What's the expected position of the second card in a deck of 108 cards

Question

Suppose we have a deck of cards from two sets of cards (104 or 108 cards).

What is the expected position of the second, let's say, spade ace ?

Any idea ?

I am trying to figure out something for a card game Gong Zhu (en.wikipedia.org/wiki/Gong_Zhu), a Hearts variant in China.

I posted this question on mathoverflow and immediately got put on hold as off-topic :( Seems a very serious math site there. Hope someone here can answer my question. Thanks.

Answer 1

Assume a $104$-card deck. If the first ace of spades is in position $n$, the expected value of the position of the second ace of spades is $n+\frac12(104-n)=52+\frac{n}2$. If $p_n$ is the probability that the first ace is in position $n$, then the expected position of the first ace is $\bar n=\sum_{k=1}^{104}p_nn$, and the expected position of the second ace is

$$\sum_{k=1}^{104}p_n\left(52+\frac{n}2\right)=52+\frac{\bar n}2\;.$$

By symmetry we must have

$$52+\frac{\bar n}2=105-\bar n\;,$$

so $\frac32\bar n=53$, $\bar n=\frac23\cdot53=\frac{106}3=35\frac13$, and the expected position of the second ace of spades is $$52+\frac{\bar n}2=52+\frac{53}3=\frac{209}3=69\frac23\;.$$

Or, more simply, it’s $105-35\frac13=69\frac23$. Note that positions $1$, $35\frac13$, $69\frac23$, and $104$ are evenly spaced.

Answer 2

Suppose the two cards are the aces of spades and the deck contains 104 cards. In order for the second ace of spades to fall at position n, the first n-1 cards must contain exactly one aces of spades and the nth card must be an ace of spades. The probability that the first n-1 cards contain exactly one ace of spades is $$\frac{\binom{2}{1} \binom{102}{n-2}} {\binom{104}{n-1}} = \frac{2 (n-1) (105-n)}{103 \cdot 104}$$ and the probability that the nth card will be an ace of spades given that the previous n-1 cards contains one ace of spades is $$\frac{1}{105-n}$$ so the probability that the nth card is the second ace of spades is $$\frac{2 (n-1) (105-n)}{103 \cdot 104} \cdot \frac{1}{105-n} = \frac {2(n-1)}{103 \cdot 104}$$ and the expected position is $$E[n] = \sum_{n=1}^{104} n \cdot \frac {2(n-1)}{103 \cdot 104} = 70$$

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[edit]A simpler derivation of the probability that the nth card will be the second ace of spades: There are $\binom{104}{2}$ ways to choose the positions of the two aces, all of which are equally likely. There are n-1 of these in which the second ace is at position n. So the probability the second ace is at position n is

$$(n-1) / \binom{104}{2} = \frac{2 (n-1)}{103 \cdot 104}$$[/edit]