Cinderella Problem. Anyone understand it?

Perhaps you could write it out so those of us not wishing to sit through a video could read it and reply.

I don't understand the video I think is the issue. I have to assume the odds of a single flip must always be 50/50 if it is a single flip. The video is 10 minutes only.

I can't get it to open. Sorry

Reply to jgill

I updated first post with YouTube video title and channel.

Reply to TiredThinker Thanks

Sleeping Beauty Problem

I still don't see how the odds are changed if the fair coin is flipped only once. If it were the lets make a deal problem of selecting twice, once initially, and again after one of the doors is removed.

After reading the Wiki piece it seems to be more a decision theory problem, and that's how they classify it. On that same page there is the following similar problem that for me is easier to understand:

Sailor's Child problem
The Sailor's Child problem, introduced by Radford M. Neal, is somewhat similar. It involves a sailor who regularly sails between ports. In one port there is a woman who wants to have a child with him, across the sea there is another woman who also wants to have a child with him. The sailor cannot decide if he will have one or two children, so he will leave it up to a coin toss. If Heads, he will have one child, and if Tails, two children. But if the coin lands on Heads, which woman would have his child? He would decide this by looking at The Sailor's Guide to Ports and the woman in the port that appears first would be the woman that he has a child with. You are his child. You do not have a copy of The Sailor's Guide to Ports. What is the probability that you are his only child, thus the coin landed on Heads (assume a fair coin)?

I agree with Tired Thinker, in the way that I am not fully seeing the problem or even "a" problem at all.

Lets say...There is no problem....Okay. Why? Because Sleeping Beauty (referred to as "her" in the following comments) volunteered...(per youtube video that was linked in OP proposing the problem)

What if only 1 option really exists but it just might occur twice sometimes?

Perhaps this is less about the numbers and more about a possible pattern in the surety of the answer, despite the outcome and if her intuition is aligned with her faith (if any) and the answer she gives.I am not sure exactly, this is just my thoughts after the first minute after watching the linked video (I stopped at the reaction of the guy in the video and was uninterested in it to be honest) about a minute in.)

Maybe the test is being done and is for measuring the subjects prediction over (or instead of) the literal probability of the odds... with faith being a component of the subjects answer and if the answers are true or false literally by guessing the result or outcome because of luck, chance or intuitively...

What If she is correct in her answer (eg.coin was flipped and landed on heads was heads and she guessed high probability on Monday making her correct on the flip and the way she answered was with excitement, calmness and acceptance...I then ponder the options further myself, asking questions like Did her faith influence her answer and did the answer come intuitively or did it come with doubt? (for example coin lands on heads, she guess low probability that it was heads therefore she was wrong about the flip because of self pity or other ways she might see her self, self doubt... eg. "no way I am that lucky" "no chance i am right" "I never guess the right answer" that would be reasonable for her to have the wrong answer to the question because without really feeling the answer it was less influenced by intuition if any...

Can there be any "hope" in the answer without the backing of faith? Is she RIGHT TO BE WRONG? And does that align with the actual outcome of the toss? Are there any patterns? Can she feel Tuesday coming or not? Hope without faith, is that relevant to the answer even if her intuition isnt the force behind the answer? Can you tell if it was? How strong of an influence does faith have on the surety of self in the answer? Is she right for doubting her self and answer perhaps? Is the problem somewhere in or about faith being questioned at all?

Why would this be a problem?

-Are the subjects with the right answer (guessing if the coin toss was heads or tails and guessing correctly) actually wrong for having the RIGHT answer?
-Or is the wrong answer in having to prove everything she is doing is right?
-Is there a problem with answers that are seemingly only stemming from faith, intuition? neither?
-Is it doubt?
-Is she wrong or right to doubt her self?
-Is she right or wrong to be so sure in her answer?
-Is it right or wrong to be either but the problem is volunteering in the first place?
-Is the problem volunteering with and having something to lose despite feelings of subject?
-Are they wrong to feel right in this? or vice versa.....

Also, how much does she believe she is going to never wake up again, is that answer a display of her hope in that? Is she showing signs of attitudes, mindset, behavior and appearance that align with the answer given, when right and when wrong literally?? (sleeping beauty meaning or whomever is the subject of study) excited or calm nervous or scared? and if the results of the flip align or follow through with the intuition, faith, display of emotion are involved in the answer maybe could that be shown in this specific study, by using a coin with 2 sides that can portray the reality of just 1 option that happens to presents itself twice?... 5050 odds that they (subject being studied and asked the question) are actually correct in the flip and the surety in the correctness of the answer...when right and when wrong....

What if its not about necessarily about the numbers but how the numbers present and align with personal feeling with or without intuition being surely relied on or of faith or belief or lack there of, these are what might tweak the possibilities of odds, not outcomes.....the attitude and tone of the answer is worth observing along with the literal 2 options 50/50 chances...the odds are HOW THE ANSWER AND , the guess, the surety in the guess and the actual turnout relate or if they do or if a pattern presents itself...

I may ask if it is possible to feel it coming? It being aware of being close to death...knowing it is happening soon? Never waking up ever again? Also I consider this.... if volunteering for this experiment is a problem...meaning one wanted to partake in the study.. then it isnt unreasonable to consider/keep in mind that those applying to do this study/volunteering might be suicidal? do they have anything to lose or feel that way (that they have nothing to lose, when that is not true or when it is...)? How can one feel that? Is the fact I just asked that last question the problem?

It feels right asking these questions even though they could all be the wrong... I am just thinking as if I were there, being studied and studying.... This is the first time I have seen this problem, I thought it was interesting.

Reply to jgill I guess the answer is 1/3 of being his only child. And it is based on a 1/2 coin flip. But what is the takeaway?

Reply to TiredThinker Looks like it could be 1/2. I used to teach stuff like this many years ago, but haven't indulged since. Could be 1/3 I guess. If you draw a diagram of paths possible, only one is favorable, with probability =1/2. @fdrake could step in here with clarifying comments. :cool:

Reply to jgill
Somewhere in the video I think he suggested that if 1/3 is correct it suggests that the multiverse might be possible. Don't know how large a leap that is.

Quoting TiredThinker

Don't know how large a leap that is

The truth table for that statement has him taking no chances.

Reply to jgill What is a truth table?

Reply to TiredThinker

Truth table

Reply to TiredThinker

This is a Veritasium video on the Cinderalla problem.

It's known as the Sleeping Beauty Problem.

I think it suggests that a fair coin flip can have odds other than 50/50

No. It "suggests" that the conditional probability of an outcome depends on any information that is obtained about that outcome. For example, if I pick a random card the probability that it is the Ace of Spades is 1/52. If I tell you it is a black card, the conditional probability is 1/26. If I tell you it is an Ace, the conditional probability is 1/4.

None of these answers changed anything. They evaluated different sets of information about the card. The issue in the Sleeping Beauty Problem is what, if any, information is gained? And two things seem confuse both the halfers and thirders described in the video:

1. Both seem to think that SB being left asleep on Tuesday, if the coin landed on Heads, is not a result of the experiment. They think this because SB can't observe it. I'm sorry, but it is a result, just one that can only be observed from the outside. If you doubt this, ask yourself if anything changes, on the days she is awakened in the video, if she is awakened on Heads+Tuesday, but not asked the question.
2. The memory wipe between Monday and Tuesday makes the observations SB can make on those two days independent to her, but not an outside observer.

MY ASSERTION: The correct solution is that SB will participate in four equally-likely outcomes that are independent to her, but not to an outside observer, because of the memory wipe. One of these outcomes is eliminated due to the fact that she observes it. So the answer is 1/3.

ONE PROOF BY DEMONSTRATION: There actually are four possible variations of the same experiment. She could be wakened always on Monday, and optionally on Tuesday; or optionally on Monday, and always on Tuesday. With either schedule, she could be wakened on the optional day after Heads, or after Tails; with this variation, she will be asked about the coin result where it is optional.

Regardless of what you think the answer is, that answer applies to each of the four variations. So, use all four at the same time. Assign a different variation to each, and use the same coin flip for them all.

On each day, three volunteers will be wakened. Two of these will be wakened both days, and one will be wakened on this day only. The question each addresses is, in essence, whether she is the "one wakening" volunteer. Yet none have any information to support the probability being different for any of the three. So the answer must be 1/3.

This is true regardless of whether each was told what combination she was assigned, and regardless of whether they can discuss it with each other (as long as they don't share assignments). So it also must be true if the other three exist, or SB merely supposes it. Which is the original problem. Essentially, the "missing" volunteer represents the "missing" observation opportunity.

+++++
Reply to jgill

Sailor's Child problem
The Sailor's Child problem, introduced by Radford M. Neal, is somewhat similar. It involves a sailor who regularly sails between ports. In one port there is a woman who wants to have a child with him, across the sea there is another woman who also wants to have a child with him. The sailor cannot decide if he will have one or two children, so he will leave it up to a coin toss. If Heads, he will have one child, and if Tails, two children. But if the coin lands on Heads, which woman would have his child? He would decide this by looking at The Sailor's Guide to Ports and the woman in the port that appears first would be the woman that he has a child with. You are his child. You do not have a copy of The Sailor's Guide to Ports. What is the probability that you are his only child, thus the coin landed on Heads (assume a fair coin)?

This is actually not the quite same thing. To see that, imagine what the answer would be if you do have a copy of The Sailor's Guide to Ports.

• If your port is the last one listed, you know that the coin landed on Tails. That is, Pr(Heads|Last)=0.
• If your port is the first one listed, you have no information about the coin. So the conditional probability is the same as the unconditional probability, Pr(Heads|First)=1/2.

But if you don't know what Q is, the answer must be between these two answers; that is, 1/2 can't be correct. So, say the fraction of the other entries that come after your port is Q, where 0<=Q<=1, then Pr(Heads|Q)=Q/(1+Q). Note that this agrees with the other two answers, and that it is 1/3 if Q=1/2.

Since you have no information about Q, you can treat it as as a uniformly distributed random variable. As the number of entries grows large, Pr(Heads|you were born) approaches 1-ln(2) ~= 0.0307,