can you elaborate?

say she bets 1 unit at even money on heads with prob p, tails with prob q, and abstains with prob 1-p-q.

then her ev in SB is (q-p)/2 units.

but say a coin is 1/3 heads. your ev is (1/3)(p-q) + (2/3)(q-p) = (q-p)/3 units.

I just woke up thinking that the reason P(Mon) needs to be 1 to make that total probability equation give 1/2 is because you are rejecting the "backwards frequentist logic" of making P(Mon) = 2/3. That is, you are essentially ignoring the fact that she could wake up on either Mon or Tues and simply looking at the fact that she will always wake up, and in fact she will always wake up on Monday. Does that make any sense to you, or should I go back to bed and wake up again?

Also, there is a difference between her not being able to form a credence for the day as PairTheBoard and cyberfish have suggested, and having P(Mon) being a nonsense undefinable quantity. If it is simply arbitrary but definable, then

P(heads) = P(heads|Mon)*P(Mon) + P(heads|Tues)*P(Tues) = 1/2

would demand that P(Mon) = 1 unless we make P(heads|Mon) equal to something other than 1/2 which I think violates the indifference principle as well as common sense because then P(tails|Mon) would be something different. If P(Mon) were nonsense, then we could just reject this equation as nonsense, but I can't imagine Jason talking about P(Mon or Tues) and P(heads|tails) if those were nonsense. But we can't say that she can't form a credence for the day, and then accept that equation which makes the credence for Monday 1. That's a contradiction.

Originally Posted by PairTheBoard
"In that case, using jason1990's notation the information of "an awakening if and only if tails" would be part of the experiment's settup information J. So the probability space going in would include P(t|J and an awakening) = 1."



Yea, my answer bothered me shortly after I posted it because it was not consistent with my explanation to Sklansky when he raised the same example - no awakening on heads. In reply to him I pointed out that IF she has an awakening then she clearly does learn something new which she didn't know going into the experiment. She didn't know beforehand if there would be at least one awakening.

You might want to take this up with jason1990. I've pasted his Theorem and following comments from post #374 below. I'll give you my thoughts and you can ask jason1990 if you're not satisfied.

Notice that in the Theorem there is no mention of ~E being something possible for Beauty to learn. Only that P(~E)>0. In the first bolded comment he explains, If "we" learn E it must have been possible for "us" to have learned not E. So in this case, if we are looking at what's happening and don't know the outcome of the coin but we see Beauty is not being awakened, "we" have learned ~E and can conclude P(t|(J and ~E)) = 0. So "we" can learn both E and ~E.

But in jason1990's second bolded comment he says the same thing about Beauty learning E. But he is working in the Original where P(There is at least 1 awakening|J) = 1 so that Beauty is always awake at some point in the experiment and available to learn her situation. So he can allow her to reason from the same perspective as "us". But in this "no awakening on heads" case, it's possible for her to not be awake and available to learn what we can learn, namely ~E. You can ask jason1990 what he says about this, but I think in this case, if Beauty learns E then it is sufficient for her to know that P(~E|J)>0 and that it would be possible for an outside observer to learn ~E even though, due to her unique perspective she would not be awake and available to learn ~E if ~E had indeed been the case.

I don't think this nullifies my other comments though. I suggest if you want to pursue the line of incorporating the Original Experiment within a New Experiment you set up the mathematics for it with the same precision as jason1990's, being careful defining the probability space, events, and conditional probabilities. Then present it to jason to see what he thinks about it. It's just getting too cumbersome to follow the logic otherwise.




PairTheBoard

You have to keep in mind that jason1990 gives Beauty only credences according to a nonindexical model. In that model there is no event of the form,

Mon = "I am awake and this is Monday"

In jason1990's model you have these kind of events.

"A Monday" = "SB has an awakening on Monday"

"A Tuesday" = "SB has an awakening on Tuesday"

So P(A Monday|J) = 1 and the event "A Monday" = (h or t)

and P(A Tuesday|J) = 1/2 and the event "A Tuesday" = t


I think there are some self locating credences Beauty can infer from this model but not all self locating credences. For example, from

P(A Tuesday|(J&h)) = 0

She can infer "If heads this is not Tuesday".

And from

P(t|(J & A Tuesday) = 1

she can infer, "If this is Tuesday the coin is tails".

But from the perspective of the nonindexical model, not only can no credence inferences be made for "This is Monday" or "This is Tuesday" but those propositions have no nonindexical meaning at all.

Of course jason1990 can speak on this better than I.


PairTheBoard

Now that I've switched to linear mode, I see why people are complaining about the envelope paradox posts being disruptive. It's a disaster. People are trying to follow the SB discussion, and every time they turn around they are interrupted by another post about the damn envelope paradox. There needs to be a feature in linear mode that allows you to skip all of the responses to any given post instead of just displaying every post chronologically. Like if someone now wanted to talk about linear mode, we certainly don't want those posts mixed in with SB and EP. Seems like a schizophrenic way to manage an internet forum; you'd think they'd have come up with something better by now. Anyway, here's another one mostly on EP with some SB content.


I'm not sure it always works as an auction because if there are more people than goods, when I have twice as many points as everyone else, I can buy 1 thing nobody can, but they can make me use up more than half my points to do it and keep me from buying anything else. Then when I have half the points as everyone else, they can keep me from buying anything. So I get an average of half of a thing, and the others get an average of the number of remaining goods to the number of remaining people which could be more than my half a thing. I think PairTheBoard's example works:

I agree, it's very interesting.

Switching will increase the average ratio of your envelope amount to each of the other's envelope amounts by a factor of 1.25. Be very careful if you ever encounter a claim like the following sentence:

"Switching will give you 1.25 times more than you started with on average".

If that is interpreted as the average of the increases is 25%, that is correct. But it doesn't mean that our average amount increases.


So did I for awhile.


You can say that the expected value of the amounts in the two envelopes are equal from the problem statement alone without regard to the distribution. Let S be the smaller amount, X be what is in your envelope, and Y be what is in the other envelope. Then

E(Y) = E(X) = 0.5*S + 0.5*2S = 1.5*S.

That's why we don't switch when we want to maximize dollars.


The original 2 envelope paradox is based on a fallacious argument that concluded that switching increases the average value of our envelope by a factor of 1.25. However, it is true even in that problem that if we do in fact switch, that

E(Y/X) = 1.25

which means that we have a 25% average increase in our envelope's value when we switch. It's just that this isn't the same as saying that the average value of our envelope increases by 25%. The average value of our envelope doesn't change. The numeraire argument sheds light on this distinction, and shows what the number 1.25 that the fallacious argument produced actually represents. It explains the paradox by taking the fallacious equation

E(Y) = 0.5*(2X) + 0.5*(X/2) = 1.25*X

and recasts it as the correct equation

E(Y') = E(Y/X) = 0.5*(2*1) + 0.5*(1/2) = 1.25

where Y' = Y/X is the value of the second envelope in units of the first envelope. That is, using the value of the first envelope as a numeraire.

But the main reason I brought it up was to introduce the numeraire concept as a way of understanding the arbitrariness of probabilities in SB. In EP, a decision changes as a result of a change in numeraire, but since it is clear that we want to maximize dollars, the numeraire is clear. But in SB by contrast, we don't even have a numeraire because nobody is having to make any decision with the goal of maximizing something. Until they do, degrees of belief don't really mean anything, and they remain mathematically undefined.


I don't think you meant to direct this at me. I didn't write what you were responding to. In fact I disagreed with it. But since you brought it up, you are essentially agreeing with Aaron's statement that there is no logic connecting frequency and degree of belief until a proper numeraire is defined, meaning a particular experiment with a pariticular thing that we want to maximize. Jason has stated that credence need not be the same as frequency for non-independent events. Arguments for 1/2 assume that the fact she can wake up on different days is irrelevant to her degree of belief. For 1/2, you would need either P(Mon) = 1, or P(tails | Mon) to not equal 1/2, or both since the answer would have to be the product of these unless you reject P(Mon) as a nonsense quantity and not merely an arbitrary quantity.

Well then that's a silly model if you can't even talk about what day of the damn week it is.

I that the paradox may be related to the hangman's paradox where we make false inferences about what we will believe in the future.

I typed that Jason allows P(heads | tails) before, I meant that he allows P(Mon | tails), apparently, and simply chooses not to set it equal to 1/2. If he allows that, he should allow P(Mon). Or is he saying not only isn't he setting it equal to anything, but that it isn't even allowable in his model? If he is disallowing it, that seems highly arbitrary since if there were some fundamental reason for disallowing it, then he shouldn't have even written it down in the first place.

Gahh come on guys. The rebuttal I gave Sklansky is saying the exact same thing as the equation Bostrom provides: if you tell her it's Monday, or that heads never results in an awakening, then the experiment is different and any inferences drawn aren't necessarily applicable and her credence (which is a subjective estimate) may be changed. And in both cases, Bostrom proves it is changed. Also, given an awakening, P of either Mon or Tuesday is 1. P (Mon) is not 1 unless she is told the coin came heads, which once again isn't the same experiment

Normalising by the 'number of bets' gives (q-p)/3 for SB.

But thats not actually what I meant. Im just saying that the change in her EV once she decides to bet heads as opposed to not betting at all, this time, before finding out the result of her bet, is the same as if she has just decided to bet on heads, this time, on a 1/3 heads coin.

Its a fairly convoluted construction meant to remove the fact that there must be a second losing bet. Its meant as a counterargument to the logic "im not going to bet heads because if im wrong yer gunna get me twice!"

Did you see the following argument?

There is a button in the center of the table. If it is pushed for the first time, an even money bet on tails is made. Otherwise nothing.

Suppose SB pushes the button p of the time. Her EV is proportional to -p + (p^2 + 2p(1-p) ) = p(1-p), so SB's optimal strategy is to push the button 1/2 of the time, not all the time.

I would like to explain, in greater mathematical detail, the reason why I said that probability theory cannot be directly applied to sentences that contain indexicals. This will be related to my brief comments in this earlier post. I will later make a follow-up, hopefully less math-heavy post, addressing the issue from a more pragmatic perspective.

Probability theory is an axiomatic system which provides a generalization of classical propositional logic. It is a mathematical set of rules that is related to, but distinct from, subjects such as statistics, game theory, decision theory, economics, gambling, and so on. All of those subjects use probability theory. But probability theory itself is just the rules of the game.

When we apply probability to propositions, we are using a formal mathematical structure which is an extension of logic. To do it rigorously, we must use the same level of rigor we would use in formal logic. First, we must write our propositions in a formal, symbolic language.

For example, we might denote the days of week (Sunday, Monday, and Tuesday) by the set {0,1,2}; the possible states of the coin (heads, tails) by {3,4}; the possible states of consciousness of Sleeping Beauty (awake, asleep) by {5,6}; the actual state of the coin by C; and Sleeping Beauty's actual state of consciousness on day n by S(n). We could add more to this if we like, but starting here, we would have the following background propositions:

"C ∈{3,4}"
"C = 3 => [(S(1) = 5) & (S(2) = 6)]"
"C = 4 => [(S(1) = 5) & (S(2) = 5)]"

After adding in details about the interviews, and so on, we would have a list of formal propositions, and our background information, J, would be the conjunction of all of these formal propositions.

To get a probability space out of this, consider the set T of all formal propositions that can be constructed involving the objects we have so far considered. Define an equivalence relation ~ on T by p ~ q if q can be proven from J & p, and p can proven from J & q. Let G = T/~ be the set of equivalence classes. The operations of conjunction, disjunction, and negation should be well-defined on G, and turn G into a Boolean algebra. (I believe this is an example of a Lindenbaum–Tarski algebra.) As long as we are considering only a finite number of propositions, Stone's theorem guarantees that there exists a measurable space (Ω,F) such that G and F are isomorphic. We can therefore identify each proposition in T, first with its equivalence class in G, and then with the corresponding element of F. In this way, whatever probabilities we assign to a proposition q in T can be transferred to the corresponding element A in F, where it is subject to the axioms of probability theory. In the end, when we have a probability measure P on the measurable space (Ω,F), we can regard it, via the isomorphism, as a function on T. In this way, we give rigorous meaning to P(q), where q is a formal proposition.

Generally, all of this happens implicitly, informally, and intuitively. It happens every time we think of ∩ as "and", and ∪ as "or". We do it all the time without all of the above rigor, and we usually do it without any problems. But I wanted to make all of this rigor explicit in order to explain my comments about indexicals. Let us add another object to our formal system. Let B denote jason1990's 50th birthday. Suppose Sleeping Beauty wakes up and wants to translate the English sentence, "Today is jason1990's 50th birthday," into the formal language in which we are operating. Indexicals such as "today", "now", "I", "here", "this", and so on, refer to a perspective or point-of-view, and are, in this sense, subjective. The word "today" is connected to Sleeping Beauty's subjective point-of-view, and is not explicitly denoted by any object in our formal system. The best that Sleeping Beauty can do toward translating this sentence is to say that she either just said "B = 1" or "B = 2", but she does not know which. There is uncertainty, but it is not uncertainty about the truth value of a formal proposition in T. Rather, it is uncertainty about the proposition itself. This is a different kind of uncertainty, and one which probability theory does not address.

I would like to write less formally now, and return to the notation from my first long post in this thread. Recall that J is the background information, and

h = "The coin lands heads."
t = "The coin lands tails."

For the sake of argument, consider a modification of the problem where an inspector arrives randomly during the experiment. The new background information is

J' = J & "An independent inspector will arrive on either Monday or Tuesday to check if Sleeping Beauty is awake or asleep."

Now suppose the inspector arrives and says to himself, "Sleeping Beauty is awake today." Or perhaps Sleeping Beauty sees the inspector and then says to herself, "I am awake today." In either case, the sentence can be understood objectively (i.e. non-indexically) by the person thinking it as saying,

E' = "Sleeping Beauty is awake on the day the inspector arrives."

In this case, it would be natural and consistent to assign probabilities that included P(h | J') = 1/2 and

P(h | E' & J') = P(h & E' | J')/P(E' | J')
= (1/2)(1/2)/[(1/2)(1/2) + (1/2)(1)]
= 1/3.

It is possible to do this, of course, because the inspector arrives only once, so "the day the inspector arrives" is a well-defined object. The inspector's arrival provides Sleeping Beauty with a way to distinguish "today" from other days in an objective manner.

But she can do something similar to this without an inspector. Let us go back to the original problem and add something different. Let

J'' = J & "Sleeping Beauty will take a 20-sided die into the experiment with her. Each time she is awakened, she will roll the die and look at the result."

Now suppose she wakes up and rolls a 5. She now has the information,

E'' = "Sleeping Beauty rolled a 5 during the experiment."

She might now calculate

P(h | E'' & J'') = P(h & E'' | J'')/P(E'' | J'')
= (1/2)(1/20)/[(1/2)(1/20) + (1/2)(1 - (19/20)²)]
= 20/59,

which is pretty close to 1/3. She would get the same number whether she rolled a 5, a 7, a 19, or any other number. If she took an n-sided die in with her, then she would calculate a probability of 1/(3 - 1/n), regardless of what number comes up. The die is giving her an objective observation that allows for the possibility of distinguishing the present awakening from any other possible awakening. It does not allow her to perfectly distinguish "today" from other days, because of the possibility of rolling the same number twice in a row, but this probability diminishes if she uses a die with more sides.

From a practical perspective, Sleeping Beauty doesn't even need a die. Perhaps she wakes up and observes that the interviewer is standing exactly 32.1 inches from her when he asks her the question. Or perhaps she observes that she has an itch on her right cheek exactly 1 minute and 18 seconds after waking. If her prior credence for these events happening on both days is much smaller than her prior credence for them happening on a given day, then her posterior credence for heads will be close to 1/3.

But of course, all of these modifications seem to violate the spirit of the problem, which is that Sleeping Beauty experiences absolutely nothing that could even possibly distinguish one awakening from another. In that case, if credence is understood in the context of formal probability theory, and if her credence for heads was 1/2 before the experiment, then it would have to be 1/2 during, since she obtains no additional, objective information.

Her credence for heads on awakening under the conditions of the OP experimental design is 1/3 before the experiment. She learns nothing new on awakening, so her credence is still 1/3. If she awakens and you tell her it is Monday, then she learns something she did not know before she slept and her credence is 1/2. If she awakens and you tell her it is Tuesday she has learned something new and her credence is 0.

Obviously the calculation is right, but I'm not sure this is a valid representation of her information. In informal terms, E'' can be true (in tails world if it happened in the other awakening) without SB knowing that it's true. So to reflect her state of knowledge, E'' should be (IMO) "is awake and looking at a 5 on a d20 she just rolled" or "is aware she rolled a 5 during the experiment". And by Bayes' theorem, this tells you nothing new of course. Using your old E'' and adding

F'' = "Beauty is aware of E''" to convert your E'' into one of mine above

P(h|E'' & F''&J'') = P(h&E''&F''|J'')/(P(F''&E''|&J'')

P(h&E''&F''|J'')=(1/2)*(1/20)*(1) = 1/40

(P(F''&E''|&J'')=P(h&E''&F''|J'') + P(t&E''&F''|J'')
=1/40 + (1/2)(39/400)(20/39) = 1/20

so P(h|E'' & F''&J'') = P(h&E''&F''|J'')/(P(F''&E''|&J'')=1/2 as expected.

Even from the perspective of somebody told about SB after the fact, and asked what the coin is, the person would say 1/2 because he has no interaction with awakenings. Then you tell him a die was rolled each time. He's like "that's nice, still 1/2". Then you tell him a 5 was rolled. He's like, "well, she's ~twice as likely to hit it with 2 rolls, so I guess it's 1/3 heads.. hey, wait, ~half the time it was tails and she rolled a 5 I wouldn't know that right now because you told me the other number instead.. still 1/2."

She rolled a 5, so E'' is true and she knows it. If there is some other objective proposition that she knows to be true, then we must identify it and condition on it. Otherwise, we can only condition on E'' & J''.

If Beauty rolls a 5 during the experiment, then Beauty is aware of it, since Beauty rolled it. So E'' => F'', which gives E'' & F'' = E''. I think you mean to define F'' to be the sentence "Beauty is aware of E'' during this awakening". Goto Post #661.

Let

E''' = "Someone randomly chooses a number that Beauty rolls during the experiment, and the chosen number is 5.",

and adjust the background information to J''' to include information about this new character. Then, yes, I agree that the most reasonable probability assignments give P(h | E''' & J''') = 1/2.

Incidentally, here are some other probabilities the die gives her. Suppose that after rolling a 5, she translates the subjective sentence, "Today is Monday," by using the objective sentence,

M'' = "Beauty rolled a 5 on Monday."

Then

P(h | M'' & E'' & J'') = P(h | M'' & J)
= P(h & M'' | J'')/P(M'' | J'')
= (1/2)(1/20)/(1/20) = 1/2,

and

P(M'' | t & E'' & J'') = P(M'' & t & E'' | J'')/P(t & E'' | J'')
= P(t & M'' | J'')/P(t & E'' | J'')
= (1/2)(1/20)/[(1/2)(1 - (19/20)²)]
= 20/39 ≈ 1/2.

I dont understand where you're going with this? What is the argument?

Jason, its hard for me to understand why her rolling a meaningless die affects her belief that it is heads. In the inspector variant, sometimes (during tuesday heads) she never sees the inspector and that affects the odds in an obvious way.

But in this variant, her roll is meaningless, that is, there is no roll that could make her belief in heads go up instead of down -- all the rolls produce the same result. For me that is just intuitively nonsensical, so I hope you can clarify this for me.

You mentioned that when you remove the possibility she can bet twice, she still bets as if it is 1/3 heads. But in this variant which appears to also strive for the same goal, it is revealed that she doesn't.

I would guess that your intuition senses a contradiction to this theorem:

I would guess that you are intuitively thinking of "the result of her roll during this awakening" as a random variable to which this theorem can be applied. But it is not a random variable. See Post #661.

How is, "she wakes up" encoded in J''?

This looks awfully Alice in Wonderlandy. She wakes up holding the 20 sided die. If she rolls the die its outcome is a random variable X uniformly distributed on {1,2,...,20}. Before rolling the die she has the credence for heads, P(h|J'') = 1/2. After she rolls the die, regardless of its outcome, her credence for heads immediatley changes to ~1/3, P(h|J'' & E''(X)] = 20/59 where E''(X) is the event E''(X) = "Beauty rolled an X during the experiment". It looks like the random value of X, whatever it turns out to be, is acting as the "numeraire" here .

And further down this rabbit hole, an objective outside observer could be provided this same E''(X) information. Suppose the experiment is run to conclusion and the outside observer is told nothing about what happened except for the infomation E''(X), "Beauty rolled an X during the experiment". As it happens, in this case he is informed E''(5), "Beauty rolled a 5 during the experiment". Should the outside observer now also change his credence for heads to 20/59?


PairTheBoard

The problem is that I think that this is what you have to say, because she knows that she's still in the experiment, and without saying that, you're not expressing her full state of knowledge. If you take the similar case where somebody walks up after the experiment, is told how it works, and asks "did she roll a 5?", then you get your numbers for sure- ~1/3 when the answer is yes, ~1/2 when the answer is no, and an overall average of .5. But even though both SB and this guy both know the information E'' and J'', I don't believe their information is identical because I don't believe this is a complete representation of their states of knowledge. They're both aware of how they learned E'' and J''. Beauty is aware that she's in the experiment and this guy is aware that he isn't (and furthermore, this guy has different information about how he learned E'' than the one from my previous post, where he's spontaneously told a 5 was rolled).

There's no way to directly incorporate her extra information of "looking at it right now" into your probability space (I assume you are imagining 20 h x 1..20 with p=1/40 each and 400 t x 1..20 x 1..20 with p=1/800 each), but if I were her, using the logic at the end of 661 I would reason: I know I'm looking at the 5 right now, but I don't know what propositions that represents if the coin is tails because I don't know what day it is. I just know I'm not looking at it right now on two different days at once. If it's Monday, then it represents this set of 20 (t,5,*), and if it's Tuesday, it represents this other set of 20 (t,*,5) (with the (t,5,5) appearing in both of course). That's convenient. Either way, I get the same answer, 1/2 (1/40 = 20/800), so I won't worry about it beyond that.

"∃n∈{1,2}: S(n) = 5".

See here. The result of this roll is not a random variable. There are, however, two natural random variables that we could define. X₁ is what is rolled on Monday. X₂ is what is rolled on Tuesday if the coin is tails, and 0 otherwise. In terms of these random variables,

E'' = {X₁ = 5} ∪ {X₂ = 5}.

As I said in my response to TomCowley,



P(h | E'' & J'') = 20/59. If an outside observer is asked for his credence for h given everything he knows, and if he knows E'' & J'', and if there is no other objective proposition that he knows to be true, then he should answer 20/59.

Well, since Beauty isn't looking for a 5, when she holds the die in her hand why does she even need to roll it? Why can't she reason as follows: "Even though my credence for heads now is P(h|J'') = 1/2, I know with certainty that after I roll the die E''(a) will be true for some a in {1,2,...,20} and I can then revise my credence for heads to P(h|J'' & E''(a)) = 20/59. Since I know with certainty that my credence for heads will be 20/59 after I roll the die, I should revise my credence for heads to 20/59 without bothering to roll the die."

For that matter, the outside observer could reason the same way. Knowing nothing about the outcome of the experiment he can reason: "J'' tells me Beauty rolled the die at least once so she must have rolled some value, say a, in {1,2,...,20}. So I know E''(a) is true for some "a" and regardless of which "a" it is I know P(h|J'' & E''(a)) = 20/59. so J'' => E''(a) for some "a", so for some "a", P(h|J'') = P(h|J'' & E''(a)) = 20/59."


PairTheBoard

Right. So she does not know which formal proposition is represented by the English sentence, "Beauty rolled a 5 today." It could be either one of:

E₁ = "Beauty rolled a 5 on Monday."
E₂ = "Beauty rolled a 5 on Tuesday."

Maybe she thinks to herself, "I know that I know either E₁ or E₂, but I don't know which one I know." She could then calculate

P(h | E₁ & J'') = 1/2,
P(h | E₂ & J'') = 0.

Then, instead of reporting her credence for heads, given everything she knows, as

P(h | (E₁ or E₂) & J'') = 20/59,

perhaps she reports that her credence is unknown, but that it is either 1/2 or 0. Maybe I can respect that. It seems a little Rumsfeldian.

I would say this is a misapplication of the following theorem:

The reason it is a misapplication is because the collection of events being considered,

{E''(a): a in {1,2,...,20}},

does not form a partition. More specifically, this collection is not pairwise disjoint.



Another point: While it is true that P(h | E''(a) & J'') = 20/59 for some a (in fact, for all a), the correct reasoning is that J'' => (E''(a) for some a), and therefore

P(h | J'') = P(h | (E''(a) for some a) & J'') = 1/2.

Note that (E''(a) for some a) can be written formally in the Boolean algebra, using disjunction, as

E''(1) or E''(2) or ... or E''(20).