## [0862] The Sleeping Yuna Problem

└ posted on Monday, 20 February 2017, by Novil

Each of those little panels with Yuna in them is drawn individually. That’s dedication to the art form that only nutcases like Bill Watterson would appreciate.

**Ye Thuza:** Rise and shine, Yuna! Because you were such an insufferable brat last week, I decided to carry out a devilish experiment on you.
**Yuna:** Really?! But I don’t have any tentacle arms!
**Ye Thuza:** It’s more of a mind game.

**Ye Thuza:** In this experiment, the mad scientist – that’s me – puts the human guinea pig – that’s you – to sleep on Sunday. Once or twice, the test subject will be awakened, interviewed, and put back to sleep with an amnesia-inducing drug that makes her forget the awakening. A coin will be tossed to determine which experimental procedure to undertake: if the coin comes up heads, the test subject will be awakened and interviewed on Monday only. If the coin comes up tails, she will be awakened and interviewed on Monday *and* Tuesday. In any case, the experiment will end on Wednesday without any further interview.

**Ye Thuza:** This means, it’s either Monday or Tuesday today. Now I ask you: what is your credence now for the proposition that the coin landed heads?gezeigt hat?
**Yuna:** That’s easy! It’s 50%!
**Ye Thuza:** Oh, is that so?

**Yuna:** No, wait…! It’s a third! … Hmm…
**Ye Thuza:** Even if I say that the coin is still lying on the kitchen table?

@ Cronek:You’re missing the most key point of the analogy. There is only one coin flip. The man has only two pockets. There are still only two possibilities.

-You get one ball, only.

-You get two balls. And I guess if we’re making it this similar to the original point, he gives you one ball, then takes it and throws it away, gives you an amnesiac drug so you forget being given the first ball, and gives you the second.

There can only be four outcomes (counting the Heads/Tuesday/No Interview) if there are four possibilities. If Ye Thuza did this two weeks in a row, that would be the case. But she only flips the coin ONCE. That means there are only two possibilities, one interview OR two interviews. It’s not one interview AND two interviews. There’s not three interviews to choose from, there’s only one, or two. And there’s no way to tell if Yuna’s in the one scenario, or the two scenario.

To go back to the analogy, the coin flip isn’t about the ball you’re given, it’s about the pocket it comes out of. In the actual problem, the coin flip isn’t about which interview Yuna is in, it’s about which branch she’s in, one interview or two interviews.

It is 50%. It doesn’t matter if there were a million wake / sleep cycles on one of the two chains.

End result is moot since neither happened and it was just a mind game to get Yuna out of bed and probably ready for school.

What makes this problem tricky is that it is temporal. Trying to make sense of a sequence of events is not easy especially when one of the events is amnesia. Let’s consider a spacial analogy.

A coin is flipped.

If the result is heads, you are given two identical sealed boxes. One contains a red ball, the other contains a blue ball.

If the result is tails, you receive two identical sealed boxes. Both boxes contain red balls.

We all agree that there is a 50-50 chance of heads or tails.

Some commenters have suggested that the rest of the scenario following the coin toss adds no further information. But it does… the fact that Yuna is being interviewed is significant. It is analogous to saying you have opened one of the boxes *and found a red ball*. Being told the ball’s colour gives extra information. (The alternative possibility of finding a blue ball is the equivalent of waking up on Tuesday without being interviewed.)

The pertinent question is this : *Given that you found a red ball*, what is the probability that the coin toss was heads?

The correct answer is 1/3.

@ Arpie:Okay, that actually was a very helpful analogy in terms of understanding why come people think the odds are 1/3.

That implies that it’s possible to find a blue ball. You can’t find a blue ball. You will never find a blue ball. From your perspective, the blue ball may as well not exist. By the same token, from Yuna’s perspective, the Tuesday in which she’s not interviewed will never occur. There’s no interview, she’s not aware of the experiment, it’s just nothing. Null.

In your analogy, it would be better to say that if the coin flip was tails, the subject would be given the two boxes containing the red balls. He’d open one box, see the red ball, then have his memory erased, then open the second box and see the second red ball, then have his memory erased again.

If the flip was heads, he’d be given the boxes containing the red ball and the blue ball. He’d open the box containing the red ball, then have his memory erased, and that’s it. He would NEVER see the blue ball. Him seeing the blue ball is not part of the experiment.

It’s a similar principle in Monty Hall, where Monty will only ever open a door and show you a goat. He’ll never open the door and show you the car.

You are describing the experiment from the perspective of an outside observer, as if this observer was given two sets of footage of Yuna’s bedroom (perv), one from Monday night and one from Tuesday night, without being told which is which, and randomly chooses one to watch. But that implies that there is a 25% chance that the observer will see footage of Yuna sleeping through the night (again, perv) with no interview.

But Yuna cannot have that perspective. Yuna’s only interaction with the experiment is being interviewed. The ball is always red. It can never be blue.

Try this analogy on for size: A coin is flipped. If it’s heads, one random person will be mailed a letter detailing the experiment. If it’s tails, two random people will be mailed the letter. Given that no recipient of the letter has any means of determining whether a second letter was mailed or not, what would they conclude the odds of the coin flip are?

It’s got to be 50%.

@ Potatamoto:Cronek is correct though.

You have to look at it this way. Yuna has no way to know what the right answer is for sure. What she does know, is that in case (HEADS) she’ll be interviewed once and in case (TAILS) she’ll be interviewed twice.

Currently Yuna is in an undesignated interview. It can be any of the three interviews. As such, the chance of it being the HEADS interview is 1/3. This is because the question has to be answered from an undefined interview and each possible interview has a 1/3 probability.

@ DysprosiumDy:That logic only works if there’s a scenario where she knows she’s not being interviewed. That can never happen.

She is in an undesignated interview, yes. She knows she’s being interviewed on THAT night. And that’s all she knows. The only TWO possibilities for HER are:

A: This is the only interview

B: There is a second interview.

There is not a third option for her, and those two options are dictated by the flip of the coin. Meaning, the odds have to be 50%.

@ Potatamoto:It suddenly occurred to me that in my statement above that somebody might argue the following:

“Since the second interview could take place either yesterday or tomorrow, isn’t that actually three possible outcomes for Yuna?” as in:

A- This is the only interview

B- The second interview will happen tomorrow

C- The second interview happened yesterday.

I’m just going to jump the gun and explain why that’s wrong now. Yuna doesn’t know what day it is. (Which, now that I think about it, is weird. Did Ye Thuza keep her asleep through the entirety of Monday and Tuesday, except for these interviews?)

But anyway, even though Yuna doesn’t know what day it is, it’s definitely either Monday or Tuesday. If it’s Monday, the second interview can only happen tomorrow, and if it’s Tuesday, the second interview can only have happened yesterday. It can’t be both options at once, it has to be one or the other. So, even though Yuna doesn’t know which day it is, she knows that there is only one possible day the second interview could take place on.

Credence is causing intermittent confusion throughout the comments. 😛

‘what is your credence now for the proposition that the coin landed heads?’

If we change the phrasing a little we can have

‘how likely do you now believe it is that the coin landed on heads?’

This thought experiment is reminisent of both the anthropomorphic principle and the ‘are we living in a simulation?’ conundrum. We are being asked whether Yuna – because she is being asked that horrible question – has been given information that affects her expectations of reality.

So what happens if you put money on it? Tell Yuna to guess ‘Heads’ or ‘Tails’ and give her five dollars if she is right, take away five dollars if she is wrong. What should she say to maximise her bank acount after the experiment is over? Well, the information she has is exactly the same whether she is being asked on Monday or Tuesday night so to simplify I shall only consider the case where she gives the same answer whether it is Monday or Tuesday.

If she always says ‘Heads’ she has a 50% chance of winning $5 and a 50% chance of losing $10

If she always says ‘Tails’ she has a 50% chance of losing $5 and a 50% chance of winning $10

Put another way, this means that the Yuna HAS been given information that influences her expectations of reality. She has been given reason to believe that it is more likely that she currently exists in the universe where the coin landed on tails.

@ Paul:Okay, I gotta admit, I’m confused by this. Your concluding sentence would indicate that you believe that Yuna’s chance of the coin landing on tails is greater than the chance it landed on heads.

But in your argument you pretty explicitly say that the odds are 50-50:

If she always says ‘Heads’ she has a *50% chance of winning $5* (heads)

*and a 50% chance of losing $10* (tails)

If she always says ‘Tails’ she has a *50% chance of losing $5* (again, heads)

*and a 50% chance of winning $10* (tails

I mean, under the rules you introduce, in one option she has a chance of winning more money if she’s right and losing less money if she’s wrong, but that doesn’t change the basic odds of her being right or wrong which are, as you said, 50%.

Potatamotowrote:The fair coin has 50-50 odds, overwise it wouldn’t be a fair coin. That is the objective probability as viewed from before the coin is flipped. If you increase the number of amnesia sleep cycles that probability of the coin landing one way or another doesn’t change.

The question, however, is based on her expectation of the coin having already landed on heads given that she is currently awake and has just had the experiment explained to her. In other words, she is being asked ‘Are you in the timeline where the coin landed heads up, or are you in the timeline where the coin landed tails up?’. This is not the same value because she experiences that question twice as often in the ‘tails’ timeline.

This is the same logic as used when considering the odds that we are in a timeline where everything is being simulated in a computer program. They posit

* that computers will continue to become better and better

* that some people will be able to, and want to simulate the 21st Century Earth experience.

Then conclude that there will be millions or billions more 21st Century Earth simulations than the one 21st Century reality that actually existed.

Given that we are currently experiencing 21st Century Earth it is, they conclude, highly likely that what we are actually experiencing is a simulation that we cannot distinguish from reality.

Let’s try that again – in succinct form.

There is a 50% chance that Yuna

onceexperiences the question and heads has been thrown,and a 50% chance that Yuna experiences the question

twiceand tails has been thrown.Therefore, I assert, if Yuna is currently experiencing the question it is twice as likely that tails has been thrown

in her timelinethan heads.Easy, 100%, she doesn’t remember it happening.

The correct answer is 40%.

The reasoning is simple enough, Ye never stated IF she have flipped the coin already, BUT she did state that it is either Monday or Tuesday. She also declared that she have decided to carry out the experiment she describes.

Therefore the experiment is underway, and there is ONLY 5 possibilities:

1. It is Monday and the coin is NOT flipped YET.

2. It is Monday and the coin turned up heads, and Yuna will be drugged tonight.

3. It is Monday and the coin turned up tails, and Yuna will be drugged tonight.

4. It is Tuesday and the coin turned up tails, and Yuna will be dugged again tonight.

5. It is Tuesday and the coin turned up heads, and Yuna will NOT be drugged tonight.

Number 5 is a bit tricky; but here is the explanation for that reasoning: The experiment only state the interviews as part of a: awake, interview, drug cycle, therefore there is a possibility that she have awakened on Tuesday, the experiment is EFFECTIVELY over, but not TECHNICALLY over, allowing for a little one-on-one talk, witch Yuna will remember, since she will not be drugged again.

The fact that “the coin is still lying on the kitchen table.” means nothing since it is not declared whether it has been flipped yet or not.

@ Zyrahtu:That is a novel way to look at the situation, but there are some holes in the logic you’re using. Option one doesn’t, can’t count as a separate option. Given the experiment’s parameters, Yuna will always been interviewed on Monday. Therefore, it doesn’t matter when the coin is flipped, right up through to the point that Ye Thuza needs to decide whether to do the Tuesday interview or not. She could flip the coin literally one second before the interview is/is not to take place, and it will not affect the odds.

Number 5 relies on information not provided by the experimenter. It’s merely stated that if the coin is heads, she will not be awakened and interviewed on Tuesday. Well, she’d being woken and interviewed now. So, it really can’t be Tuesday with a flip of heads. If you look at the chart, the heads branch is just Monday, with nothing happening on Tuesday at all.

Options 2-4 actually kind of illustrate something a little tricky about the question that I think are throwing a lot of people off. If the question was, what day is it, Monday or Tuesday, then you’d have to say that there was a 75% chance of it being Monday, and a 25% chance of it being Tuesday. This is because there would be two random selections, both 50%. The initial coin flip, to determine which branch of the experiment we’re on, and then the selection of the day. Which then results in what effectively counts as four options, as will always happen with two 50-50 random selection.

The tricky part is: if the initial coin flip is heads, when you make the second selection, both options are Monday, since with a flip of heads, there is no interview on Tuesday. So if that second choice were a coin flip, heads would be Monday, and Tails would also be Monday.

With a flip of tails, the second selection has a 50% chance of choosing Monday, and a 50% chance of choosing Tuesday. So, out of four options, Monday is 3/4 and Tuesday is 1/4.

A lot of people are taking that line of reasoning, which is sound, and applying it to the odds of the initial coinflip, which is not. Odds don’t change. You can play with them afterwards, but if the odds of a single coin flip are 50-50, those odds will always be 50-50, no matter what the consequences of the flip are. Yuna knows she’s being interviewed on this night, and she knows that a coin flip will determine whether this is the only interview she will have, or whether a second interview will take place/has already taken place.

A coin flip can only result in two possible outcomes. If there are more, then a second random selection has taken place, like choosing which particular day it is. But we’re not concerned with that, we’re only concerned with the first coin flip. It isn’t more likely that it’s tails because that results in two interviews, because there’s only two interviews if the flip was tails. If it’s heads, there’s only one interview. There’s not three interviews to choose from, there’s only one, OR two. Is this Yuna’s only interview? Or is there another? That is the question put to her, and the odds are 50%.

The odds that the coin came up heads are 50/50.

The question is whether the information that she is being interviewed changes her estimation of what is true.

What differentiates this from a Monte Haul problem is that she’s been asked what the odds are. She’s NOT being asked to predict what the result was.

If she were being asked to guess the result, then if she had any sort of scoring where every interview’s answer was worth the same amount, the fact that tails gets two interviews weights the value of tails higher than the value of heads. The weighting of two interviews to one, where each interview is of equal value, would lead to her maximal reward function being to always guess Tails, because her expected value is twice as high (2/3 vs 1/3) for making that guess.

If, instead, she were being graded on her answer, but her final grade would be a ratio of number of correct guesses to number of interviews, then the weighting is removed, and she again has an expected reward of 50% of the maximum possible reward.

But, again, she’s not being asked to guess the result. She’s being asked to give the odds, given that she’s being interviewed, that the coin came up heads.

She isn’t maximizing a reward function. She’s giving odds that the coin came up heads. These are two different propositions. The odds the coin came up “heads” are the same, when she has no way of knowing if this was the single interview she would get for “heads” or one of the two interviews she would get for “tails.”

The Monty Haul problem works because it is essentially saying that, if a coin is “heads” and you guess it right, you get $5, but if a coin is “tails” and you guess it right, you get $10. With no prior knowledge, you should guess “tails” because it maximizes your expected value for reward. This doesn’t actually change the odds that the coin came up heads or tails.

And it’s the latter question she’s being asked.

TL;DR: If she were being asked to make a guess, and that she would get a reward for every interview she is right, she should guess “tails,” because the expectation value is 2/3 of the maximum reward for going for “tails.” This is because there are two chances for reward on “tails,” weighting “tails” as the more valuable result.

Since she is being asked WHAT THE ODDS ARE, and guessing the coin’s sidedness is irrelevant to her reward, the odds the coin landed “heads” are 50/50. This is because there is no reward either way.

@ Paul:This is kind of a rehash of what I said in the above post, but anywho…

The odds of a particular random outcome don’t change. If the odds of a coin flip are 50-50, those odds will remain 50-50, with two possible outcomes. If any additional outcomes emerge, it’s because a second random selection was made. But we’re not interested in any additional random outcomes, we’re only interested in the initial coin flip.

You said it yourself, again.

That 50% chance is the coin flip, and it’s all we’re interested in. The selection of which particular day she’s in is another random selection, and if looked at correctly, it still ends up with a 50% chance of heads or tails. Like so:

There is a 50% chance (1/2) that the coin flip is heads. If it’s heads, there’s a 100% chance that it’s Monday (1/1) since if the flip is heads, there’s no interview on Tuesday. So, the odds that it’s Monday, with a flip of heads, is (1/2)*(1/1)= (1/2) Thus, a 50% chance that it’s Monday with a flip of heads.

There’s a 50% chance (1/2) that the coin flip is tails. If it’s tails, there is then a 50% chance (1/2) that the interview is taking place on Monday or Tuesday. So, the overall odds of it being Monday with a flip of tails is 25% (1/2)*(1/2)=(1/4). The same holds true for Tuesday with tails.

So, overall, there is a 50% chance of it being Monday, with a flip of heads. (1/2)*(1/1)= (1/2)

And the odds of it being being Monday or Tuesday with a flip of tails are both 25% respectively. (1/2)*(1/2)=(1/4) And (1/4)+(1/4)=(2/4) So there’s a 50% chance that the flip was tails as well.

So no matter how you play with the outcomes of the initial coin flip, the odds of the flip will always be 50%.

50%

The coin was tossed only once, in the night from Sunday to Monday, to determine whether or not the subject was to be woken and interrogated only on Monday or on Monday and Tuesday. Since there is only one coin toss, the probability is 50%

@ r4m0n:But what if ou don’t remember that you set up this protocol? What if you accidentally scratch your arm? What if…[Infinite series detected. Program terminated.]

All of these comments could serve as a big thought baloon over Yuna’s head. 🙂 I’m reading them and going, “And THIS is why she is going to go nuts”.

Mind you, it’s interesting to see how so many people are starting their posts with “X is the correct answer”, or “So and so is correct”. So much certainty going around, even when calling out very different results, and all of that certainty backed up by long explanations. And it’s still a sandstorm, it still feels like three blind men describing an elephant. Rather fascinating.

Personally, I think the 50% and 1/3 arguments are BOTH right, depending on how you choose to interpret the question. I also think statistics and probabilty are sufe-fire ways to bugger up very simple things in very strange ways, so I don’t much trust them anyway.

@ Peter Piers:Better to say that both can be right depending on the observer, or maybe, the parameters. ^_^

Probability and statistics are interesting. It’s a branch where there are indeed solid, concrete answers (not really math, otherwise 😛 ) and they are verifiable…if you run the numbers often enough. But that’s hard to translate for a single person. Like with Monty Hall, if I tell you that you’ve got a 2/3 chance of winning if you switch, you’re picking the MORE likely option if you do, but there’s still a 1/3 chance that you should stay, which isn’t negligible. I mean, if you played 20 times, you’d win more often than you’d lose, but you probably wouldn’t appear on a game show more than once!

Actually, the same principle that makes Monty Hall interesting is the exact same principle that makes this one interesting. Eliminating potential possibilities. not gonna go exhaustively into Monty Hall, but if you do the math, the fact that when Monty opens the door with a 100% chance of showing a goat, and a 0% chance of showing the car is what causes your odds of winning if you switch to be 2/3. If, when you picked a door with a goat behind it, he had an even chance of opening the door with the car and the door with the other goat, your odds of winning when you’d switch would be 50-50. But, that would also mean that in the third of all games, he’d show the contestant the car when he opened the door, and the game would end. That’s outside the scope of the scenario, which states that the door he opens will ALWAYS have a goat behind it. Plus, it’s just bad television to do it the other way. 😉

In this problem, nothing happens to Yuna on Tuesday night if Ye Thuza rolls heads. No interview, nothing. Tuesday night with heads rolled isn’t part of the experiment. There’s even a blank spot on the drawing up there. So when we figure the odds, we know that if heads are rolled, there’s a 100% chance that the interview is taking place on Monday, as opposed to a 50-50 split of it being Monday or Tuesday if tails comes up. I got the math for this sketched up a couple posts up, and it works out to 50% odds for Yuna.

Now, as I said a little further up, if we got an outside observer and handed him two sets of footage, one for Monday and one for Tuesday night and had him randomly watch one, that’d be different. If the footage he watched was from an interview, there’d be a 2/3 chance it came from a tail flip, and a 1/3 chance it came from a head flip. BUT, he’d also have a 25% chance of seeing footage of Yuna sleeping through the night with no interview at all. Which totally gives away how the coin flip ended up. See, it’s different for him because when he sees an interview, he knows a possibility has been eliminated. He sees an interview, and he knows he’s just eliminated 25% of the possible outcomes. But that’s a perspective Yuna can’t ever have. There aren’t any other possibilities for her, there’s just the interviews. Just like Monty opening the door and showing the car isn’t part of the experiment, Tuesday night with a flip of heads isn’t part of the experiment.

If we wanted our outside observer’s experience to match Yuna’s, we’d play it different. First, we’d roll the dice and perform the interview(s). If it was heads, we’d give the observer the footage of Monday’s sole interview to watch. If it was tails, we’d flip another coin, and give him either Monday or Tuesday’s footage based on that. 100% chance of showing him Monday with a flip of heads, 50% chance of showing him either Monday or Tuesday with a flip of tails. So he’s got a 50% chance of seeing footage of a head flip scenario, and 25% + 25%= 50% odds of seeing footage of a tail flip scenario.

I know that statistics and probability kind of feels innately squishy or spooky, but it’s still math, and math always has concrete answers. You just gotta make sure you’ve got all the ducks in a row, and it all works out.

Like I said, it comes down to what question you’re answering.

Yuna wasn’t asked whether she thinks it came up heads or tails, with a reward (even as miniscule as “being right during this interview”) for a correct guess. If she had been asked THAT question, then she has an expectation value of 2/3 the maximum reward if she chooses Tails, and an expectation value of 1/3 the maximum reward if she chooses Heads, assuming all interviews are weighted equally (rather than weighting maximum possible reward to be equal for all coin-flip results).

Yuna was asked what she thinks the probability that it came up Heads or Tails was. This probability doesn’t change based on the knowledge that she is being interviewed. She has no weight associated with each interview, because the question doesn’t reward her for a correct prediction of the result. The question rewards her for a correct estimate of the probability that the coin came up “Heads.”

Because the default assumption in these problems is that it’s a fair coin, assigning a 50/50 probability to “heads” is by definition correct. Even if it came up Tails in this experiment, she is correct for saying it is 50% likely to have come up Heads.

Essentially, Potatamoto is right about what information “I am being interviewed” gives her: none. The Monty Haul problem works because of expectation values of reward. If this were formed with a reward for correct guessing that weighted all interviews equally, it would mean she should guess “Tails” is what came up to maximize her expected reward. But that’s not what’s going on.

Segevwrote:Why

assumethe interviews are weighted equally when we know that isn’t the case? We know that the interview ‘heads-monday’ is 50%, and that weights of ‘tails-monday’ plus ‘tails-monday’ add to 50%. Yuma is asked credence of the coin being heads. The number of interviews she receives if the coin is tails doesn’t effect the probability of heads.Adding a reward introduces an interesting twist. If the reward is the intangible satisfaction of being correct, the reward is taken away by the amnesia potion. Yuma can only remember being right one time, regardless of how many interviews.

If Yuma was asked to ‘guess’ heads or tails and at the end of the experiment she would receive 2 cookies for every correct guess, she can change the expected return even though the odds remain the same. If Yuma has no way of knowing which interview it is, her logic and guess would be the same each time, so. If she guesses heads, the expected value is 1 cookie: she receives 0 cookies per interview if tails (50% chance) and 2 cookies if heads (50% chance).

If she guesses tails the expected value is 2 cookies: she receives 2 cookies per interview if tails (50% chance of 4 cookies) and 0 cookies if heads (50% chance).

Agreed

The answer is: “Go to the kitchen and look at the coin.”

[quote]it would mean she should guess “Tails” is what came up to maximize her expected reward.[/quote]

Her ‘expected reward’ is being right. You know, like in pointless internet arguments. Saying ‘Tails’ gives the maximum expectation of being right because if she says Tails and is right then she is right twice as much as if she says Heads and is right. while if she says Tails and is wrong she is not wrong as much as if she says Heads and is wrong.

Because she is being asked about her belief in the coin’s state, ‘being right’ is what it’s all about.

Paulwrote:This is absolutely true. Adding or worrying about a ‘reward’ of any kind has no bearing whatsoever on the state of the coin. A reward or prize isn’t part of the experiment, and even if it was, it wouldn’t change the odds in the slightest. The question of any kind of reward is a different question entirely.

But, Paul, your line of reasoning runs something like this, I imagine…there are three potential interviews. Heads/Monday, Tails/Monday, Tails/Tuesday. All three are equally likely to occur, and two happen if the coin flip is tails. Therefore, it’s twice as likely that a given interview is happening with the coin being on tails.

There is a serious flaw in that reasoning. Namely…three interviews are not taking place. There’s no scenario in which there are three interviews to pick from. You either have to pick from one interview OR two interviews. This is a very important distinction, because it means that all the interviews have a chance of occurring, but NOT an equal chance.

We both have agreed that we’re dealing with a fair coin, with a 50% chance of landing on heads or tails respectively. So then, our options look like this:

50%- Heads: Heads/Monday

50%- Tails: Tails/Monday, Tails/Tuesday

So if the coin lands on heads, there’s a 100% chance of Heads/Monday, simply because there’s no alternative. By contrast, if the coin lands on tails, there’s a 50% chance each of Tails/Monday and Tails/Tuesday. Probability is multiplicative. So, 50% times 100% is 50%. (1/2)*(1/1)=(1/2) And 50% times 50% is 25%. (1/2)*(1/2)= (1/4).

So yes, if the coin lands on tails, there are twice as many interviews as if the coin lands on heads. But, looked at collectively, each one of those individual interviews are half as likely to occur as the single interview that happens if the coin lands on heads. Taken together, they equal the odds of that one heads flip interview. And since Yuna has no means of determining whether one interview or two has taken place, the odds for the coin flip from her perspective must always be 50%.

This whole experiment is very easy to model in real life. Just flip a coin (or use another random generator of your choice) for every head, give yourself an A. For every tail, flip another coin. Give yourself a B for heads and a C for tails. Do it, I dunno, 20 or 30 times. Then add up your Bs and Cs, and compare them to your number of As.

Potatamotowrote:There is a problem with this view (that often returns (in slight variations) in your explanations). You assume that the probability of heads/tails the fair 50/50 is. But you also (quite covertly) assume that Yuna is being interviewed.

Then you conclude that the probability must be 50/50. (which you actually already assumed; so you’re circlereasoning).

I won’t be repeating what I allready posted, but be sure to take a look at it.

Rich314wrote:No it is not. An interview on heads-monday won’t necessarily be 50%. (the statement does have its charms, but that doesn’t garantee anything 😉 ).

Try stating your successfull cases and all cases for the percentages your noting, that makes things easier.

Remember, the whole point of pointing out that the reward version requires that Yuma be asked whether it came up heads or tails. Because that is not the case, and she is not being asked what she should predict, the whole scenario that could make a 2/3 tails argument have any weight goes out the window.

The question she is asked is what the probability is that the coin landed “heads.” This probability is 50%. She has no information gained from knowing she is being interviewed. She only knows she could be in any of three interviews, and that those are divided into the “heads” set and the “tails” set. Each set has a 50% chance of occurring, and she has no means of telling which set she is in.

With no information gained from knowledge that she is being interviewed, the odds that it came up heads, based on her knowledge, are 50/50.

I think I figured it out. the answer is 100%

Given – Yuna is put to sleep after the interview

Given – Yuna has not been put to sleep

Given – The day is monday OR tuesday

If – The coin was tails; Then – Yuna is put to sleep on Monday AND Tuesday

Since – Yuna is NOT put to sleep AND it is Monday OR Tuesday

Therefore – The coin was NOT tails

If – The coin was heads; Then – Yuna is put to sleep on Monday ONLY

Since – Yuna is NOT put to sleep AND it is Monday OR Tuesday

Therefore – The coin was heads

@ Yannick:I’m not covertly assuming she’s being interviewed…she is, in fact, being interviewed. The key difference between our arguments is that you are giving probabilistic weight to a scenario where she’s not being interviewed. I go over that in the post here:

@ Potatamoto:No, she’s not sleeping through the night during this interview. But is that an option for her? I mean, of course, if the coin flip is heads, she will sleep through the night on Tuesday. But what kind of option is that, from Yuna’s perspective? She’ll be asleep, unconscious, with no knowledge of the experiment, or anything else. She won’t know she’s not being interviewed. Furthermore, when she is being interviewed (because she MUST be interviewed at least once,) nothing changes for her whether there’s a second interview or not.

I used this analogy once before, but I still think it’s the best fit. Imagine you run an experiment where you flip a coin. If you flip heads, you mail a letter detailing the experiment to a random person. If you flip tails, you mail the letter to two random people. No recipient of the letter will be able to tell if a second letter was sent or not, so from the perspective of a person who received a letter, what are the odds the coin was heads or tails. It’s pretty obviously 50%.

As noted online in various places, a coin flip does

nothave 50:50 odds.I like how Yuna doesn’t even question whether her mom would actually experiment on her own daughter like this.

Ok guys, you really get this game wrong.

-From the statistical point of view the only thing that matter is the coin toss, that has a 50/50 chance of being either head or tails. All the other elements are mialeading data that have no influence on the outcome of the coin toss.

-If you want to discuss about the phrasing of the question, my interpretation is that you should say what you believe is the result of the coin toss. Again, the toss has already happened, and you have no knowledge about its result, so you can choose as you wish and have always a 50/50 chance of getting the right answer

Of course, the real point here is that ALL the information is misleading, and the question is deliberatly ambiguous, so you are trapped in an unsolvable enigma.

The devilish experiment, “kind of a mind game”, is not the one described, but the act itself of posing the enigma.

The only way to win is not playing.

@ Beregorn:Precisely. Boiled down, this is what I have said. Since Yuna is guaranteed to experience at least one interview, and her experience will be identical whether she has one or two, the fact that she’s being interviewed gives no information that will make her expectations of the coin flip anything but 50-50.

But if you flip a coin three times…

I’m getting Zero Time Dilemma vibes here

@Beregorn That is not really the case. As I understand it, it is more of conditional probability problem. What is the chance of the coin landing heads assuming you know I am asking you this question.

In the end, it really depends on how you want to count success. If you want to be right as many time as possible in infinite attempts (at limit), then the correct answer would be 1/3. If you only care about being right once, then it is 1/2. IMO.

My opinion: 0%.

My reasoning: it’s Monday. There never was any experiment. It was just a mind game. Dun dun dun!

Really, the actual probability that the coin landed on is less than 50%. In no place does her mother dictate the point in time that coin is flipped in the procedure. The coin must only be flipped before the morning of Tuesday, when Yuna would be awoken on a tails flip. My initial gut feeling is something of about 25% chance that a coin flip occurred and resulted in heads

Now a better question. What is the probability that the coin has landed on

Tails.There’s five fundamental answers. Yes. No. I don’t know. I don’t care. Wrong question.

The correct answer to the question as formulated is “I don’t care.” Let’s fix that.

Say instead of heads vs. tails, the mad scientist has stolen all the milk you need for your morning tea. Dastardly. Or they haven’t. You could get milk before work, but you will be late.

Let’s start with the case where you will be hit with the amnesia potion if they haven’t stolen your milk. Clearly, the correct choice is not to get any. If you choose to get milk, you might be late twice, in a row, for no reason. If you don’t choose to get milk, you will miss out on at most one lovely tea session.

With the other case, clearly the correct choice is to go. If you’re only awakened once and unnecessarily get more milk, probably your boss will let it go. If you’re late twice, you can explain that some fiendish criminal stole your milk, so it was for a good cause. And look: no job is worth giving up tea twice.

I also like Nick Bostrom’s limit-taking. What if, instead of being awakened twice on tails, you’re awakened a million times. I’ll add this: what if you gain a dollar every time you guess correctly, and lose a dollar every time you guess wrong. What is the best guess, strategically?

Or, imagine you gain a dollar every time you correctly guess the day of the week. Should you guess Monday, Tuesday, or does it not matter?

Interesting wrinkle: if you guess heads half the time and tails half the time, you will be right half the time against a normal coin. If the sleeping Yuna experiment is repeated enough times, you will also be right half the time, whether we use Bostrom’s modification or not.

However, if you guess tails, unlike a normal coin, you will be right more than half the time.