Card draws and two players game

Question

Let's consider a deck of $52$ cards $(A,K,Q,J,10,9,8,7,6,5,4,3,2)$. Two players each draw $5$ cards without putting them back into the deck.

1. Calculate the probability that the first player draws at least one Ace and at least one King

2. Calculate the probability that the second player draws exactly one King knowing that the first player has drawn the King of Hearts and exactly one 2 among his five cards.

Now suppose that the two players carry out a series of draws one after the other (the first one goes first) and the winner is the one who draws the King of Spades first.

3. Calculate the probability that at least four cards must be revealed before the King of Spades is revealed, in the case in which reinsertion is carried out and in the case in which it is not carried out.

4. If the players decide to put the cards back into the deck after each draw (shuffling), is the first player more likely to win the game?

My attempt is the following:

1. If $A$ is the requested event, then $A^C$ is the event in which neither Aces nor Kings are extracted.

$$\mathbb{P}(A)=1-\mathbb{P}(A^C)=1-\frac{\binom{44}{5}}{\binom{52}{5}}\approx 0.58$$

2. If $B$ is the requested event, then

$$\mathbb{P}(B)=\frac{\binom{3}{1}\binom{44}{4}}{\binom{47}{5}}\approx 0.27$$

3. I have that, if $N$ is the variable that counts the number of cards drawn.

EXTRACTION WITH REINSERTION:

$$\mathbb{P}(N\geq 4)=1-\mathbb{P}(N<4)=\frac{1}{52}+\frac{51}{52}\cdot \frac{1}{52}+\left(\frac{51}{52}\right)^2\cdot \frac{1}{52}=$$

$$=1-\frac{1}{52}\left(1+\frac{51}{52}+\left(\frac{51}{52}\right)^2\right)\approx 0.943$$

EXTRACTION WITHOUT REINSERTION:

$$\mathbb{P}(N\geq 4)=1-\mathbb{P}(N<4)=1-\left(\frac{1}{52}+\frac{51}{52}\cdot\frac{1}{51}+\frac{51}{52}\cdot \frac{50}{51}\cdot \frac{1}{50}\right)=$$

$$=1-\frac{3}{52}=\frac{49}{52}\approx 0.942$$

4. No, the probability is always the same.

I don't understand if is it correct that in 3. the probabilities are the same and if 4. is a correct deduction. Any suggestions?

Answer 1

In scenario $4$, let $p$ be the probability that the first player wins, so that the probability that the second player wins is $1-p$.

Assume that the first player does not draw the King of Spades on the first draw. That happens with probability $\frac{51}{52}$. If that does happen (so that the game continues), then the players have effectively switched positions so the conditional probability that the second player wins is $p$.

In other words, $1-p=\frac{51}{52}p$, so $p=\frac{52}{103}$.

The result is different if the players are drawing without replacement, because if the first player fails to draw the King of Spades on the first draw, the second player now has fewer cards to draw from and therefore has an increased chance to draw the King of Spades. The game without replacement is exactly the same as just permuting the entire deck, with each permutation having equal probability. Player $1$ wins if the King of Spades is in an odd position and Player $2$ wins if the King of Spades is in an even position. This shows the game without replacement is a $50$-$50$ game.

The probabilities are not the same in scenario $3$. Both of your calculations are correct, but the difference in the fractions is too small to show up in the first two significant figures.

Answer 2

Your answer to Q1 is wrong, because you are assuming that the question is equivalent to asking for the the complement of the probability that the person did not draw any Ace or King.

The person can fail to simultaneously draw at least one Ace and at least one King by drawing (for example) 4 Kings and a deuce. This alternative method of failing the event is (in effect) omitted from your computation of $~\displaystyle 1 - \dfrac{\binom{44}{5}}{\binom{52}{5}}.$

The problem can be attacked by Inclusion-Exclusion. See this article for an introduction to Inclusion-Exclusion. Then, see this answer for an explanation of and justification for the Inclusion-Exclusion formula.

Following the syntax in the second link above, let $~S~$ denote the set of all possible combinations of $~5~$ cards from the $~52~$ card deck. Let $~S_1~$ denote the subset of $~S~$ where no King was drawn. Let $~S_2~$ denote the subset of $~S~$ where no Ace was drawn.

Then, you want

$$\frac{1}{\binom{52}{5}} \times \left\{ ~|S| - |S_1 \cup S_2| ~\right\}. \tag1 $$

Let $~\displaystyle T_0 = |S| = \binom{52}{5}.~$

Let $~T_1 = |S_1| + |S_2|.~$
By symmetry, $~\displaystyle T_1 = 2 \times \binom{48}{5}.$

Let $~\displaystyle T_2 = |S_1 \cap S_2| = \binom{44}{5}.$

Then, by Inclusion-Exclusion theory, the expression in (1) above is equivalent to

$$= \frac{1}{\binom{52}{5}} \times \left\{ ~T_0 - T_1 + T_2 ~\right\}$$

$$= \frac{1}{\binom{52}{5}} \times \left\{ ~\binom{52}{5} - \left[ ~2 \times \binom{48}{5} ~\right] + \binom{44}{5} ~\right\}.$$