The 1⁄3 solution makes the assumption that the probability of heads given an awakening is:
lim N-> infinity n1/(n1+n2+n3)
But, we have a problem here. N does not equal n1+n2+n3, it is equal to n1+n2.
Why is that a problem? Why would N have to be equal to n1+n2+n3? Only because it does in your other example?
(ETA)I’m not sure of where you’re formula “lim N-> infinity n1/(n1+n2+n3)” comes from—as the third example shows, it just doesn’t work in all cases. That doesn’t mean that your alternative formula is better in the sleeping beauty case.
Because this, lim N-> infinity n1/(n1+n2+n3), is p1 if the counts are from independent draws of a multinomial distribution.
We have outcome-dependent sampling here. Is lim N-> infinity n1/(n1+n2+n3) equal to p1 in that case? I’d like to see the statistical theory to back up the claim. It’s pretty clear to me that people who believe the answer is 1⁄3 pictured counts in a 3 by 1 contingency table, and applied the wrong theory to it.
(ETA) The formula “lim N-> infinity n1/(n1+n2+n3)” is what people who claim the answer is 1⁄3 are using to justify it. The 1⁄2 solution just uses probability laws. That is, P(H)=1/2. P(W)=1, where W is the event that Beauty has been awakened. Therefore, P(H|W)=1/2.
It’s pretty clear to me that people who believe the answer is 1⁄3 pictured counts in a 3 by 1 contingency table, and applied the wrong theory to it.
I’ll have to disagree with that—there is a pretty clear interpretation in which 1⁄3 is a “correct” answer: if the sleeping beauty is asked to bet X dollars that heads came up, and wins $60 if she’s right, for up to which values X should she accept the bet? (if the coin comes up tails, she gets the possibility to bet twice)
In that scenario, X=$20 is the right answer, which corresponds to a probability of 1⁄3. Do you agree with that? (I haven’t read all the threads, you probably adressed this somewhere)
See, here I’m not using any “lim N-> infinity n1/(n1+n2+n3)” , so I feel you’re being unfair to 1/3rders.
I don’t agree, because the question is about her subjective probability at an awakening. The betting question you described is a different one.
For example, suppose I flip a coin and tell you you will win $60 if heads came up, but I require that you make the bet twice if tails came up? You’d be willing to bet up to $30, but that doesn’t mean you think heads has probability 1⁄3. If Beauty really thinks heads has probability 1⁄3, she’d be willing to accept the bet up to $30 even if we told her that we’d only accept one bet (of course, we wouldn’t tell her that she’s already made a bet on Tuesday. Payout would be on Wed).
The wikipedia page for the sleeping beauty problem says:
Suppose this experiment were repeated 1,000 times. We would expect to get 500 heads and 500 tails. So Beauty would be awoken 500 times after heads on Monday, 500 times after tails on Monday, and 500 times after tails on Tuesday. In other words, only in a third of the cases would heads precede her awakening. So the right answer for her to give is 1⁄3.
That’s why I think people are picturing counts in a contingency table when they come up with the 1⁄3 answer.
I don’t agree, because the question is about her subjective probability at an awakening. The betting question you described is a different one.
She would also bet at an awakening. If you ask her to bet when she just broke up, it would seem weird that she would say “my subjective probability for heads is 1⁄2, but I’ll only willing to bet up to $20 − 1⁄3 of the winnings if it’s heads.”
It seems even weirder in the Xtreme Sleeping Beauty, where she’s awakened a thousand times : “my subjective probability for heads is 1⁄2, but I’m only willing to bet up to 6 cents”.
Yes, you get a different result if you change the betting rules where only one bet per “branch” counts, but I don’t see why that’s closer to the problem as originally stated.
It seems even weirder in the Xtreme Sleeping Beauty, where she’s awakened a thousand times : “my subjective probability for heads is 1⁄2, but I’m only willing to bet up to 6 cents”.
I guess I don’t see why it’s weird. The number of times she will bet is dependent on the outcome. So, even though at each awakening she thinks probability of heads is 1⁄2, she knows if it’s tails she’ll have to bet many more times than if heads. We’re essentially just making her bet more money on a loss than on a win.
In that case, what does it even mean to say “my subjective probability for heads is 1/2”? Subjective probability is often described in terms of bettings—see here.
Seems to me this is mostly a quarrel of definitions, and that when you say “people who believe the answer is 1⁄3 pictured counts in a 3 by 1 contingency table, and applied the wrong theory to it.”, you’re being unfair. They’re just using a different definition of “subjective probability”
“Subjective probability” is a basic term in decision theory and economics, though. If you want to roll your own metric, surely you should call it something else—to avoid much confusion.
Why is that a problem? Why would N have to be equal to n1+n2+n3? Only because it does in your other example?
(ETA)I’m not sure of where you’re formula “lim N-> infinity n1/(n1+n2+n3)” comes from—as the third example shows, it just doesn’t work in all cases. That doesn’t mean that your alternative formula is better in the sleeping beauty case.
Because this, lim N-> infinity n1/(n1+n2+n3), is p1 if the counts are from independent draws of a multinomial distribution.
We have outcome-dependent sampling here. Is lim N-> infinity n1/(n1+n2+n3) equal to p1 in that case? I’d like to see the statistical theory to back up the claim. It’s pretty clear to me that people who believe the answer is 1⁄3 pictured counts in a 3 by 1 contingency table, and applied the wrong theory to it.
(ETA) The formula “lim N-> infinity n1/(n1+n2+n3)” is what people who claim the answer is 1⁄3 are using to justify it. The 1⁄2 solution just uses probability laws. That is, P(H)=1/2. P(W)=1, where W is the event that Beauty has been awakened. Therefore, P(H|W)=1/2.
I’ll have to disagree with that—there is a pretty clear interpretation in which 1⁄3 is a “correct” answer: if the sleeping beauty is asked to bet X dollars that heads came up, and wins $60 if she’s right, for up to which values X should she accept the bet? (if the coin comes up tails, she gets the possibility to bet twice)
In that scenario, X=$20 is the right answer, which corresponds to a probability of 1⁄3. Do you agree with that? (I haven’t read all the threads, you probably adressed this somewhere)
See, here I’m not using any “lim N-> infinity n1/(n1+n2+n3)” , so I feel you’re being unfair to 1/3rders.
I don’t agree, because the question is about her subjective probability at an awakening. The betting question you described is a different one.
For example, suppose I flip a coin and tell you you will win $60 if heads came up, but I require that you make the bet twice if tails came up? You’d be willing to bet up to $30, but that doesn’t mean you think heads has probability 1⁄3. If Beauty really thinks heads has probability 1⁄3, she’d be willing to accept the bet up to $30 even if we told her that we’d only accept one bet (of course, we wouldn’t tell her that she’s already made a bet on Tuesday. Payout would be on Wed).
The wikipedia page for the sleeping beauty problem says:
That’s why I think people are picturing counts in a contingency table when they come up with the 1⁄3 answer.
She would also bet at an awakening. If you ask her to bet when she just broke up, it would seem weird that she would say “my subjective probability for heads is 1⁄2, but I’ll only willing to bet up to $20 − 1⁄3 of the winnings if it’s heads.”
It seems even weirder in the Xtreme Sleeping Beauty, where she’s awakened a thousand times : “my subjective probability for heads is 1⁄2, but I’m only willing to bet up to 6 cents”.
Yes, you get a different result if you change the betting rules where only one bet per “branch” counts, but I don’t see why that’s closer to the problem as originally stated.
I guess I don’t see why it’s weird. The number of times she will bet is dependent on the outcome. So, even though at each awakening she thinks probability of heads is 1⁄2, she knows if it’s tails she’ll have to bet many more times than if heads. We’re essentially just making her bet more money on a loss than on a win.
In that case, what does it even mean to say “my subjective probability for heads is 1/2”? Subjective probability is often described in terms of bettings—see here.
Seems to me this is mostly a quarrel of definitions, and that when you say “people who believe the answer is 1⁄3 pictured counts in a 3 by 1 contingency table, and applied the wrong theory to it.”, you’re being unfair. They’re just using a different definition of “subjective probability”
Don’t you think so?
Based on my interaction with people here, I think we all are talking about the same thing when it comes to subjective probability.
I agree that you can use betting to describe subjective probability, but there are a lot of possible ways to bet.
“Subjective probability” is a basic term in decision theory and economics, though. If you want to roll your own metric, surely you should call it something else—to avoid much confusion.
That is why I’d rather talk in terms of bets than subjective probability—they don’t require precise technical definitions.