Since I now know that I will soon rationally infer that this note was written during an H-awakening with probability 1/3 (on the basis of no new information), I can already infer this right now. — Pierre-Normand

Then before the experiment starts the thirder will say "since I now know that I will soon rationally infer that the coin will have landed heads with probability 1/3 (on the basis of no new information), I can already infer this right now, before the coin is tossed."

But I think this is wrong.

Making n large makes Sleeping Beauty's epistemic situation on Wednesday, when she receives a note, nearly identical to her situation when she wrote the note, since the Bayesian updating she can perform on the basis of the note being unique is negligible. — Pierre-Normand

They're not nearly identical. On Wednesday she knows that she only had the opportunity once. When she wrote the note she didn't know that it was her only opportunity. So contrary to the above, there is new information on Wednesday.

Note that when Sleeping Beauty doesn't receive a note on Wednesday, her credence P(H) = 1/2 doesn't merely differ in value from her credence P(H) = 1/3 during awakenings; the predicates P() also have different meanings. During awakenings, P(H) refers to the odds that her current awakening episode is occurring during a coin toss that landed heads. On Wednesday, P(H) refers to the odds that the experimental run she is exiting from was an H-run. While in each case the biconditionals "I am now in an H-awakening iff I am now (and will be) in an H-run" or (on Wednesday) "I was in an H-awakening iff I am now in an H-run" hold, the probabilities don't necessarily match due to the two-to-one mapping between T-awakenings and T-runs. — Pierre-Normand

I address this here.

the prior probability for any event is based set of all possibilities that could occur — JeffJo

And her being woken a second time if the coin lands heads can't occur, which is why its prior probability is 0, not $1\over4$.

And that is important here, because you are insisting that a two-day collection of events (I'll call your two passes Monday and Tuesday since only the order matters to anything). You are calling Monday+Tails and Tuesday+Tails the same event. But to SB, who can only observe one at a time, they are distinct events that each have half the prior probability that you assign to the combination. — JeffJo

There aren't two days in my example.

When I will read again the note that I am currently writing, on Wednesday, I will be able to rationally infer that it is twice as likely that this note was written by me on the occasion of a T-awakening. — Pierre-Normand

That depends on the probability that you will be given the opportunity to write a note. If that probability is 1/2 then it won't be rational on Wednesday to infer that it is twice as likely that this note was written by me on the occasion of a T-awakening.

That was only in the specific case where n = 2. As n grows larger, P(H) tends towards 1/3. — Pierre-Normand

And as it grows smaller, P(H) tends to 1. I don't understand the relevance of any of these three answers.

Why is the correct answer given by any of these situations, let alone by the situation where n is arbitrarily large?

Yes, I got the maths wrong there (and deleted my post before you replied; apologies).

Though I don't see why I should accept your claim that if "she receives a single note on Wednesday, Sleeping Beauty comes to be causally and epistemically related to the coin result in the exact same manner as she was when she originally wrote the note."

As you said yourself, if the probability of her writing a note is 1/2 then if she finds exactly one note on Wednesday then her credence in Heads is 1/2.

As the occasions to write a note become rarer (e.g. 1/n with n >> 1), the frequency of those overlapping notes become negligible (n times as many single notes are received as double notes) and Sleeping Beauty's epistemic state (i.e. the value of her credence) approaches asymptotically her epistemic state as she was writing the note. And, as I had suggested in my previous post, this is because when she receives a single note on Wednesday, Sleeping Beauty comes to be causally and epistemically related to the coin result in the exact same manner as she was when she originally wrote the note. — Pierre-Normand

If heads and n = 100 then the probability of writing a note is 1/100

If tails and n = 100 then the probability of writing exactly one note is 1/100.

So if she finds exactly one note on Wednesday then her credence in heads is 1/2.

@Pierre-Normand

These cannot all be true:

1. Credence "is a statistical term that expresses how much a person believes that a proposition is true"
2. My current interview is a heads interview iff I have been assigned one heads interview
3. The fraction of interviews which are heads interviews is $1\over3$
4. The fraction of experiments which have one heads interview is $1\over2$
5. My credence that my current interview is a heads interview is equal to the fraction of interviews which are heads interviews
6. My credence that I have been assigned one heads interview is equal to the fraction of experiments which have one heads interview

Propositions like 5 and 6 might usually be true, but they are not true by definition.

Given 1 and 2, my credence that my current interview is a heads interview is equal to my credence that I have been assigned one heads interview.

Therefore given 1, 2, 3, and 4, one or both of 5 and 6 is false.

So how will your reasoning let you choose between 5 and 6 without begging the question?

I accept 6 and reject 5. My credence that my current interview is a heads interview isn't equal to the fraction of interviews which are heads interviews.

My argument is:

P1. If I am certain that A is true if and only if B is true then the degree to which I believe that A is true is equal to the degree to which I believe that B is true
P2. I am certain that my current interview is my only interview if and only if I have been assigned only one interview
C1. Therefore the degree to which I believe that my current interview is my only interview is equal to the degree to which I believe that I have been assigned only one interview (from P1 and P2)
P3. I am certain that if I have been assigned at random by a fair coin toss either one or two interviews then the probability that I have been assigned only one interview is $1\over2$
P4. I am certain that I have been assigned at random by a fair coin toss either one or two interviews
C2. Therefore I am certain that the probability that I have been assigned only one interview is $1\over2$ (from P3 and P4)
P5. The degree to which I believe that I have been assigned only one interview is equal to what I am certain is the probability that I have been assigned only one interview
C3. Therefore the degree to which I believe that I have been assigned only one interview is $1\over2$ (from C2 and P5)
C4. Therefore the degree to which I believe that my current interview is my only interview is $1\over2$ (from C1 and C3)
P6. I am certain that my current interview is my only interview if and only if the coin landed heads
C5. Therefore the degree to which I believe that the coin landed heads is $1\over2$ (from P1, C4, and P6)

Suppose we update the protocol so that on rare occasions, which present themselves with equal probability on each awakening episode, Sleeping Beauty is able to write down a note saying "I have now been awakened and interviewed." She can retain this note and read it again on Wednesday. Upon rereading the note on Wednesday, she can reason that it is twice as likely that such a note was produced if the coin landed tails since she would have been twice as likely to write it during such an experimental run. — Pierre-Normand

Not necessarily.

Assume a probability of 1/2 each time. The probability of writing it if the coin landed heads is 1/2. The probability of writing it (at least once) if the coin landed tails is 3/4. It is 3/2 times as likely to have been tails.

Assume a probability of 2/3 each time. The probability of writing it if the coin landed heads is 2/3. The probability of writing it (at least once) if the coin landed tails is 8/9. It is 4/3 times as likely to have been tails.

But notice that as the probability of writing a note each time approaches 1 the "greater likelihood" of it having been tails gets smaller, approaching 1.

Also, apply the principle of this reasoning to this.

A fair coin toss is equally likely to be heads as tails, she is less likely (not guaranteed) to wake if tails, therefore if she does wake then she reasons that it's less likely to be tails. In waking she rules out TTT.

If the answer to this problem is 4/7 then the answer to the normal problem is 1/2. If the answer to the normal problem is 1/3 then the answer to this problem is 1/2.

You are asking if it has occurred when you know it hasn't — JeffJo

Sleeping Beauty doesn't know that it hasn't occurred. She has amnesia.

"Prior" refers to before information revealed — JeffJo

So after waking, and before new information is revealed, the prior probability that the coin landed heads and that she is being woken for a second time is 1/4? Then what new information is revealed after waking that allows her to rule out that prior probability? It can’t be “being awake” because that isn’t new information. And it can’t be “being asked her credence” because if she has just woken then she knows with certainty that she is about to be asked her credence.

The simplest answer is the correct one. The prior probability that the coin will or did land heads and that she is being or will be woken for a second time is and always was 0. That’s just a rule of the experiment.

Nothing is ruled out when woken or asked her credence that wasn’t already ruled out before the experiment started.

Even Elga understood this:

Before being put to sleep, your credence in H was 1/2. I’ve just argued that when you are awakened on Monday, that credence ought to change to 1/3. This belief change is unusual. It is not the result of your receiving new information

The prior probabilities, for an awakened SB, are 1/4 for each. — JeffJo

It makes no sense to say that when she wakes there is then a prior probability that she’s “asleep” of $1\over4$ that is immediately ruled out. If some prior probability is ruled out when she wakes then it must be that the prior probability is established before she wakes. But the prior probability that she will wake a second time if the coin lands heads is 0, not $1\over4$.

If she’s awake then it is just the case that either the coin hasn’t been tossed or it landed tails.

I can set out an even simpler version of the experiment with this in mind:

1. Sleeping Beauty is given amnesia
2. She is asked her credence that a coin has been tossed
3. A coin is tossed
4. If the coin lands tails then:
4A. She is given amnesia
4B. She is asked her credence that a coin has been tossed

The prior probability that step 2 will happen is 1 and the prior probability that step 4B will happen is $1\over2$.

When Sleeping Beauty is given amnesia she knows that she is in either step 1 or step 4A. When asked her credence she knows that she is in either step 2 or step 4B.

No prior probability is ruled out when she is given amnesia or asked her credence.

You have to say that in step 2 her credence is $1\over3$. I have to say that in step 2 her credence is $1\over4$. Of note is that neither of us can just apply the principle of indifference and say that in step 2 her credence is $1\over2$.

Thirder reasoning is that because step 2 is twice as likely to occur as step 4B then I am twice as likely to be in step 2 as step 4B.

Halfer reasoning is that because step 2 is twice as likely to occur as step 4B and that because if 4B will occur then I am equally likely to be in step 2 as step 4B then I am three times as likely to be in step 2 as step 4B.

There is no theory of when prior probabilities are established. But if there were, it would be fom the start, not before the start. — JeffJo

So when is this alleged P(X) = 1/4 prior established if not before the experiment starts?

It cannot be when she is asked her credence as you’ve said that in being asked her credence this prior is reduced to 0.

But it also cannot be before because there is no “current interview” before she’s asked her credence.

So this alleged prior just makes no sense.

But this is the entire controversy behind the Sleeping Beauty Problem. One that I have shown can be trivially removed. And that is why you ignore it. — JeffJo

I'm not ignoring it. I'm showing you that your version with the P(HH) = 1/4 prior that is ruled out when asked is not the same as the normal problem because the normal problem doesn't have an equivalent P(X) = 1/4 prior that is ruled out when asked.

And the prior probability that the current waking, is a step-1 waking, is 1/2. — JeffJo

Prior probabilities are established before the experiment starts, so there is no “current waking” prior because there is no “current waking” before the experiment starts.

We’re talking about prior probabilities, i.e the probabilities as established before the experiment starts.

The prior probability that step 1 will happen is 1.
The prior probability that step 2 will happen is 1.
The prior probability that step 3 will happen is 1/2.
The prior probability that step 4 will happen is 1/2.

There is no prior probability equal to 1/4.

When she is asked her credence she cannot rule out any of these prior probabilities, and she is being asked her credence that step 3 occurs.

Your question "what is your credence the coin will/did land on Heads" is asking SB to distinguish between the cases where your coin will/did land on Heads, and will/did land on Tails. So cases A and B, which depend on the same distinction, must be distinct outcomes to SB. — JeffJo

She’s being asked her credence that step 3 happens.

Step 1 just isn’t two events with a prior probability of 1/4 each. It’s one event with a prior probability of 1. Step 2 has a prior probability of 1 and steps 3 and 4 each have a prior probability of 1/2.

And the reason for the shopping example is pointing out that the four parts that I highlighted and labeled A, B, C, and D each have a prior of 1/4. — JeffJo

No they don’t. Your “A or B” isn’t two separate things but one thing with prior probability 1. C and D each have a prior probability of 1/2; C will happen if the coin lands heads and D will happen if the coin lands tails, and the prior probability that a coin will land heads is 1/2.

And why do you count being sent home in step 3 as part of the probability space but not being sent home in step 4 as part of the probability space”? You’re being inconsistent.

That 1/4 chance that she would have been taken shopping. — JeffJo

I’m not asking about your shopping example. I’m asking about this example:

1. Sleeping Beauty is given amnesia and asked her credence that the coin will or did land heads
2. The coin is tossed
3. If the coin lands heads then she is sent home
4. If the coin lands tails then she is given amnesia, asked her credence that the coin will or did land heads, and sent home

there is a "prior P(X) = 1/4 that becomes P(X) = 0 when she’s asked her credence" in your experiment. — JeffJo

Which is what? What prior P(X) = 1/4 becomes P(X) = 0 when she’s asked her credence?

In your experiment the prior probability P(HH) = 1/4 becomes P(HH) = 0 when she’s asked her credence. But there is no prior P(X) = 1/4 that becomes P(X) = 0 when she’s asked her credence in my simplified form of the experiment.

Hence your experiment is not equivalent and your solution doesn’t apply.

My claim is simply regarding what the word means.

Credence "or degree of belief is a statistical term that expresses how much a person believes that a proposition is true."

Therefore, if I believe that A iff B and if my credence in A is 1/2 then my credence in B is 1/2.

What it means for one's credence to be 1/2 rather than 1/3 is a secondary matter. My only point here is that it cannot be that each of these is true:

1. I believe that A iff B
2. My credence in A is 1/2
3. My credence in B is 1/3

Either both my credence in A and my credence in B is 1/2 or both my credence in A and my credence in B is 1/3.

So now we have an issue:

1. I believe that H-run iff H-interview
2. My credence in H-run is equal to my credence in H-interview
3. 1/2 of all runs are H-runs
4. 1/3 of all interviews are H-interviews

Given (2) it cannot be that my credence in H-run is equal to the fraction of all runs that are H-runs and that my credence in H-interview is equal to the fraction of all interviews that are H-interviews.

It may be that one's credence is a reflection of the ratio in some reference class, but given the above, (3) and/or (4) are the wrong reference class to use.

So we go back to my previous argument:

P1. If I have been assigned at random either one heads interview or two tails interviews then the probability that I have been assigned one heads interview is $1\over2$
P2. I have been assigned at random either one heads interview or two tails interviews
C1. Therefore my credence that I have been assigned one heads interview is $1\over2$
P3. My current interview is a heads interview iff I have been assigned one heads interview
C2. Therefore my credence that my current interview is a heads interview is $1\over2$

P1, P2, and P3 are true, C1 follows from P1 and P2, and C2 follows from C1 and P3. Therefore C2 is true.

My credence that my current interview is a heads interview is equal to the fraction of runs assigned one heads interview, not the fraction of interviews which are heads interviews.

Consider an alternative experiment setup where Sleeping Beauty awakens less often rather than more often when the coin lands tails. For instance, we could eliminate Tuesday awakenings altogether and ensure that Sleeping Beauty awakens once on Monday when the coin lands heads, and only half the time when it lands tails (by tossing a second coin, say). — Pierre-Normand

I consider a similar situation here.

Halfer reasoning gives an answer of 4/7 and thirder 1/2. I think 4/7 is more reasonable.

I’m less likely to wake if tails, heads and tails are equally likely, and so if I do wake my credence is that it’s less likely to be tails.

Even though in the long run the number of heads- and tails-awakenings are equal.

Pr(Heads) = Pr(Heads&First Time) + Pr(Heads&Second Time) — JeffJo

The prior P(Heads & Second Time) = 0 as established by the rules of the experiment. She will never be asked a second time if the coin lands heads. So there's nothing for her to later rule out when she's asked her credence.

When she is asked the second time, the "prior probability" of heads is ruled out. — JeffJo

No it's not. She doesn't know that she's being asked a second time. She can't rule out heads.

This does not implement the original problem. She is wakened, and asked, zero tomes or one time. — JeffJo

She’s asked once in step 1 and then, optionally, again in step 4.

No prior probability is ruled out when asked.

You know 50% is a ratio, right? — Srap Tasmaner

Pierre-Normand is saying that P(X) refers to the ratio of Xs to non-Xs in some given reference class.

I'm saying that P(X) refers to the degree to which I believe X to be true.

If P(X) refers to the degree to which I believe X to be true, and if I believe that A iff B, then P(A) = P(B).

The conclusion doesn't follow because, while the biconditional expressed in P3 is true, this biconditional does not guarantee a one-to-one correspondence between the set of T-interviews and the set of T-runs (or "T-interview sets"). Instead, the correspondence is two-to-one, as each T-run includes two T-interviews. — Pierre-Normand

There is a one-to-one correspondence between the set of H-interviews and the set of H-runs.

H-interview iff H-run
P(H-run) = 1/2
Therefore, P(H-interview) = 1/2

Credences … can be thought of as ratios — Pierre-Normand

One’s credence is the degree to which one believes a thing to be true. Often one’s credence is determined by the ratios but this is not necessary, as shown in this case.

If one believes that A iff B and if one is 50% sure that A is true then one is 50% sure that B is true. That just has to follow.

So:

P1. If I have been assigned at random either one heads interview or two tails interviews then the probability that I have been assigned one heads interview is $1\over2$
P2. I have been assigned at random either one heads interview or two tails interviews
C1. Therefore my credence that I have been assigned one heads interview is $1\over2$
P3. My current interview is a heads interview iff I have been assigned one heads interview
C2. Therefore my credence that my current interview is a heads interview is $1\over2$

You would have to argue:

P4. If I have been assigned at random a heads interview, a first tails interview, or a second tails interview then the probability that I have been assigned a heads interview is $1\over3$
P5. I have been assigned at random a heads interview, a first tails interview, or a second tails interview
C3. Therefore my credence that my current interview is a heads interview is $1\over3$

But P5 is false.

I think you have chosen the wrong reference class. Given the manner in which the experiment is conducted, as shown by this Venn diagram, the reference class of all interviews is of no use to Sleeping Beauty. With respect to individual interviews the only relevant reference classes are "heads", "tails", "first", and "second".

Turn coin C2 over, to show its opposite side — JeffJo

OK, I understand your argument now, and it's what I said to you before: in your experiment the prior probability P(HH) = 1/4 is ruled out when woken; in the normal experiment where I am woken once if heads and twice if tails there is no prior probability P(X) = 1/4 that is ruled out when woken.

To make this fact clearer the experiment will be conducted in the simplest possible form:

1. Sleeping Beauty is given amnesia and asked her credence that the coin will or did land heads
2. The coin is tossed
3. If the coin lands heads then she is sent home
4. If the coin lands tails then she is given amnesia, asked her credence that the coin will or did land heads, and sent home

It said race could not be considered a reason to permit or deny admission into college under the Constitution. — Hanover

Unless it's a military academy, in which case they can.

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What I said in that post you dissected applies to one pass only, and the intent was to have two passes where, if there was no question in the first, there would be in the second. — JeffJo

So if the coin combination is HH then the participant will be asked their credence during the second pass? If so then you are wrong when you said "she also knows that the fact that she is awake eliminates (H,H) as a possibility."

And in the original, on Tuesday after Heads, you are also not asked for a credence. — JeffJo

What matters is the probability that you will be asked for your credence at least once during the experiment.

In the normal problem you are certain to be asked at least once during the experiment.

In your problem you are not certain to be asked at least once during the experiment.

Your problem isn’t equivalent and so the answer to your problem is irrelevant.

The subject in my implementation is always asked. — JeffJo

No they're not. I'll quote you:

1. Two coins will be arranged randomly out of your sight. By this I mean that the faces showing on (C1,C2) are equally likely to be any of these four combinations: HH, HT, TH, and TT.

2. Once the combination is set, A light will be turned on.

3. At the same time, a computer will examine the coins to determine if both are showing Heads. If so, it releases a sleep gas into the room that will render you unconscious within 10 seconds, wiping your memory of the past hour. Your sleeping body will be moved to a recovery room where you will be wakened and given further details as explained below.

4. But if either coin is showing tails, a lab assistant will come into the room and ask you a probability question. After answering it, the same gas will be released, your sleeping body will be moved the same way, and you will be given the same "further details."

The lab assistant only asks for my credence if the coin combination isn't HH.

If you take away this condition and so I am always asked my credence then the answer is 1/2.

The passer-by sees all of the flashes and does not know the genetic status of the fireflies producing them. This is analogous to Sleeping Beauty experiencing all of her awakenings but not knowing if they're unique (generated by a coin having landed heads) or one of a series of two (generated by a coin having landed tails). — Pierre-Normand

The passer-by can see a mix of single and double-flashing fireflies. Sleeping Beauty can't. She either sees one firefly flash once or she sees one firefly flash twice (forgetting if she's seen the first flash).

All you've imagined here is that we collect several iterations of the Sleeping Beauty experiment, mix up all the interviews, and then have Sleeping Beauty "revisit" them at random. That's not at all equivalent. This is equivalent to a sitter being assigned a room and I already accepted that for the sitter the answer is 1/3.

In your scenario there are a bunch of flashes going off in a forest and me, a passer-by, randomly sees one of them. This is comparable to a sitter being assigned a room.

But this isn't how things work for Sleeping Beauty. It's not the case that her interview is randomly selected from the set of all interviews. It's the case that her interview set is randomly selected from the set of all interview sets. That's just how the experiment works and so is how she should reason.

If we apply your reasoning to the example here then we conclude that P(Heads|Awake) = 1/2, which I think is wrong.

I'm less likely to wake if tails and so if I do wake it's less likely to be tails, and so P(Heads|Awake) > P(Tails|Awake).

This is satisfied only if we use halfer reasoning and conclude that P(Heads|Awake) = 4/7.

In your example being asked your credence isn't certain. In Sleeping Beauty's it is. That's why your example isn't equivalent.

Pierre-Normand also tried to explain this to you here.

I find it unusual that you maintain that when faced with a potential outcome O in a situation S, your credence P(O) should only reflect the intrinsic propensity of an object to generate O, disregarding how O affects the likelihood of you being in this situation. — Pierre-Normand

That's not what I said.

In the Sleeping Beauty problem I am guaranteed to wake up at least once if tails and guaranteed to wake up at least once if heads. The coin toss does not determine the likelihood of me waking up. It only determines the number of times I'm woken up. But this frequency is irrelevant given the guarantee.

Only if there is no guarantee is the frequency relevant. Again, see this and this.

What matters is that in the Sleeping Beauty problem the prior probability of being asked one's credence at least once is 1 and the prior probability of being asked one's credence at least once if heads is 1, and in your problem the prior probability of being asked one's credence at least once is 3/4 and the prior probability of being asked one's credence at least once if the first coin is heads is 1/2.

Because of this your problem is not equivalent. The answer to your problem is 1/3 only because of those prior probabilities. The Sleeping Beauty problem has different prior probabilities and so a different answer of 1/2.

This ratio is also the relevant one for her to predict from which wing she would likely exit from if she had a chance to escape during any given awakening episode.

If on each day the chance to escape is 1/2 then the prior probability of being given at least one chance to escape if the dice rolls 1-5 is 1/2 and if the dice rolls 6 is 63/64. Given the prior probability that I'm more likely to be given at least once chance to escape if the dice rolls 6 it is reasonable to infer that if I am given a chance to escape that the dice most likely rolled 6.

This is the same reasoning used here to infer that one's credence in heads should be greater in experiment 1 than in experiment 2 and used here to infer that one's credence should favour heads.

However, what does not logically follow is that P'(not-'six') = 5/6, if we interpret this to mean that in five out of six potential awakening episodes, she finds herself in not-'six' episodes. The relevant ratio in this context is P'(not-'six') = 6/11. — Pierre-Normand

This has nothing to do with credence.

I am asked to place two bets on a single future coin toss. If the coin lands heads then only the first bet is counted. What is it rational to do? Obviously to bet on tails, even though my credence isn't that tails is more likely. The same principle holds in the Sleeping Beauty experiment where I'm put to sleep and woken up either once or twice depending on a coin toss.

In both cases the fact that there are twice as many T-wins as H-wins has nothing to do with the likelihood of the coin having landed tails. There are twice as many T-wins as H-wins because the experimental setup says that if it's heads then you get one bet and if it's tails then you get two bets, and heads and tails are equally likely.

That in many cases there are twice as many A-outcomes as B-outcomes because A is twice as likely isn't that in every case if there are twice as many A-outcomes as B-outcomes then A is twice as likely. There are plenty of other factors that can contribute to an outcome ratio that doesn't reflect the likelihood of each outcome. The Sleeping Beauty problem is one such example.

I have indeed conceded that the inference is valid (as are the applications of Bayes' theorem predicated on it) as long as we avoid equivocating the meaning of P(). — Pierre-Normand

I think there's only one meaning of P(), and it is such that P(HInterview) = P(HRun) = 1/2.

It is a mistake to reason that P(X) should reflect the ratio of X to not-X, as explained above.

In #2, A knows she will be wakened and that the coin is irrelevant. So Pr(Heads|Awake)=Pr(Heads)=1/2. B knows that she will only be wakened if Heads. So Pr(Heads/Awake)=1. — JeffJo

Neither participant knows if they are A or B.

In my version, there is one subject who knows she will be wakened, just not how many times. — JeffJo

She doesn't know that. If both coins land heads then she's not asked her credence.