Saturday, 28 February 2015

The Objections of chrysics - Part 2

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.

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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?