Thread: Basic question on chance

I left school in the '60s, and I've forgotten most of the statistics I ever knew, so I expect this question is easily answered.

You have one of those spinning discs that has a red, a green and a blue sector (thus producing white). If you throw a dart at it, you have a 1 in 3 chance of hitting red, so over a large number of throws, say 9000, you can expect to score 3000 hits on red. But suppose you divide them into 1000 groups of 9 consecutive throws, and score each session - 3/9, 2/9, 5/9, etc. Two questions:

1. If 3/9 is your average score, does that mean it is also your most common score?

2. Will 2/9 and 4/9 occur with equal frequency?

Originally Posted by Terpsichore

I left school in the '60s, and I've forgotten most of the statistics I ever knew, so I expect this question is easily answered.

You have one of those spinning discs that has a red, a green and a blue sector (thus producing white). If you throw a dart at it, you have a 1 in 3 chance of hitting red, so over a large number of throws, say 9000, you can expect to score 3000 hits on red. But suppose you divide them into 1000 groups of 9 consecutive throws, and score each session - 3/9, 2/9, 5/9, etc. Two questions:

1. If 3/9 is your average score, does that mean it is also your most common score?

2. Will 2/9 and 4/9 occur with equal frequency?

The the number X of reds in nine throws is binomially distributed like so:



Therefore, the answer to your first question is yes, and the answer to your second question is no.

Speaking of averages: in other cases, the average (or the expected value) need not even be a possible outcome of the experiment (like, say, when throwing dice only once, you get an expected value of 3.5 which is not a possible outcome).
So, you cannot expect the average (or, rather, the expected value) to be the most likely outcome. But for your particular experiment it really is.

Okay, thanks for that. The answer is not as simple as I thought.