Money from trafficking.

here's a good one to make a few dollars as you drive through traffic on your way to the casino


offer your passenger a bet at an even $5 that in the next 20 cars you see there will be two whose car number plates end in the same last two numbers


if the car's number is abc123 the 23 is the same as that of a car with 423xyz as its number


there are 100 possible numbers, from 00 to 99, so he will think he has the best of it in only 20 cars


what are your chances of winning each bet?


how many cars would make this a fair game?

Let P(n) be the probability the first n cars have distinct numbers .


So P(1)=1


P(2)=99/100,


P(n)= 100(99)(98).... 101-n)/100^n.


Now my calculator is at home, but maybe someone else can punch in the numbers.

You win 87% of the time with 20 cars. 12 cars would make it a fair game.


This is just like the birthday problem. I believe that out of 20 people in a room, it is more likely than not that 2 people will have the same birthday.


One I found awhile back is that if you give a full deck of cards to each person at a full holdem table and have them each draw a card, two of them will draw the same card over half the time.

BTW, before anyone tries this in California, you only win 52% of the time with 9 players. With 10 players you win 60%.

of course, BruceZ has it spot on


the trick is to calculate the chance that the same number plate (or birthday) does NOT appear


(either all numbers are different or at least two are the same)


find this number and subtract it from 1 to give you the chance that it DOES appear


with two cars the chance the second car has a different number is 99 times out of 100, or 100/100 x 99/100


with three it is 100/100 x 99/100 x 98/100


with four, 100/100 x 99/100 x 98/100 x 97/100


and so on

You must have read the same book I just did. Some British author, can't remember the name. Good stuff, though.

i got it from john haigh's "taking chances"


there were some other good ones that i'll try to post soon