But the average roll is 3.5

Probability Level 4

Suppose I have a line of squares, labelled $0,1,2,3,4, \ldots$ and so on. I place a counter on the square 0. On every turn, I roll a fair die (with the numbers 1 to 6) and move forward as follows: if I was on square $P$ before the roll and I get a $x$ then I move to square $P+x$ . Let $X_n$ be the chance that I land on the square labelled $n$ . Find

$\text{max}(X_n)? \space\space ( n \in \mathbb{N})$

and submit your answer as $n$ .

There is no maximum but

X_n

isn't strictly increasing 6

X_n

is strictly increasing 7

1 solution

Varsha Dani
Oct 6, 2018

$X_n$ = probability of landing on $n$ . In order to land on $n$ one must first land on one of $n-6, n-5, \dots n-1$ and then roll the appropriate number. Thus

$X_n = \frac{1}{6} \sum_{j=1}^{6} X_{n-j}$

Also, since the counter starts on 0, the base of the recurrence is $X_{i} = 0$ for $i<0$ and $X_0 =1$ . Clearly $X_n$ is increasing on 1 through 6. Thereafter, for each $n$ , since $X_n$ averages the previous six values, $X_n \le \max\{X_{n-j} | 1\le j\le 6\}$ . It follows that the maximum values of $X_n$ over windows of size 6 are decreasing. Therefore $X_n$ is maximized at $n=6$ .

By the way I found your question to be somewhat confusingly phrased. You may wish to rephrase it as:

Find $\mathrm{argmax}_{n\in \mathbb{N}} (X_n)$ .

Find the value of $n$ that maximizes $X_n$ .

But the average roll is 3.5

1 solution

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