Question

I am going to play a card game. I get 5 chances to draw an Ace from a 52 card deck. Each time I draw, I do not place the drawn card back into the deck. The game is over when either I draw an Ace, or I have drawn 5 cards without drawing an Ace. It costs me $1 to play. If I draw the Ace of spades, I win $50, if I draw any other Ace I win $10. What is the probability I will draw an Ace (other than spades) and the probability I will draw the Ace of spades.

Answer 1

Question: I am going to play a card game. I get 5 chances to draw an Ace from a 52 card deck. Each time I draw, I do not place the drawn card back into the deck. The game is over when either I draw an Ace, or I have drawn 5 cards without drawing an Ace. It costs me $1 to play. If I draw the Ace of spades, I win $50, if I draw any other Ace I win $10. What is the probability I will draw an Ace (other than spades) and the probability I will draw the Ace of spades.

Solution:

By classical definition of probability let's find probabilities $\mathsf{P}(\mathsf{A}_1),\mathsf{P}(\mathsf{A}_2)$ of events $\mathsf{A}_1,\mathsf{A}_2$ ..

$A_{1} =$ "Draw an A at the $1^{\mathrm{st}}$ turn", $\mathsf{P}(\mathsf{A}_1) = \frac{1}{52}$

$A_{2} =$ "Draw an A at the $2^{\text{nd}}$ turn" $\equiv$ "draw no ace at the first turn and then draw A at the second one", $\mathsf{P}(\mathsf{A}_2) = \frac{48}{52} \cdot \frac{1}{51} = \frac{4}{221}$ .

By analogy,

$A_{3} =$ "Draw an A at the $3^{\mathrm{rd}}$ turn", $\mathsf{P}(\mathsf{A}_3) = \frac{48}{52} \cdot \frac{47}{51} \cdot \frac{1}{50} = \frac{94}{5525}$ .

$A_{4} =$ "Draw an A at the $4^{\text{th}}$ turn", $\mathsf{P}(\mathsf{A}_4) = \frac{48}{52} \cdot \frac{47}{51} \cdot \frac{46}{50} \cdot \frac{1}{49} = \frac{4324}{270725}$ .

$A_{5} =$ "Draw an A at the $5^{\text{th}}$ turn", $\mathsf{P}(\mathsf{A}_5) = \frac{48}{52} \cdot \frac{47}{51} \cdot \frac{46}{50} \cdot \frac{45}{49} \cdot \frac{1}{48} = \frac{3243}{216580}$ .

Since $A =$ "Draw an $A \spadesuit$ " = $A_1 \cup A_2 \cup A_3 \cup A_4 \cup A_5$

is the union of mutually exclusive events $A_{1}, A_{2}, A_{3}, A_{4}, A_{5}$

then

$\mathsf{P}(\mathsf{A}) = \mathsf{P}(\mathsf{A}_1) + \mathsf{P}(\mathsf{A}_2) + \mathsf{P}(\mathsf{A}_3) + \mathsf{P}(\mathsf{A}_4) + \mathsf{P}(\mathsf{A}_5) = \frac{1}{52} +\frac{4}{221} +\frac{94}{5525} +\frac{4324}{270725} +\frac{3243}{216580} = \frac{4618}{54145}\approx 0.09.$

Let $B$ be the event "draw no ace". Its probability is

$\mathsf{P}(\mathsf{B}) = \frac{48}{52}\cdot \frac{47}{51}\cdot \frac{46}{50}\cdot \frac{45}{49}\cdot \frac{44}{48} = \frac{35673}{54145}$

Consider event $C =$ "Draw an ace, but not $A \spadesuit$ " = "Draw any ace" \ "Draw $A \spadesuit$ " = (U \B)A.

Here U is the sample space. Therefore, according to additivity of probability of mutually exclusive events we have

$\mathsf{P}(\mathsf{C}) = 1 - \mathsf{P}(\mathsf{B}) - \mathsf{P}(\mathsf{A}) = 1 - \frac{35673}{54145} -\frac{4618}{54145} = \frac{13854}{54145}\approx 0.26.$

Answer: $\mathsf{P}(\text{"Draw an ace, but not } \mathsf{A} \spadesuit") = \frac{13854}{54145} \approx 0.26$ ;

$\mathsf{P}(\text{"Draw A} \spadesuit") = \frac{4618}{54145} \approx 0.09.$