.

[AngusThermopyle]

3 with A6, by far.

Two AA: About 1 in 6016, from earlier threads.

Given 3 players has Ax ( guestimate around 1/25 ).
Player 1 has A6 .. about 1/13
Player 2 also has A6 ... about 1/17
Player 3 also has A6 ... about 1/25

Ballpark 1 in 125,000

Will try to think of a more "intuitive" and less math reason.

[Siegmund]

My own guess was that it'd be close - 3 vs 2 hands, but much easier-to-be-dealt hands.

3xA6 is rarer... so it comes down to what your mental image of "by far" is, I guess.

I make the chance 10C3 * 16/1326 * 9/1225 * 4/1128, for 1 in 26508.

[MickeyHoldem]

I think this term is a wee bit high... it over counts the times there are 4 hands of A6 by a factor of 4.

Four A6:
Q4 = ((16*9*4*1)/24) * 44c12 * 11!! = 5261703498291240

Three A6:
Q3 = ((16*9*4)/6) * 46c14 * 13!! - 4 * Q4 = 3111921783275104800

# deals:
D = 52c20 * 19!! = 82492346176096206475125

so exactly 3 A6 = 3111921783275104800 / 82492346176096206475125 = 0.000037468627243
= ~1 in 26689

atleast 3 A6 = (5261703498291240 + 3111921783275104800) / 82492346176096206475125 = 0.000037532411385 = ~1 in 26644

you answer is the first part of Q3 / D

((16*9*4)/6) * 46c14 * 13!! / (52c20 * 19!!) = ~1 in 26508

[Siegmund]

You're right, I forgot to subtract the tiny chance of four A6s being out.

GRB

[Slacker13]

I played at the Taj one day, I hit the nut straight on the flop, three of us betting & raising to the end, we all 3 flipped over 56 suited.