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I'm working Matthew M. Conroy's "A Collection of Dice Problems". Question 14 is the following:

Suppose we roll a fair die 10 times. What is the probability that the sequence of rolls is non-decreasing (i.e., the next roll is never less than the current roll)?

First I considered what happens when you only perform two rolls. Let $d_i$ be the value of the $i^{\text{th}}$ roll. Then for two rolls, we have three possibilities:

  • $d_1 < d_2$
  • $d_1 = d_2$
  • $d_1 > d_2$

From here we note that $P(d_1 < d_2) = P(d_1 > d_2)$ and $P(d_1 = d_2) = 6 \times \frac{1}{6} \times \frac{1}{6} = \frac{1}{6}$. We can combine the above to conclude that $$1 = P(d_1 < d_2) + P(d_1 > d_2) + P(d_1 = d_2) \implies P(d_1 < d_2) = \frac{5}{12}$$ And thus for two rolls, the probability that you roll a non-decreasing sequence is $P=\frac{5}{12} + \frac{1}{6}=\frac{7}{12}$.

Now if I consider three rolls. First we will perform two rolls and have a probability of $\frac{7}{12}$ of having a non-decreasing sequence. Then we roll the third time, again we have three situations.

  • $d_2 < d_3$
  • $d_2 = d_3$
  • $d_2 > d_3$

I wish to assert that $P(d_2 < d_3) = P(d_2 > d_3)$. But I am not sure if they are equal. Since I have forced $d_2$ to be from a non-decreasing sequence, the value of $d_2$ is skewed towards larger values (for example, there is only one way $d_2=1$, but three ways $d_2 = 2$). If $P(d_2 < d_3) = P(d_2 > d_3)$, Then I can just solve it the same have i did for two rolls, and thus for 10 rolls. But again, that would require the assertion $P(d_2 < d_3) = P(d_2 > d_3)$. Is such a assertion valid?

user21820
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WaterDrop
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  • $P(d_2 < d_3) = P(d_2 > d_3)$ in the $6^3$ potential three dice rolls (by symmetry/exchangeability) but this need not apply as a conditional probability, either in an overall non-decreasing sequence in which case $P(d_2 < d_3) >P (d_2>d_3)=0$, or just conditioned on $d_1 \le d_2$ in which case $P(d_2 < d_3) <P (d_2>d_3)$ since $d_2$ is likelier to be high. – Henry Sep 28 '24 at 20:27

2 Answers2

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The problem can be addressed much more simply.

  • Count the number of ways $10$ balls can be placed in $6$ bins marked $1-6$, using stars and bars

  • Note that each of the $\binom{10+6-1}{6-1} = 3003$ results thus obtained can yield only one non-decreasing sequence.

  • A result of $\;\;\fbox{2}\fbox{0}\fbox{3}\fbox{1}\fbox{2}\fbox{2}\;$, e.g. means a sequence of $1-1-3-3-3-4-5-5-6-6$

  • Thus $Pr = \dfrac{3003}{6^{10}}$

true blue anil
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An easier approach is to count the possibilities of the differences $g_i=d_i -d_{i-1}$ all being at least $0$.

To generalise this, let's suppose you roll an $k$-sided die $n$ times. We will also take the fictional $d_0=1$ and $d_{n+1}=k$ since these would be associated with non-negative differences. So we need the $n+1$ differences (including the fictional ones at the beginning and end) to add up to $k-1$, and a stars and bars argument will tell us there are ${n+k-1 \choose n}$ ways of doing this. Divide this count by $k^n$ to get a probability.

To take your example with $k=6$,

  • For $n=2$ dice rolls, this gives $\frac{21}{36} =\frac7{12}$ as you found
  • For $n=3$ dice rolls, this gives $\frac{56}{216} =\frac7{27}$
  • For $n=4$ dice rolls, this gives $\frac{126}{1296} =\frac7{72}$
  • For $n=5$ dice rolls, this gives $\frac{252}{7776} =\frac7{216}$

though sadly the pattern of $7$ being the reduced numerator does not quite continue

  • For $n=6$ dice rolls, this gives $\frac{462}{46656} =\frac{77}{7776}$
  • For $n=7$ dice rolls, this gives $\frac{792}{279936} =\frac{11}{3888}$
  • For $n=8$ dice rolls, this gives $\frac{1287}{1679616} =\frac{143}{186624}$
  • For $n=9$ dice rolls, this gives $\frac{2002}{10077696} =\frac{1001}{5038848}$
  • For $n=10$ dice rolls, this gives $\frac{3003}{60466176} =\frac{1001}{20155392} \approx 0.00005$
Henry
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