Coin Puzzle: Predict the Other's Coin
pratikpoddarcse.blogspot.in
Coin Puzzle: Predict the Other's Coin
1–10 of 50 posts
Re: Coin Puzzle: Predict the Other's Coin
#2Anyone find a solution? I have an idea but want to check.
Re: Coin Puzzle: Predict the Other's Coin
#3[deleted]
Re: Coin Puzzle: Predict the Other's Coin
#4Humm let me see if I get this right
- Assuming everybody is honest
The chance of A getting the guess right is 0.5, same for B, so the joint probability should be 0.25 (or 1/4)
You're out of 2 dollars (3 - the 1 you get) for something with a chance of 1/4 (in the other 3/4 you're being payed $1) SO
Looks like it's worth playing.
Expected losses: 2 * 1/4 Edit: see comment below
Edit 2, see the other comment below where the probability of prediction can be of 1/2, in this case: Expected losses: 2* 1/2, wins: 1/2 so you should not play
Re: Coin Puzzle: Predict the Other's Coin
#5The team of A and B can increase the chance from ¼ to ½ by consistently taking their output from step 2 and using it as the guess for step 3.
Same thing if they both agree to guess the inverse of step 2.
Re: Coin Puzzle: Predict the Other's Coin
#6Anyone find a solution? I have an idea but want to check.
[deleted]
Re: Coin Puzzle: Predict the Other's Coin
#7Both A and B guess the same as the results of their own toss. Their answers are correct if the tosses are HH and TT, 50% chance.
So no, C should not play this game.
Re: Coin Puzzle: Predict the Other's Coin
#8The team of A and B can increase the chance from ¼ to ½ by consistently taking their output from step 2 and using it as the guess for step 3. Same thing if they both agree to guess the inverse of step 2.
[deleted]
Re: Coin Puzzle: Predict the Other's Coin
#9If they both agree to use their own coins result as the guess of the OTHER person's coin, they should get it right 50% of the time instead of 25%.
Possible tosses:
TT win
HT lose
TH lose
HH win
Counter-intuitive though isn't it.
Why does using your own result improve the odds of winning the game!?
Re: Coin Puzzle: Predict the Other's Coin
#10Humm let me see if I get this right - Assuming everybody is honest The chance of A getting the guess right is 0.5, same for B, so the joint probability should be 0.25 (or 1/4) You're out of 2 dollars (3 - the 1 you get) for something with a chance of 1/4 (in the other 3/4 you're being payed $1) SO Looks like it's worth playing. Expected losses: 2 * 1/4 Edit: see comment below Edit 2, see the other comment below where…
> You're out of 3 dollars for something with a chance of 1/4 (in the other 3/4 you're being payed $1)
You're out $2 (since you get the $1 from the other players at the beginning, no matter what)