She knew from the start that she is twice as likely to be asked when it is tails. So, her estimate of the chances of her being awakened facing tails should be bigger from the beginning.
Thank you, your explanation for the 1⁄3 answer makes sense to me. I’m still a bit confused about it, but i think i feel like i might be changing my mind.
I’ll try to figure out what would happen if SB makes a bet on the coin flip at each interview. Suppose she guesses heads each time, then:
Given that the result was heads, then she is interviewed once, and she is right once.
Given that the result was tails, then she is interviewed twice, and she is wrong twice.
… meaning that if the experiment is repeated several times, the guess “heads” will be correct for one out of three guesses. Just like you said.
(Perhaps it’s important to realize that, if the coin lands on tails, then she’s guaranteed to wake up once on Monday, and also guaranteed to wake up once on Tuesday. Now that i read your other comment again, i see your meaning when you say that p(heads) and p(tails) for each day is the same.)
I couldn’t decide exactly you meant by “twice as likely to be asked (woken) when it’s tails” either. I’m going to guess that you’re averaging evenly over Monday and Tuesday, in which case I agree. After marginalizing over M/T, P(wake|heads)=1/2 and P(wake|tails)=1.
“She knew from the start that she is twice as likely to be asked when it is tails. ”
The probability that she would be asked is 1, regardless of the outcome of the coin. Her estimate of the chances of her being awakened should have been 1.
She knew from the start that she is twice as likely to be asked when it is tails. So, her estimate of the chances of her being awakened facing tails should be bigger from the beginning.
Thank you, your explanation for the 1⁄3 answer makes sense to me. I’m still a bit confused about it, but i think i feel like i might be changing my mind.
I’ll try to figure out what would happen if SB makes a bet on the coin flip at each interview. Suppose she guesses heads each time, then:
Given that the result was heads, then she is interviewed once, and she is right once.
Given that the result was tails, then she is interviewed twice, and she is wrong twice.
… meaning that if the experiment is repeated several times, the guess “heads” will be correct for one out of three guesses. Just like you said.
(Perhaps it’s important to realize that, if the coin lands on tails, then she’s guaranteed to wake up once on Monday, and also guaranteed to wake up once on Tuesday. Now that i read your other comment again, i see your meaning when you say that p(heads) and p(tails) for each day is the same.)
I couldn’t decide exactly you meant by “twice as likely to be asked (woken) when it’s tails” either. I’m going to guess that you’re averaging evenly over Monday and Tuesday, in which case I agree. After marginalizing over M/T, P(wake|heads)=1/2 and P(wake|tails)=1.
“She knew from the start that she is twice as likely to be asked when it is tails. ”
The probability that she would be asked is 1, regardless of the outcome of the coin. Her estimate of the chances of her being awakened should have been 1.
Yes: her estimate of the chances of her being awakened is indeed 1.