## Saturday, 28 February 2015

### The Objections of chrysics - Part 2

Please note that since I wrote this article, I have been persuaded that the argument it relates to is wrong (meaning that chrysics and Mathematician and irishsultan and ChalkboardCowboy were all right from the start and I should have listened to them rather than arguing with them).  Fortunately, I didn't because, for me at least, this little intellectual journey has been far more interesting than it would have otherwise been.

The correct answer for the scenario as it is worded is not 1/2 but rather 1/3 (meaning that the likelihood of winning as a consequence of staying is 2/3).

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The discussion goes on between chrysics and me.  Please note that this has been written with one reader in mind, so I haven't provided much in the way of context.  There are, however, links below to other relevant posts which might be of assistance to other readers.  Note that the comments below that I am responding to are from /r/math over at reddit.

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First my preceding comment as context:

(quoted) chrysics: And if we say that there's a Red Mary game where the contestant has picked Red and White: your argument is again correct, there are two equally likely possibilities (MAC and MCA). The contestant has a 50% chance of winning by switching.

me (as wotpolitan): That's been my point right from The Reverse Monty Hall Problem, although it might be more clearly stated in Marilyn Gets My Goat.
As I said on my blog, everything else has been a diversion, likely due to me not making myself absolutely crystal clear. I agree that the argument is not valid when we have a Red Mary game in which Red Mary is not revealed, but this would then an isotropic White Ava or Green Ava game.

I appreciate your analysis of why, overall, the likelihood of winning from switching across multiple iterations, is 1/3 (or 2/3 in the classic Monty Hall Problem). However, ChalkboardCowboy argues that this is in contravention of the Law of Large Numbers. Would you have a response to that? (Note, my suspicion is that ChalkboardCowboy is misapplying the Law of Large Numbers.)

PS: I know that we have been at this for a long time, but could you please take another look at the original Reverse Monty Hall Problem article and see if I somehow failed to describe what I meant to describe.

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And then chrysics’ post that I am responding to:

Perhaps my last comment was unclear. I am not saying that the whole of your argument is valid, and the contestant will deduce that they have a 50% chance of winning by switching. It is emphatically not the case that the contestant has a 50% chance of winning by switching, if there are no further qualifiers applied to that statement. Your argument about equally likely options existing, and thus providing a 50% chance of winning, is valid only in specific circumstances (the contestant selected Red and Green; the contestant selected Red and White), and only to somebody with knowledge that is unavailable to the contestant (the location of Mary, even if she is not revealed).

There is also an additional scenario in which the argument does not apply. Your definition of a Red Mary game as being any game in which the Red Door holds Mary puts no constraints on which doors are selected by the contestant. As such, it requires that you consider the additional scenario (the contestant does not pick the Red Door at all. They can never win by switching, as that gets them the Red Door which by definition holds Mary) as equally likely to each of the others. The probability of winning by switching, given that Mary is behind the Red Door, is thus reduced from 50% to 1/3.

What I'm saying is that if there exists an outside observer who knows:

1. The contestant's choice of doors, and

2. The location of Mary (assumed to be red, for the sake of simplicity)
then that observer will, in some but not all scenarios, deduce that the contestant has a 50% chance of winning. In another scenario, that observer will deduce that the contestant has a 0% chance of winning by switching.

The contestant themselves can never (correctly) deduce that switching provides them with a 50% chance of winning. They do not have the same information available to them. The contestant is asked, once the door is opened and they find themselves in a Revealed Red Mary (or a Revealed Green/White Ava, as the case may be), to evaluate the probability that they win by switching. To do that, they must weigh up all of the possible ways in which they could potentially arrive at the scenario they now find themselves in, and must also consider how likely each of those ways is to actually produce the scenario they find themselves in. In doing so, they follow a procedure equivalent to that I outlined in my earlier post.

Your argument is valid only in the scenario that the Red Door is one of the contestant's two chosen doors. Your argument does not show that the contestant has a 50% chance of winning a Red Mary game. It shows that the contestant has a 50% chance of winning a Red Mary game if and only if they have selected the Red Door as one of their two. Which is true only 2/3 of the time.

The analysis I'm providing - let me make it very clear, once again - is entirely independent of how many times you play the game. It is not an analysis exclusively of the probabilities you get when playing repeatedly. Nor is it an analysis exclusively of the probabilities when playing a single iteration, but it can be treated as such if that is what interests you. Because the probabilities deduced by the contestant are in no way dependent upon how many times the game is played.

My response to ChalkboardCowboy's statement would be that ChalkboardCowboy is exactly right. I'd be interested to know why you think the law of large numbers cannot be applied here (or why you think it applies but in a different way than ChalkboardCowboy says, if that's the case).

(quoted) wotpolitan: could you please take another look at the original Reverse Monty Hall Problem article and see if I somehow failed to describe what I meant to describe.

I believe you've described exactly the same process throughout, with the exception of the one time you said that the host is committed to open the Red Door if it holds Mary. I'll accept this was an error as you've otherwise been consistent both before and after in saying the host will choose randomly if both selected doors hold goats. Your initial blog post seems perfectly clear to me. You ask for the likelihood, as deduced by the contestant, that switching will win the car. There's not really any room for ambiguity here. We can narrow it down to, for example, only considering the possibilities in which Mary is behind the Red Door, and perform the analysis that way, and this does not affect the result as it is one of many equally likely scenarios which each produce the same set of probabilities.

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First off, I have to reiterate my rather strange sounding claim that, overall, with The Reverse Monty Hall Problem, the contestant will benefit from a policy of staying, winning 2/3 of the time, if the game is played repeatedly.  I’ve not denied that.

Secondly, I’ll quote myself from The Reverse Monty Hall Problem (I keep linking to it because my original words are there and I think that many people are responding to what they think I said and not to what I actually said, I don’t feel particularly responsible for arguing for what they think I said):

One day you decide to go out to buy a new puzzle book at the massive Honty Mall.  When you enter, however, you are confronted by three doors and a rather dishevelled amateur philosopher who swiftly talks you into trying out a variation of an old game show puzzle (you obviously like puzzles, so it was an easy task).

The puzzle is put to you as briefly yet comprehensively as possible:

·         There are three doors, there is a goat behind two of the doors and behind the third is a car.

·         If, at the end of the game, you open the door with the car behind it, you win the car.

·         First, you select two doors (not the one door of the Classic Monty Hall Problem).

·         The philosopher will then open one of the doors you selected, revealing a goat.

·         You then have the option to switch from your remaining selected door or stay.

·         Before being allowed to open a door, you must provide the likelihood that a switch will win you the car (even if you choose to stay).

·         The placement of the goats and car is randomised.

Do you switch or stay, and what is the likelihood of winning from a switch?

In the exact situation in which the contestant finds herself, two doors have been selected (later described in Marilyn Gets My Goat as Red-White, Red-Green or White-Green, and specified as Red-Green) and one door has been opened (specified as the Red Door in Marilyn Gets My Goat).  While I do discuss the specific example of Mary revealed behind the Red Door (referred to as “Red Mary”) after the Red and Green Doors were selected, if I have correctly understood the term this situation is isomorphic with:

Red Mary & Red-White

White Mary & Red-White

White Mary & White-Green

Green Mary & Red-Green

Green Mary & White-Green

Red Ava & Red-White

Red Ava & Red-Green

White Ava & Red-White

White Ava & White-Green

Green Ava & Red-Green

Green Ava & White-Green

These constitute all of the situations in which the contestant might find herself after selecting two doors and after the host has opened a door to reveal a goat in a single iteration, one shot instance of the Reverse Monty Hall Problem.

You (chrysics) said:

(neopolitan's) argument about equally likely options existing, and thus providing a 50% chance of winning, is valid only in specific circumstances (the contestant selected Red and Green; the contestant selected Red and White), and only to somebody with knowledge that is unavailable to the contestant (the location of Mary, even if she is not revealed).

This sort of misses the point, but you seemed to have got the point earlier when you wrote:

If we say that there's a Red Mary game where the contestant has picked Red and Green:

your argument is correct, there are two possibilities (MAC and MCA), each of which is equally likely. The contestant has a 50% chance of winning by switching.

Because all the situations in which the contestant finds herself after the door has been opened are isomorphic, no matter which doors she selected and no matter which door was opened to reveal which goat, she will have a 50% chance of winning by switching.

---

With regard to the objections of ChalkboardCowboy and the Law of Large Numbers, this law applies to “the result of performing the same experiment a large number of times”.  Remember that in the situation in which the contestant must assess the likelihood of winning as a result of switching, she has picked two doors, one door has opened and one goat has been revealed.  This opened door and revealed goat tells the contestant which particular subset of mini-games the mini-game that she is playing belongs to (in our example, MAC or MCA).  If we don’t care to distinguish between goats, then simply the opened door tells us (in our example ggC or gCg).

If we try to apply the Law of Large Numbers the way that ChalkboardCowboy is implying, then we don’t get to know which subset of mini-games apply with each iteration, we just know that it is one of the set [ACM, AMC, CAM, CMA, MAC, MCA] (or [Cgg, gCg, ggC] if we don’t care about the goat’s identity).  This means that we are not “performing the same experiment”.

It would be akin to walking into a room with three (fair) gambling machines, one with an average payout of 1/2, one with an average payout of 1/10 and one with an average payout of 1/100, selecting one at random and expecting to get a payout of about 1/5 from a single machine.  However, if you go in and play the 1/2 machine 1,000 times, then you will get close to an average payout of about 1/2.  Do it a million times and you’ll be even closer to 1/2.  (This example is not entirely analogous to what is going on with Reverse Monty Hall Problem, I know that.  I am just highlighting the fact that the Law of Large Numbers won’t work if you are playing different games.)

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So, the question I have is: did I somehow fail to make perfectly clear that I was talking about a decision made by the contestant after the door was opened (and hence after doors were selected) – a situation which, by your calculation, gives the contestant a 1/2 likelihood of winning as a consequence of switching?

And a further question is: how was my scenario substantively different to the original question raised by Craig F. Whittaker?

## Friday, 27 February 2015

### The Objections of chrysics - Part 1

Please note that since I wrote this article, I have been persuaded that the argument it relates to is wrong (meaning that chrysics and Mathematician and irishsultan and ChalkboardCowboy were all right from the start and I should have listened to them rather than arguing with them).  Fortunately, I didn't because, for me at least, this little intellectual journey has been far more interesting than it would have otherwise been.

The correct answer for the scenario as it is worded is not 1/2 but rather 1/3 (meaning that the likelihood of winning as a consequence of staying is 2/3).

---

Over on Reddit, I’ve been engaged in discussions with many “dumb repeaters” and a small cadre of intelligent, highly educated people who have provided useful feedback in the way of objections to what I’ve been writing about recently.

Because I find Reddit particularly unwieldy for more in-depth discussions, I will try to respond to a longer comment here.  It’s not intended that these be comprehensible for all readers, in fact it is pretty much directed at one person, but the challenge was made publically, so the response is presented publically.  If you have not already been engaged in the furore, I suggest looking back at earlier articles here and traipsing over to reddit.com to check things out.  Try r/badmathematics where I am in the running for a Golden Goat Award (I’d like to thank my family …), while you are there, you can soak up the arguments against my position, which I strongly recommend that you do.

The comment is from u/chrysics but this person has also commented on this blog:

neopolitan – 100% of the time when a Red door has been opened to reveal Mary will the Red door have been opened to reveal Mary

chrysics – Nobody is disputing that. But this can occur in two ways:

1.    Red Mary can be the only choice the host was allowed to make

2.    The host had a choice between Red Mary and Green Ava

Now, the probability of either arrangement of goats is equal. But the arrangement of goats required for case two only gives you a 50% chance of a Red Mary game, while case 1. gives you a 100% chance. Two thirds of the time that Red Mary shows up, it's because you were in case 1.

See Q10 of Marilyn Gets My Goat, this just says the same thing.  Also, I address this issue at Monty Two Face.

neopolitan – It's a single iteration, one-shot scenario,

chrysics – That has nothing to do with anything.

Naturally I disagree.  I very specifically wrote The Reverse Monty Hall Problem to make it a real situation in which there is a single iteration of the game.  I peopled the scenario with real people (and tried to make that even clearer in Marilyn Gets My Goat).

neopolitan – Personally, I don't really care if the goats are interchangeable or not,

chrysics – Well you should, because - unless you're using the term 'interchangeable' to mean something different from its established meaning in probability - it changes the outcome of the game.

I’m probably not using the term 'interchangeable' in a strict sense.  All I mean is that in a real version of the game, there will be two goats and these goats are not the same goat and thus these real goats are not interchangeable.  (Equally, there is not 2/3 of a goat behind each door and 1/3 of a car.)

neopolitan – I can build my argument on them being the same goat hidden behind two doors, if that is what you prefer.

chrysics – It's not. I don't care about the goats. There is one door with a car and two losing doors. If both losing doors are selected, the host decides which to open randomly; otherwise the host opens the only losing door selected. That is all that is important. The losing doors can hold separate goats, the same goat, nothing, a guitar in one and a performance art troupe behind the other, it's completely and utterly irrelevant.

I agree.  But some people arguing wanted to stick with hypothetical goats, strangely divisible goats, even when I was trying to talk about a notionally real situation.

neopolitan – Which do you actually want me to argue? That the goats are interchangeable or not interchangeable?

chrysics – The interchangeability of the goats is a function of the rules of the game. You have consistently described a set of rules in which the goats are interchangeable. But you keep saying they're interchangeable, and on that one occasion you also described a version of the game in which the rules differ in a subtle-but-important way from your other descriptions, which just so happens to make the goats non-interchangeable.

I don't want you to argue that the goats are interchangeable, nor do I want you to argue that the goats are not interchangeable. I want you to clarify which set of rules you want to play by. Is the host required to open Red Mary whenever that is a possibility, or is the host permitted to open Green Ava even when Mary is behind the red door? If the latter, then I also want you to explain why you insist that the goats are not interchangeable.

Actually, this simply isn’t true.  When I first mentioned the concept of “interchangeability”, I wrote (emphasis added):

(as wotpolitan) It is easy enough to overcome logically. I say my scenario is a single instance, one-shot game. Then someone says, aha, but he could have opened a different door. But my point is that he didn't. He opened a specific door revealing a specific goat and it is this specific scenario that the contestant has to deal with. Some say that the goats are interchangeable, and I say that might be the case with a purely imaginary game, but I put the contestant in a shopping centre with me (the dishevelled amateur philosopher), a relatively real person, conducting the experiment with goats that I keep trying to tell people are real, and patently not interchangeable. They say, if you run it multiple times you'll get 2/3, I agree and say that it's different if you run it just once. I show how it works with conditional probability, they say that that would be fine, but conditional probability doesn't apply - perhaps they are right about that, but it seems to apply fine from what I can see.

I am fine using whichever you prefer, but if you have no preference, let’s go with non-interchangeable goats called Mary and Ava, because it prevents the problem mentioned above.

In the scenario I describe at both The Reverse Monty Hall Problem and Marilyn Gets My Goat, there is no obligation on the part of the host to open the Red Door if Mary is behind it, unless the car is behind the other selected door in a Red-White or Red-Green scenario.  However, if the host has opened the Red Door to reveal Mary, then there is an obligation to consider only scenarios in which the host could open the Red Door to reveal Mary for that particular instance of the game (which is the only instance as I formulated the scenario).

neopolitan – I think your comment there is simply irrelevant because the host makes one decision and one decision only to open one door

chrysics – In potential Red Mary games, the host only make a decision half of the time - the other half of the time, the host has no choice at all. And half of those decisions result in the outcome of the game not being Red Mary. Of the potential Red Mary games, only 3/4 are actual Red Mary games. In 2/3 of those 3/4, you win the car by sticking with your initial choice. The other 1/4 are irrelevant, as they're not Red Mary games at all.

This looks to be based on a misunderstanding of what a Red Mary game is.  I take responsibility if I was insufficiently clear, but a Red Mary game is a game in which Mary is behind the Red door, irrespective of whether she is revealed by the host as being behind the Red Door.

However, if we call these Revealed Red Mary games, you are right in that the host makes the decision whether to open the Red Door or nor half the time and is obliged to open the other half of the time.  I address the problem highlighted here at Monty Two Face.

chrysics – Let me break it into stages, so you can tell me where you think I go wrong. (We'll assume for the sake of consistency and simplicity that the contestant's chosen doors are Red, the leftmost door, and Green, the rightmost door (neopolitan comment – agreed, this is consistent with the scenario as laid out in Marilyn Gets My Goat)

1.    A Red Mary game denotes a game in which the host opens the Red door, revealing the goat Mary to be behind it (neopolitan comment – not agreed, this is a subset of the set of games I call Red Mary games, we could call this a Revealed Red Mary game, I will update your text below to reflect this, updates in brackets)

2.    The goats and the car are arranged behind the doors at random

3.    A (Revealed) Red Mary game may occur when the goats and car are arranged MAC or MCA

4.    Both of those arrangements are equally likely:
P(MAC) = P(MCA)

5.    No other arrangement can produce a (Revealed) Red Mary game:
P((
Revealed)RedMary| (!MCA && !MAC) ) = 0

6.    The contestant has no knowledge of the arrangement until the host opens the door

7.    At that point, the contestant has no knowledge beyond what is revealed (Mary is behind the Red door) and anything that can be deduced from that

8.    If the arrangement is MAC, there is a 100% chance that the host will open the Red Door, producing a (Revealed) Red Mary game

9.    If the arrangement is MAC, there is no possible result other than the one given above:
P((
Revealed)RedMary|MAC) = 1

10. If the arrangement is MCA, there is a 50% chance that the host will open the Red Door, producing a (Revealed) Red Mary game

11. If the arrangement is MCA, there is a 50% chance that the host will open the Green Door, producing a (Revealed) Green Ava game

12. If the arrangement is MCA, there is no possible result other than the two given above:
P((
Revealed)RedMary|MCA) + P((Revealed)GreenAva|MCA) = 1, and they are equally likely:
P((
Revealed) RedMary|MCA) = P((Revealed)GreenAva|MCA)

13. As MAC and MCA are equally likely arrangements,
P((
Revealed)RedMary|MAC) = P((Revealed)RedMary|MCA) + P((Revealed)GreenAva|MCA)

14. Using:
P((
Revealed)RedMary|MCA) = P((Revealed)GreenAva|MCA), and
P((
Revealed)RedMary|MAC) = P((Revealed)RedMary|MCA) + P((Revealed)GreenAva|MCA), we arrive at:
P((
Revealed)RedMary|MAC) = P((Revealed)RedMary|MCA) + P((Revealed)RedMary|MCA)

15. Using:
P((
Revealed)RedMary| (!MCA && !MAC) ) = 0, it can be deduced that:
P((
Revealed)RedMary) = P((Revealed)RedMary|MCA) + P((Revealed)RedMary|MAC)

16. Using:
P((
Revealed)RedMary|MAC) = P((Revealed)RedMary|MCA) + P((Revealed)RedMary|MCA), this may be rewritten as:
P((
Revealed)RedMary) = P((Revealed)RedMary|MCA) + [P((Revealed)RedMary|MCA) + P((Revealed)RedMary|MCA)] (the square brackets denote nothing other than where the equation has been re-written)

17. The re-written equation P((Revealed)RedMary) = P((Revealed)RedMary|MCA) + P((Revealed)RedMary|MCA) + P((Revealed)RedMary|MCA) accounts for all (Revealed) Red Mary games, and consists of three equal contributions.

18. The first of these contributions accounts for the (Revealed) Red Mary games arising with the arrangement MCA.

19. The second and third contributions - each of which is equal individually to the first - account for all remaining (Revealed) Red Mary games, in which the arrangement is MAC. Collectively, this MAC. Collectively, this accounts for two thirds of (Revealed) Red Mary games

20. For any given (Revealed) Red Mary game, there is a 2/3 probability that the arrangement is MAC.

I am not well versed enough in this notation to confidently identify the precise step in which the error (as I see it) creeps in.  However, the issue is that we are talking about a situation in which there is a Revealed Red Mary, meaning that the only options are that we are playing an MAC game or an MCA game.  I agree that we will only see the Revealed Red Mary half the time when we are playing a Red Mary game of the form MCA (where Red and Green were selected).  But I disagree that this means that we can eliminate the MCA games that we would not see.  This is the point I am trying to make at Monty Two Face.  I think that we need to compare Red Mary games, not Revealed Red Mary games.

By rejecting Red Mary games that are also Revealed Green Ava games (ie 50% of the MCA games), which is effectively what you are doing, we skew the result.

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Update:

As mentioned in the comment below, I was not seeing the wood for the trees when reviewing chrysics post.  I was convinced that his core argument was that I was wrong, so I completely overlooked that his conclusion was that the probability of a Red Mary being revealed as a result of a MAC distribution of goats and car is 2/3 - a point I made in Marilyn Gets My Goat (see Q10 and also Q16).  Therefore, I actually agree with him.  No wonder that I could not find any error (a fact that was a little disconcerting, since it all seemed pretty straightforward).

That said, my last few comments above still stand, I just wasn't properly addressing what chrysics was saying.

### Monty Two Face

Please note that since I wrote this article, I have been persuaded that the argument it relates to is wrong.  The correct answer for the Reverse Monty Hall Problem is not 1/2 but rather 1/3 (meaning that the likelihood of winning as a consequence of staying is 2/3).  Note also that I has already accepted in this article that I had been wrong, but even in that acceptance of being wrong, I was wrong.  But I'm right now ... just not in what is written below.

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When I wrote this article, I thought 1) that it was right and 2) the scenario describedt was analogous with the Reverse Monty Hall Problem.  I was certainly wrong with 1) and if so, I might well be with 2).  I will need to think carefully though to see if it the below is analogous with the Reverse Monty Hall Problem, because if it is (which I have come to doubt) then I am wrong about the Reverse Monty Hall Problem.

Interestingly, I came to believe that I am wrong about what follows via conditional probability which I am told is not applicable to the Monty Hall Problem - so this is an argument in support of 2) being wrong.

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Say that Monty has two coins, one is a standard fair coin with a head and a tail.  The other is a fair coin, in that it is balanced so as to land on either side with equal likelihood, but it has a head on both sides.

Say that Monty selects a coin at random and he tosses it and he records the result.  He repeats this process nineteen times to get the following result (random tosses generated by a spreadsheet):

 Table 1

Now, say that we want to look only at the instances in which there might be confusion as to which coin, Monty tossed.  We would have to eliminate all the times that tails was tossed.  So we have:

 Table 2

What is the likelihood that, given that Monty tossed a coin and got heads that it was the HH coin?  The answer is 2/3.  In the relative small sample we have here, it looks like this (with HH in red):

 Table 3

Heads appeared 15 times, and the HH coin was responsible for that appearance 9 times.  The fact that each of these events were independent and the sample size is small make it quite likely that the HH coin would have been observed somewhere between 8 and 12 times.  With a sample size of 500, the result was still closer to 0.7 than to 0.66.

However, if Monty stopped on the first run, what would have been the likelihood that the HH coin was used, if a heads had resulted?  Sure the result was a heads, but I’m not eliminating the possibility of there having been a tails.  I’m just forcing the scenario to one in which there was a heads.

In this scenario, we only know that Monty selected the coin at random from a choice of two.  We must conclude that the likelihood that he picked that HH coin is 1/2.  This is because the sample space we would need to consider, from multiple iterations is that in Table 1, rather than that in Table 2.

An analogous situation applies in the Reverse Monty Hall Problem.  In the Red Mary scenario (as introduced in Marilyn Gets My Goat), HH is equivalent to the car being behind the Green Door and HT is equivalent to Ava being behind the Green Door.  In the latter case, Monty could have chosen to open the Green Door to reveal a Green Ava, but he didn’t.

This is presented in order to explain how, in a single run, we can get a 1/2 result while getting a 2/3 result over multiple runs.  Those arguing for a 2/3 result in a single iteration, one shot instance version of the game do so by incorrectly dealing with the possibility that Monty could have opened the other door if two goats have been selected by the contestant.

---

Basically, I overreached.  I was trying desperately to think of a scenario that could help explain a feature of the Reverse Monty Hall Problem, I thought that I had found one and I burst into print without mulling it over sufficiently.

The likelihood of the coin being a HT coin is given by Pr(HT|H) = Pr(HT ∧ H) / Pr(H) where:

Pr(HT|H) = the likelihood of the coin being an HT coin given that we see a heads

Pr(HT ∧ H) = the likelihood of the coin being an HT and our seeing a heads = 1/4

Pr(H) = the likelihood of seeing a heads = 3/4

So Pr(HT|H) = 1/3 and Pr(HH|H) = 2/3

We can calculate Pr(HH|H) directly too:

Pr(HH|H) = the likelihood of the coin being an HH coin given that we see a heads

Pr(HH ∧ H) = the likelihood of the coin being an HT and our seeing a heads = 2/4

Pr(H) = the likelihood of seeing a heads = 3/4

And so, Pr(HH|H) = 2/3

---

ChalkboardCowboy has a comment below.  Here is a table that is relevant to my response:

Note that MC = Magic Coin (which is simulated to provide Heads with a likelihood of 1/10^12, using an American trillion) and NC = Normal Coin (which is simulated to provide Heads with a likelihood of 1/2).

If I see a heads, I am going to be pretty confident that we have a normal coin and not ChalkboardCowboy's magic coin.

(Note while I was wrong above, ChalkboardCowboy was also being a little silly with his magic coin example.  That said, no harm done, I found my error eventually - largely thanks to "drip" who forced me to go over the workings that I should have done from the start.)

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The dice do actually model something rather close to the Reverse Monty Hall Problem, so not only was I wrong, I was wrong about why I was wrong.  Since I now conform with the majority opinion, if I am now wrong again about being wrong about why I was wrong, I am at least wrong in the same way as almost everyone else.