Daily Maths Challenges

195 readers

1 users here now

Share your cool maths problems.

Complete a challenge:

Post your solution in comments, if it is exactly the same as OP's solution, let us know.
Have fun.

Post a challenge:

Doesn't have to be original, as long as it is not a duplicate.
Challenges not riddles, if the post is longer than 3 paragraphs, reconsider yourself.
Optionally include solution in comments, let it be clear this is not a homework help forums.
Tag [unsolved] if you don't have a solution yet.
Please include images, if your question includes complex symbols, attach a render of the maths.

Feel free to contribute to a series by DMing the OP, or start your own challenge series.

founded 7 months ago

MODERATORS

[email protected]

Coin-flipping game (lemmy.world)

submitted 6 months ago* (last edited 6 months ago) by [email protected] to c/[email protected]

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?

you are viewing a single comment's thread
view the rest of the comments

[–] el_psd 3 points 6 months ago (1 children)

I read the other answers in the thread, and I definitely would have gotten this wrong without that. This answer stands a chance at least

Independence

A and C are independent, A and B are not. Since you have a chance to win on turn 4 that you do not have on turn 1, P(A|B) = 1/2 + 1/16 + 1/64... = 7/12 != P(A). Conversely, P(A|C) = 1/4 + 1/16 + 1/64... = 1/3 = P(A)

[–] [email protected] 3 points 6 months ago (1 children)

response

This is the correct answer, although P(A|B) should actually be 2/3 rather than 7/12 - I think you meant 1/2 + 1/8 + 1/32 + ...?

The reasoning is good, either way. Since past flips won't affect future flips, if the player has made it to turn 5, an odd turn, then their future prospects are no different than they were on turn 1, another odd turn - so A and C are independent. Similarly, A and B are dependent because your chances of winning and losing effectively flip: If you've made it to an even turn, then you now win if it takes an odd number of flips from there to get Heads.

So it should be an almost paradoxical-seeming situation: You win 1/3 games overall, you win 2/3 games that make it to turn 2, you win 1/3 games that make it to turn 3, you win 2/3 games that make it to turn 4, 1/3 that make it to turn 5, etc.

[–] el_psd 3 points 6 months ago

response

Thanks for the catch! Once I noticed that the probability of winning on the initial flip was the difference between the two, I stopped thinking about the other terms...