Card hand probability?

Edit #3 -- M3, I'm pretty confident in my answer, since I used one straight-forward method and one brutish method which both agree. I would probably be willing to place a wager on my answer, since I am a gambling man ;)

My first method works because only one person can get >1 ace. That's why it doesn't over-count anything when I multiply by the 3 players, so there's no need to worry about overlap or inclusion-exclusion. My 2nd method works because it explicitly counts all combos and partitions of the 15 cards dealt to the opponents (though I omitted the C(9,4) in the numerator and denominator since they cancel).

Edit #4 -- Aha, good thing you are kind and didn't accept the wager.

There are not 51 cards left in the deck, there are 47, since as you reiterated, Ms X has exactly 1 ace meaning her other cards are known. What threw me is that they're not known specifically. But they are known to be non-aces, so that still subtracts from the deck for our purposes here.

1 - 3*[C(3,2)*C(44,3) + C(44,2)] / C(47,5) = 995/1081 = 92.0444%

And my 2nd method becomes:

Pr(someone has 2 aces) =

C(3,2) * C(44,13) * C(13,3) + .

C(44,12) * C(3,2) * C(12,3)

/ C(47,15) / C(14,4)

Pr(someone has 3 aces) = C(44,12) * C(12,2) / C(47,15) / C(14,4)

** EDIT #5 final bleeping edit I hope! **

Yes changed the 48's to 44's int he 2nd method but forgot to in the 1st. Fixed. It took me a minute to know what the heck you were talking about with sauce and geese lol. What's good for the denominator is good for the numerator indeed.

Well,

this is what i propose :

let be X1, X2, X3, X4 the random variables representing the number of aces in the hands of players N0 1, 2, 3, 4

Mr X is called player 1, he has X1 = 1 ace in hand : P(X1 = 1) = 1

let's compute the complementary event E2 : one player hes more aces than player 1 :

we can notice that, as there are 4 aces, only 1 among the 3 can have more aces:

(E2 | X1 = 1) = ...

= (X2 = 2 | X1 = 1) υ (X3 = 2 | X1 = 1) υ (X4 = 2 | X1 = 1) υ (X2 = 3 | X1 = 1) υ (X3 = 3 | X1 = 1) υ (X4 = 3 | X1 = 1)

the events are "exclusive" (french vocabulary : A ∩ B = ∅ ==> P(A υ B) = P(A) + P(B) )

and :

P(X2 = 2 | X1 = 1) = P(X3 = 2 | X1 = 1) = P(X4 = 2 | X1 = 1)

and :

P(X2 = 3 | X1 = 1) = P(X3 = 3 | X1 = 1) = P(X4 = 3 | X1 = 1)

then for every ace:

P(X2 = 2 | X1 = 1) = 4*P(X2 = 2 | X1 = ace of heart)

with : in the hand of X2 : 2 aces among 3 times 2 "any" cards among 52 - 4 aces = 48

possibilities on 5 cards among 52 - 1 ace (in player_1's hand) = 51

P(X2 = 2 | X1 = ace of heart) = 3C2 * 48C3 / 51C5 and 51C5 = 51*50*49*48*47/(5*24)=51*10*49*47

= 3 * 48 * 47 * 46 /3 / (51*49*47*10)

# 0,088355342...

so : times 4, so to cover all cases

==> P(X2 = 2 | X1 = 1) # 0,35342

same procedure for :

P(X2 = 3 | X1 = 1) = 4*P(X2 = 3 | X1 = ace of heart)

P(X2 = 3 | X1 = ace of heart) = 3C3 * 48C2 / 51C5

= 1 * 48 * 47 /2 / (51*49*47*10)

# 0000960384

==> P(X2 = 3 | X1 = 1) = 0,003841537

# 0,3573

therefore :

P(X2 has more aces | X1 = 1) # 0,35342 + 0,003841537

# 0,3573

and finally :

P(no player has two aces or more | X1 = 1) # 0,6427 = 64.27%

hope it' ll help !!