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Shuffle an ordinary deck of 52 playing cards containing four aces. Then turn up the cards from the top until the first ace appears. On the average how many cards are required to be turned before producing the first ace?

The question is answered here using symmetry. I have seen other such questions.

Is there a name for this method? Specifically, is there a name for the type of symmetry being used.

Note: I don't want a solution. There are many solutions already available. I am asking for a name for the symmetry used.

Starlight
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  • (no known name ; a generic name for this kind of questions is "runs"). A remark : the accepted answer is a little "handwaving" : happily there is the comment by Roy ! – Jean Marie Mar 26 '24 at 10:41

1 Answers1

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Expectation is about averages, so one way to understand this problem is to imagine the $52$ cards make $52$ "cuts" on a $0-1$ line, so are evenly placed at $1/53, 2/53, ...k/53,...52/53$

Likewise, the aces would make "cuts" to be evenly placed at $1/5,2/5/...5/5$

I can't say what type of symmetry this is, it is just symmetry of evenly placed elements on an average

Then the first ace will be found at the point so that $1/5 = k/53 \Rightarrow k = 53/5 = 10.6$

Another more formal way to get the same result is to use indicator random variables. Consider the $4$ aces together in conjunction with one non-ace, and define an indicator variable $X_i$ for the non-ace $i$ to have $Pr=1$ if it is in front of all the aces, $0$ otherwise

so $P(X_i) = \frac15$, since the non-ace is equally likely to be at any of the $5$ positions

Now the expectation of an indicator variable is just the probability of the event it indicates, thus

$\mathbb{E}[X_i] = P(X_i) = \frac15$

and by linearity of expectation, which operates even when the random variables are not independent,

$\mathbb{E}[X] = \Sigma{ \mathbb{E}[X_i]} = \frac{48}5 = 9.6$

and the first ace is expected to be at position $9.6+1 =10.6$

PS

If "how many cards are required to be turned before producing the first ace" is interpreted to exclude the first ace, the answer will be one less, $9.6$

true blue anil
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