User talk:Coffee2theorems

Welcome!

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Request for edit summary

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probit graph

Hi, I propose updating the probit graph. The probit is a function with domain [0,1] and range on the real line. The axies should be labeled as such. Were this an article on statistics (which it isn't it's an article on the function) your labels would be great. Pdbailey 00:51, 4 October 2006 (UTC)[reply]

I removed labels altogether, making it a simple plot of

\Phi ^{-1}:[0;1]\to (-\infty ;+\infty )\!

without any units. The graph in logit doesn't have any labels either. -- Coffee2theorems | Talk 15:00, 12 January 2007 (UTC)[reply]

Okay, that's fine, but here's a question for you. If zero is in the domain, what value does it take on in the range? Pdbailey 19:35, 7 April 2007 (UTC)[reply]

Indeed the domain isn't really [0,1] but (0,1) unless you're willing to say that it maps 0 and 1 to negative and positive infinities using extended reals, but what I meant was really that there are labels "0.0" and "1.0" in the plot instead of "0.0001" and "0.9999". I thought you were just poking a bit of fun at my sloppy notation here and decided a response wasn't necessary, but then I just noticed that the probit article at the moment says that the range of the sigmoid function is [0,1] (!). I'll change that to (0,1).. -- Coffee2theorems (talk) 15:09, 1 May 2008 (UTC)[reply]

Inductive Logic

Hey, I am with you on the inductive logic edit. The last sentence added makes it look redundant. If you need any support on that, just let me know. Pjwerner (talk) 12:29, 25 October 2008 (UTC)[reply]

Oh, it was reverted, I didn't notice. I'm not really interested in edit warring, but the reason for the reversion only attempts to address half of the reason for the original edit, so I think I'll prod it once more, this time to only drop the strange allusion to Russell's teapot. -- Coffee2theorems (talk) 13:07, 25 October 2008 (UTC)[reply]

Minimum of exponential variables

Hi. I reverted a change to the exponential distribution article about the minimum of exponential variables, and then noticed that it was your change, and that you are a statistician, and I am not. This prompts me to notify you of my change, in case I am mistaken. I think that you ran afoul of the notation in the article, where the rate $\lambda$ is used in the pdf like this: $\lambda e^{-\lambda x}$ , instead of the alternative definition of $e^{-x/\lambda }/\lambda$ . -- Coffee2theorems (talk) 16:07, 8 January 2009 (UTC)[reply]

In fact the change I made assumed that what was there already was OK (which i didn't think about) and was to show the result of the complicated formula for a special case where the formula is simple ... in the correct version the rate goes up by a factor of n while the mean is divided by n. Your correction to treat the rate parameter properly gives a much simpler formula for the rate which means that the speical case of equal rates us probably not required. Anyway what is there now seems to be correct. Melcombe (talk) 17:03, 8 January 2009 (UTC)[reply]

You maybe interested in the Article Rescue Squadron

I notice that you are part of Category:Inclusionist_Wikipedians. I would like you to consider joining the Article Rescue Squadron. Rescue Squadron members are focused on rescuing articles for deletion, that might otherwise be lost forever. I think you will find our project matches your vision of Wikipedia.

Ikip (talk) 00:53, 21 February 2009 (UTC)[reply]

MHP Mathematical formulation.

Hello, I have responded to your comments on the proposed Math. Form. for the MHP. Thanks for taking your time to reply, if you care to. glopk (talk) 17:18, 1 July 2010 (UTC)[reply]

Monty Hall problem

Please note that formal mediation is re-starting. Your name has been mentioned as a recently active editor [1]. Let us know if you wish to be a particpant. Sunray (talk) 00:40, 15 August 2010 (UTC)[reply]

I'll pass on this one. I tried for quite a while to get some sense into the MHP discussions, but it does not seem to be going anywhere and it's an endless time sink. People there literally produce reams upon reams of talk page text, much of which is quite difficult to interpret in the way the writer intended (particularly Gerhardvalentin and Glkanter). Looking at the mediation page, it seems to me that nothing has changed there. I do not think Wikipedia MHP talk is an efficient (i.e. good effect/effort ratio) use of time. Maybe it will become less of a quagmire in a few years. -- Coffee2theorems (talk) 13:20, 20 August 2010 (UTC)[reply]

Bayes Theorem

Hi, I've reverted your recent deletion. I think the text was necessary to distinguish between the technical sense of "prior" and the more familiar meaning of the word. The justification you've given in the edit summary doesn't make sense to me. Happy to discuss further in article Talk, MartinPoulter (talk) 19:48, 30 January 2011 (UTC)[reply]

Invitation to comment at Monty Hall problem RfC

Because of your previous participation at Monty Hall problem, I am inviting you to comment on the following RfC:

Talk:Monty Hall problem#Conditional or Simple solutions for the Monty Hall problem?

--Guy Macon (talk) 16:53, 8 September 2012 (UTC)[reply]

No email?

Any particular reason you don't have email enabled? Just curious. -- Rick Block (talk) 04:58, 30 September 2012 (UTC)[reply]

I think I reasoned that it's unnecessary, and that it's better to keep Wikipedia matters on Wikipedia. I also don't check e-mail that often anymore, because most online communication tends to happen using methods nowadays. -- Coffee2theorems (talk) 10:55, 30 September 2012 (UTC)[reply]

I suppose this doesn't have to be private. I just wanted to thank you for your efforts on the MHP. -- Rick Block (talk) 15:57, 30 September 2012 (UTC)[reply]

Oh. Thanks! :) -- Coffee2theorems (talk) 18:01, 30 September 2012 (UTC)[reply]

Devlin

On the talk page of MHP this thresad is left unsolved, that's why I bring it again to your attention:

Sorry, Coffee2theorems, often people come with alternative descriptions as if they would make the problem more understandable. As if the original setting is not simple enough as it is. Anyway, if you like to use an alternative setting, like yours, with tiny doors, etc, you firstly have to show it's equivalent to the MHP, and guess ... it isn't. That's often the point, someone gives an alternative description, from which they believe it is equivalent to the MHP, and then proves their different way of solving with this alternative description. Why are people that reluctant to admit Devlin's combining is nonsense? Nothing particularly difficult there. I will spell it out again. Devlin reasons: 1) the not chosen doors 2 and 3 have together probability 2/3 on the car (what he omits to mention, or simply doesn't realize, is that each has probability 1/3); 2) the opened door 3 has probability 0 on the car. To me this is a contradiction. And to you?

Please comment. Nijdam (talk) 09:02, 11 October 2012 (UTC)[reply]

OK, so you don't like a model with indistinguishable doors. I can't say I'm a big fan of that approach either. Nevertheless, there are many possible ways of interpreting the MHP, whatever you or I may think of them. If you attempt to justify one interpretation as The One True Way with any seriousness, you'll probably end up conceding that it is a subjective notion, and that other more or less reasonable people may well disagree.

You can make a Devlin-type solution work with distinguishable doors, too. You may call it something else, if you dislike Devlin's article. Maybe you'd call it "Gill's combined doors solution" or something. Call it whatever you wish; I call it a Devlin-type solution, to distinguish it from an Adams-type solution (both with combined doors). The "-type" suffix is a concession to avoid confusion; I previously called the same thing simply "Devlin's solution". A complete Devlin-type solution would look like this:

If there is to be a definite solution at all, the answer must be the same no matter how the doors are numbered in the problem statement. In particular, the probability of winning by switching has to be the same whether the host opened door 2 or door 3. The opened door number therefore must be irrelevant (the answer does not depend on it), and we may pretend that the player only knows that one of the doors was opened, but not which. One can e.g. imagine a blind player, who is told that the host opened one of his doors to reveal a goat, but the number wasn't said.

When the host opens one of his doors (door 2 or door 3) to reveal a goat, no new information is revealed: the player already knew the host would do that. As no new information is revealed, the chances of the car being behind one of the host's doors are still 2/3. The car cannot be behind the opened door, so the chances of it being behind the remaining door must be 2/3. The player should switch to the remaining door.

It is not necessary to assume that a definite solution exists; it can be shown. In the Krauss and Wang version, the host picks randomly. One can simulate the game using e.g. a fair coin, and so estimate the probability of winning. That means that a definite probability exists for any particular way to number the doors. Further, it cannot matter whether the doors are numbered 1, 2, 3 or 1, 3, 2, because choosing one of 2, 3 by a fair coin is just the same thing as choosing one of 3, 2 by a fair coin. It's the same simulation either way, so the probabilities are the same, and a definite probability (one that does not depend on the numbering scheme) therefore exists. The problem does have a definite solution, and by the previous argument, it must be 2/3.

There shouldn't be any problem with this, except that it's rather long, and maybe could be written a bit more clearly. I think the third paragraph probably ought to be omitted if another complete proof is presented (surely one is enough). I don't know about the first paragraph.

The car is behind each of the doors door 1, door 2 and door 3 with probability 1/3, yes. However, it is behind the remaining door with probability 2/3, and behind the opened door with probability 0. You don't know which door each one is (it could be either door 2 or door 3), so there is no contradiction. -- Coffee2theorems (talk) 07:36, 12 October 2012 (UTC)[reply]

Unless, as according to Devlin, the opened door is door Number 3. Don't you see the basic idea of Devlin is wrong from the start and many people accept it as a solution. I read in one article even that the probability of the opened door 3 jumped to the remaining door 2. I'm not trying to push any particular point of view of mine, I'm fighting all the nonsense. I'm happy with any presentation of the problem and any correct solution, as long as it is made clear what they mean. Nijdam (talk) 09:01, 12 October 2012 (UTC)[reply]

I'm not attached to anyone's particular wording of a Devlin-type solution, and I doubt anyone else is either. I agree that clarity is good, but I'm not sure what that exactly means. Clarity to whom? Littering a simple argument with a thousand proofs of the obvious can obscure it. If the things proved are not so obvious after all, then I think clarity depends on whether you noticed that and whether you could fill in the gaps. Instead of seeking clarity to everyone, leaving it at "clear to many people" is also good, is it not? If one complete and clear argument is presented, then you can always rely on that if the others were unclear.

The probability does jump from door 3 to door 2; any conditional proof for a model with distinguishable doors shows that in the end, including the one above. The subjective probabilities of the blind player do not jump like that, true. But consider a friend in the audience who does see. For them, the probability does jump from door 3 to door 2. The blind player doesn't know that, but he does know that for his friend, the probability jumped from door 3 to door 2, or from door 2 to door 3. It jumps from the open door to the remaining door, the player simply does not know which is which. Now, if you stop pretending that you're blind (or take off the blindfold) at this point, your probability jumps as well; you "become your friend". I think that's the basic intuition used in these arguments. I'm not saying that people explicitly think of blindness or of an imaginary companion observer like that, merely that the underlying intuition works similarly. -- Coffee2theorems (talk) 13:30, 13 October 2012 (UTC)[reply]

Maybe the basic intuition is easier to see this way: Imagine that you, the analyst, are on the phone with the player. He says that he picked door 1. His car belief distribution is now (1/3, 1/3, 1/3). He tells you that the host opened a door. His car distribution must now be either (p, 1-p, 0) or (p, 0, 1-p), if there is to be a definite probability answer to the problem. The player knew in advance that this would happen – he knew he would come to believe p for the first door any which way – so p=1/3 is the only consistent possibility by the Crystal Ball Theorem. So his distribution is either (1/3, 2/3, 0) or (1/3, 0, 2/3). Suppose that the host opened door 3; in that case, the probability jumps from door 3 to door 2, and the distribution is (1/3, 2/3, 0). The player should switch. -- Coffee2theorems (talk) 14:30, 13 October 2012 (UTC)[reply]

I have the idea you do not see what actually is wrong with the combined doors (Devlin's) argumentation. So: he reasons: the combined doors have together probability 2/3 on the car. After the host has opened door 3 this door must have probability 0 on the car. But we know - as Devlin reasons - that doors 2 and 3 together have chance 2/3, hence (!?) door 2 now must have probability 2/3. How come you do not see this is nonsense? In this reasoning different probabilities are mixed. In formula: the original probabilities on the car are p1=p2=p3=1/3. Hence (combining doors) p2+p3=2/3. Now the host opens door 3, the new probability q3=0. Now comes the flaw: p2+p3=2/3, but we do not know q2+q3. Of course I know how you have to solve the MHP. But this does not work. The combining argument: p2+p3=2/3, is not helpful in any way, does in no way contribute to the solution, and hence not to the understanding. Only in a misleading way, by suggesting that the combined probability p2+p3 is a constant. Nijdam (talk) 10:24, 14 October 2012 (UTC)[reply]

I think I have probably complained about the same thing a long time ago. Yes, if you see it as mixing the probabilities, then it's obviously nonsense. The point is that there's a step of p2+p3 = q2+q3 in the middle (or equivalently, p1 = q1). I read "Your choice of door A has a chance of 1 in 3 of being the winner. I have not changed that." as the assertion that p1 = q1. Why read it nonsensically?

Some (e.g. Martin) claim that p1 = q1 is so obvious that it does not need to be shown. You and I disagree with that, yes. Yet I'd be fine with an explanation that says "it can be shown that p1 = q1" or "if there is to be a definite answer at all, p1 = q1 because...", or something like that, so that the argument does not appear to go through when it shouldn't. I'm not so sure I'd even object to simply stating that "p1=q1" like Devlin basically does in the above quote. It's not like it would be the only non-obvious step in a proof that ever existed. If there's one explanation where every step is simple and easy to follow, then isn't that enough?

As to why p1 = q1 seems obvious to some, it's probably because they don't feel that the host is giving anything away when he opens one of his doors. Ningauble's analogy was pretty good: someone accused of hiding a card up their sleeves and then rolling up one sleeve to "show" that they're not hiding anything does not end up convincing anyone; it's "obvious" that they can roll up one sleeve without giving away anything (except that it's not that sleeve). It may not be so obvious when you start thinking about it, but it does seem obvious if you don't think, and in this case the intuition isn't actually wrong (for exchangeable "sleeves"). -- Coffee2theorems (talk) 18:39, 16 October 2012 (UTC)[reply]

Glad to hear. I know all this, and I'm not willing to "fix" the solution. As I said: combining, i.e. stating that p1+p2=2/3, does not contribute to any understanding. And I'm quite convinced that most people - including Devlin - just don't (din't) notice the difference between p1+p2 and q1+q2. That's why it is so appealing, and that's why I consider the combining argument highly misleading. Nijdam (talk) 22:17, 16 October 2012 (UTC)[reply]

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