Question

Peter rolls an 8 die and then Kate gets three chances to roll a 6-die and get greater number than whatever Pter rolled. What is the probability that at least one of Kate’s rolls is greater than what Peter's rolls?

Answer 1

Let X be the number which Peter got.

• For X = 6, 7, 8, Kate can never roll a bigger number than X.
  Therefore, P(Greater | X = 6, 7, 8) = 0
• For X = 5,
  P(Greater | X = 5)
  = Probability that there is at least one 6 in three rolls by Kate
  = 1 - Probability that the number on all three rolls is <= 5
  = 1 - (5/6)³ = 1 - (125/216) = 91/216
  P(Greater | X = 5) =  91/216
• Similarly for other values of X, we get the probabilities here as:
  P(Greater | X = 4) = 1 - (4/6)³ = 1 - (8/27) = 19/27
  P(Greater | X = 3) = 1 - (3/6)³ = 7/8
  P(Greater | X = 2) = 1 - (2/6)³ = 26/27
  P(Greater | X = 1) = 1 - (1/6)³ = 215/216

Therefore the probability that at least one of Kate’s rolls is greater than what Peter's rolls is computed here as:

= (1/8)*(Sums of all the conditional probabilities computed above)

= (1/8)*(0 + (91/216) + (19/27) + (7/8) + (26/27) + (215/216) )

= 0.4948

Therefore 0.4948 is the required probability here.