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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.

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  • $\begingroup$ If you know the expected position of the first ace, the expected position of the second (there being two spade aces) will be that same position when the cards are taken in reverse order. $\endgroup$ – Mark Bennet Oct 3 '13 at 17:03
  • $\begingroup$ For single card in 108 card deck, the expected position is the mean of a uniform distribution. so (a+b)/2 = (1+108)/2 = 54.5. So you think the expected position for the second card is also 54.5 ? That seems odd as two cards can not be at the same position... $\endgroup$ – whoji Oct 3 '13 at 17:16
  • $\begingroup$ The point is if there are two cards the expected distance of the first from the top is equal to the expected distance of the second from the bottom. $\endgroup$ – Mark Bennet Oct 3 '13 at 17:25
  • $\begingroup$ That's a good clue. Haven't thought about that. Thanks Mark. $\endgroup$ – whoji Oct 3 '13 at 17:28
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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.

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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 $$E[n] = \sum_{n=1}^{104} n \cdot \frac {2(n-1)}{103 \cdot 104} = 70$$


[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]

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