1.1 Deriving the PDF and Expected Value for Max Roll

The max roll after d rolls is a discrete random variable with 6 possible outcomes. To find it’s probability distribution function (pdf) we need to find the probability of getting one of these outcomes.

Let’s first look at the concrete case of getting a max roll of 4 after d=3 rolls. First, we know that all of our rolls need to be under 5, which we can model as \((\frac{4}{6})^3\).

Next we need to know, given that all our rolls are under 5, the probability that at least one was a 4. We can model this as 1 minus the probability that none of the rolls were 4, or \(1-(\frac{3}{4})^3\). Note that the denominator is a 4, since we are calculating based on the condition that all our rolls were under 5.

We can combine these probabilities as \((\frac{4}{6})^3\times[1-(\frac{3}{4})^3]\). This comes out to

(4/6)^3*(1-(3/4)^3)

[1] 0.1712963

or 17.1%. We can follow this model for any desired max roll and for any number of dice rolls. The generic formula becomes \(P(Max Roll = \textit{k}|\textit{d } rolls) = (\frac{k}{6})^d\times[1-(\frac{k-1}{k})^d]\).

Distibuting the outside into the rest, we get \(P(Max Roll = \textit{k}|\textit{d } rolls) = (\frac{k}{6})^d-(\frac{k-1}{6})^d\). Using this formula we can write a function to generate a pdf for max roll

pdfMaxRoll = function(d) {
  k = 1:6
  (k/6)^d*(1-((k-1)/k)^d)
}
pdfMaxRoll(3)

[1] 0.00462963 0.03240741 0.08796296 0.17129630 0.28240741 0.42129630

As you can see, when we input 3 into this function, the 4th value of the output matches our calculation for a max of 4 given 3 dice rolls. We can modify the function slightly to allow for generation of a pdf table for all dice rolls from 1 to 5.

pdfMaxRoll = function(d=0) {
  if(d==0) 
    return(data.frame(1:6,pdfMaxRoll(1),pdfMaxRoll(2),pdfMaxRoll(3),pdfMaxRoll(4),pdfMaxRoll(5)))
  k = 1:6
  (k/6)^d*(1-((k-1)/k)^d)
}
pdfMaxRoll()

In the 3rd column we can see the same data for 3 dice rolls.

Now we can calculate the expected value for max dice roll. We can get the expected value of a discrete random variable with the formula \(EV = \sum\limits_{k=1}^n k\times p_k\). In our case, k is a value for max dice roll, p_k is the probability of getting that value, and n is 6.

So the expected value for max roll, given d dice rolls would be \(EV = 1*[(\frac{1}{6})^d-(\frac{0}{6})^d] + 2*[(\frac{2}{6})^d-(\frac{1}{6})^d] + 3*[(\frac{3}{6})^d-(\frac{2}{6})^d] + ...\).

As you can see this series cascades and terms cancel out. At the end we are left with \(EV = 6-(\frac{5}{6})^d-(\frac{4}{6})^d-(\frac{3}{6})^d-(\frac{2}{6})^d-(\frac{1}{6})^d\). We can write this as a function

expMaxRoll = function(d=1:5) {
  6 - (5/6)^d - (4/6)^d - (3/6)^d - (2/6)^d - (1/6)^d
}
expMaxRoll()

[1] 3.500000 4.472222 4.958333 5.244599 5.430941

The default input for this function is the range from 1 to 5, so the output we get is the expected value for max roll for 1 to 5 dice rolls.