Probability & Statistics

Aces

Thirteen cards each to four players. Placing the aces one at a time, the second avoids the first holder with probability 39/51, the third both holders with 26/50, the fourth 13/49, so every hand holds an ace with probability 2197/20825, about 0.105.

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A 52-card deck is dealt evenly to four players, 13 cards each. What is the probability that every player ends up with exactly one of the four aces?

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#Place the aces in turn

Forget the other cards and drop the four aces into the 52 slots one at a time. The first ace lands anywhere and marks a hand. The second must miss that hand, so it has 39 good slots of the 51 left. The third must reach one of the two untouched hands, 26 of 50. The fourth must hit the last hand, 13 of 49.

39/51second ace avoids the first hand
26/50third ace takes a new hand
13/49fourth ace fills the last hand
Place the aces one at a time, each into a fresh hand. The chances multiply to 2197/20825, about 10.5 percent.

#Multiply

P=3951⋅2650⋅1349=13351⋅50⋅49=219720825≈0.105.(1)\PP = \frac{39}{51}\cdot\frac{26}{50}\cdot\frac{13}{49} = \frac{13^3}{51\cdot 50\cdot 49} = \frac{2197}{20825} \approx 0.105. \tag{1}

So all four players hold an ace a little more than one time in ten.