1.2 Calculating Max Roll Probabilities for 5v4

In a 5v4 matchup how likely is it that the player with more dice will win?

First let’s get the pdf’s for max roll for 1<=rolls<=5

pdf = pdfMaxRoll()
pdf

We can visualize this data as such:

plot(pdf[,6],col=5,type="l",xlab="Max Roll",ylab="P(occurence)")
for(i in 2:5)
  lines(pdf[,i],col=i-1)

legend(1,0.6,1:5,text.col=1:5,title="# Rolls")

We also have the expected value for Max Roll for each 1<=rolls<=5

ev = expMaxRoll()
ev

[1] 3.500000 4.472222 4.958333 5.244599 5.430941

Now let’s plot these expected values

plot(ev,type='l',xlab='Dice Rolls',ylab='Expected Maximum Roll')

The first calculation we want to do is \(P([Max Roll | 5 rolls] > [Max Roll | 4 rolls])\).

We should split this up into independent sub-probabilities, namely:

• \(P(Max|5 = 6)*P(Max|4 < 6)\)
• \(P(Max|5 = 5)*P(Max|4 < 5)\)
• \(P(Max|5 = 4)*P(Max|4 < 4)\)
• \(P(Max|5 = 3)*P(Max|4 < 3)\)
• \(P(Max|5 = 2)*P(Max|4 < 2)\)

We can get the required probabilities simply by looking at our PDF table

pdf

So the first calculation is

pdf[6,6]*sum(pdf[1:5,5])

[1] 0.2884464

We can automate to do all the calculations

total = 0
for(i in 6:2)
  total = total + pdf[i,6]*sum(pdf[1:i-1,5])
total

[1] 0.3484327

The probability of a tie would be \(P(Max|5 = Max|4) = P(Max|5 = 6)*P(Max|4 = 6) + P(Max|5 = 5)*P(Max|4 = 5) + ...\) , which we can calculate with

sum(pdf[1:6,6]*pdf[1:6,5])

[1] 0.4015745

So the probability of \(P(Max|5 < Max|4)\) would be

1-.4015745-.3484327

[1] 0.2499928

So, in a 5v4 the player with the extra die has a 34.8% chance of winning, 25.0% chance of losing, and 40.2% chance of tying on the max roll.

When we move on to \(P([2nd Highest Roll | 5 rolls] > [2nd Highest Roll | 4 rolls])\) the issue we run into is that the possible values for the 2nd highest roll are affected by the value for the max roll. This makes solving this very non-trivial.