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For practice, I decided to invent a variation on the problem of the expected number of cards to be drawn until the first ace is found. I would appreciate some feedback and clarification on certain details of my solution, which I describe at the end.

Problem: Given a deck of $52$ standard playing cards, what is the expected value of the highest card seen before drawing the first ace?* (Where we set $J = 11, Q = 12, K = 13$).

Solution: Let $H$ be the random variable corresponding to the value of the highest card. We note that the four aces divide the deck into five "boxes". Then, we may observe that:

$$\mathbb{P}(H \le n) = \mathbb{P}( \text{no card of value >} n \text{ is before the first } A )$$

For brevity, I will write this as $\mathbb{P}(S_n)$. We can imagine any permutation of the deck (ignoring permutation of the aces) as inserting cards into the gaps in between previously placed cards (including the aces).

Then, by considering this framing:

$$\mathbb{P}(S_n) = \prod_{k=0}^{4(13-n)-1} \mathbb{P}(\text{highest } (k+1) \text{ cards placed after } A \; | \text{ highest } k \text{ cards placed after } A)$$

$$ = \prod_{k=0}^{4(13-n)-1} \frac{4+k}{5+k} = \frac{4}{4(14-n)} = \frac{1}{14-n}$$

Then, noting that $\mathbb{P}(H = n) = \mathbb{P}(S_n) - \mathbb{P}(S_{n-1})$,

$$\mathbb{E}[H] = \sum_{n=2}^{13} n \bigg(\frac{1}{14-n} - \frac{1}{15-n}\bigg) \approx 10.74$$

*The solution ignores the probability that an ace is the first card, which is taken to be negligible.

Questions

1. Why is the probability space above is equivalent to the uniform distribution on all shuffled decks? It seems obvious, but is there a way to justify it more rigorously?
2. Why can we ignore the order of the cards when we insert them in the way described above?

The second question in particular may seem inane, but it's still quite difficult for me to determine if and when it's necessary to account for permutations, and I would appreciate an intuitive explanation if there is one.

For practice, I decided to invent a variation on the problem of the expected number of cards to be drawn until the first ace is found. I would appreciate some feedback and clarification on certain details of my solution, which I describe at the end.

Problem: Given a deck of $52$ standard playing cards, what is the expected value of the highest card seen before drawing the first ace? (Where we set $J = 11, Q = 12, K = 13$, and we say that the value is $0$ if the first card is an ace.)

Solution: Let $H$ be the random variable corresponding to the value of the highest card. We note that the four aces divide the deck into five "boxes". Then, we may observe that:

$$\mathbb{P}(H \le n) = \mathbb{P}( \text{no card of value >} n \text{ is before the first } A )$$

For brevity, I will write this as $\mathbb{P}(S_n)$. We can imagine any permutation of the deck (ignoring permutation of the aces) as inserting cards into the gaps in between previously placed cards (including the aces).

Then, by considering this framing:

$$\mathbb{P}(S_n) = \prod_{k=0}^{4(13-n)-1} \mathbb{P}(\text{highest } (k+1) \text{ cards placed after } A \; | \text{ highest } k \text{ cards placed after } A)$$

$$ = \prod_{k=0}^{4(13-n)-1} \frac{4+k}{5+k} = \frac{4}{4(14-n)} = \frac{1}{14-n}$$

Then, noting that $\mathbb{P}(H = n) = \mathbb{P}(S_n) - \mathbb{P}(S_{n-1})$,

$$\mathbb{E}[H] = \sum_{n=2}^{13} n \bigg(\frac{1}{14-n} - \frac{1}{15-n}\bigg) \approx 10.74$$

Questions

1. Why is the probability space above is equivalent to the uniform distribution on all shuffled decks? It seems obvious, but is there a way to justify it more rigorously?
2. Why can we ignore the order of the cards when we insert them in the way described above?

The second question in particular may seem inane, but it's still quite difficult for me to determine if and when it's necessary to account for permutations, and I would appreciate an intuitive explanation if there is one.

For practice, I decided to invent a variation on the problem of the expected number of cards to be drawn until the first ace is found. I would appreciate some feedback and clarification on certain details of my solution, which I describe at the end.

Problem: Given a deck of $52$ standard playing cards, what is the expected value of the highest card seen before drawing the first ace?* (Where we set $J = 11, Q = 12, K = 13$).

Solution: Let $H$ be the random variable corresponding to the value of the highest card. We note that the four aces divide the deck into five "boxes". Then, we may observe that:

$$\mathbb{P}(H \le n) = \mathbb{P}( \text{no card of value >} n \text{ is before the first } A )$$

For brevity, I will write this as $\mathbb{P}(S_n)$. We can imagine any permutation of the deck (ignoring permutation of the aces) as inserting cards into the gaps in between previously placed cards (including the aces).

Then, by considering this framing:

$$\mathbb{P}(S_n) = \prod_{k=0}^{n-1} \mathbb{P}(\text{first } (k+1) \text{ cards placed before } A \; | \text{ first } k \text{ cards placed before } A)$$

$$ = \prod_{k=0}^{n-1} \frac{4+k}{5+k} = \frac{4}{4(14-k)} = \frac{1}{14-k}$$

Then, noting that $\mathbb{P}(H = n) = \mathbb{P}(S_n) - \mathbb{P}(S_{n-1})$,

$$\mathbb{E}[H] = \sum_{n=2}^{13} n \bigg(\frac{1}{14-n} - \frac{1}{15-n}\bigg) \approx 10.74$$

*The solution ignores the probability that an ace is the first card, which is taken to be negligible.

Questions

1. Why is the probability space above is equivalent to the uniform distribution on all shuffled decks? It seems obvious, but is there a way to justify it more rigorously?
2. Why can we ignore the order of the cards when we insert them in the way described above?

The second question in particular may seem inane, but it's still quite difficult for me to determine if and when it's necessary to account for permutations, and I would appreciate an intuitive explanation if there is one.

For practice, I decided to invent a variation on the problem of the expected number of cards to be drawn until the first ace is found. I would appreciate some feedback and clarification on certain details of my solution, which I describe at the end.

Problem: Given a deck of $52$ standard playing cards, what is the expected value of the highest card seen before drawing the first ace?* (Where we set $J = 11, Q = 12, K = 13$).

Solution: Let $H$ be the random variable corresponding to the value of the highest card. We note that the four aces divide the deck into five "boxes". Then, we may observe that:

$$\mathbb{P}(H \le n) = \mathbb{P}( \text{no card of value >} n \text{ is before the first } A )$$

For brevity, I will write this as $\mathbb{P}(S_n)$. We can imagine any permutation of the deck (ignoring permutation of the aces) as inserting cards into the gaps in between previously placed cards (including the aces).

Then, by considering this framing:

$$\mathbb{P}(S_n) = \prod_{k=0}^{4(13-n)-1} \mathbb{P}(\text{highest } (k+1) \text{ cards placed after } A \; | \text{ highest } k \text{ cards placed after } A)$$

$$ = \prod_{k=0}^{4(13-n)-1} \frac{4+k}{5+k} = \frac{4}{4(14-n)} = \frac{1}{14-n}$$

Then, noting that $\mathbb{P}(H = n) = \mathbb{P}(S_n) - \mathbb{P}(S_{n-1})$,

$$\mathbb{E}[H] = \sum_{n=2}^{13} n \bigg(\frac{1}{14-n} - \frac{1}{15-n}\bigg) \approx 10.74$$

*The solution ignores the probability that an ace is the first card, which is taken to be negligible.

Questions

1. Why is the probability space above is equivalent to the uniform distribution on all shuffled decks? It seems obvious, but is there a way to justify it more rigorously?
2. Why can we ignore the order of the cards when we insert them in the way described above?

The second question in particular may seem inane, but it's still quite difficult for me to determine if and when it's necessary to account for permutations, and I would appreciate an intuitive explanation if there is one.