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.
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.