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Coin-flipping game (self.dailymaths)

submitted 7 months ago* (last edited 7 months ago) by zkfcfbzr to c/dailymaths

12 comments fedilink hide all child comments

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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[–] [email protected] 4 points 7 months ago* (last edited 7 months ago) (2 children)

Are all games played to completion? Assuming yes,

answer

P(A) = Sum[1/4^n,{n,1,∞}] = 0.333...
P(B) = 1/8
P(C) = 1/16

One of A,B or A,C is dependent, as it'd flip the polarity of the remaining games, the other is independent, as it wouldn't. I'm not too worried about which is which, as it's just an off-by-one error. xD

Assuming no, You'd have to know the distribution of when games were abandoned, I'd think.

[–] zkfcfbzr 5 points 7 months ago* (last edited 7 months ago) (1 children)

Games are always played to completion, though if you wanna make it (barely) more challenging you can add in a 5% chance for both players to get bored and give up on each round (before flipping), leading to a loss. Though it seems unlikely - after flipping a quarter 20 times and getting Tails every time, I'd be inclined to keep flipping if anything.

response

These are correct. It is possible to reason out which of B and C is independent of A without going into the numbers.

[–] [email protected] 3 points 7 months ago

after flipping a quarter 20 times and getting Tails every time, I'd be inclined to keep flipping if anything.

It's a sunk cost fallacy, it's never too late to get bored and give up. ;)

[–] [email protected] 2 points 7 months ago

A supertask.