Yet one more variant. On my view it’s structurally and hence statistically equivalent to Iterated Sleeping Beauty, and I present an argument that it is. This one has the advantage that it does not rely on any science fictional technology. I’m interested to see if anyone can find good reasons why it’s not equivalent.
The Iterated Sleeping Beaty problem (ISB) is the original Standard
Sleeping Beauty (SSB) problem repeated a large number N of times. People always seem to want to do this anyway with all the variations, to use the Law of Large Numbers to gain insight to what they should do in the single shot case.
The Setup
As before, Sleeping Beauty is fully apprised of all the details ahead of time.
The experiment is run for N consecutive days (N is a large number).
At midnight 24 hours prior to the start of the experiment, a fair coin is tossed.
On every subsequent night, if the coin shows Heads, it is tossed again; if it shows Tails, it is turned over to show Heads.
(This process is illustrated by a discrete-time Markov chain with transition matrix:
[1/2 1/2] = P
[ 1 0 ]
and the state vector is the row
x = [ Heads Tails ],
with consecutive state transitions computed as x * P^k
Each morning when Sleeping Beauty awakes, she is asked each of the following questions:
“What is your credence that the most recent coin toss landed Heads?”
“What is your credence that the coin was tossed last night?”
“What is your credence that the coin is showing Heads now?”
The first question is the equivalent of the question that is asked in the Standard Sleeping Beauty problem. The second question corresponds to the question “what is your credence that today is Monday?” (which should also be asked and analyzed in any treatment of the Standard Sleeping Beauty problem.)
Note: in this setup, 3) is different than 1) only because of the operation of turning the coin over instead of tossing it. This is just a perhaps too clever mechanism to count down the days (awakenings, actually) to the point when the coin should be tossed again. It may very well make a better example if we never touch the coin except to toss it, and use some other deterministic countdown mechanism to count repeated awakenings per coin toss. That allows easier generalization to the case where the number of days to awaken when Tails is greater than 2. It also makes 3) directly equivalent to the standard SB question, and also 1) and 3) have the same answers. You decide which mechanism is easier to grasp from a didactic point of view, and analyze that one.
After that, Beauty goes on about her daily routine, takes no amnesia drugs, sedulously avoids all matter duplicators and transhuman uploaders, and otherwise lives a normal life, on one condition: she is not allowed to examine the coin or discover its state (or the countdown timer) until the experiment is over.
Analysis
Q1: How should Beauty answer?
Q2: How is this scenario similar in key respects to the SSB/ISB scenario?
Q3: How does this scenario differ in key respects from the SSB/ISB scenario?
Q4: How would those differences if any make a difference to how Beauty should answer?
My answers:
Q1: Her credence that the most recent coin toss landed Heads should be 1⁄3. Her credence that the coin was tossed last night should be 1⁄3. Her credence that the coin shows Heads should be 2⁄3. (Her credence that the coin shows Heads should be 1⁄3 if we never turn it over, only toss, and 1/K if the countdown timer counts K awakenings per Tail toss.)
Q2: Note that Beauty’s epistemic state regarding the state of the coin, or whether it was tossed the previous midnight, is exactly the same on every morning, but without the use of drugs or other alien technology. She awakens and is asked the questions once every time the coin toss lands Heads, and twice every time it lands tails. In Standard Sleeping Beauty, her epistemic state is reset by the amnesia drugs. In this setup, her epistemic state never needs to be reset because it never changes, simply because she never receives any new information that could change it, including the knowledge of when the coin has been tossed to start a new cycle.
Q3: In ISB, a new experimental cycle is initiated at fixed times—Monday (or Sunday midnight). Here the start of a new “cycle” occurs with random timing. The question arises, does the difference in the speed of time passing make any difference to the moments of awakening when the question is asked? Changing labels from “Monday” and “Tuesday” to “First Day After Coin Toss” and “Second Day After Coin Toss” respectively makes no structural change to the operation of the process. Discrete-time Markov chains have no timing, they have only sequence.
In the standard ISB, there seems to be a natural unit of replication: the coin toss on Sunday night followed by whatever happens through the rest of the week. Here, that unit doesn’t seem so prominent, though it still exists as a renewal point of the chain. In a recurrent Markov chain, the natural unit of replication seems to be the state transition. Picking a renewal point is also an option, but only as a matter of convenience of calculation; it doesn’t change the analysis.
Q4: I don’t see how. The events, and the processes which drive their occurence haven’t changed that I can see, just our perspective in looking at them. What am I overlooking?
Iteration
I didn’t tell you yet how N is determined and how the experiment is terminated. Frankly, I don’t think it matters all that much as N gets large, but let’s remove all ambiguity.
Case A: N is a fixed large number. The experiment is terminated on the first night on which the coin shows Heads, after the Nth night.
Case B: N is not fixed in advance, but is guaranteed to be larger than some other large fixed number N’, such that the coin has been tossed at least N’ times. Once N’ tosses have been counted, the experiment is terminated on any following night on which the coin shows Heads, at the whim of the Lab Director.
Q5: If N (or N’) is large enough, does the difference between Case A and B make a difference to Beauty’s credence? (To help sharpen your answer, consider Case C: Beauty dies of natural causes before the experiment terminates.)
Note that in view of the discussion under Q3 above, we are picking some particular state in the transition diagram and thinking about recurrence to and from that state. We could pick any other state too, and the analysis wouldn’t change in any significant way. It seems more informative (to me at any rate) to think of this as an ongoing prcess that converges to stable behavior at equilibrium.
Extra Credit:
This gets right to the heart of what a probability could mean, what things can count as probabilities, and why we care about Sleeping Beauty’s credence.
Suppose Beauty is sent daily reports showing cumulative counts of the nightly heads/tails observations. The reports are sufficiently old as not to give any information about the current state of the coin or when it was last tossed. (E.g., the data in the report are from at least two coin tosses ago.) Therefore Beauty’s epistemic state about the current state of the coin always remains in its initial/reset state, with the following exception. Discuss how Beauty could use this data to--
corroborate that the coin is in fact fair as she has been told.
update her credences, in case she accrues evidence that shows the coin is not fair.
For me this is the main attraction of this particular model of the Sleeping Beauty setup, so I’m very interested in any possible reasons why it’s not equivalent.
Sorry I was slow to respond .. busy with other things
My answers:
Q1: I agree with you: 1⁄3, 1⁄3, 2⁄3
Q2. ISB is similar to SSB as follows: fair coin; woken up twice if tails, once if heads; epistemic state reset each day
Q3. ISB is different from SSB as follows: more than one coin toss; same number of interviews regardless of result of coin toss
Q4. It makes a big difference. She has different information to condition on. On a given coin flip, the probability of heads is 1⁄2. But, if it is tails we skip a day before flipping again. Once she has been woken up a large number of times, Beauty can easily calculate how likely it is that heads was the most recent result of a coin flip. In SSB, she cannot use the same reasoning. In SSB, Tuesday&heads doesn’t exist, for example.
Consider 3 variations of SSB:
Same as SSB except If heads, she is interviewed on Monday, and then the coin is turned over to tails and she is interviewed on Tuesday. There is amnesia and all of that. So, it’s either the sequence (heads on Monday, tails on Tuesday) or (tails on Monday, tails on Tuesday). Each sequence has a 50% probability, and she should think of the days within a sequence as being equally likely. She’s asked about the current state of the coin. She should answer P(H)=1/4.
Same as SSB except If heads, she is interviewed on Monday, and then the coin is flipped again and she is interviewed on Tuesday. There is amnesia and all of that. So, it’s either the sequence (heads on Monday, tails on Tuesday), (heads on Monday, heads on Tuesday) or (tails on Monday, tails on Tuesday). The first 2 sequences have a 25% chance each and the last one has a 50% chance. When asked about the current state of the coin, she should say P(H)=3/8
The 1⁄2 solution to SSB results from similar reasoning. 50% chance for the sequence (Monday and heads). 50% chance for the sequence (Monday and tails, Tuesday and tails). P(H)=1/2
If you apply this kind of reasoning to ISB, where we are thinking of randomly selected day after a lot of time has passed, you’ll get P(H)=1/3.
I’m struggling to see how ISB isn’t different from SSB in meaningful ways.
Perhaps this is beating a dead horse, but here goes.
Regarding your two variants:
1 Same as SSB except If heads, she is interviewed on Monday, and then the
coin is turned over to tails and she is interviewed on Tuesday. There is
amnesia and all of that. So, it’s either the sequence (heads on Monday,
tails on Tuesday) or (tails on Monday, tails on Tuesday). Each sequence
has a 50% probability, and she should think of the days within a sequence
as being equally likely. She’s asked about the current state of the
coin. She should answer P(H)=1/4.
I agree. When iterated indefinitely, the Markov chain transition matrix is:
acting on state vector [ H1 H2 T1 T2 ], where H,T are coin toss outcomes and 1,2 label Monday,Tuesday. This has probability eigenvector [ 1⁄41⁄4
1/4 1⁄4 ]; 3 out of 4 states show Tails (as opposed to the coin having been tossed Tails). By the way, we have unbiased sampling of the coin toss outcomes here.
If the Markov chain model isn’t persuasive, the alternative calculation is to look at the branching probability diagram
and compute the expected frequencies of letters in the result strings at each leaf on Wednesdays. This is
0.5 * ( H + T ) + 0.5 * ( T + T ) = 0.5 * H + 1.5 * T.
2 Same as SSB except If heads, she is interviewed on Monday, and then the
coin is flipped again and she is interviewed on Tuesday. There is amnesia
and all of that. So, it’s either the sequence (heads on Monday, tails on
Tuesday), (heads on Monday, heads on Tuesday) or (tails on Monday, tails
on Tuesday). The first 2 sequences have a 25% chance each and the last one
has a 50% chance. When asked about the current state of the coin, she
should say P(H)=3/8
I agree. Monday-Tuesday sequences occur with the following probabilities:
HH: 1/4
HT: 1/4
TT: 1/2
Also, the Markov chain model for the iterated process agrees:
to compute expected frequencies of letters in the result strings,
0.25 * ( H + H ) + 0.25 * ( H + T ) + 0.5 * ( T + T ) = 0.75 * H + 1.25 * T
Because of the extra coin toss on Tuesday after Monday Heads, these are biased observations of coin tosses. (Are these credences?) But neither of these two variants is equivalent to Standard Sleeping Beauty or its iterated variants ISB and ICSB.
The 1⁄2 solution to SSB results from similar reasoning. 50% chance for the sequence (Monday and heads). 50% chance for the sequence (Monday and tails, Tuesday and tails). P(H)=1/2
(Sigh). I don’t think your branching probability diagram is correct. I don’t know what other reasoning you are using. This is the diagram I have for Standard Sleeping Beauty
And this is how I use it, using exactly the same method as in the two examples above. With probability 1⁄2 the process accumulates 2 Tails observations per week, and with probability 1⁄2 accumulates 1 Heads observation. The expected number of observations per week is 1.5, the expected number of Heads observations per week is 0.5, the expected number of Tails observations is 1 per week.
0.5 * ( H ) + 0.5 * ( T + T ) = 0.5 * H + 1.0 * T
Likewise when we record Monday/Tuesday observations per week instead of Heads/Tails, the expected number of Monday observations is 1, expected Tuesday observations 0.5, for a total of 1.5. But in both of your variants above, the expected number of Monday observations = expected number of Tuesday observations = 1.
Thanks for your response. I should have been clearer in my terminology. By “Iterated Sleeping Beauty” (ISB) I meant to name the variant that we here have been discussing for some time, that repeats the Standard Sleeping Beauty problem some number say 1000 of times. In 1000 coin tosses over 1000 weeks, the number of Heads awakenings is 1000 and the number of Tails awakenings is 2000. I have no catchy name for the variant I proposed, but I can make up an ugly one if nothing better comes to mind; it could be called Iterated Condensed Sleeping Beauty (ICSB). But I’ll assume you meant this particular variant of mine when you mention ISB.
You say
Q3. ISB is different from SSB as follows: more than one coin toss; same number of interviews regardless of result of coin toss
“More than one coin toss” is the iterated part. As far as I can see,
and I’ve argued it a couple times now, there’s no essential difference between SSB and ISB, so I meant to draw a comparison between my variant and ISB.
“Same number of interviews regardless of result of coin toss” isn’t correct. Sorry if I was unclear in my description. Beauty is interviewed once per toss when Heads, twice when Tails. This is the same in ICSB as in Standard and Iterated Sleeping Beauty. Is there an important difference between Standard Sleeping Beauty and Iterated Sleeping Beauty, or is there an important difference between Iterated Sleeping Beauty and Iterated Condensed Sleeping Beauty?
Q4. It makes a big difference. She has different information to condition
on. On a given coin flip, the probability of heads is 1⁄2. But, if it is
tails we skip a day before flipping again. Once she has been woken up a
large number of times, Beauty can easily calculate how likely it is that
heads was the most recent result of a coin flip.
We not only skip a day before tossing again, we interview on that day too!
I see how over time Beauty gains evidence corroborating the fairness of the
coin (that’s exactly my later rhetorical question), but assuming it’s a fair coin, and barring Type I errors, she’ll never see evidence to change her initial credence in that proposition. In view of this, can you explain how she can use this information to predict with better than initial accuracy the likelihood that Heads was the most recent outcome of the toss? I don’t see how.
In SSB, Tuesday&heads doesn’t exist, for example.
After relabeling Monday and Tuesday to Day 1 and Day 2 following the coin toss, Tuesday&Heads (H2) exists in none of these variants. So what difference is there?
Q1: I agree with you: 1⁄3, 1⁄3, 2⁄3
Good and well, but—are these legitimate credences? If not, why not? And
if so, why aren’t they also in the following:
Standard Iterated Sleeping Beauty is isomorphic to the following Markov
chain, which just subdivides the Tails state in my condensed variant into
Day 1 and Day 2:
[1/2, 1/2, 0]
[0, 0, 1]
[1/2, 1/2, 0]
operating on row vector of states [ Heads&Day1 Tails&Day1 Tails&Day2 ],
abbreviated to [ H1 T1 T2 ]
When I say isomorphic, I mean the distinct observable states of affairs are
the same, and the possible histories of transitions from awakening to next awakening are governed by the same transition probabilities.
So either there’s a reason why my 2-state Markov chain correctly models my
condensed variant that allows you to accept the 1⁄3 answers it computes,
that doesn’t apply to the three-state Markov chain and its 1⁄3 answers
(perhaps you came to those answers independently of my model), or else
there’s some reason why the three-state Markov chain doesn’t correctly model
the Iterated Sleeping Beauty process. Can you help me see where the difficulty may lie?
I’m struggling to see how ISB isn’t different from SSB in meaningful ways.
I assume you are referring to my variant, not what I’m calling Iterated Sleeping Beauty. If so, I’m kind of baffled by this statement, because under similarities, you just listed
fair coin
woken twice if Tails, once if Heads
epistemic state reset each day
With the emendation that 2) is per coin toss, and in 3) “each day” = “each awakening”, you have just listed three essential features that SSB, ISB and ICSB all have in common. It’s exactly those three things that define the SSB problem. I’m claiming that there aren’t any others. If you disagree, then please tell me what they are. Or if parts of my argument remain unclear, I can try to go into more detail.
Yet one more variant. On my view it’s structurally and hence statistically equivalent to Iterated Sleeping Beauty, and I present an argument that it is. This one has the advantage that it does not rely on any science fictional technology. I’m interested to see if anyone can find good reasons why it’s not equivalent.
The Iterated Sleeping Beaty problem (ISB) is the original Standard Sleeping Beauty (SSB) problem repeated a large number N of times. People always seem to want to do this anyway with all the variations, to use the Law of Large Numbers to gain insight to what they should do in the single shot case.
The Setup
As before, Sleeping Beauty is fully apprised of all the details ahead of time.
The experiment is run for N consecutive days (N is a large number).
At midnight 24 hours prior to the start of the experiment, a fair coin is tossed.
On every subsequent night, if the coin shows Heads, it is tossed again; if it shows Tails, it is turned over to show Heads.
(This process is illustrated by a discrete-time Markov chain with transition matrix:
and the state vector is the row
with consecutive state transitions computed as x * P^k
Each morning when Sleeping Beauty awakes, she is asked each of the following questions:
“What is your credence that the most recent coin toss landed Heads?”
“What is your credence that the coin was tossed last night?”
“What is your credence that the coin is showing Heads now?”
The first question is the equivalent of the question that is asked in the Standard Sleeping Beauty problem. The second question corresponds to the question “what is your credence that today is Monday?” (which should also be asked and analyzed in any treatment of the Standard Sleeping Beauty problem.)
Note: in this setup, 3) is different than 1) only because of the operation of turning the coin over instead of tossing it. This is just a perhaps too clever mechanism to count down the days (awakenings, actually) to the point when the coin should be tossed again. It may very well make a better example if we never touch the coin except to toss it, and use some other deterministic countdown mechanism to count repeated awakenings per coin toss. That allows easier generalization to the case where the number of days to awaken when Tails is greater than 2. It also makes 3) directly equivalent to the standard SB question, and also 1) and 3) have the same answers. You decide which mechanism is easier to grasp from a didactic point of view, and analyze that one.
After that, Beauty goes on about her daily routine, takes no amnesia drugs, sedulously avoids all matter duplicators and transhuman uploaders, and otherwise lives a normal life, on one condition: she is not allowed to examine the coin or discover its state (or the countdown timer) until the experiment is over.
Analysis
Q1: How should Beauty answer?
Q2: How is this scenario similar in key respects to the SSB/ISB scenario?
Q3: How does this scenario differ in key respects from the SSB/ISB scenario?
Q4: How would those differences if any make a difference to how Beauty should answer?
My answers:
Q1: Her credence that the most recent coin toss landed Heads should be 1⁄3. Her credence that the coin was tossed last night should be 1⁄3. Her credence that the coin shows Heads should be 2⁄3. (Her credence that the coin shows Heads should be 1⁄3 if we never turn it over, only toss, and 1/K if the countdown timer counts K awakenings per Tail toss.)
Q2: Note that Beauty’s epistemic state regarding the state of the coin, or whether it was tossed the previous midnight, is exactly the same on every morning, but without the use of drugs or other alien technology. She awakens and is asked the questions once every time the coin toss lands Heads, and twice every time it lands tails. In Standard Sleeping Beauty, her epistemic state is reset by the amnesia drugs. In this setup, her epistemic state never needs to be reset because it never changes, simply because she never receives any new information that could change it, including the knowledge of when the coin has been tossed to start a new cycle.
Q3: In ISB, a new experimental cycle is initiated at fixed times—Monday (or Sunday midnight). Here the start of a new “cycle” occurs with random timing. The question arises, does the difference in the speed of time passing make any difference to the moments of awakening when the question is asked? Changing labels from “Monday” and “Tuesday” to “First Day After Coin Toss” and “Second Day After Coin Toss” respectively makes no structural change to the operation of the process. Discrete-time Markov chains have no timing, they have only sequence.
In the standard ISB, there seems to be a natural unit of replication: the coin toss on Sunday night followed by whatever happens through the rest of the week. Here, that unit doesn’t seem so prominent, though it still exists as a renewal point of the chain. In a recurrent Markov chain, the natural unit of replication seems to be the state transition. Picking a renewal point is also an option, but only as a matter of convenience of calculation; it doesn’t change the analysis.
Q4: I don’t see how. The events, and the processes which drive their occurence haven’t changed that I can see, just our perspective in looking at them. What am I overlooking?
Iteration
I didn’t tell you yet how N is determined and how the experiment is terminated. Frankly, I don’t think it matters all that much as N gets large, but let’s remove all ambiguity.
Case A: N is a fixed large number. The experiment is terminated on the first night on which the coin shows Heads, after the Nth night.
Case B: N is not fixed in advance, but is guaranteed to be larger than some other large fixed number N’, such that the coin has been tossed at least N’ times. Once N’ tosses have been counted, the experiment is terminated on any following night on which the coin shows Heads, at the whim of the Lab Director.
Q5: If N (or N’) is large enough, does the difference between Case A and B make a difference to Beauty’s credence? (To help sharpen your answer, consider Case C: Beauty dies of natural causes before the experiment terminates.)
Note that in view of the discussion under Q3 above, we are picking some particular state in the transition diagram and thinking about recurrence to and from that state. We could pick any other state too, and the analysis wouldn’t change in any significant way. It seems more informative (to me at any rate) to think of this as an ongoing prcess that converges to stable behavior at equilibrium.
Extra Credit:
This gets right to the heart of what a probability could mean, what things can count as probabilities, and why we care about Sleeping Beauty’s credence.
Suppose Beauty is sent daily reports showing cumulative counts of the nightly heads/tails observations. The reports are sufficiently old as not to give any information about the current state of the coin or when it was last tossed. (E.g., the data in the report are from at least two coin tosses ago.) Therefore Beauty’s epistemic state about the current state of the coin always remains in its initial/reset state, with the following exception. Discuss how Beauty could use this data to--
corroborate that the coin is in fact fair as she has been told.
update her credences, in case she accrues evidence that shows the coin is not fair.
For me this is the main attraction of this particular model of the Sleeping Beauty setup, so I’m very interested in any possible reasons why it’s not equivalent.
Sorry I was slow to respond .. busy with other things
My answers:
Q1: I agree with you: 1⁄3, 1⁄3, 2⁄3
Q2. ISB is similar to SSB as follows: fair coin; woken up twice if tails, once if heads; epistemic state reset each day
Q3. ISB is different from SSB as follows: more than one coin toss; same number of interviews regardless of result of coin toss
Q4. It makes a big difference. She has different information to condition on. On a given coin flip, the probability of heads is 1⁄2. But, if it is tails we skip a day before flipping again. Once she has been woken up a large number of times, Beauty can easily calculate how likely it is that heads was the most recent result of a coin flip. In SSB, she cannot use the same reasoning. In SSB, Tuesday&heads doesn’t exist, for example.
Consider 3 variations of SSB:
Same as SSB except If heads, she is interviewed on Monday, and then the coin is turned over to tails and she is interviewed on Tuesday. There is amnesia and all of that. So, it’s either the sequence (heads on Monday, tails on Tuesday) or (tails on Monday, tails on Tuesday). Each sequence has a 50% probability, and she should think of the days within a sequence as being equally likely. She’s asked about the current state of the coin. She should answer P(H)=1/4.
Same as SSB except If heads, she is interviewed on Monday, and then the coin is flipped again and she is interviewed on Tuesday. There is amnesia and all of that. So, it’s either the sequence (heads on Monday, tails on Tuesday), (heads on Monday, heads on Tuesday) or (tails on Monday, tails on Tuesday). The first 2 sequences have a 25% chance each and the last one has a 50% chance. When asked about the current state of the coin, she should say P(H)=3/8
The 1⁄2 solution to SSB results from similar reasoning. 50% chance for the sequence (Monday and heads). 50% chance for the sequence (Monday and tails, Tuesday and tails). P(H)=1/2
If you apply this kind of reasoning to ISB, where we are thinking of randomly selected day after a lot of time has passed, you’ll get P(H)=1/3.
I’m struggling to see how ISB isn’t different from SSB in meaningful ways.
Perhaps this is beating a dead horse, but here goes. Regarding your two variants:
I agree. When iterated indefinitely, the Markov chain transition matrix is:
acting on state vector [ H1 H2 T1 T2 ], where H,T are coin toss outcomes and 1,2 label Monday,Tuesday. This has probability eigenvector [ 1⁄4 1⁄4 1/4 1⁄4 ]; 3 out of 4 states show Tails (as opposed to the coin having been tossed Tails). By the way, we have unbiased sampling of the coin toss outcomes here.
If the Markov chain model isn’t persuasive, the alternative calculation is to look at the branching probability diagram
[http://entity.users.sonic.net/img/lesswrong/sbv1tree.png (SB variant 1)]
and compute the expected frequencies of letters in the result strings at each leaf on Wednesdays. This is
I agree. Monday-Tuesday sequences occur with the following probabilities:
Also, the Markov chain model for the iterated process agrees:
acting on state vector [ H1 H2 T1 T2 ] gives probability eigenvector [ 1⁄4 1⁄8 1⁄4 3⁄8 ]
Alternatively, use the branching probability diagram
[http://entity.users.sonic.net/img/lesswrong/sbv2tree.png (SB variant 2)]
to compute expected frequencies of letters in the result strings,
Because of the extra coin toss on Tuesday after Monday Heads, these are biased observations of coin tosses. (Are these credences?) But neither of these two variants is equivalent to Standard Sleeping Beauty or its iterated variants ISB and ICSB.
(Sigh). I don’t think your branching probability diagram is correct. I don’t know what other reasoning you are using. This is the diagram I have for Standard Sleeping Beauty
[http://entity.users.sonic.net/img/lesswrong/ssbtree.png (Standard SB)]
And this is how I use it, using exactly the same method as in the two examples above. With probability 1⁄2 the process accumulates 2 Tails observations per week, and with probability 1⁄2 accumulates 1 Heads observation. The expected number of observations per week is 1.5, the expected number of Heads observations per week is 0.5, the expected number of Tails observations is 1 per week.
Likewise when we record Monday/Tuesday observations per week instead of Heads/Tails, the expected number of Monday observations is 1, expected Tuesday observations 0.5, for a total of 1.5. But in both of your variants above, the expected number of Monday observations = expected number of Tuesday observations = 1.
Thanks for your response. I should have been clearer in my terminology. By “Iterated Sleeping Beauty” (ISB) I meant to name the variant that we here have been discussing for some time, that repeats the Standard Sleeping Beauty problem some number say 1000 of times. In 1000 coin tosses over 1000 weeks, the number of Heads awakenings is 1000 and the number of Tails awakenings is 2000. I have no catchy name for the variant I proposed, but I can make up an ugly one if nothing better comes to mind; it could be called Iterated Condensed Sleeping Beauty (ICSB). But I’ll assume you meant this particular variant of mine when you mention ISB.
You say
“More than one coin toss” is the iterated part. As far as I can see, and I’ve argued it a couple times now, there’s no essential difference between SSB and ISB, so I meant to draw a comparison between my variant and ISB.
“Same number of interviews regardless of result of coin toss” isn’t correct. Sorry if I was unclear in my description. Beauty is interviewed once per toss when Heads, twice when Tails. This is the same in ICSB as in Standard and Iterated Sleeping Beauty. Is there an important difference between Standard Sleeping Beauty and Iterated Sleeping Beauty, or is there an important difference between Iterated Sleeping Beauty and Iterated Condensed Sleeping Beauty?
We not only skip a day before tossing again, we interview on that day too! I see how over time Beauty gains evidence corroborating the fairness of the coin (that’s exactly my later rhetorical question), but assuming it’s a fair coin, and barring Type I errors, she’ll never see evidence to change her initial credence in that proposition. In view of this, can you explain how she can use this information to predict with better than initial accuracy the likelihood that Heads was the most recent outcome of the toss? I don’t see how.
After relabeling Monday and Tuesday to Day 1 and Day 2 following the coin toss, Tuesday&Heads (H2) exists in none of these variants. So what difference is there?
Good and well, but—are these legitimate credences? If not, why not? And if so, why aren’t they also in the following:
Standard Iterated Sleeping Beauty is isomorphic to the following Markov chain, which just subdivides the Tails state in my condensed variant into Day 1 and Day 2:
operating on row vector of states [ Heads&Day1 Tails&Day1 Tails&Day2 ], abbreviated to [ H1 T1 T2 ]
When I say isomorphic, I mean the distinct observable states of affairs are the same, and the possible histories of transitions from awakening to next awakening are governed by the same transition probabilities.
So either there’s a reason why my 2-state Markov chain correctly models my condensed variant that allows you to accept the 1⁄3 answers it computes, that doesn’t apply to the three-state Markov chain and its 1⁄3 answers (perhaps you came to those answers independently of my model), or else there’s some reason why the three-state Markov chain doesn’t correctly model the Iterated Sleeping Beauty process. Can you help me see where the difficulty may lie?
I assume you are referring to my variant, not what I’m calling Iterated Sleeping Beauty. If so, I’m kind of baffled by this statement, because under similarities, you just listed
fair coin
woken twice if Tails, once if Heads
epistemic state reset each day
With the emendation that 2) is per coin toss, and in 3) “each day” = “each awakening”, you have just listed three essential features that SSB, ISB and ICSB all have in common. It’s exactly those three things that define the SSB problem. I’m claiming that there aren’t any others. If you disagree, then please tell me what they are. Or if parts of my argument remain unclear, I can try to go into more detail.