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Sir,

Three containers hold 100 balls each, numbered from 00 to 99... we are to extract 1 ball from each container... prior to this I wrote on a piece of paper two, two digit numbers... example, I chose 34 and 97

The numbers extracted from the containers were 34, 12 and 97... I was succesful in pairing both my numbers with two balls from the three extracted...

What were my odds of doing so? 1 in x?

I would really appreciate you help in putting a number on x, for this is very important an ongoing project and I cant find the person to help me.

Thank you.

Well, assume you have your two numbers X and Y written down. Then you pick three numbers A, B, C at random also (from these bins).

The probability that you match both numbers depends a bit on how you choose X and Y.

If you are choosing your own picks at random, there is the question of whether or not you are allowed to pick the same number twice -- and if so, do you have to match it twice, or just once?

I'm going to assume X and Y have to be different, even though A, B, and C could be the same.

Because X and Y are different and A, B, and C are random, it really doesn't matter which X and Y you pick -- any pair X and Y have the same chances of winning (assuming X=Y is not allowed).

So whatever X and Y are, they are chosen at random.

The chance of A=X is 1/100, so the chance of not matching is 99/100.

But there are three ways to get X, either X=A, X=B, or X=C. So the chances of not getting it three times in a row is (99/100)×(99/100)×(99/100).

This means the chance that matching at least one of the A's is 1 - (99/100)^3.

So that's a smaller part of a larger problem. It's a bit harder to account for matching A and B simultaneously. So here's the idea: There are four outcomes:

1. You match neither A nor B.

2a. You match A only.

3a. You match B only.

4. You match A and B.

You can also have these outcomes:

2b. You match A (and maybe B, maybe not).

3b. You match B (and maybe A, maybe not).

Because 1, 2a, 3a, and 4 are exactly all the possible outcomes, you know:

P(1) + P(2a) + P(3a) + P(4) = 1

Also, 2b and 3b are just:

P(2b) = P(2a) + P(4)

P(3b) = P(3a) + P(4)

But we don't have much other info. But because of these three algebraic relations, it might be easier to solve for P(4) by solving for the other stuff instead. Because X and Y aren't really any different, we also know P(2b)=P(3b) and P(2a)=P(3a). So to sum up what we know:

1. You match neither A nor B.

2a. You match A only.

3a. You match B only.

4. You match A and B.

You can also have these outcomes:

2b. You match A (and maybe B, maybe not).

3b. You match B (and maybe A, maybe not).

Because 1, 2a, 3a, and 4 are exactly all the possible outcomes, you know:

P(1) + 2 P(2a) + P(4) = 1

P(2b) = P(2a) + P(4)

We know P(2b) already, we have P(2b) = 1 - (1-p)^3, where p=1/100.

In a similar way, we can get P(1) = (1-2p)^3.

If we know these two, we can get:

P(1) + 2 ( P(2b) - P(4) ) + P(4) = 1

P(1) + 2 P(2b) - P(4) = 1

P(4) = P(1) + 2 P(2b) - 1

P(4) = [ (1-2p)^3 ] + 2 [ 1 - (1-p)^3 ] - 1 = 6p^2(1-p) ≈ 0.000594

So about 1/20th of one percent chance -- very low probability indeed.

I can answer all questions up to, and including, graduate level mathematics. I do not have expertise in statistics (I can answer questions about the mathematical foundations of statistics). I am very much proficient in probability. I am not inclined to answer questions that appear to be homework, nor questions that are not meaningful or advanced in any way.

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