One could argue that if the coin is flicked and comes up tails then we have both “Tails&Monday” and “Tails&Tuesday” as both being correct, sequentially.
Yes, it is a commonplace occurrence that “Today is Monday” and “Today is Tuesday” can both be true, on different days. This doesn’t ordinarily prevent people from assigning probabilities to statements like “Today is Monday”, when they happen to not remember for sure whether it is Monday or not now. And the situation is the same for Beauty—it is either Monday or Tuesday, she doesn’t know which, but she could find out if she just left the room and asked some passerby. Whether it is Monday or Tuesday is an aspect of the external world, which one normally regards as objectively existing regardless of one’s knowledge of it.
All this is painfully obvious. I think it’s not obvious to you because you don’t accept that Beauty is a human being, not a python program. Note also that Beauty’s experience on Monday is not the same as on Tuesday (if she is woken). Actual human beings don’t have exactly the same experiences on two different days, even if the have memory issues. The problem setup specifies only that these differences aren’t informative about whether it’s Monday or Tuesday.
What do you mean by “probability”.
I’m using “probability” in the subjective Bayesian sense of “degree of belief”. Since the question in the Sleeping Beauty problem is what probability of Heads should Beauty have when awoken, I can’t see how any other interpretation would address the question asked. Note that these subjective degree-of-belief probabilities are intended to be a useful guide to decision-making. If they lead one to make clearly bad decisions, they must be wrong.
Consider two very extreme cases of the sleeping beauty game: - Guess wrong and you die! (GWYD) - Guess right and you live! (GRYL)
If we look at the GWYD and GRYL scenarios you describe, we can, using an “outside” view, see what the optimal strategies are, based on how frequently Beauty survives in repeated instances of the problem. To see whether Beauty’s subjective probability of Heads should be 1⁄2 or 1⁄3, we can ask whether after working out an optimal strategy beforehand, Beauty will change her mind after waking up and judging that P(Heads) is either 1⁄2 or 1⁄3, and then using that to decide whether to follow the original strategy or not. If Beauty’s judgement of P(Heads) leads to her abandoning the optimal strategy, there must be something wrong with that P(Heads).
For GWYD, both the strategy of deterministically guessing Heads on all awakenings and the strategy of deterministically guessing Tails on all awakenings will give a survival probability of 1⁄2, which is optimal (I’ll omit the proof of this).
Suppose Beauty decides ahead of time to deterministically guess Tails. Will she change her mind and guess Heads instead when she wakes up?
Suppose that Beauty thinks that P(Heads)=1/3 upon wakening. She will then think that if she guesses Heads, her probability of surviving is P(Heads)=1/3. If instead, she guesses Tails, she thinks her probability of surviving is P(Tails & on other wakening she also guesses Tails), which is 2⁄3 if she is sure to follow the original plan in her other wakening, and is greater than 1⁄3 as long as she’s more likely than not to follow the original plan on her other wakening. So unless Beauty thinks she will do something perverse on her other wakening, she should think that following the original plan and guessing Tails is her best action.
Now suppose that Beauty thinks that P(Heads)=1/2 upon wakening. She will then think that if she guesses Heads, her probability of surviving is P(Heads)=1/2. If instead, she guesses Tails, she thinks her probability of surviving is P(Tails & on other wakening she also guesses Tails), which is 1⁄2 if she is sure to follow the original plan in her other wakening, and less than 1⁄2 if there is any chance that on her other wakening she doesn’t follow the original plan. Since Beauty is a human being, who at least once in a while does something strange or mistaken, the probability that she won’t follow the plan on her other wakening is surely not zero, so Beauty will judge her survival probability to be greater if she guesses Heads than if she guesses Tails, and abandon the original plan of guessing Tails.
Now, if Beauty thinks P(Heads)=1/2 and then always reasons in this way on wakening, then things turn out OK—despite having originally planned to always guess Tails, she actually always guesses Heads. But if there is a non-negligible chance that she follows the original plan without thinking much, she will end up dead more than half the time.
So if Beauty thinks P(Heads)=1/3, she does the right thing, but if Beauty thinks P(Heads)=1/2, she maybe does the right thing, but not really reliably.
Turning now to the GRYL scenario, we need to consider randomized strategies. Taking an outside view, suppose that Beauty follows the strategy of guessing heads with probability h, independently each time she wakes. Then the probability that she survives is S=(1/2)h+(1⁄2)(1-h*h) - that is, the probability the coin lands Heads (1/2) times the probability she guesses Heads, plus the probability that the coin lands Tails time the probability that she doesn’t guess Heads on both awakenings. The derivative of S with respect to h is (1/2)-h, which is zero when h=1/2, and one can verify this gives a maximum for S of 5⁄8 (better than the survival probability when deterministically guessing either Heads of Tails).
This matches what you concluded. However, Beauty guessing Heads with probability 1⁄2 does not imply that Beauty thinks P(Heads)=1/2. The “guess” here is not made in an attempt to guess the coin flip correctly, but rather in an attempt to not die. The mechanism for how the guess influences whether Beauty dies is crucial.
We can see this by seeing what Beauty will do after waking if she thinks that P(Heads)=1/2. She knows that her original strategy is to randomly guess, with Heads and Tails both having probability 1⁄2. But she can change her mind if this seems advisable. She will think that if she guesses Tails, her probability of survival will be P(Tails)=1/2 (she won’t survive if the coin landed Heads, because this will be her only (wrong) guess, and she will definitely survive if the coin landed Tails, regardless what she does on her other awakening). She will also compute that if she guesses Heads, she will survive with probability P(Heads)+P(Tails & she guesses Tails on her other wakening). Since P(Heads)=1/2, this will be greater than 1⁄2, since the second term surely is not exactly zero (it will be 1⁄4 if she follows the original strategy on her other awakening). So she will think that guessing Heads gives her a strictly greater chance of surviving than guessing Tails, and so will just guess Heads rather than following the original plan of guessing randomly.
The end result, if she reasons this way each awakening, is that she always guesses Heads, and hence survives with probability 1⁄2 rather than 5⁄8.
Now lets see what Beauty does if after wakening she thinks that P(Heads)=1/3. She will think that if she guesses Tails, her probability of survival will be P(Tails)=2/3. She will think that if she guesses Heads, her probability of survival will be P(Heads)+P(Tails & she guesses Tails on her other wakening). If she thinks she will follow the original strategy on her other wakening, the this is (1/3)+(2/3)*(1/2)=2/3. Since she computes her probability of survival to be the same whether she guesses Heads or Tails, she has no reason to depart from the original strategy of guessing Heads or Tails randomly. (And this reinforces her assumption that on another wakening she would follow the original strategy.)
So if Beauty is a Thirder, she lives with probability 5⁄8, but if she is a Halfer, she lives with lower probability of 1⁄2.
We are in complete agreement about how beauty should strategize given each of the three games (bet a dollar on the coin flick with odds K, GWYL and GWYD). The only difference is that you are insisting that “1/3” is Beauty’s “degree of belief”. (By the way I am glad you repeated the same maths I did for GWYD, it was simple enough but the answer felt surprising so I am glad you got the same.)
In contrast, I think we actually have two quantities:
“Quobability”—The frequency of correct guesses made divided by the total number of guesses made.
“Srobability”—The frequency of trials in which the correct guess was made, divided by the number of trials.
Quabability is 1⁄3, Scrobability is 1⁄2. “Probability” is (I think) an under-precise term that could mean either of the two.
You say you are a Bayesian, not a frequentist. So for you “probability” is degree of belief. I would also consider myself a Bayesian, and I would say that normally I can express my degree of belief with a single number, but that in this case I want to give two numbers, “Quobability = 1⁄3, Scrobability =1/2″. What I like about giving two numbers is that typically a Bayesian’s single probability value given is indicative of how they would bet. In this case the two quantities are both needed to see how I would bet given slightly varies betting rules.
I was still interested in “GRYL”, which I had originally assumed would support the thirder position, but (for a normal coin) had the optimal tactic being to pick at 50⁄50 odds. I just looked at biased coins.
For a biased coin that (when flicked normally, without any sleep or anmesia) comes up heads with probability k. (k on x-axis), I assume Beauty is playing GRYL, and that she is guessing heads with some probability. The optimal probability for her strategy to take is on the y-axis (blue line). Overall chance of survival is orange line.
You are completely correct that her guess is in no way related to the actual coin bias (k), except for k=0.5 specifically which is an exception not the rule. In fact, this graph appears to be vaguely pushing in some kind of thirder position, in that the value 2/3rds takes on special significance as beyond this point beauty always guesses heads. In contrast when tails is more likely she still keeps some chance of guessing heads because she is banking on one of her two tries coming up tails in the case of tails, so she can afford some preparation for heads.
“Quobability”—The frequency of correct guesses made divided by the total number of guesses made.
“Srobability”—The frequency of trials in which the correct guess was made, divided by the number of trials.
Quabability is 1⁄3, Scrobability is 1⁄2. “Probability” is (I think) an under-precise term that could mean either of the two.
I suspect that the real problem isn’t with the word “probability”, but rather the word “guess”. In everyday usage, we use “guess” when the aim is to guess correctly. But the aim here is to not die.
Suppose we rephrase the GRYL scenario to say that Beauty at each awakening takes one of two actions—“action H” or “action T”. If the coin lands Heads, and Beauty takes action H the one time she is woken, then she lives (if she instead takes action T, she dies). If the coin lands Tails, and Beauty takes action T at least one of the two times she is woken, then she lives (if she takes action H both times, she dies).
Having eliminated the word “guess”, why would one think that Beauty’s use of the strategy of randomly taking action H or action T with equal probabilities implies that she must have P(Heads)=1/2? As I’ve shown above, that strategy is actually only compatible with her belief being that P(Heads)=1/3.
Note that in general, the “action space” for a decision theory problem need not be the same as the “state space”. One might, for example, have some uncertain information about what day of the week it is (7 possibilities) and on that basis decide whether to order pepperoni, anchovy, or ham pizza (3 possibilities). (You know that different people, with different skills, usually make the pizza on different days.) So if for some reason you randomized your choice of action, it would certainly not say anything directly about your probabilities for the different days of the week.
Maybe we are starting to go in circles. But while I agree the word “guess” might be problematic I think you still have an ambiguity with what the word probability means in this case. Perhaps you could give the definition you would use for the word “probability”.
“In everyday usage, we use “guess” when the aim is to guess correctly.” Guess correctly in the largest proportion of trials, or in the largest proportion of guesses? I think my “scrob” and “quob” thingies are indeed aiming to guess correctly. One in the most possible trials, the other in the most possible individual instances of making a guess.
“Having eliminated the word “guess”, why would one think that Beauty’s use of the strategy of randomly taking action H or action T with equal probabilities implies that she must have P(Heads)=1/2?”—I initially conjectured this as weak evidence, but no longer hold the position at all, as I explained in the post with the graph. However, I still think that in the other death-scenario (Guess Wrong you die) the fact that deterministically picking heads is equally good to deterministically picking tails says something. This GWYD case sets the rules of the wager such that Beauty is trying to be right in as many trials as possible, instead of for as many individual awakenings. Clearly moving the goalposts to the “number of trials” denominator.
For me, the issue is that you appear to take “probability” as “obviously” meaning “proportion of awakenings”. I do not think this is forced on us by anything, and that both denominators (awakenings and trials) provide us with useful information that can beneficially inform our decision making, depending on whether we want to be right in as many awakenings or trials as possible. Perhaps you could explain your position while tabooing the word “probability”? Because, I think we have entered the Tree-falling-in-forest zone: https://www.lesswrong.com/posts/7X2j8HAkWdmMoS8PE/disputing-definitions, and I have tried to split our problem term in two (Quobability and Srobability) but it hasn’t helped.
Perhaps you could give the definition you would use for the word “probability”.
I define it as one’s personal degree of belief in a proposition, at the time the judgement of probability is being made. It has meaning only in so far it is (or may be) used to make a decision, or is part of a general world model that is itself meaningful. (For example, we might assign a probability to Jupiter having a solid core, even though that makes no difference to anything we plan to do, because that proposition is part of an overall theory of physics that is meaningful.)
Frequentist ideas about probability being related to the proportion of times that an event occurs in repetitions of a scenario are not part of this definition, so the question of what denominator to use does not arise. (Looking at frequentist concepts can sometimes be a useful sanity check on whether probability judgements make sense, but if there’s some conflict between frequentist and Bayesian results, the solution is to re-examine the Bayesian results, to see if you made a mistake, or to understand why the frequentist results don’t actually contradict the Bayesian result.)
If you make the right probability judgements, you are supposed to make the right decision, if you correctly apply decision theory. And Beauty does make the right decision in all the Sleeping Beauty scenarios if she judges that P(Heads)=1/3 when woken before Wednesday. She doesn’t make the right decision if she judges that P(Heads)=1/2. I emphasize that this is so for all the scenarios. Beauty doesn’t have to ask herself, “what denominator should I be using?”. P(Heads)=1/3 gives the right answer every time.
Another very useful property of probability judgements is that they can be used for multiple decisions, without change. Suppose, for example, that in the GWYD or GRYL scenarios, in addition to trying not to die, Beauty is also interested in muffins.
Specifically, she knows from the start that whenever she wakes up there will be a plate of freshly-baked muffins on her side table, purchased from the cafe down the road. She knows this cafe well, and in particular knows that (a) their muffins are always very delicious, and (b) on Tuesdays, but not Mondays, the person who bakes the muffins adds an ingredient that gives her a stomach ache 10 minutes after eating a muffin. Balancing these utilities, she decides to eat the muffins if the probability of it being Tuesday is less than 30%. If Beauty is a Thirder, she will judge the probability of Tuesday to be 1⁄3, and refrain from eating the muffins, but if Beauty is a Halfer, she will (I think, trying to pretend I’m a halfer) think the probability of Tuesday is 1⁄4, and eat the muffins.
The point here is not so much which decision is correct (though of course I think the Thirder decision is right), but that whatever the right decision is, it shouldn’t depend on whether Beauty is in the GWYD or GRYL scenario. She shouldn’t be considering “denominators”.
One could argue that if the coin is flicked and comes up tails then we have both “Tails&Monday” and “Tails&Tuesday” as both being correct, sequentially.
Yes, it is a commonplace occurrence that “Today is Monday” and “Today is Tuesday” can both be true, on different days. This doesn’t ordinarily prevent people from assigning probabilities to statements like “Today is Monday”, when they happen to not remember for sure whether it is Monday or not now. And the situation is the same for Beauty—it is either Monday or Tuesday, she doesn’t know which, but she could find out if she just left the room and asked some passerby. Whether it is Monday or Tuesday is an aspect of the external world, which one normally regards as objectively existing regardless of one’s knowledge of it.
All this is painfully obvious. I think it’s not obvious to you because you don’t accept that Beauty is a human being, not a python program. Note also that Beauty’s experience on Monday is not the same as on Tuesday (if she is woken). Actual human beings don’t have exactly the same experiences on two different days, even if the have memory issues. The problem setup specifies only that these differences aren’t informative about whether it’s Monday or Tuesday.
What do you mean by “probability”.
I’m using “probability” in the subjective Bayesian sense of “degree of belief”. Since the question in the Sleeping Beauty problem is what probability of Heads should Beauty have when awoken, I can’t see how any other interpretation would address the question asked. Note that these subjective degree-of-belief probabilities are intended to be a useful guide to decision-making. If they lead one to make clearly bad decisions, they must be wrong.
Consider two very extreme cases of the sleeping beauty game: - Guess wrong and you die! (GWYD) - Guess right and you live! (GRYL)
If we look at the GWYD and GRYL scenarios you describe, we can, using an “outside” view, see what the optimal strategies are, based on how frequently Beauty survives in repeated instances of the problem. To see whether Beauty’s subjective probability of Heads should be 1⁄2 or 1⁄3, we can ask whether after working out an optimal strategy beforehand, Beauty will change her mind after waking up and judging that P(Heads) is either 1⁄2 or 1⁄3, and then using that to decide whether to follow the original strategy or not. If Beauty’s judgement of P(Heads) leads to her abandoning the optimal strategy, there must be something wrong with that P(Heads).
For GWYD, both the strategy of deterministically guessing Heads on all awakenings and the strategy of deterministically guessing Tails on all awakenings will give a survival probability of 1⁄2, which is optimal (I’ll omit the proof of this).
Suppose Beauty decides ahead of time to deterministically guess Tails. Will she change her mind and guess Heads instead when she wakes up?
Suppose that Beauty thinks that P(Heads)=1/3 upon wakening. She will then think that if she guesses Heads, her probability of surviving is P(Heads)=1/3. If instead, she guesses Tails, she thinks her probability of surviving is P(Tails & on other wakening she also guesses Tails), which is 2⁄3 if she is sure to follow the original plan in her other wakening, and is greater than 1⁄3 as long as she’s more likely than not to follow the original plan on her other wakening. So unless Beauty thinks she will do something perverse on her other wakening, she should think that following the original plan and guessing Tails is her best action.
Now suppose that Beauty thinks that P(Heads)=1/2 upon wakening. She will then think that if she guesses Heads, her probability of surviving is P(Heads)=1/2. If instead, she guesses Tails, she thinks her probability of surviving is P(Tails & on other wakening she also guesses Tails), which is 1⁄2 if she is sure to follow the original plan in her other wakening, and less than 1⁄2 if there is any chance that on her other wakening she doesn’t follow the original plan. Since Beauty is a human being, who at least once in a while does something strange or mistaken, the probability that she won’t follow the plan on her other wakening is surely not zero, so Beauty will judge her survival probability to be greater if she guesses Heads than if she guesses Tails, and abandon the original plan of guessing Tails.
Now, if Beauty thinks P(Heads)=1/2 and then always reasons in this way on wakening, then things turn out OK—despite having originally planned to always guess Tails, she actually always guesses Heads. But if there is a non-negligible chance that she follows the original plan without thinking much, she will end up dead more than half the time.
So if Beauty thinks P(Heads)=1/3, she does the right thing, but if Beauty thinks P(Heads)=1/2, she maybe does the right thing, but not really reliably.
Turning now to the GRYL scenario, we need to consider randomized strategies. Taking an outside view, suppose that Beauty follows the strategy of guessing heads with probability h, independently each time she wakes. Then the probability that she survives is S=(1/2)h+(1⁄2)(1-h*h) - that is, the probability the coin lands Heads (1/2) times the probability she guesses Heads, plus the probability that the coin lands Tails time the probability that she doesn’t guess Heads on both awakenings. The derivative of S with respect to h is (1/2)-h, which is zero when h=1/2, and one can verify this gives a maximum for S of 5⁄8 (better than the survival probability when deterministically guessing either Heads of Tails).
This matches what you concluded. However, Beauty guessing Heads with probability 1⁄2 does not imply that Beauty thinks P(Heads)=1/2. The “guess” here is not made in an attempt to guess the coin flip correctly, but rather in an attempt to not die. The mechanism for how the guess influences whether Beauty dies is crucial.
We can see this by seeing what Beauty will do after waking if she thinks that P(Heads)=1/2. She knows that her original strategy is to randomly guess, with Heads and Tails both having probability 1⁄2. But she can change her mind if this seems advisable. She will think that if she guesses Tails, her probability of survival will be P(Tails)=1/2 (she won’t survive if the coin landed Heads, because this will be her only (wrong) guess, and she will definitely survive if the coin landed Tails, regardless what she does on her other awakening). She will also compute that if she guesses Heads, she will survive with probability P(Heads)+P(Tails & she guesses Tails on her other wakening). Since P(Heads)=1/2, this will be greater than 1⁄2, since the second term surely is not exactly zero (it will be 1⁄4 if she follows the original strategy on her other awakening). So she will think that guessing Heads gives her a strictly greater chance of surviving than guessing Tails, and so will just guess Heads rather than following the original plan of guessing randomly.
The end result, if she reasons this way each awakening, is that she always guesses Heads, and hence survives with probability 1⁄2 rather than 5⁄8.
Now lets see what Beauty does if after wakening she thinks that P(Heads)=1/3. She will think that if she guesses Tails, her probability of survival will be P(Tails)=2/3. She will think that if she guesses Heads, her probability of survival will be P(Heads)+P(Tails & she guesses Tails on her other wakening). If she thinks she will follow the original strategy on her other wakening, the this is (1/3)+(2/3)*(1/2)=2/3. Since she computes her probability of survival to be the same whether she guesses Heads or Tails, she has no reason to depart from the original strategy of guessing Heads or Tails randomly. (And this reinforces her assumption that on another wakening she would follow the original strategy.)
So if Beauty is a Thirder, she lives with probability 5⁄8, but if she is a Halfer, she lives with lower probability of 1⁄2.
We are in complete agreement about how beauty should strategize given each of the three games (bet a dollar on the coin flick with odds K, GWYL and GWYD). The only difference is that you are insisting that “1/3” is Beauty’s “degree of belief”. (By the way I am glad you repeated the same maths I did for GWYD, it was simple enough but the answer felt surprising so I am glad you got the same.)
In contrast, I think we actually have two quantities:
“Quobability”—The frequency of correct guesses made divided by the total number of guesses made.
“Srobability”—The frequency of trials in which the correct guess was made, divided by the number of trials.
Quabability is 1⁄3, Scrobability is 1⁄2. “Probability” is (I think) an under-precise term that could mean either of the two.
You say you are a Bayesian, not a frequentist. So for you “probability” is degree of belief. I would also consider myself a Bayesian, and I would say that normally I can express my degree of belief with a single number, but that in this case I want to give two numbers, “Quobability = 1⁄3, Scrobability =1/2″. What I like about giving two numbers is that typically a Bayesian’s single probability value given is indicative of how they would bet. In this case the two quantities are both needed to see how I would bet given slightly varies betting rules.
I was still interested in “GRYL”, which I had originally assumed would support the thirder position, but (for a normal coin) had the optimal tactic being to pick at 50⁄50 odds. I just looked at biased coins.
For a biased coin that (when flicked normally, without any sleep or anmesia) comes up heads with probability k. (k on x-axis), I assume Beauty is playing GRYL, and that she is guessing heads with some probability. The optimal probability for her strategy to take is on the y-axis (blue line). Overall chance of survival is orange line.
You are completely correct that her guess is in no way related to the actual coin bias (k), except for k=0.5 specifically which is an exception not the rule. In fact, this graph appears to be vaguely pushing in some kind of thirder position, in that the value 2/3rds takes on special significance as beyond this point beauty always guesses heads. In contrast when tails is more likely she still keeps some chance of guessing heads because she is banking on one of her two tries coming up tails in the case of tails, so she can afford some preparation for heads.
CODE
By “GWYL” do you actually mean “GRYL” (ie, Guess Right You Live)?
Yes I do, very good point!
I think we actually have two quantities:
“Quobability”—The frequency of correct guesses made divided by the total number of guesses made.
“Srobability”—The frequency of trials in which the correct guess was made, divided by the number of trials.
Quabability is 1⁄3, Scrobability is 1⁄2. “Probability” is (I think) an under-precise term that could mean either of the two.
I suspect that the real problem isn’t with the word “probability”, but rather the word “guess”. In everyday usage, we use “guess” when the aim is to guess correctly. But the aim here is to not die.
Suppose we rephrase the GRYL scenario to say that Beauty at each awakening takes one of two actions—“action H” or “action T”. If the coin lands Heads, and Beauty takes action H the one time she is woken, then she lives (if she instead takes action T, she dies). If the coin lands Tails, and Beauty takes action T at least one of the two times she is woken, then she lives (if she takes action H both times, she dies).
Having eliminated the word “guess”, why would one think that Beauty’s use of the strategy of randomly taking action H or action T with equal probabilities implies that she must have P(Heads)=1/2? As I’ve shown above, that strategy is actually only compatible with her belief being that P(Heads)=1/3.
Note that in general, the “action space” for a decision theory problem need not be the same as the “state space”. One might, for example, have some uncertain information about what day of the week it is (7 possibilities) and on that basis decide whether to order pepperoni, anchovy, or ham pizza (3 possibilities). (You know that different people, with different skills, usually make the pizza on different days.) So if for some reason you randomized your choice of action, it would certainly not say anything directly about your probabilities for the different days of the week.
Maybe we are starting to go in circles. But while I agree the word “guess” might be problematic I think you still have an ambiguity with what the word probability means in this case. Perhaps you could give the definition you would use for the word “probability”.
“In everyday usage, we use “guess” when the aim is to guess correctly.” Guess correctly in the largest proportion of trials, or in the largest proportion of guesses? I think my “scrob” and “quob” thingies are indeed aiming to guess correctly. One in the most possible trials, the other in the most possible individual instances of making a guess.
“Having eliminated the word “guess”, why would one think that Beauty’s use of the strategy of randomly taking action H or action T with equal probabilities implies that she must have P(Heads)=1/2?”—I initially conjectured this as weak evidence, but no longer hold the position at all, as I explained in the post with the graph. However, I still think that in the other death-scenario (Guess Wrong you die) the fact that deterministically picking heads is equally good to deterministically picking tails says something. This GWYD case sets the rules of the wager such that Beauty is trying to be right in as many trials as possible, instead of for as many individual awakenings. Clearly moving the goalposts to the “number of trials” denominator.
For me, the issue is that you appear to take “probability” as “obviously” meaning “proportion of awakenings”. I do not think this is forced on us by anything, and that both denominators (awakenings and trials) provide us with useful information that can beneficially inform our decision making, depending on whether we want to be right in as many awakenings or trials as possible. Perhaps you could explain your position while tabooing the word “probability”? Because, I think we have entered the Tree-falling-in-forest zone: https://www.lesswrong.com/posts/7X2j8HAkWdmMoS8PE/disputing-definitions, and I have tried to split our problem term in two (Quobability and Srobability) but it hasn’t helped.
Perhaps you could give the definition you would use for the word “probability”.
I define it as one’s personal degree of belief in a proposition, at the time the judgement of probability is being made. It has meaning only in so far it is (or may be) used to make a decision, or is part of a general world model that is itself meaningful. (For example, we might assign a probability to Jupiter having a solid core, even though that makes no difference to anything we plan to do, because that proposition is part of an overall theory of physics that is meaningful.)
Frequentist ideas about probability being related to the proportion of times that an event occurs in repetitions of a scenario are not part of this definition, so the question of what denominator to use does not arise. (Looking at frequentist concepts can sometimes be a useful sanity check on whether probability judgements make sense, but if there’s some conflict between frequentist and Bayesian results, the solution is to re-examine the Bayesian results, to see if you made a mistake, or to understand why the frequentist results don’t actually contradict the Bayesian result.)
If you make the right probability judgements, you are supposed to make the right decision, if you correctly apply decision theory. And Beauty does make the right decision in all the Sleeping Beauty scenarios if she judges that P(Heads)=1/3 when woken before Wednesday. She doesn’t make the right decision if she judges that P(Heads)=1/2. I emphasize that this is so for all the scenarios. Beauty doesn’t have to ask herself, “what denominator should I be using?”. P(Heads)=1/3 gives the right answer every time.
Another very useful property of probability judgements is that they can be used for multiple decisions, without change. Suppose, for example, that in the GWYD or GRYL scenarios, in addition to trying not to die, Beauty is also interested in muffins.
Specifically, she knows from the start that whenever she wakes up there will be a plate of freshly-baked muffins on her side table, purchased from the cafe down the road. She knows this cafe well, and in particular knows that (a) their muffins are always very delicious, and (b) on Tuesdays, but not Mondays, the person who bakes the muffins adds an ingredient that gives her a stomach ache 10 minutes after eating a muffin. Balancing these utilities, she decides to eat the muffins if the probability of it being Tuesday is less than 30%. If Beauty is a Thirder, she will judge the probability of Tuesday to be 1⁄3, and refrain from eating the muffins, but if Beauty is a Halfer, she will (I think, trying to pretend I’m a halfer) think the probability of Tuesday is 1⁄4, and eat the muffins.
The point here is not so much which decision is correct (though of course I think the Thirder decision is right), but that whatever the right decision is, it shouldn’t depend on whether Beauty is in the GWYD or GRYL scenario. She shouldn’t be considering “denominators”.