# The Most Likely Position

### Problem

### Remark

This is problem 6 from the Wrangle contest at the Joint Mathematics Meetings 2018. Its formulation seems to be open to interpretations. There are at least two:

On many, many trials, what is the expected position of the first ace?

For what position of the first ace the probability is the highest?

We already have an answer to the first interpretation. The second question appears to be obvious to experts, with an immediate answer. The answer is still surprising to non-experts. So, this is the interpretation will be solving below:

A standard deck of the cards has $52$ cards, of which four are aces. When the deck is shuffled, for what position of the first ace the probability is the highest?

Solution 1 below is by twitter.com/@BKKeconcode, Solution 2 is by Amit Itagi.

### Solution 1

Let $p(k)$ be the probability for the first ace to appear at position $k.$ Then

$\displaystyle \begin{align} &p(1)=\frac{4}{52}\\ &p(2)=\frac{48}{52}\cdot\frac{4}{51}\\ &p(3)=\frac{48}{52}\cdot\frac{47}{51}\cdot\frac{4}{50}\\ &p(4)=\frac{48}{52}\cdot\frac{47}{51}\cdot\frac{46}{50}\cdot\frac{4}{49}\\ &\ldots\\ &p(k)=\left(\prod_{j=0}^{j=k-2}\frac{48-j}{52-j}\right)\cdot\frac{4}{53-k}. \end{align}$

Note that

$\displaystyle\begin{align} \displaystyle \frac{p(k+1)}{p(k)}&=\left(\frac{49-k}{53-k}\cdot\frac{4}{52-k}\right)\cdot\left(\frac{53-k}{4}\right)\\ &=\frac{49-k}{52-k}\lt 1, \end{align}$

implying $p(k+1)\lt p(k),$ so that $p(1),$ however small in itself, is the largest of all the other probabilities.

### Solution 2

If the first ace is at position $n,$ the number of ways of placing the remaining three aces is $\displaystyle {52-n\choose 3}.$ Having chosen these four positions, the number of permutations of cards is $4!48!.$ Thus, the probability of the first ace being at position $n$ is

$\displaystyle P(n)={52-n\choose 3}\cdot\frac{4!48!}{52!}=\frac{\displaystyle {52-n\choose 3}}{\displaystyle {52\choose 4}}.$

$\displaystyle {52-n\choose 3}$ is a decreasing function of $n.$ Thus, the probability is maximized for $n=1.$

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