Question

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?

Answer 1

Answer:

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