AJJ board, probability of Jx vs 6 players?

[DavidC]

Hey there.

I have Ax on a board of AJJ vs 6 players.

What is the chance (given random cards for starters) that someone has a J.

The X in my hand is not a J.

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My guess:

Cards left: 52-2-3 = 47.

Hands remaining: 47*46/2 = 1081.

JX hands = 2*12 (JX) + 1 (JJ) = 25?

P(Jx++) = 25/1081 =~ 0.023127

P(No one Jx++) = (1-P(Jx++))^6 = 0.8690191

Therefore the probability of someone having Jx++ is about 13%?

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edit: revised guess

JX hands = 2*45+1 = 91
P(J+) = 91/1081 = 0.084181
P(Not J+) = 0.9158
P(everyone not J+) = p(not j+)^6 = 0.59
P(someone has J+) = therefore 41%

[Cobra]

There are 47 unknown cards - 2 jacks and 45 other cards.

The probability that one or more of your six opponents was dealt at least one jack is 1- (47c12/45c12) = 45%

Cobra

[jay_shark]

Lets say there are 6 players with random cards like you said .

The probability that there are no jacks amongst 12 cards is 45c12/47c12 . This is simple , since there are 45 non jack cards left in the deck and 47 unknown cards . Now we take the complement which is 1-45c12/47c12=44.9% which is the probability that at least one jack is out there .

[DavidC]

It's been a while since I did permutations and combinations, would you mind writing out 47c12 in factorial or just straight up fractions so I know what that means?

i.e. is 47c12: 47*46*45...*36?

[jay_shark]

47c12 is 47!/[(47-12)!12!]

Think of a string of 47 letters with 12 a's and 35 b's. The number of distinct combination of letters is 47c12 which is 47!/[35!12!]

Now for each letter label a1,a2,a3,...a12 and b1,b2,b3,...b35 . Note that a1a2 is the same as a2a1 so we must not over count . The total number of ways of arranging these letters with disregard to order is 47! . However , we divide by 35!*12! since the indices for the a's and b's can be arranged 12! and 35! different ways respectively .

does that make any sense ?