## Friday, 27 February 2015

### 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.

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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

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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.

1. By your logic, the following argument is also correct; do you agree?

I have two coins. The first is a normal fair coin that lands heads half the time and tails half the time. The other is a magic coin that has a one in a trillion chance of landing heads, and lands tails the rest of the time. The coins are otherwise indistinguishable.

I choose one of the two coins at random, flip it, and see that it has landed heads, then I stop.

In this scenario, we only know that I selected the coin at random from a choice of two. We must conclude that the likelihood that I picked the magic coin is 1/2. 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.

1. What? This is a bizarre response on your part. If you plug the results from your magic coin into my tables, you don't get anything close to 1/2 result. Until you run quite a few iterations, you are unlikely to get any heads at all (although it remains possible). I have my suspicions as to why your reaction here is so bizarre, so please don't think I am having a go at you because I have put your "magic coin" into an additional table above, along with a little commentary.

2. I'm not talking about an iterated sequence of throws. This is a one-time event!

Specifically, I'm using the argument from your sixth paragraph ("However, if Monty stopped on the first run..."). Suppose the first toss is heads, and I stop. What is the chance that it was the magic coin that I flipped? I claim that your argument says it's 1/2, but maybe I'm misunderstanding you.

3. I think you are misunderstanding me. I'm glad that, finally, someone is willing to accept that as at least a possibility!

Does the table not indicate that you were misunderstanding me? I thought it was pretty clear and I am wondering how I could make it more clear.

4. No you're not misunderstanding him ChalkboardCowboy, according to neopolitan's argument the answer is 1/2, and that is based on this statement :
"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 NC (changed from original) coin is 1/2"

With your MC it's a trillion times more likely heads resulted from the normal coin, with neopolitan's 2 coins it is twice as likely heads resulted from the HH coin - but in both games there are still just 2 coins and according to neopolitan MOnty selected the coin at random, making the answer 1/2 in both games (It's nonsense of course, but at least your argument is consistent nonsense)

5. In my last post I didn't mean your argument ChalkboardCowboy was nonsense, but your example was consistent with neopolitan's example, which was of course nonsense..

6. Marley52, perhaps you might want to read things a little more closely. I quite clearly explained why I don't agree with ChalkboardCowboy's extrapolation of the example. It seems that you are just trying to play the game with the big boys without having much of a clue about the rules of the game. It's amusing that, in another comment here, you accuse me of trolling.

If you want me to take that post seriously and address it, I suggest you apologise. At the moment, it simply seems to not be worth my while.

7. Bo neopolitan you didn;t explain (you never do) you just said "If I see a heads, I am going to be pretty confident that we have a normal coin and not ChalkboardCowboy's magic coin." Why? What;s the difference between Table1 and Table4 (ChalkboardCowboy's table)?

Similarly I say "If I see a heads, I am going to be pretty confident that we have a HH coin and not a TH coin", and that's because it's twice as likely I'll see heads from a HH coin as a TH coin, therefore I'm twice as confident it's a HH coin and the TH coin.

8. Hi neopolitan, I think you misunderstand ChalkboardCowboy's first post. I believe he was trying to provide a counterexample to the argument in the sixth paragraph. Specifically, in the argument of your sixth paragraph, the answer of 1/2 is independent of the parameter p, which is the probability of landing a heads on the biased coin. ChalkboardCowboy simply substituted p with a ridiculously low value of one in a trillion to show you that if the argument in your sixth paragraph is correct, then the answer should still be 1/2. The fact that you don't agree that the answer is 1/2 shows that the argument in the sixth paragraph is wrong.

9. On reflection, I agree that when you fiddle with the coins, the argument for there being a simple 1/2 likelihood based only on the fact that there was a selection from two coins is shown to be insufficient.

However, this was not actually my main thrust. My main argument (one which ChalkboardCowboy inadvertently made far more strongly and effectively than I did) was that we cannot ignore the distribution of Tails when trying to make an assessment of the likelihood of having picked the HT coin for tossing.

If you look at Table 1, it contains all the outcomes. If you look at Table 4, it contains all the outcomes. In both instances, we use ALL the outcomes to assist us in assessing the likelihood having used the HT coin in a single toss. If we get a heads, it's 1/2, because 25% of the time we expect to see a heads from an HT coin and 50% of the time we expect to see a heads from the HH coin and 25% of the time we expect to see a tails from the HT coin.

With the magic coin, we don't have the same situation. We expect to see the 50-50 H-T split with the natural coin and a 1-999,999,999,999 H-T split with the the magic coin. Therefore, if we see a heads, the likelihood is very low that it's the magic coin.

But, and I can't stess this enough, the example is given only to show that with the Reverse Monty Hall Problem, given that it is a single iteration, one shot instance of the game, we cannot ignore the fact that, with any Red May game, there is a possibility that Ava will be revealed rather than Mary. (There isn't with a *Revealed* Red Mary game, because Red Mary has been revealed, but as I've said elsewhere what revealing Red Mary tells you is that the Mary is behind the Red Door and, therefore, you now know that you are not playing any of the games in which Ava or the Car are behind the Red Door.)

10. Hi neopolitan, could you elaborate on how you got the value of 1/2 from this part?

"If we get a heads, it's 1/2, because 25% of the time we expect to see a heads from an HT coin and 50% of the time we expect to see a heads from the HH coin and 25% of the time we expect to see a tails from the HT coin."

I would like to know how the 25%, 50% and the 25% come in and result in 1/2.

I am genuinely curious as to how you obtained 1/2, because it still seems to me that the answer of 1/2 from the tables and you calculations are independent of the probability of landing a heads on a biased coin.

I would also like to know whether you can assign an exact numerical value to the probability of getting a magic coin given that a heads was landed, based on your algorithm.

Many thanks.

11. Hi drip, thanks for asking this question. It forced me to go back and revisit what wrote here, during which I found an error in my thinking. I've updated accordingly, acknowledged my error and I now realise that this was not a good example for showing what I wanted to show (perhaps this approach was flawed anyway).

However, I still think that it is possible to end up with 1/2 with the Reverse Monty Hall problem because, inter alia, I am not convinced that my equation at "The Monty Hall Problem and Conditional Probability" is inapplicable. I'll avoid going into depth at this time because I want to ponder it some more (in other words, I want to make sure that I am not deep in a hole before I decide whether to keep digging).

12. Oh and you wanted an answer on the magic coin:

Pr(MC|H) = Pr(MC ∧ H) / Pr(H) = 1/10^12 / ~1/4 = ~4/10^12

In other words it's rather unlikely to be the magic coin if you see a heads, and you'll see tails very close to 3/4 of the time.

2. neopolitan, I mentioned on your other blog that you don't understand conditional probability, and here you are once again demonstrating that very fact. Other posters here (on your blogs) and on reddit have repeatedly shown where you are going wrong in your analysis and yet still you persist. If you're not trolling you must be delusional.

Anyway it's time to put this ridiculous nonsense of yours to rest once and for all but before I do that, did you realise that if you add a TT coin to your "Monty Two Face" (making "Monty Three Face"?) you have replicated Bertrand's Box Paradox (BBP) which was published by the French mathematician Joseph Bertrand (1822–1900) in his 1889 text Calcul des ProbabititÃ©s. It is ironic that you are invoking BBP (perhaps unwittingly) to support your position because in reality it does the exact opposite,

Anyway to the mathematical proof that your conclusion "..... that the likelihood that he picked that HH coin is 1/2" is terminally and fatally flawed.

Let P(HH) = Probability that the HH coin is picked
Let P(TH) = Probability that the TH coin is picked

We want to calculate P(HH|Heads) - the (conditional) probability that the HH coin was picked given Heads is observed.

So. P(Heads|HH) = 1 (Probability Heads is observed given the HH coin is picked)
And P(Heads|TH) = 1/2 (Probability Heads is observed given the TH coin is picked)
And P(HH) = P(TH) = 1/2 (each coin is equally likely to picked)

So plugging these values into Bayes equation gives:

P(HH|Heads) = [1 x 1/2]/[(1 x 1/2) + (1/2 x 1/2)] = [1/2]/[3/4] = 2/3.

(You can apply a similar mathematical proof to your RMHP to get an answer of 2/3 for staying)
Give it a rest, you were incorrect when you first posted your RMHP problem and you're still wrong now.

1. This comment is currently not being addressed due to the abusive nature of the commenter (and the fact that there's nothing new in it).

2. There's only a mathematial proof that the answer is 2/3, that's all. Of course if you don't believe in probability theory then I suppose there's nothing in it for you.
Why don''t YOU give a mathematical proof that the answer is 1/2 instead of making baseless assertions? You can't, because one doesn't exist.

3. You were right Marley and so I apologise, the answer is 2/3. That said, I would have actually read further if you weren't being a dick about it. Fortunately, drip took a better approach and that led me to find the error myself. Only just now have I noticed that you were doing the conditional probability thing - which, if I had read it earlier, would have saved me some grief.

Oh well. You said below that I was being a clown and, in this article, it is indeed true that I was. I now need to ponder the rest to see if that is the same there too - which is by no means automatically the case merely because I screwed up here.

4. Since the RHMP and MHP are merely variants of Bertrand's Box Paradox (as is your coin example) you'll find the answer is the same. The Bayes equation for RHMP and MHP is almost identical to the one above.
I apologise if I came over as a bit of a dick - I'm sure you haven't got enormously large feet :)

5. I shouldn't complain, I suppose, I almost certainly come across as a bit of a dick. I have been thinking about the coin example and the extent to which is it analogous with RMHP and MHP (while I should have been sleeping). I think the problem is that the coins artificially group the doors if we are going to consider them as an analogy.

I think that what it might effectively do (in analogous form) is force the contestant to select between two options, A and B, which represent either a single door or a combination of doors, where the combination of doors is artificially limited. The host knows which is which, but the contestant doesn't - although the contestant is given enough information to work out the limitation associated with the combinations, albeit in this coin game format (so it's not blindingly obvious what that limitation actually is). Then the host opens a door revealing a goat (without telling the contestant which group the opened door was from - equivalent to limiting the game to those tosses which result in a heads). Then the contestant has to guess whether it is better to switch or stay (noting that, at this point, the contestant doesn't know which of the remaining two doors is "hers").

It may be possible to work through this to arrive at the precise likelihood of benefitting from a switch, but it's a little mind-bending and not quite the MHP or the RMHP, but rather a blend of the two. But we know that the answer would be 2/3 (because we indicated that the limitation to the available combinations was analogous with the coin toss game).

3. Over on reddit, I said this: "I haven't actually heard you say it out loud, but you do realize that with this statement you are unambiguously claiming that the Law of Large Numbers is false, right?"

Your reply was, "I'm suggesting, at worst, that you are misapplying it. A new article is on the way to explain precisely this point."

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I don't think you've attempted in this article to explain how I'm misapplying the Law of Large Numbers. I recognize that you haven't had formal mathematical training, so I assume that your understanding of the LoLN is based on an English-language paraphrase of it, such as the following from Wikipedia:

"Borel's law of large numbers, named after Ã‰mile Borel, states that if an experiment is repeated a large number of times, independently under identical conditions, then the proportion of times that any specified event occurs approximately equals the probability of the event's occurrence on any particular trial; the larger the number of repetitions, the better the approximation tends to be."

I believe that's an excellent description of of the law, and this is NOT one of those cases where natural language is just not suited to convey the mathematical content of a statement. In other words, I don't believe your lack of mathematical training should be an obstacle to your understanding of the LoLN nor keep you from explaining exactly how you think I'm misapplying it.

Showing me how I've misapplied the LoLN is fundamentally NOT the same as giving a new (or repeating an old) argument of your own, wouldn't you agree? Where exactly is my error in claiming that your arguments directly contradict the LoLN?

1. Someone on reddit asked neopolitan what it would take for him to admit he was wrong, I don't recall what his response was but it clearly isn't;
a) Evidence. Statistical data shows that answer is 2/3, (and he even accepts this but perists with his assertion that in a single game the answer is a 1/2 even though this mathematically impossible)

b) Reason. Numerous commentators have provided simple logical proofs that in a single trial the answer is 2/3 (and even he admits that you are twice as likely to observe 'heads' from a HH coin than a TH coin)

c) Maths. As in b) mathematics proves the answer is 2/3, he waves his arms and says "nothing to see here" as id maths is matter of opinion.

What's left; magic, intuition, wishful thinking? I suppose we could try neopolitan's own tactic and just make a baseless assertion: "the answer is 2/3 because I SAY IT IS!"

Seriously though, he says he's not troling, then what is he doing? He's certainly not making sense by any known definition of the word.

2. ChalkboardCowboy,

I can certainly see how, taken out of context, what I wrote could be taken as meaning that I had some interest in proving that you are misapplying the Law of Large Numbers. However, the strand that we were commenting on was started by OmarDiamond who struggles with the idea that the probability of an single instance of the Reverse Monty Hall Problem might you a 1/2 likelihood of benefitting from switching while multiple instances give you a 2/3 likelihood of benefitting from staying. That's the point that I was intending to address. I don't have any abiding interest in the Law of Large Numbers, nor in investigating your relationship to it.

That said, I've talked about how in the Reverse Monty Hall Problem, you are actually playing one of six mini-games and you don't know which one (three mini-games if you don't distinguish between goats, or if you just open the losing door to see the same empty room). Once you open a door, you eliminate all but two of the mini-games. The Law of Large Numbers in this instance would be including all six (or three) mini-games, while a post door opening single iteration, one shot instance of the Reverse Monty Hall Problem would only involve two mini-games. So, I think you are looking at different things and so the Law of Large Numbers doesn't apply.

3. neopolitan,

We are all struggling with "the idea that the probability of an single instance of the Reverse Monty Hall Problem might [give] you a 1/2 likelihood of benefitting from switching while multiple instances give you a 2/3 likelihood of benefitting from staying." The reason we struggle with it--okay, this isn't the right word after all--the reason we reject it is because it directly contradicts a mathematical theorem, the Law of Large Numbers.

I'm not expecting you to have an "abiding interest" in the LoLN for its own sake--after all, you aren't a mathematician. But you do manifestly have an abiding interest in making this argument of yours, and since the LoLN contradicts it, I would at least expect you to be interested in it on that basis.

Your second paragraph doesn't make much sense to me. The situation under consideration here is, "the door has just been opened in a Revealed Red Mary game", and the event whose probability we're considering is, "the player can win by staying". The LoLN says that, if we see that situation happen a large number of times, then the proportion of times the player can win by staying is the SAME as the probability that the player can win by staying on any single instance considered in isolation.

When you say, "Once you open a door, you eliminate all but two of the mini-games. The Law of Large Numbers in this instance would be including all six (or three) mini-games, while a post door opening single iteration, one shot instance of the Reverse Monty Hall Problem would only involve two mini-games," it sounds like you don't think the LoLN can be applied to the exact same situation as the one-off single instance situation. But there's absolutely no reason we can't.

4. Like Omar Diamond I struggle with how the 'mathematically impossible' (Average[1/2 +1/2+.,.1/2] = 2/3) is supposedly proven by your HH and TH coin example.

1) Take a set of random coin flips of a HH coin and a TH coin and record the frequency of which coin was flipped AND the outcome of the flip (Outcome: 50:50 for distribution of HH/TH coins, and 75:25 for distribution of 'Heads'/'Tails')

2) Take a subset of the above data including only those coins flips which landed 'Heads' (the only flips we're interested in) and note the frequency of which coin produced those 'Heads' (Outcome: 2:1 distrubtion for HH/TH)

3) Take the frequency of which coin was flipped from Set1 and compare that with the frequency of which coin produced 'Heads' from t(he completely different) Set2, wave your magic wand and hey presto!-, there's your "proof". Absolutely laughable, even a 10 year old wouldn't be fooled by that.

When you said you were one of "the big boys" were you referring to the size of your feet, because you're a real clown.

Feel free to comment, but play nicely!

Sadly, the unremitting attention of a spambot means you may have to verify your humanity.