If you have two chances and each chance has a 1 in 3 chance of success...?

Question

What would be you're expected success rate? Would it be something other than 1 in 3?

In this particular case, say I'm throwing a die and I want to get a 1 or a 2. I'm going to throw the die a maximum of 2 times and I only care that it's successful one of the two times. Obviously the second roll is completely independent of the first.

kieranhr: I was wondering if that was the correct way to do this. Thanks for the responses.

Answer 1

It's easier to look at it the other way. What's your chance of failing?
You have a 2/3rds chance of failing each time. To find the odds of failing both times you multiply them together. So you have a 4/9ths chance of failing both times.
Your chance of succeeding at least once then is (1 - 4/9) = 5/9ths, or a little better than 1 in 2.

Edit: That's the right way of doing it alright Justin, but you don't have to take my word for it. If you had some time to spare, you could get two die and start rolling them and recording answers. You'd soon see the success rate converge to 5/9.

Answer 2

2 in 3.

Answer 3

The odds of success would be 2 in 3. The odds of both being successful would be 1 in 9.

Answer 4

1/9

Answer 5

If you are selecting from three cards, one being an ace, and you get two tries, your first try gives 1 in 3. the second, since you have two left to draw from gives 1 in two. The two give 1/3 + 1/2 or 5/6
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KrazyKyngeKorny(Krazy, not stupid)
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Answer 6

Hi. Not if the outcome were the same. Failing the first has no influence on the second outcome.

Answer 7

nope

Answer 8

edit: misread question

Answer 9

no