Who is asking: Teacher
Question: The odds of winning in a game are 3 out of 10. How many times would one have to play the game in order to be sure they would win?
Since the probability of winning the game if you play it once is 3/10, there is no guarantee you will ever win the game no matter how many times you play (I am assuming that losing or winning the first time you play does not affect the probability of winning or losing the next time, that is the events are independent). Thus every time you play the game the probability that you win is 3/10 and you can never be sure that you will win.
You can, however, determine the number of times you would need to play to come within some acceptable limit of certainty of winning, such as 99% sure you would win.
This may be calculated as follows:
Let n be the number of times you need to play
The probability of not winning the game 1 - 3/10 = 7/10.
The probability of not winning the game in n attempts is equal to the probability of not winning the game in the first attempt AND not winning the game in the second attempt AND ... AND not winning the game in the nth attempt which is (7/10)n since the probability of the intersection of n independent events is the product of the probabilities.
So the probability of winning at least once in n attempts is 1 - (7/10)n
Since we want to be 99% sure of winning, we require 1 - (7/10)n to be greater than or equal to 0.99.
Rearranging this we find we need (7/10)n to be less than or equal to 0.01. Trying some values for n you can see that (7/12)12 = 0.0138413 and (7/10)13 = 0.0096889 and so we should play the game 13 times in order to be 99% sure of winning at least once.
Hope this helps,