4 cards are shuffled and placed face down. Hidden faces display 4 suits: spades, clubs, diamonds, hearts. You turn over cards until win or lose

Question

Problem: 4 cards are shuffled and placed face down. Hidden faces display 4 suits: spades, clubs, diamonds, hearts. You turn over cards until win or lose. You win if you turn over spades AND clubs. You lose if you turn over hearts. Find P(win)

I can't seem to find what's wrong with my approach.

Law of total probability:

$P(lose) = P(lose | H_1) * P(H_1) + P(lose | H_2) * P(H_2) + P(lose | H_3) * P(H_3) + P(lose | H_4) * P(H_4)$ where $Hx$ denotes the Hearts card being in the $x^{th}$ position

$P(lose) = 1/4 * 1 + 1/4 * 1 + 1/4 * (1/4)^2 + 1/4 * 0 = 0.515625$

If Hearts is in either 1st or 2nd position, we always lose. If Hearts is in 3rd position, we lose if Spades & Clubs are exactly in 1st and 2nd position, which happens with $P = (1/4)^2$, in the last scenario we cannot lose

$P(win) = 1-P(lose) = 0.484375$ But the official answer is $2/3$, so I don't know where I went wrong. Please also explain the correct answer like I'm 5.

Thanks for your help!

Answer 1

Think about where the diamond may be when given that the heart is somewhere.

We lose if the heart is in position 1 or 2, or if the heart is in position 3 and the diamond is not in position 4.

$$\begin{align}\mathsf P(L)&=\mathsf P(H_1\cup H_2\cup(H_3\cap D_4^{\small\complement}))\\[1ex]&=\mathsf P(H_1)+\mathsf P(H_1)+\mathsf P(H_3)\;\mathsf P(D_4^{\small\complement}\mid H_3)\end{align}$$

When given that the heart is in position 3, the probability that the diamond is not in position 4 is: $2/3$.

$$\begin{align}\mathsf P(L)&=\tfrac 14+\tfrac 14+\tfrac 14\cdot\tfrac 23\\[1ex]&=\tfrac{2}{3}\end{align}$$

Answer 2

I think your problem is that your tree of possibilities needs pruning. You also are missing a conditional probability: The probability of spades and clubs being the first two given the heart being 3rd is not 1/16 but rather 1/3. This problem is more easily done by drawing the tree.

Answer 3

If Hearts is in either 1st or 2nd position, we always lose.

Correct. This happens with probability $\frac{1}{2}$.

If Hearts is in 3rd position, we lose if Spades & Clubs are exactly in 1st and 2nd position.

This is not correct. You lose if Diamonds is in either 1st or 2nd position. This happens with probability $\frac{1}{4}\times \frac{2}{3}=\frac{1}{6}$.

The probability of losing is therefore $\frac{1}{2}+\frac{1}{6}=\frac{2}{3}.$