Y is 10, therefore the sample space for X is [5, 10] — Michael

I know it seems that way to you. That is your subjective view of the situation. But we are told no such thing. We are not told X is selected by some random process or even by a choice. There is no such thing, so far as we know, as a "sample space for X".

Here's an analogy. Suppose a friend tells me he's going to flip either his red/blue coin or his yellow/green coin. What's my expectation of red? Is it 1/4? How would I know that? He didn't tell me how he was going to choose which coin he was going to flip. We do have — unlike in the envelope problem — a space known to have four elements: [[red, blue], [yellow, green]]. Maybe what he meant was that he was
always going to flip the red/blue coin. That counts as flipping either. Maybe he meant he would randomly choose one, maybe some other method. I have no way of knowing. I should probably treat the two possible spaces indifferently, because what else can I do?

In our situation, I know it's tempting to say we should average the chance that our sample space is [10, 20] and the chance that it's [5, 10], but we can't. Even if we choose to treat the two spaces indifferently, we do not know there are two such spaces. This is the difference.

@andrewk I think argued that we have to treat X as a random variable, and that we have to treat any value we cannot exclude as a possible value for X. I can see the appeal of that, I really can. But it clearly leads to the wrong answer. If I present envelopes containing either £10 or £20 to a hundred switchers and let them open one, half of them will think there might be envelopes with £5 out there and half of them will think there might be envelopes with £40 out there. Half of them will trade their £10 for the £20, and half of them, sad to say, will trade their £20 for £10. Those with £10 did not risk losing only £5. Those with £20 had no chance of getting £40.

I know it seems that way to you. That is your subjective view of the situation. But we are told no such thing. We are not told X is selected by some random process or even by a choice. There is no such thing, so far as we know, as a "sample space for X". — Srap Tasmaner

But your own argument made this same assumption.

$\small P(U=2X \mid Y=10)(2X)$ is the same as $\small P(X=10)(2X)$

$\small P(U=X \mid Y=10)(X)$ is the same as $\small P(X=5)(X)$

Edit: It's not even an assumption. It simply follows from a) Y = 10 and b) Y is X or 2X. If Y is 10 and Y is X then X is 10. If Y is 10 and Y is 2X then X is 5.

But your own argument made this same assumption. — Michael

That's not an argument, it's just a calculation, and it didn't. It weights the possible values known to be in the sample space for $\small U$, namely $\small X$ and $\small 2X$, by the probability of the event of $\small U$ containing the larger of the two values, given that the value of one envelope is known to be 10, and the probability of the event of $\small U$ containing the smaller of the two values, given that one envelope is known to contain 10.

Knowing that one envelope contains 10 does not tell you whether $\small U$ is the larger or the smaller envelope, and does not tell you whether 5 or 20 are in the sample space. Not knowing whether 5 or 20 are in the sample space for $\small U$ we cannot use them. We could use 10, but 10 is the value of the envelope we've already got.

That's not an argument, it's just a calculation, and it didn't. It weights the possible values known to be in the sample space for $U$, namely $X$ and $2X$, by the probability of the event of $U$ containing the larger of the two values, given that the value of one envelope is known to be 10, and the probability of the event of $U$ containing the smaller of the two values, given that one envelope is known to contain 10. — Srap Tasmaner

To ask if U is the larger of the two, given that Y is 10, is to ask if X is 10 and U is 20.

To ask if U is the smaller of the two, given that Y is 10, is to ask if X is 5 and U is 5.

So if there's a probability of 0.5 that U is larger than Y, where Y is 10, then there's a probability of 0.5 that U is 20, and if there's a probability of 0.5 that U is smaller than Y, where Y is 10, then there's a probability of 0.5 that U is 5. So 0.5 * 20 + 0.5 * 5.

Knowing that one envelope contains 10 does not tell you whether UU is the larger or the smaller envelope, and does not tell you whether 5 or 20 are in the sample space. Not knowing whether 5 or 20 are in the sample space for UU we cannot use them. We could use 10, but 10 is the value of the envelope we've already got. — Srap Tasmaner

We derive the sample space from:

1. Y = 10
2. U = 2X or U = X.

If U = 2X then Y = X. As Y = 10, 2X = 20.
If U = X then Y = 2X. As Y = 10, X = 5.

And yet my use of X and 2X leads to the correct conclusion, while your substituting Y and Y/2 leads to the wrong conclusion. Since that is the only difference, you ought to conclude that you cannot substitute this way.

((This is fun, but I have to go work on the Tractatus. I'll check back here later today.))

In this expression

$\small P(U=2X \mid Y=X)(2X)$

$\small (2X)$ is a dollar value — Srap Tasmaner

Yes. And that dollar value is twice the value of $\small Y$, so if $\small Y=10$ then $\small 2X=20$. How can you possibly disagree with this? It's right there in your own equation: $\small Y = X$. So $\small 2X = 2Y$.

You can't condition on $\small Y = X$ and then multiply by $\small 2X$ where $\small 2X \ne 2Y$. That's just a contradiction.

And the same issue on the other side. You can't condition on $\small Y = 2X$ and then multiply by $\small X$ where $\small X \ne \frac12 Y$. That's just a contradiction. If $\small Y = 10$ then $\small X = 5$.

You can't condition on Y=X and then multiply by 2X where 2X≠2Y. That's just a contradiction. — Michael

It's really not.

There are two different sorts of things here. One is the conditional probability of an event occurring, one is the value in our sample space for U that U will have if that event occurs; by abuse of notation we're actually calling that event "U = 2X". The way the game is set up, P(U = 2X | Y = X) = 1 and P(U = X | Y = 2X) = 1. No assertion is made here about the event "Y = X" actually occurring, or about its probability.

Maybe you're imagining we're talking about Y here the same way we're talking about R in this coin toss:

$\small \begin{align} E(R) &= P(R = \mathrm H)(10) + P(R = \mathrm T)(100) \\ &= \frac12(10) + \frac12(100) \\ &= 55\end{align}$

But we're not. Here we're talking about the probability of the events "R = H" and "R = T" occurring, and weighting the values from our sample space, 10 and 100, by those probabilities.

In my expectation calculations, we're talking about U = ... occurring, not Y = ..., so you need to read them as

$\small \begin{align} E(U \dots) &= P(U = 2X \dots)(2X) + P(U = X \dots)(X) \end{align}$

We're just matching up events of U taking a value from the sample space with that value.

no matter what you imagine it might be or how much you know about the envelopes. — Jeremiah

Well, with full knowledge of the situation there's a 100 % chance that one envelope contains $ 10,-- and the other $ 20,-- and there's no need to invoke probablility. The reason we invoke probability at all is simply because we have incomplete knoweldge of the situation. And that's why a bet makes sense at all. I don't understand how can say that knowledge doesn't matter. It's the entire point of it.

The expected value of U is the probability that event A is the case multiplied by the value of U if event A is the case plus the probability that event B is the case multiplied by the value of U if event B is the case.

Event A is "U is the larger envelope, given that Y is £10". If event A is the case then the value of U is £20. Your own calculation accepts that the probability that event A is the case is 0.5. So it's 0.5 * 20.

Event B is "U is the smaller envelope, given that Y is £10". If event B is the case then the value of U is £5. Your own calculation accepts that the probability that event B is the case is 0.5. So it's 0.5 * 5.

I honestly cannot understand which step here you disagree with.

The sample space for U has two values in it. One of them is 10, and that's taken by Y. That leaves exactly one slot open in the sample space for either 5 or 20, and it certainly will be one of those, but we don't know which. If you know for a fact that one of those two values cannot be a value of U, because it cannot be in the sample space from which the values of U are drawn, then you cannot use both of those values when calculating the expected value of U. One of them has to go.

Since we don't know which one has to go, the only thing to do is continue, reluctantly perhaps and with some disappointment, using X and 2X. And what you find is that your expectation for U, in terms of X, does not change when you learn that Y = 10. Knowing only one of the two values is just not enough information.

That it is not enough information should be obvious. Look again at this example:

If I present envelopes containing either £10 or £20 to a hundred switchers and let them open one, half of them will think there might be envelopes with £5 out there and half of them will think there might be envelopes with £40 out there. Half of them will trade their £10 for the £20, and half of them, sad to say, will trade their £20 for £10. Those with £10 did not risk losing only £5. Those with £20 had no chance of getting £40.

There is a perfect symmetry here: half were lucky in the draft but unlucky in the trade; the other half were unlucky in the draft but lucky in the trade. Not one of them actually had any idea which they were, whether they saw £10 in their envelope or £20. And they all traded for no reason at all.

Since we don't know which one has to go, the only thing to do is continue, reluctantly perhaps and with some disappointment, using X and 2X. And what you find is that your expectation for U, in terms of X, does not change when you learn that Y = 10. Knowing only one of the two values is just not enough information. — Srap Tasmaner

If you describe your expectation for U as 2X and if Y is known to be 10 then you're describing your expectation for U as £20. If you describe your expectation for U as X and if Y is known to be 10 then you're describing your expectation for U as £5.

That it is not enough information should be obvious. Look again at this example: — Srap Tasmaner

We're talking about the expected value of U if Y is 10. Situations where Y is 20 (or 5) aren't relevant to this consideration.

So you could say that the envelope sets are all £10 and £20 and everyone with £10 gains £10 by switching or you could say that the envelope sets are all £5 and £10 and everyone with £10 loses £5 by switching, or you could say that half the envelope sets are £10 and £20 and half are £5 and £10 and half the people with £10 gain £10 by switching and half lose £5 by switching.

There are a number of different real life scenarios (conditioned on Y = 10) that we can consider all with different values of U and gains/losses for switching. But what we're discussing right now is your calculation:

$\small \begin{align} E(U \mid Y=10) &= P(U=2X \mid Y=10)(2X) + P(U=X \mid Y=10)(X) \\ &=\frac12 2X + \frac12 X \\ &= \frac32 X \end{align}$

You are accepting that the probability that U is larger than 10 is 0.5 and that the probability that U is smaller than 10 is 0.5.

But if U is larger than 10 then U is 20 and if U is smaller than 10 then U is 5.

So it's 0.5 * 20 + 0.5 * 5.

And if we were to condition on X being either 5 or 10 then we'd come to two different conclusions (neither of which are your (3/2)X):

$\small \begin{align} E(U \mid Y=10, X=5) &= P(U=2X \mid Y=10, X=5)(2X) + P(U=X \mid Y=10, X=5)(X) \\ &=0 \times 2X + 1 \times X \\ &=X \\ &=5 \end{align}$

$\small \begin{align} E(U \mid Y=10, X=10) &= P(U=2X \mid Y=10, X=10)(2X) + P(U=X \mid Y=10, X=10)(X) \\ &=1 \times 2X + 0 \times X \\ &= 2X \\ &= 20 \end{align}$

But none of what you've said here addresses the calculation as explained here, so please point out exactly which step it is you disagree with.

I've just been hit by the consequences of some of the issues raised in my Bayesian note. I didn't notice them before, but they're in there.

The first is that, unless the player adopts an assumed distribution for X, they have no probability distribution by which they can calculate expected gains, so it's simply meaningless to ask what their expected gain is. We can ask what the host expects the gain to be if we like. Since the host knows what X is, the answer to that is:

-X (a certain loss) if the host knows the player has the envelope with 2X
+X (a certain gain) if the host knows the player has the envelope with X
0 (an even chance of a gain or loss of X) if the host doesn't know which envelope the player has, but believes the choice of which envelope to put the doubled amount in was random with even odds.

But what the host expects doesn't bear on the question, which is 'what should the player do?' (assuming the player's utility function is the identity function).

Stripped of all the mathematics, my simplest answer to that (more sophisticated alternatives are possible) is: the player should estimate L, the maximum amount she thinks the game would be prepared to pay out. If Y<L/2 she should switch, otherwise she should not.

This is consistent with the player assuming that X can be any amount between 0 and L/2, with uniform probability. That is the assumption used in @Srap Tasmaner's PHP program, with num_trials = L/2, if I've interpreted the PHP correctly (I may not be remembering the variable name exactly, but you'll know the one I mean).

With that assumed distribution, the player will expect a zero gain if she switches regardless of the value of Y. But if she switches only if Y<L/2 the expected gain will be Y/2 for those cases.

I predict that, if that sim is run with a strategy that only switches when Y<num_trials / 2, it will converge on a gain that averages L/4 across those trials and 3L/16 across all trials.

Interestingly, the Bayesian note doesn't side completely with either prior 'side' of the argument. It shows that there exists a switching strategy that gives the player a strong expectation of gain over the Y they have seen, but it also shows that, if she has adopted an assumption of the distribution of X, and uses a strategy that switches regardless of the value of Y, her expectation of gain from switching is zero. The way we reconcile that with the fact that by switching she'll either gain Y or lose Y/2 is to observe that the probabilities of those two outcomes are not equal. The probabilities must be inferred from where the observed Y sits in the player's prior distribution for Y, which is inferred from her prior distribution for X. The higher Y is, the greater is the probability that it is the doubled amount, so the greater is the probability of losing Y/2 rather than gaining Y by switching.

For detail, see the note. The simplest example I can think of is where the player assumes that X is either 10,000 or 20,000. Then her assumed distribution for Y is 10,000 (prob = 1/4), Y=20,000 (prob=1/2), Y=40,000 (prob=1/4). If she sees Y=40,000 she knows she has the doubled amount (given her prior assumed distribution of X), so she shouldn't switch. Otherwise she should. That strategy gives an expected gain of (1/4) . 10,000 - (1/4) . 10,000 + (1/4) . 20,000= 5,000. That gain is equal to the expected loss that arises if she also switches when Y=40,000 (lose 40,000 / 2 with probability 1/4, ie lose 5,000). But using a Bayesian switching strategy, she sets up an expected gain of 5,000.

This also demonstrates why looking in the envelope makes a difference. If you don't look in the envelope, you can't use the above strategies, as they decide whether to switch based on the value of Y.

If Y<L/2 she should switch, otherwise she should not. — andrewk

That's exactly what my program does (although I switch if Y is less than or equal to L/2). In my case, L is the highest 2X envelope seen so far and so doesn't require guessing some upper bound.

I predict that, if that sim is run with a strategy that only switches when Y<num_trials / 2, it will converge on a gain that averages L/4 across those trials and 3L/16 across all trials. — andrewk

It's a .25 gain in my program.

if I've interpreted the PHP correctly. — andrewk

Do you know any programming languages? I could try writing my program in something else so you can better see what it's doing in case the PHP is too confusing.

Most of my languages are very old - Fortran, BASIC, Pascal, C, Simula, LISP, Prolog. These days I mostly use R, and am toying with Python in case that turns out to be useful if I want to get into 'data science'. I think I managed to understand srap's PHP program, so I'll try to work through yours as well as part of a self-education exercise.

It seems weird how variable names start with $. I suppose there's some carefully thought-out reason for that.

It's a .25 gain in my program. — Michael

That's a different metric than the ones I quoted. Mine is in dollars while yours, if I'm reading the PHP correctly, is the dollars divided by the total of all Ys observed, minus 1.

Measured on the same basis as yours, my model gives 0.5. That is more because mine uses L from the beginning, rather than progressively adapting it based on observations. My suggestion is that we set L to be the budget of the game show, because we know they can't pay more than that. If we don't know that we could use a suitably small proportion of the net assets of the company running the game show. Or if that's unknown, a suitably small proportion of the GDP of the country it comes from (or of the world, if we don't know the country).

Which approach is appropriate may depend on the exact setup of the repetitions. Yours appears to be a serial one, where the info from previous trials is available to the player in subsequent trials. Alternatively, we could have a parallel one, where a team of 10,000 players do one trial each simultaneously, and aim to maximise their team winnings.

There's also the question of whether we use the assumed prior distribution of X to model the possible values of X. To not do that requires having an additional distribution that models the uncertainty in our prior distribution, which is a bit too meta- for me at this time of night. I'll sleep on it. If we start with L estimated as above, in the serial case we could update it (improve it) in a Bayesian fashion based on each new observation of 2X.

Here's my attempt at R:


The gain is .25.

That's a different metric than the ones I quoted. Mine is in dollars while yours, if I'm reading the PHP correctly, is the dollars divided by the total of all Ys observed, minus 1. — andrewk

It's the amount you walk away with by following the switching strategy divided by the amount you walk away with by never switching, minus 1.

If I was just showing the amounts then it would be:

Switching strategy: £3,786,812
Never switch: £3,020,844

The switching strategy gives you 1.25 times the amount from never switching (which is just what the division and minus 1 is done to show).

I did not say you have full knowledge. All of the concerns you are expressing have already been addressed by multiple people multiple times, and I don't feel like going in circles over this any more.

I summed this up towards the start of the thread:

You have two envelopes, A and B.

There are two possible amounts for the contents of the two envelopes either X or 2X. You don't know which is which.

We'll call amount X case R, and and amount 2X case S.

You are handed envelope A.

Before you open it, you know that A could be case R or it could be case S. You have no clue which it is, so as an uninformative prior you give each case a 50% likelihood.

Now for the tricky part, you open envelope A and see it has 10 bucks in it.

Intuitively this seems like new information which would call for you to update your prior; however, it still does not tell you if are you in case R or case S. You don't know if that is 10=X or 10=2X. So in truth you can't really update your prior, as your prior was based on the uncertainty of being in case R or case S.

So your prior stands.

You consider swapping. You look at envelope B and realize it could be in case R or case S. You don't know which, so just as before you give it an uninformative prior and you give each case a 50% likelihood.

Now you could swap, but really probabilistically it changes nothing as your initial uncertainty would still stand. You would just trade the uncertainty of A for the uncertainty of B which is the same uncertainty. And the math? Well the math is indifferent to all options. So the decision to swap really comes down to if you feel lucky or not. — Jeremiah

I mostly use R, and am toying with Python in case that turns out to be useful if I want to get into 'data science' — andrewk

There is no reason to put data science in quotations; it is a real branch of modern science.
The advancements of computers and the age of big data is what created a need for specialist in analyzing data.The two most common forms of data scientists are statisticians and computer scientists; however, there are other flavors as well.

R is mostly used by statisticians for data analysis, my impression is that Python is mainly used in machine learning, which would be the computer science side. From what I have read, it is not too hard for a statistician to cross over into machine learning, as many of the machine learning concepts come from statistical theory and practice. If your goal is simply to engage data analysis, and you already know R that is well enough, and more preferable if you wish to go the statistics path. However, being a statistician requires that you know both Classical and Bayesian methods, as many of the model assumptions in Classical statistics persist over into Bayesian statistics.

so please point out exactly which step it is you disagree with — Michael

You are taking a weighted average of one value which is not only possible but actual, and one value that is not possible. You cannot calculate an expected value for a variable using values not in the sample space for that variable.

You are taking a weighted average of one value which is not only possible but actual, and one value that is not possible. — Srap Tasmaner

You saying that there's a 50% chance that U is 2X given that Y is £10 is you saying that there's a 50% chance that U is £20.

You saying that there's a 50% chance that U is X given that Y is £10 is you saying that there's a 50% chance that U is £5.

The pro switchers never really gave a good reason how to reconcile the fact that their model has the potential to make impossible predictions. That is a valid reason for concern and if a model which makes no new assumptions upon seeing Y only provides possible outcomes then that is a definitive reason to favor that model.

Until they fix that problem, that will stand against their methods of approach.

The error here is making new assumptions upon seeing Y.

You saying that there's a 50% chance that U is 2X given that Y is £10 is you saying that there's a 50% chance that U is £20.

You saying that there's a 50% chance that U is X given that Y is £10 is you saying that there's a 50% chance that U is £5. — Michael

I'm going to meet you half-way, Michael.

In essence what I'm saying is that P(U = 2X | Y = 10) = P(U = 2X), which entails that (U = 2X ∩ Y = 10) = ∅. The trouble here, which we should have fixed long ago, is just abuse of notation. I can look at that and understand it as saying that the values loaded into the envelopes are independent of my choice of envelope. But it also suggests that the value of U and the value of Y are unrelated, which is plainly false. It's much more cumbersome to do it the long way, but I guess there's nothing for it.

We'll label the envelopes L and R. There are two independent events:
(1) [L = x, R = x], the loading of the envelopes; and
(2) [Y = L, Y = R]. the choosing of an envelope by the player.
(There are several equivalent ways of defining these.) We're also going to assign probabilities Y = L = 1/2 = Y = R, and L = x = 1/2 = R = x. (The defense for this is that since the player cannot deliberately choose L even if she wants to, she is indifferent, and since she is indifferent then there is no point in the loader of the envelopes not being indifferent. There is no gamesmanship attached to which indistinguishable envelope is which.)

The blind expectation looks like this:

$\small \begin{align} E(U) &= P(U=x)(x) + P(U=2x)(2x) \\ &= P(U=L)P(L=x)(x) + P(L=R)P(R=x)(x) + P(U=L)P(L=2x)(2x) + P(U=R)P(R=2x)(2x) \\ &= \frac12\cdot\frac12(x) + \frac12\cdot\frac12(x) + \frac12\cdot\frac12(2x) + \frac12\cdot\frac12(2x) \\ &= \frac32x\end{align}$

What happens when we have selected and opened an envelope with value 10?

There are two possibilities: Y = L & L = 10 or Y = R & R = 10. It seems safe to treat these as equally likely. But now we have complications. Y = L and L = 10 are independent; the sort of events we'll be conditioning on this pair are U = R and R = x, which are also independent. But Y = L and U = R are not independent and L = 10 and R = x are not independent.

Case 1: Y = L & L = 10
$\small \begin{align} E(U\mid C_1) &= P(U=x\mid C_1)(x) + P(U=2x\mid C_1)(2x) \\ &= P(U=R\mid Y=L)P(R=x\mid L=10)(x) + P(U=R\mid Y=L)P(R=2x\mid L=10)(2x) \\ &= 1\cdot\frac12(x) + 1\cdot\frac12(2x) \\ &= \frac32x \end{align}$

Obviously Case 2 will be the same and E(U) = 3X/2. This looks a whole lot like what I did in this post.

But is it right? I know what you're going to say: (x) should be 5 and (2x) should be 20. After all, if L = 10 and R = x, then R = 5.

But consider this: I dropped some terms from the expansion, namely all the terms in which U = L, because we're doing the case in which Y = L. All those terms just go to 0, so I dropped them. What if it turns out x = 10? Then P(R = x | L = 10) would also go to 0, and we'd end up with U = 20, which is fine.

Your way would have the term containing (5) disappear when x = 10. Does that make sense?

*** ADDED: Yeah, actually it does. That's what makes this so misleading. There's no avoiding the sample space issue. ***

We could even, if we wanted, go ahead and do more cases in which we set x = 10, now that we know that's a possibility. The only problem with that is that we cannot do a complementary case that's x =/= 10. What would you plug in for that?

So now we're back to the same issue: there is one unknown value in our sample space for L and R. It's either 5 or 20, but two values don't go in one slot. And you cannot calculate an expected value using values not known to be in your sample space. So we're stuck with x.

((Actually we don't need x. I'll be posting a version in which the loading sample space it's just [L = 2R, R = 2L], and our expected value is just (L + R)/2.))

We'll label the envelopes L and R. — Srap Tasmaner

Please Nooooooo! Let's not do that. I reserved the symbol L many pages ago for the maximum possible value ('Limit') in the player's prior distribution for 2X. If we start using it for something else, like one of the envelopes, we'll end up a terrible muddle. :joke:

I'd be interested in your thoughts on this post, and whether you've tried doing a simulation using the simple strategy outlined there. The discussion has been very fluid but I think I'm agreeing with you that the player's expected gain from blind switching is zero, but her expected gain from switching based on comparing the value of Y to L/2 is strongly positive (here L has the original meaning of Limit, not Left).

Thank you for that. Now I can check my progress in PHP interpretation by comparing the two.

Here's another slightly different version with a little twist.

As above, we'll have an envelope loading event, this time [L = 2R, R = 2L].

I'm just going to do the case for Y = L = 10. In addition to dropping all the terms that include U=L, I'm going to leave U=R (=1) out of the terms I keep.

E(U|Y=L=10) = P(L=2R | L=10)(R) + P(R=2L | L=10)(R)

Now the question is, can I do this:

= P(R=5 | L=10)(R) + P(R=20 | L=10)(R)
= R

That is, maybe there's no harm in doing a little algebra in how you describe the events whose probabilities you're looking at, but all that stays inside the parentheses-- you still don't get to touch (R) because there's no (R | L=10) or something.

I feel a little weird about it, but maybe there's room for a partially subjective description here, so long as you're careful not to average values that are not known to be in the sample space.

What do you think @Jeremiah? Is this okay?

Nooooooo — andrewk

Sorry, man.

the player's expected gain from blind switching is zero, but her expected gain from switching based on comparing the value of Y to L/2 is strongly positive. — andrewk

I think a more complete, realistic answer is around here, yes. Still, for a single trial, your guess would have to be awfully lucky to be any help.

for a single trial, your guess would have to be awfully lucky to be any help. — Srap Tasmaner

Why? It's not about helping, it's about her expected gain, which is a pure calculation and nothing to do with luck. And my analysis is for a single trial.