If I take a 1/30 chance two separate times, what are my chances of winning once?

Question

If I take a chance that is 1/30 two separate times in a row, what are my chances of winning once? (Trying to settle a debate, They say its 1/15, I believe it to be slightly less than that because the first chance has no affect on the second, so its 1/30 twice, not 1/15, I just don't know the equation to show this.)

Answer 1

I'm assuming you mean what are the chances of winning at least once?

If so, look at the chances of losing both times, then subtract that from 1

You'll lose the first time with probability 29/30 and win 1 out of 30 times

Same with the 2nd try. 29/30 chance of losing

Multiply them: 29/30 x 29/30 = 841/900 which equals 841/30 over 30 = 28.0333.../30

subtract that from 1 to get 1.9666..../30 which is slightly less than 2/30 or 1/15

1.9666.... is very close to 2, but slightly less.

The error in just adding them 1/30 + 1/30 = 2/30 = 1/15 can be seen better if you consider doing it 30 times.

you wouldn't add 1/30 30 times or you'd get 30/30 = 1 or a certainty that you'd win. It's not certain. It's very likely, but you'd never get certainty just by doing it 30 times. Your expected number of wins after doing it 30 times is 1, but that's different from the probability of a win.

Answer 2

Your chances of winning once are the same as your chances of winning on the first OR winning on the second.

If P(A) = 1/30 and P(B) = 1/30, then P(A or B) = P(A) + P(B)

P(A ∪ B) = P(A) + P(B)

= 1/30 + 1/30

= 2/30

= 1/15