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I am having some trouble solving the following.

You draw cards from a standard 52-card deck without replacement until you get an ace. You draw from what is remaining until you get a 2. You continue on with 3. What is the expected number you will be on after the entire deck runs out?

It was natural to let

• $T_i = \text{first position of card whose value is }i$
• $U_i = \text{last position of card whose value is }i$

So the problem essentially amounts to figuring out the probability that you will be on $k$ when the deck runs out, namely:

$$Pr(T_1<\cdots<T_k \cup U_{k+1} < T_k)$$

I can see that

$$Pr(T_1<\cdots<T_k) = 1/k! \\ \text{and} \\ Pr(U_{k+1} < T_k) = 1/70$$

but could not get any further...

I am having some trouble solving the following.

You draw cards from a standard 52-card deck without replacement until you get an ace. You draw from what is remaining until you get a 2. You continue on with 3. What is the expected number you will be on after the entire deck runs out?

It was natural to let

• $T_i = \text{first position of card whose value is }i$
• $U_i = \text{last position of card whose value is }i$

So the problem essentially amounts to figuring out the probability that you will be on $k$ when the deck runs out, namely:

$$Pr(T_1<\cdots<T_k \cap U_{k+1} < T_k)$$

I can see that

$$Pr(T_1<\cdots<T_k) = 1/k! \\ \text{and} \\ Pr(U_{k+1} < T_k) = 1/70$$

but could not get any further...

I am having some trouble solving the following.

You draw cards from a standard 52-card deck without replacement until you get an ace. You draw from what is remaining until you get a 2. You continue on with 3. What is the expected number you will be on after the entire deck runs out?

It was natural to let

$T_i = \text{first position of card whose value is }i$

$U_i = \text{last position of card whose value is }i$

So the problem essentially amounts to figuring out

$Pr(T_1<\cdots<T_k \cup U_{k+1} < T_k) = \text{probability that you will be on $k$ when the deck runs out}$

I can see that

$Pr(T_1<\cdots<T_k) = 1/k!$ and $Pr(U_{k+1} < T_k) = 1/70$

but could not get any further...

I am having some trouble solving the following.

You draw cards from a standard 52-card deck without replacement until you get an ace. You draw from what is remaining until you get a 2. You continue on with 3. What is the expected number you will be on after the entire deck runs out?

It was natural to let

• $T_i = \text{first position of card whose value is }i$
• $U_i = \text{last position of card whose value is }i$

So the problem essentially amounts to figuring out the probability that you will be on $k$ when the deck runs out, namely:

$$Pr(T_1<\cdots<T_k \cup U_{k+1} < T_k)$$

I can see that

$$Pr(T_1<\cdots<T_k) = 1/k! \\ \text{and} \\ Pr(U_{k+1} < T_k) = 1/70$$

but could not get any further...