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Coin-flipping game (lemmy.world)

submitted 6 months ago* (last edited 6 months ago) by zkfcfbzr@lemmy.world to c/dailymaths@lemmy.world

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We're playing a game. I flip a coin. If it lands on Tails, I flip it again. If it lands on Heads, the game ends.

You win if the game ends on an even turn, and lose otherwise.

Define the following events:

A: You win the game

B: The game goes on for at least 4 turns

C: The game goes on for at least 5 turns

What are P(A), P(B), and P(C)? Are A and B independent? How about A and C?

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[–] zkfcfbzr@lemmy.world 4 points 6 months ago* (last edited 6 months ago)

In probability, two events are said to be independent if one event happening has no effect on the probability of another event happening. So coin flips, as an example, are independent - because when you flip a coin and get Tails, that doesn't affect the probability of the next coin flip also coming up Tails.

So in this context, asking if A and B are independent is asking: Does knowing the game lasts at least four turns change the probability of winning? And similarly for A and C. Does knowing the game lasts at least 5 turns change the probability that the game will end in a victory?

Rest of response

To be clear about your other answers, saying P(B) = 1/16 and P(C) = 1/32 are not correct - I was saying if you adjusted your formula, from 1/2^n to 1/2^(n-1), then your answers would be correct. So the probability of the game lasting at least four turns is 1/2^(4-1) = 1/2^3 = 1/8, and the probability of it lasting at least 5 turns is 1/2^(5-1) = 1/2^4 = 1/16.

But I think you were more asking about why this didn't affect your win rate - that's because there's a subtle difference between "making it to the nth round" and "having the game end on the nth round" - and that difference is that once you make it to a round, you then have a 1/2 probability to end the game - which makes the probability of ending on the nth round the 1/2^n you used.