Betting Probability

**ryanhorne** · October 12th 2009, 11:10 AM

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!

**CaptainBlack** · October 12th 2009, 08:44 PM

Originally Posted by ryanhorne

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!

The probability that a car has the last two digits equal is $1/10$ (explain why).

The number of cars with the last two digits equal in a sample of $n$ has a binomial distribution $B(n,0.1)$ , so the probability of $N$ having the last two digits equal is:

$p(N)= b(N;n,0.1)={n \choose N}0.1^N0.9^{n-N}$

Then the probability of two or more having the last two digits equal is:

$p(N\ge2 )=1-p(N<2)=1-p(N=0)-p(N=1)$

Now for your bet to be fair you need to find an $n$ such that $p(N\ge 2)=0.5$ , but there is no $n$ that makes this bet fair, 16 seems to be the closest (maybe I have gone wrong somewhere I will have to check this again)

CB

**ryanhorne** · October 13th 2009, 12:56 AM

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!

**CaptainBlack** · October 13th 2009, 01:23 AM

Originally Posted by ryanhorne

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!

Lets just call the last two digits the cars number

So we have that the bet is that two or more cars in the next $n$ have the same number.

The probability that all the cars have different numbers is:

$p(0)=\frac{100!}{(100-n)!}\times\frac{1}{100^n}$

The probability that the first and second cars have the same number and all the others are different is:

$p(1^*)=\frac{100!}{(100-n+1)!}\times\frac{1}{100^{n-1}}\times \frac{1}{100}$

This is the same as the probability that the $i$ and $j$ -th cars have the same number and all the rest are different. There are $n(n-1)/2$ such pairs so:

$p(1)=\frac{100!}{(100-n+1)!}\times\frac{1}{100^{n-1}}\times \frac{n(n-1)}{200}$

If the above is correct (and I am issuing no guarantees at present) we need to find $n$ such that $p(0)+p(1)=0.5$

Now the nearest I can get to $0.5$ is $0.4967$ so presumably this is still not quite right (with $n=19$ ) !?

CB

**ryanhorne** · October 13th 2009, 01:55 AM

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....

**CaptainBlack** · October 13th 2009, 02:13 AM

Originally Posted by CaptainBlack

Lets just call the last two digits the cars number

So we have that the bet is that two or more cars in the next $n$ have the same number.

The probability that all the cars have different numbers is:

$p(0)=\frac{100!}{(100-n)!}\times\frac{1}{100^n}$

The probability that the first and second cars have the same number and all the others are different is:

$p(1^*)=\frac{100!}{(100-n+1)!}\times\frac{1}{100^{n-1}}\times \frac{1}{100}$

This is the same as the probability that the $i$ and $j$ -th cars have the same number and all the rest are different. There are $n(n-1)/2$ such pairs so:

$p(1)=\frac{100!}{(100-n+1)!}\times\frac{1}{100^{n-1}}\times \frac{n(n-1)}{200}$

If the above is correct (and I am issuing no guarantees at present) we need to find $n$ such that $p(0)+p(1)=0.5$

Now the nearest I can get to $0.5$ is $0.4967$ so presumably this is still not quite right (with $n=19$ ) !?

CB

Wrong interpretation (I was calculating the probability than there be no more than one repeat number), the probability that we want is 1 minus the probability that all the cars have different numbers and if this is $0.5$ then the probability that they all have different numbers is $0.5$ , that is $p(0)=0.5$ , so we want n such that

$p(0)=\frac{100!}{(100-n)!}\times\frac{1}{100^n}=0.5$

Then $n=12$ gives a probability of $0.503$ for $p(0)$ , still not exact, but close.

CB

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