Assumption:
- Each player draws two cards at once.
- There are two players.
Notation rule:
C(n, r)
is nCr
or n combination r
.
Among 52 different unique cards, the probability that the King of Spades is one of the two cards:
52/C(52,2) = 2/51
a := 2/51
The goal: to calculate the probability of the first player's winning.
Let's say I'm the first player.
Case 1
Answer is P{I win at the first round} + P{I lose, he lose, I win} + P{I lose, he lose, I lose, he lose, I win} + ..., where there are the 52 cards at each round
Thus
a + (1-a)^2*a + (1-a)^4*a + ...
= a ( 1 + r + r^2 + r^3 + ... ), where r:= (1-a)^2
= a ( 1 / (1-r) ), because r < 1
= a ( 1 - (1-a)^2)
= 1/(2-a)
= 51/100
= 0.51
Case 2
Answer is P{I win at the first round} + P{I lose, he lose, I win} + P{I lose, he lose, I lose, he lose, I win} + ... + P{(I lose, he lose) * 12 times, I win, and two cards left on the ground}, where there are 2 less cards at each round than the previous round
Thus
The probability of winning at each round (where round is from 0 to 12)
round 0: 52/C(52, 2)
round 1: (1 - 52/C(52, 2)) * (1 - 50/C(50, 2)) * 48/C(48, 2)
round 2: (1 - 52/C(52, 2)) * (1 - 50/C(50, 2)) * (1 - 48/C(48, 2)) * (1 - 46/C(46, 2)) * 44/C(44, 2)
...
round 12: (1 - 52/C(52, 2)) * (1 - 50/C(50, 2)) * (1 - 48/C(48, 2)) * (1 - 46/C(46, 2)) * ... (1 - 6/C(6, 2)) * 4/C(4, 2)
Answer
= sum { prob(round 0) to prob(round 12)}
= 52/C(52, 2) + sum{i=1 to 12, product{j=0 to 2i-1, 1 - (52-2j)/C(52-2j,2)} * (52-4i)/C(52-4i, 2)}
Although the expression looks complicating, it can be simplified.
For a given 1 <= i <= 12, the prob(round i)
= product{j=0 to 2i-1, 1 - (52-2j)/C(52-2j,2)} * (52-4i)/C(52-4i, 2)
= product{j=0 to 2i-1, (49-2j)/(51-2j)} * 2/(51-4i)
= {49/51 * 47/49 * ... * (51-4i)/(53-4i)} * 2/(51-4i)
= 2/51
Note:
n/C(n, 2) = 2/(n-1)
1 - n/C(n, 2) = (n-3)/(n-1)
Thus prob(round i) = 2/51, for i = 0, 1, ..., 12
Thus answer = 2/51 * 13 = 0.50980392156...
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