I'm trying to work out if random variance in dice rolls is more likely to influence a given situation in a game rather than the overall expected values of those dice rolls being significant. The game is a common table-top miniature game, where one must roll certain dice in succession but only if you've previously scored a success.

To give frame to the question, let's assume the following:

  1. The initial number of dice to be rolled is 20
  2. If a roll is 'successful' then that die is used in the next round of rolling, and if it is 'unsuccessful' then it is removed
  3. There are 3 rounds of rolling
  4. Round 1 success is determined by a roll of 4, 5, or 6
  5. Round 2 success is determined by a roll of 3, 4, 5, or 6
  6. Round 3 success is determined by a roll of 5 or 6

Now the issue I have is that quite often people will look at the overall expected value of the dice rolling game and make assumptions based on this. In our scenario, with 20 dice and consecutive probabilities of success of 1/2, 2/3, and 1/3, the overall expected value of our final successful dice at the end of round 3 would be 2.22 (if not rounding to whole numbers).

However I'd imagine that the individual variance (if I'm using the correct word here) of any given set of dice rolls would play a far more important role than that final expected value given the relatively small sample space of only 20 initial dice.

So my question is two-fold:

  1. How many dice would you have to roll to be relatively certain of getting close to your expected value of final successes?
  2. Given the above example of rolling 20 dice, what certainty would you have in getting that expected value?

For argument's sake, let's say that I'd like to be 90% certain of getting within 1 either side of my expected value for question 1 - and feel free to use the above numbers to illustrate if it's easier.

Happy to answer any clarifications as required!

Thanks for your help!


Let’s take my problem to an extreme. With the probabilities for success as above (50%, 66.6%, then 33.3%), if I were to roll 1 die my ‘expected’ outcome mathematically to get three successful rolls would be 0.11. However, by the end of the three rolls I will either have a value of 0 successes, or 1. Both are distant from my expected value.

Even if we increase the number of dice to 10, with a now increased expected value of final successes of 1.11, it’s plausible that I could end up with 6 successes overall, which again would be distant from my expected value.

However, if I increased the number of dice to 10,000,000, there’s a high probability that my final successes would be close to my expected value of 1,111,111 given the likelihood that many outlying rolls would not confer statistical significance.

So, at what ‘number of dice’ rolled does the probability of my final results falling within, say, one standard deviation on either side of my expected value, become 90%? How many dice must I roll before the chances of ‘randomness’ affecting my overall result is statistically reduced to less than 10% ie. I am 90% likely to achieve final successes within one standard deviation on either side of my expected value?

Hope that clarifies it somewhat!

  • 1
    $\begingroup$ I don't understand how this works. Can you explain how exactly you estimated that expected value? What values are you considering? What do you mean that a dice is "removed"? $\endgroup$ Oct 20, 2020 at 12:38
  • $\begingroup$ The expected value I calculated was using the example numbers above ie. 20 dice, a 50% chance of succss in round 1, a 66.6% chance of success in round 2, and a 33.3% chance of success in round 3: 20 x 0.5 x 0.66 x 0.33 = 2.22. $\endgroup$
    – Jase
    Oct 21, 2020 at 13:31

2 Answers 2


You are interested in the distribution of number of successes, that is how many dice out of an initial $x$ will be left in the game after three rolls, given the rules. You estimated $E[X]=np=20 \cdot 0.11=2.2$, this is the expected number of dies that remain in the game at the end. The variance is $Var[X]=np(1-p)=20 \cdot 0.11(1-0.11)=1.958$.

Your question, at the least the last one, is how many dice would you have to roll, such that you decrease the variance enough that 90 % of outcomes are within one standard deviation of the expected value. So if the expected value is 2.2, 90 % of outcomes would have to be between 1 and 3.

This cannot be done. In fact, what you will find is that, by increasing the number of dice, while keeping the probability the same, the variance actually increases (so you would have to decrease the number of dice to decrease the variance). I think you got this idea from the distribution of the mean, where increasing the sample size decreases the variance of the distribution, that is the distribution gets narrower.

In your case the only way you can decrease the variance is by changing the probability of success (which is related to changing the number of rolls, which is the question in your title but not in the actual text). This is actually a rule which is sometimes used when estimating the sample size, the variance is greatest at probability 0.5, and then decreases as you move towards the extremes 0 and 1.

  • $\begingroup$ Right, I see what you're saying and thanks for the answer. I think I just realised my question is somewhat nonsensical... whoops! $\endgroup$
    – Jase
    Oct 22, 2020 at 10:48

Perhaps the simplest thing to do here is just to compute the probability of every possible outcome at the end of the three-stage process. For the sake of broader interest, let me generalise your description of the problem by supposing we have a $K$-round process with $n$ initial dice, with $\boldsymbol{\theta} = (\theta_1, \theta_2, ..., \theta_K)$ as the respective success probabilities for each round of the process. We will also assume that the successes for each die over each round are independent with these fixed probabilities.

One simpler way of looking at this is to suppose that you roll all dice in all rounds, but a die is only a success at the end of the process if it rolled successfully in all rounds. This is equivalent to your stipulated process, where dice are removed if they do not roll successfully in a given round. To facilitate the analysis, let $X_{i,k}$ denote the indicator variable for a successful outcome with the $i$th die on the $k$th round. Then you have independent indicators with distribution $X_{i,k} \sim \text{Bern}(\theta_k)$. Let $Y_i$ denote the indicator variable for a successful outcome with the $i$th die over all rounds. Then you have independent indicators with distribution:

$$Y_i = \min (X_{i,1},...,X_{i,K}) \sim \text{Bern} \Bigg( \prod_{k=1}^K \theta_k \Bigg).$$

Consequently, the distribution for the total number of successful outcomes $Y$ for the process is:

$$Y \sim \text{Bin} \Bigg( n, \prod_{k=1}^K \theta_k \Bigg).$$

The mean and variance of the total number of successes is:

$$\mathbb{E}(Y) = n \prod_{k=1}^K \theta_k \quad \quad \quad \mathbb{V}(Y) = n \Bigg( \prod_{k=1}^K \theta_k \Bigg) \Bigg( 1-\prod_{k=1}^K \theta_k \Bigg).$$

I will not address your other question regarding the number of initial dice required to get within a certain bound of the expected value with a certain probability. That is not a particularly interesting question and so I will leave it to you to compute from the present functions if it is something of interest. Once you have the probabilities of every possible outcome for any initial number of dice, you can get a good sense of what is likely to happen in this process.

Example: In your example you have $n=20$ initial dice and you have a process with $K=3$ rounds with success probabilities $\boldsymbol{\theta} = (\tfrac{1}{2}, \tfrac{2}{3}, \tfrac{1}{3})$. We can compute and plot the probabilities for the number of successes at the end of the process as follows.

#Set the parameters
n     <- 20
probs <- c(1/2, 2/3, 1/3)

#Compute the success probabilities
PROBS <- dbinom(0:n, size = n, prod(probs))
names(PROBS) <- 0:n

#Plot the success probabilities
barplot(PROBS, col = 'blue',
        main = 'Successes at End of Process',
        xlab = 'Number of Successes', ylab = 'Probability')

In this particular example the mean number of successes is $\mathbb{E}(Y) = 20/9 = 2.2222$ and the variance is $\mathbb{V}(Y) = 160/81 = 1.9753$. The standard deviation of the number of successes is $\mathbb{S}(Y) = \sqrt{160/81} = 1.4055$. (Your intuition about the mean of the process is accurate.) You can see from the barplot that you are most likely to get two successes from this process, slightly less likely to get one or three successes, and so on.

enter image description here

  • $\begingroup$ I still can't figure out what the OP is trying to ask. Could you provide an explicit statement of your interpretation of the question? $\endgroup$
    – whuber
    Oct 21, 2020 at 13:09
  • $\begingroup$ Thanks for the great answer Ben! $\endgroup$
    – Jase
    Oct 21, 2020 at 13:33
  • $\begingroup$ @whuber: I'm also not entirely sure of the question OP is asking, though it is evidently some aspect of the behaviour of the final number of successes. I think giveing the full distribution and moments probably covers most aspects of interest. $\endgroup$
    – Ben
    Oct 22, 2020 at 1:09
  • $\begingroup$ I think I might have confused the issue by accidentally using a term, variance, that I didn't mean mathematically but rather linguistically. What I was trying to ask is at what point does our expected result actually become a highly probable actual result? Is 20 dice too low, such that the natural randomness of dice-rolling ends up playing a bigger part of our actual result rather than what our expected value is? If we instead use 2,000,000 dice, would we be more likely to get our expected result because randomness is less impactful? Or am I misunderstanding something fundamentally? $\endgroup$
    – Jase
    Oct 22, 2020 at 10:30

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