Assume that the last two digits on a car number plate are equally likely to be any of the 100 outcomes 00, 01, ..., 99. Jacques bets Jules at even money, that at least two of the next n cars seen have the same last two digits. What value of n makes this bet fair?
I am pretty sure the answer is 12, as the only way this is fair is when all the number plates are different, which is at 1*0.99*0.98*0.97 etc, and I did this until it got to 0.5, which was after 12 cars.
However, is there a better way of writing this? This is not a very mathematical answer!
Yes that is the route I went down, however, I showed this to my maths teacher and the question really means this:-
What makes the bet even, when there are n cars, that the same digits will turn up twice eg 67 88 92 67.
However, my maths teacher has alot of problems in expressing his english! (Punch)
According to the book the answer is 12, however there are no calculations!
The way I see it at the moment is that if we see it from the others guys point of view, he wins every time the car is not the same. So, at what point when every car is different do his odds become even.
So with n = 2
His odds are, 1*0.99 = .99
So hw will win 99 time out of a hundred. This is because it makes no difference what the first car is, and the sceond car can be any other 99 from that 100.
Then, lets say n=5, the his odds of winning are
1*0.99*0.98*0.97*0.96 = 0.903
Now, when n=12, we get
1*.99*.98......*0.89 = 0.503
So at 12 the odds are pretty much even.
To start off writing this mathemtaically is pretty diffcult.... all i have so far is...
(1*(1-0*0.01)) * (1*(1-1*0.01)) * (1*(1-2*0.01)) * etc....(Crying)