Is this probability calculation right? (99% chance of winning a 1 in 20,000 random game)

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The odds of winning in a game are 1 out of 20,000. How many times would one have to play the game in order to be sure they would win?

Let n be the number of times you need to play

The probability of not winning the game 1 - 1/20000= 19999/20000.

Using this probability calculation below:

1 - (19,999/20,000)ⁿ

As I understand I should solve for n and equal something like ~0.009 to represent a 99% of winning.

So what I found is n = 94,000

Is it fair to say if there are 94,000 players of the game with a 1 in 20,000 chance of winning (no skill involved) that there is a 99% chance that someone wins?

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1 Answer

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"Is it fair to say if there are 94,000 players of the game with a 1 in 20,000 chance of winning (no skill involved) that there is a 99% chance that someone wins?"

It would be better to say there is 99% that at least one person wins.

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