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Given a standard deck of cards our goal is to pick the King of Spades to win and we draw two cards in each round. What would the probability of winning given these two cases: 1. We draw the first card with replacement. 2. We draw the first card with no replacement.

I think the probability of the first one is: $$ \dfrac{1}{52} + \dfrac {1}{52}$$ and the same thing for the second one, but I feel that it's wrong. Thank you in advance.

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    $\begingroup$ One does not simply add probabilities in most scenarios. By your logic if we were to try $104$ times in a row to draw a king of spades with replacement then we would have a probability of $1/52 + 1/52 + \dots + 1/52$ where we add $1/52$ a total of $104$ times which would come out to a final total of $2$... which is larger than $1$... which is impossible. Simply adding like this does not give a probability... it gives an expected value, the expected number of times that it should happen. Drawing $104$ cards we expect to have seen a king of spades twice. $\endgroup$
    – JMoravitz
    Commented Apr 15, 2019 at 21:49
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    $\begingroup$ You may have been told somewhere that "or" means to add the probabilities, that $Pr(A\cup B)=Pr(A)+Pr(B)$ and that "and" means to multiply, that $Pr(A\cap B)=Pr(A)\times Pr(B)$. This is incorrect. Those only work in the very restrictive scenarios where they are mutually exclusive (and so you can add like this) or when they are independent (and so you can multiply like this). The correct things to do that you can do whenever are instead $Pr(A\cup B) = Pr(A)+Pr(B)\color{red}{-Pr(A\cap B)}$ and $Pr(A\cap B)=Pr(A)\times Pr(B\color{red}{|A})$ $\endgroup$
    – JMoravitz
    Commented Apr 15, 2019 at 21:51
  • $\begingroup$ As indicated by the previous comments, your first computation of $~\dfrac{1}{52} + \dfrac{1}{52} ~$ counts twice the situation where you draw the King of Spades on both draws. This is why you (in effect) must subtract $~\color{red}{\text{Pr}(A \cap B)}.$ $\endgroup$
    – user2661923
    Commented Jul 15 at 18:47

2 Answers 2

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$$P(\text{King of spades is drawn}) = P(\text{King of spades is drawn in the first try}) + P(\text{King of spades is drawn in the second try and not in the first try})$$

Case 1:

$$P(\text{King of spades is drawn in the first try}) = 1/52$$

$$P(\text{King of spades is drawn in the second try and not in the first try}) = 51/52*1/52$$

Case 2:

$$P(\text{King of spades is drawn in the first try}) = 1/52$$

$$P(\text{King of spades is drawn in the second try and not in the first try}) = 51/52*1/51$$

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  • $\begingroup$ I think my definition of the event was a little confusing so have edited my answer to make it clearer. Thanks $\endgroup$
    – Vizag
    Commented Apr 15, 2019 at 22:35
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Assumption:

  1. Each player draws two cards at once.
  2. 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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