Assume that the last two digits on a car number plate are equally likely to be any of the one hundred outcomes {00, 01, 02, ... 98, 99}.

1. Peter bets Paul, at even money, that at least 2 of the next n cars seen will have the same last two digits. Does n=16 favour Peter or Paul?

2. What value of n would make this a pretty fair bet?

1) Probability that Peter wins = PeW = 48.53%, while the probability that Paul wins = PaW = 1-PeW = 51.47% (Peter si favoured)

2) The bet will be pretty fair at n=16 or 17

Step-by-step explanation:

Since each car number plate is independent of the others, then the random variable X=number of cars that have the same last two digits follows a binomial distribution. Thus the probability distribution is

P (X) = n! / ((n-x) !*x!) * p^n * (1-p) ^ (n-x)

where

p = probability that a car has the same last 2 digits = {00, 11, ...,99}/{00, 01, 02, ... 98, 99} = 10/100 = 1/10

n = number of cars = 16

Then the probability that Peter wins PeW is:

PeW=P (X≥2) = 1 - P (X<2) = 1 - F (X=2) = 1 - 0.5147 = 0.4853 = 48.53%

therefore

Probability that Peter wins = PeW = 48.53%

Probability that Paul wins = PaW = 1-PeW = 51.47% (Peter si favoured)

for n=17, PeW = 0.518, PaW = 0.482

for n=15, PeW = 0.450, PaW = 0.548

then the bet will be pretty fair at n=16 or 17