Throwing dice: Compute the probability of strictly increasing values in the first $n$ throws

Question

Let a die be thrown $m$ times. What is the probability of strictly increasing values in the first $n< m$ throws? In other words, compute the probability $P(X=n)$ with $X\,\widehat{=}$"The first $n$ face values are strictly increasing".

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The problem here is that I can't seem to find a rule which is common to how the probabilities of $P(X=1),P(X=2),\ldots$ seem to form in order to arrive at a general rule. How do I approach this?

Answer 1

Consider the set of all rolls of $n$ dice, each die having $N$ sides. There are $N^n$ such sequences. How many of these sequences are strictly increasing?

Every possible sequence of $n$ strictly increasing die rolls can be obtained by selecting $n$ distinct values, without replacement, from the set of the die's $N$ faces, and then sorting them into a strictly increasing order. There are $N \choose n$ distinct ways to do this.

Thus, the probability of a sequence of $n$ rolls being strictly increasing is $$ p = \frac{1}{N^n} { N \choose n}. $$

For $N = 6$, these values work out to be:

• $n = 1$: $p = 1$
• $n = 2$: $p = 5/12$
• $n = 3$: $p = 5/54$
• $n = 4$: $p = 5/432$
• $n = 5$: $p = 1/1296$
• $n = 6$: $p = 1/46656$