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each Roll is the sum of six six-sided dice ( therefore a value of 6 to 36). How many times would I have to roll (and take the sum of 6d6) before I would have a 90% expectation of seeing three 6d6 sums that are sequentially less than or equal to each other.

Example...roll 1 is 27. Roll 2 is 30. Roll 3 is 17. Roll 4 is 15. Roll 5 is 14.

Rolls 3, 4, and 5 are consecutive in the series and each <= the previous roll sum.

I know all of the possible outcomes for every 6d6, and can find the likelihood of rolling each value. So I assume for any value of my current roll r ( the sum of 6d6 ), I would add up the probabilities of rolling every number <= r. Then....I don’t know. Thank you!

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  • $\begingroup$ Could you please explain what you mean by "in a row," given you state you roll all six dice at once? What do you mean by "less than or equal to each other" when describing a set of three values?? $\endgroup$
    – whuber
    May 5, 2020 at 17:28
  • $\begingroup$ Are you referring to the totals of 6d6? Thus, the $i$-th roll consists of rolling 6d6 and recording the total as $x_i$, and you are asking how many such rolls you need to do to have a 90% chance that there exists an $i$ with $x_i\geq x_{i+}\geq x_{i+2}$? $\endgroup$ May 5, 2020 at 20:47
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    $\begingroup$ Sorry, to clarify...I roll 6d6 and total the values of the six dice. I continue to roll six dice, tracking the total for each set of six. I’m trying to find how many rolls to be 90% likely of hitting a sequence of three 6d6 totals that are less than or equal to each other in sequence....not necessarily related to the original roll total. $\endgroup$
    – MoeFaux
    May 6, 2020 at 7:59

1 Answer 1

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TL;DR: 11 (possibly 10). Plus 2.

I got this by simulation, and to be honest, I think a closed formula will be quite painful to derive. Here is my thinking. Note that if the first three rolls already satisfy the condition, then I will record this as stopping after step 1, so if you are interested in the total number and would count this as step 3, you need to add 2 to all numbers below.

First, we can quite easily derive the probability mass function (PMF) $(p_k)$ for the sum of 6d6 by convolution. In R:

n_dice <- 6
probs <- rep(1/6,6)
for ( ii in 2:n_dice ) probs <- convolve(probs,rep(1/6,6),type="open")
names(probs) <- n_dice:(6*n_dice)

Now, it's not overly hard to derive the probability that the first three roll totals $R_1, R_2, R_3$ fulfill your condition, i.e., the probability $q_1$ that your dice rolling stops after 1 step:

$$ \begin{align*} q_1 = \; & P(R_1\geq R_2\geq R_3) \\ = \; & \sum_{k=6}^{36} P(R_1\geq k)P(R_2=k)P(R_3\leq k) \\ = \; & \sum_{k=6}^{36}\bigg(\sum_{j=k}^{36}p_j\bigg)p_k\bigg(\sum_{j=6}^k p_j\bigg) \\ \approx \; & 0.2016245. \end{align*} $$

In R:

q_1 <- sum(sapply(seq_along(probs),function(kk)sum(probs[kk:length(probs)])*probs[kk]*sum(probs[1:kk])))

We can still find the probability $q_2$ that you stop after the second step:

$$ \begin{align*} q_2 = \; & P(R_1< R_2\geq R_3\geq R_4) \\ = \; & \sum_{k=6}^{36} P(R_1<R_2)P(R_2\geq k)P(R_3=k)P(R_4\leq k) \\ = \; & \sum_{k=6}^{36}\bigg(\sum_{j=k}^{36}\Big(\sum_{i=6}^{j-1}p_i\Big)p_j\bigg)p_k\bigg(\sum_{j=6}^k p_j\bigg) \\ \approx \; & 0.1409793. \end{align*} $$

In R again:

q_2 <- 0
for ( kk in 6:length(probs) ) {
    q_2 <- q_2 +
        sum(sapply(kk:length(probs),function(jj)sum(probs[1:(jj-1)])*probs[jj])) *
        probs[kk] *
        sum(probs[1:kk])
}
q_2

Interestingly, the precise same calculation gives us $q_3$, since the condition to stop at step 3 is $R_2<R_3\geq R_4\geq R_5$, and $R_1$ doesn't matter. So since all rolls are iid,

$$ q_3 = P(R_2<R_3\geq R_4\geq R_5) = P(R_1< R_2\geq R_3\geq R_4) = q_2 \approx 0.1409793. $$

Unfortunately, the problem starts about here. To calculate $q_4$, we have to get a handle on why the rolling didn't stop in steps 1-3. We know that $R_3<R_4\geq R_5\geq R_6$ (because we are stopping in step 4), but this relationship imposes a condition on our previous rolls: given this condition, the third roll is more likely to be lower, and since we didn't stop in step 1, this changes the conditional distributions of $R_1$ and $R_2$. The cases get complicated quickly.

So unless you are happy with a stopping probability $<q_1+q_2+q_3\approx 0.483583$ (for which the formulas above give the exact probabilities), the best you can do is likely to simulate. Here is a simulation of the stopping probabilities in R:

n_sims <- 1e6
results <- rep(0,n_sims)
pb <- winProgressBar(max=n_sims)
for ( ii in 1:n_sims ) {
    setWinProgressBar(pb,ii,paste(ii,"of",n_sims))
    set.seed(ii)
    roll_1 <- sum(sample(1:6,n_dice,replace=TRUE))
    roll_2 <- sum(sample(1:6,n_dice,replace=TRUE))
    roll_3 <- sum(sample(1:6,n_dice,replace=TRUE))
    counter <- 1
    while ( roll_1<roll_2 | roll_2<roll_3 ) {
        roll_1 <- roll_2
        roll_2 <- roll_3
        roll_3 <- sum(sample(1:6,n_dice,replace=TRUE))
        counter <- counter+1
    }
    results[ii] <- counter
}
close(pb)
hist(results,breaks=seq(0.5,max(results)+0.5),col="grey",freq=FALSE)

histogram

Note how the first three tabulated frequencies are close to the $q_1, q_2, q_3$ we calculated above:

> table(results)/n_sims
results
       1        2        3        4        5        6        7        8        9       10       11       12       13       14       15       16       17       18       19       20       21       22       23       24       25       26 
0.201307 0.140820 0.141095 0.104467 0.087303 0.066984 0.053460 0.042390 0.033982 0.026595 0.021161 0.016777 0.013222 0.010485 0.008337 0.006501 0.005204 0.004071 0.003291 0.002639 0.002096 0.001638 0.001300 0.001002 0.000833 0.000668 
      27       28       29       30       31       32       33       34       35       36       37       38       39       40       41       42       43       44       45       46       47       48       49       50       51       52 
0.000500 0.000372 0.000298 0.000217 0.000204 0.000156 0.000124 0.000110 0.000072 0.000068 0.000047 0.000040 0.000036 0.000027 0.000024 0.000019 0.000009 0.000013 0.000005 0.000005 0.000007 0.000006 0.000002 0.000002 0.000002 0.000001 
      53       54       58       61       66       69 
0.000001 0.000001 0.000001 0.000001 0.000001 0.000001 

Finally, to get the number you are looking for, we look at the cumulative frequencies and note where this exceeds 90%:

> cumsum(table(results)/n_sims)
       1        2        3        4        5        6        7        8        9       10       11       12       13       14       15       16       17       18       19       20       21       22       23       24       25       26 
0.201307 0.342127 0.483222 0.587689 0.674992 0.741976 0.795436 0.837826 0.871808 0.898403 0.919564 0.936341 0.949563 0.960048 0.968385 0.974886 0.980090 0.984161 0.987452 0.990091 0.992187 0.993825 0.995125 0.996127 0.996960 0.997628 
      27       28       29       30       31       32       33       34       35       36       37       38       39       40       41       42       43       44       45       46       47       48       49       50       51       52 
0.998128 0.998500 0.998798 0.999015 0.999219 0.999375 0.999499 0.999609 0.999681 0.999749 0.999796 0.999836 0.999872 0.999899 0.999923 0.999942 0.999951 0.999964 0.999969 0.999974 0.999981 0.999987 0.999989 0.999991 0.999993 0.999994 
      53       54       58       61       66       69 
0.999995 0.999996 0.999997 0.999998 0.999999 1.000000 

So our result is 11 (but because of sampling variability, there is a small chance that it's actually 10).

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  • $\begingroup$ (+1) You need to add $2$ to your answers because you aren't counting the first two rolls. For confirmation of your answer, note that a slight overestimate can be obtained by assuming each roll is a uniform random value in $[0,1],$ for which an exact solution can be found (giving $15$ as the upper bound). $\endgroup$
    – whuber
    Jul 7, 2020 at 13:18
  • $\begingroup$ @whuber: thanks, good points. I edited the +2 in. $\endgroup$ Jul 7, 2020 at 13:21

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