andrewducker | Monty Hall

Cannot believe no one told me the million doors version when trying to explain it to me. It would have taken me way less time to 'get' it if they had.

Yes... the million doors version makes it obvious doesn't it.

I confess I got it wrong when I first encountered it but I had had four or five pints.

It does seem to be exposing something in the human brain that doesn't quite match up to reality. Why my brain could see it properly for a million doors, but not for three doors is a mystery to me.

Humans are just really really really bad at statistics.

Even pigeons do better.

I prefer to think of it as humans having a wide range of abilities at statistics.

I first heard of this one when I attend the RI Christmas lectures given by Ian Stewart; my Mother (a maths teacher) didn't believe his proof (which had been framed for the, fairly young but keen, audience) although I did.

It's very irritating to explain to people in pubs, much easier with pen and paper :-p

I actually find it easier to explain in pubs cos you can use upturned empty glasses and a penny to illustrate it. Then people can see where the prize is and so they can see why they should switch - much like the diagram in the "decision tree" section of Andy's explanation link above.

Mind you, 1/3 is a really high probability anyway. You'll get the goat if you switch enough times that you'll be convinced it's random, because you're human

though I got it right - it was for 2 wrong reasons

1. [instant thought] It wouldn't be a 'problem' if the instantly obvious answer (doesn't matter) were true.
2. [immediately following thought] My brain rephrased it as a 1 in 2 (50% chance) once you have 2 doors left and know one of them is a goat and one is a car

I failed yesterday to satisfactorily explain to Steven the 'correct' (66%) ratio, so I'd be interested in this "million doors" explanation.....? Link?

I must ask my sister this one ans see what she says - she is numerically ultra-sharp and a statistician (insurance risk analyst) by trade (well at least til she went management...)

The link is part of the wikipedia article in the post.

For how I internalised it, see my response to

dpolicar below.

Shouldn't the poll really have the "It doesn't make any difference if I switch or not" option, given that's the answer most folks give to the MHP?

That would be "I do not believe that switching is beneficial"

I should maybe have clarified: "I do not believe that switching is beneficial" includes both, "I believe that switching is harmful (<50/50)" and, "It doesn't matter (50/50)" which is two of the three possible answers lumped together.

Maybe Andy intended the irony in lumping 2/3 of the wrong answers together in a problem where the right answer gets you the prize 2/3 of the time - after all, we all know how he likes to polarise his polls :)

(I still think that "It doesn't matter" should be separate given it's such a popular conclusion).

Yup. I've never heard anyone claim that not switching gave an advantage, so including it as an option didn't seem worthwhile.

Neither answer quite covers it.

I "believe" that switching is beneficial because of a combination of peer pressure, having found some argument that seems compelling that suggests it, and having sat down for an hour with a deck of cards and run Monty Hall simulations and counted the results.

But it has never "clicked" for me, which makes it a very unstable belief.

Then again, I have a lot of beliefs like that.

I should add that I also "believe" it on the philosophical grounds that probabilities (at least macroscopic linear ones) are quantifications of ignorance, not facts about the world, so it ought not surprise me that manipulating my ignorance will make things more or less probable.

But really internalizing that is hard.

I wrote some code this afternoon to prove it to myself.

What I discovered was that either I had chosen the prize with my initial guess, or I had not. Opening doors that I hadn't chosen did not change the correctness of my initial guess. And therefore, there was always a one in three chance that my original guess was correct.

Therefore, as there was only one other door to choose, that door must have a two in three chance of being correct.

I don't like the answers because "I believe" is not correct. I *know* and can *trivially prove* that switching is beneficial. There is no belief required.

There is belief involved, but there is no faith. One can believe true things. In fact, on my philosophy courses the best definition that I encountered for knowledge was "justified, true, belief".

I still don't like the term used here, as "believe" can also be in something untrue.

Yup. But as people on both sides would presumably think that their belief was true, using anything stronger would just have made people on the wrong side even more wrong :->

I probably would have worded it as "switching is beneficial" versus "switching is not beneficial" versus "monty who?"

Yeah, that works better.

Bayesian statistics uses the phrase "prior belief" all the time.

The book I told you which had the solution? Recommended you should always change. I voted you shouldn't because I was all for going with my book's answer, but I remembered wrong. I did remember it wasn't what you expected, so I double bluffed myself!

Although, it still feels uncomfortable, and anyways, its a travesty against goats...

Exactly - why won't anyone think of the goats?

I've heard the million doors explanation before and yet people still go, "Ah, I get it for a million doors and the odds are 999,999 to 1 but if there are only three doors, doesn't it come back to 50/50?" because they can't make the leap from a million to three. I always tried to explain it with just the three to avoid confusion.

I also wonder if people don't get it because the choice seems to be between "car" or "goat" and there's an inherent 50/50 feel to that. If the three doors hide a car, a goat and a sheep, then (unless you're of a particular wool-loving persuasion) there are now three different outcomes rather than just cars and goats.

Some explanations assume you pick door 1 with either a goat or a car behind it. In my head that also adds another level of complexity because, if you don't spot the symmetry in the problem then the next question is: "But what if I start with door 2?"

By differentiating between sheep and goat, you can also do away with the door numbers in explaining it and keep it simple:

It then comes down to:
If you picked the goat first, the host reveals the sheep. If you switch you WIN.
If you picked the sheep first, the host reveals the goat. If you switch you WIN.
If you picked the car first, the host reveals one of the animals. If you switch you LOSE.

You don't know which one you picked but you will always be in one of the three scenarios above. In two thirds of the scenarios switching wins you the car, in the other one it doesn't, ergo switching is good 2/3 of the time. Job done.

Maybe it's just me but that seems simple and straightforward in my head. Simpler than trying to explain why "If you picked a goat then the host reveals the other goat" has to be counted twice for the two combinations of goats.

Edited 2011-08-17 16:55 (UTC)

The explanation I always like to use is that you're not picking between your first choice and the one remaining door. You're picking between your first choice and *the best of all the other possible choices*.

As a statistician, I couldn't really afford to get this wrong...

PS:

You run Jurassic Park 3.0. Your lead scientist tells you he has just bred two new Velociraptor babies. You fire him immediately for breeding Velociraptors, but now you have a problem: Those are billion-dollar killing machines, so you can't afford to just write them off if it's possible you can keep them.

If the new babies are female, you can safely put them in your tank and they can't breed, because you only keep female Velociraptors. If they're male, you'll have to eat the loss and sell them to the Roast Dino Hut franchise, or something.

You ask the newly-promoted Head Scientist if the babies are female. "Yes," she says, "I've checked the first one, and it's female!".

What are the odds that, when she checks, the second one will also be female?

(assume each baby has a 50/50 chance of being female)

50%.

There are four possible worlds I am living in:
MM
MF
FM
FF
Because she has proven that I do not live in either of the first two worlds, I am left with two possible worlds, one with a male second raptor, one with a female. And thus a 50% chance that I am sharing it with two female raptors.

Do I win a goat to feed to my raptors?

Interestingly, this is a situation where getting it wrong and winning a goat is actually slightly beneficial!

And you do, because you haven't proven you're not in universe #2. After all, the first one she checked could have been either one.

More particularly: There's no way to tell if the first one she checked was in either position. So all you know is that they're not both male.

It's irrelevent. If you are given a fact that one is female, and assuming there are no hidden facts like eggs from the same litter have to be the same gender, then it is a 50-50 that the other one is female.

Nope! The trick is that you're not determining the sex of a random baby, you're determining the sex of a random baby from a set of two random babies.

The possible sets of two random babies are:

MM
MF
FM
FF

All of those two-baby sets are equally likely.
You know that one of the two babies is female, but not which one of the two - your Head Scientist's slightly misleading "first one" is "the first one she checked" and, if the original comment could be edited, it would be by now.

But!

All you know is that you have four equally-likely possibilities... and you're DEFINITELY not in possibliity #1, "MM".

This leaves you with three possible combinations:

MF
FM
FF
where are least one baby is female.

And in 2/3 of the cases, the second baby is male.

(This is a classic nonintuitive result, like Monty Hall's problem.)

Wrong.

You are not weighting your chances. Your options are MF FM or FF but the chance of having a male are 50-50. You then have a 50% chance of the Male being in place 1 or 2. Thus you have 25% chance of MF and FM and a 50% chance of FF.

Intriguing argument.

Imagine that Monty Hall has three goats and a car, behind four doors. The car is equally likely to be behind each door, such that your odds are:

1: 25%
2: 25%
3: 25%
4: 25%

Monty opens door #1 to reveal no car.

*I* say that your odds are now
1: 0%
2: 33%
3: 33%
4: 33%

*You* say that your odds are now:
1: 0%
2: 25%
3: 25%
4: 50%

I am wondering why you think this is.

You chose a door before you were shown a goat in one of the other 3. The host KNOWS that that door didn't have the car behind it before offering you a swap. He revealed extra information about the system of 3 doors after chosen. The important distinction is that this information was not a given at the point you chose the door as that the host knows for certain one of those doors was with a goat inside.

You chose a door before you were shown a goat in one of the other 3.

Very true, and the fact of that choice was why the odds acted in a nonintuive way.

In this problem, he's doing the opposite. He's *not* adding information by telling you that the car is not behind door #1, so the odds of the car being behind the other three doors should be equal, right?

Okay let me try it this way and see if I can distinguish between the Monty Hall problem and not.

You have 3 doors. I tell you that door number 1 has a goat behind it, but a car hides behind door number 2 or 3. The odds in thsi scenario are:-

Door 1 0%
Door 2 50%
Door 3 50%

------------

In the Monty Hall Problem you are told a car is behind one of the 3 doors and you are told to pick. For arguments sake we will say you chose Door number 3. The odds are thus:

Door 1 33%
Door 2 33%
Door 3 33%

Monty then knows that door number 1 has a goat behind it and shows you it. He asks if you want to swap. The odds of it behind behind either door number 1 or 2 remain 66%, but as he has ruled out the first door the odds are now:

Door 1 0%
Door 2 66%
Door 3 33%

If door number 1 contained the car, he wouldn't open it, he would instead open door number 2. If door number 3 contained the car then it wouldnt matter which one he opened. It all revolves around him giving new information at this point. The information is dependant upon your choice and THAT is why it changes things.

That's exactly right. You have the problem defined perfectly and you have the correct answer.

In the baby raptors situation, imagine "FF" to be your "car", and the other three possible results to be "goats".

All four results are equally likely, which corresponds to Monty having four doors, right?

*Without* you picking, Monty eliminates one goat, the "MM" goat, which leaves three doors. Each of those three doors now has a 1/3 chance of being the "FF" car, right?

Wrong. You are introducing the notion that the position is or importance and then confusing the odds.

Either you do know that the female is in the 1st slot or you don't. If you do the options are:

F* M
F* F

If you don't then the options are:

F* M
F* F
M F*
F F*

Where F* is the one that was identified. You may give these all equal weighting now. You are mixing up the 2 solutions are deciding that the position of the female matters only if there is a Male. Wrong.

So you're once again arguing that

1: 25%
2: 25%
3: 25%
4: 25%

compresses to
1: 0%
2: 25%
3: 25%
4: 50%

once you know that "it's not 1"?

Although if you wish to apply the above to the solution, you need to add % chances :-

Before you know one is female the options are:
MM 25%
MF 25%
FM 25%
FF 25%

AFTER you know one is female the options are:
MM 0%
MF 25%
FM 25%
FF 50%

Why does the lack of MM make the MF possibilities less likely?

It doesn't. There is a 50% chance of FF or MF/FM. The location of Male or Female doesn't change the fact that each option is 50% as a whole. There is a 50-50 that the Male is in the 1st or 2nd slot so that is half of the 50% (25%) for each option.

Your method is not a Monty Hall problem. That is when you are given NEW information AFTER a decision has been made. This example gives informations BEFORE a decision has been made.

Nope. you want to know if you have two females.

Your scientist chooses one and checks it. You have information. You now want to know the odds of the other being female. That's a re-evaluation of the probability after you've been given information.

That doesn't seperate out the system.

Let me apply this to dice. I roll 2 dice. I reveal that one of them is a 6 but don't reveal which dice. The options are 1-6 2-6 3-6 4-6 5-6 6-1 6-2 6-3 6-4 6-5 6-6. You will notice that 6-6 only appears once in the list. Do you honestly believe that I have less chance of rolling a 12 than rolling a 7?

If you don't say which die is the 6 then you have to let both dice vary completely in your options. It should be:

1-6 2-6 3-6 4-6 5-6 6-6
6-1 6-2 6-3 6-4 6-5 6-6

not

1-6 2-6 3-6 4-6 5-6
6-1 6-2 6-3 6-4 6-5 6-6

6-6 has to appear twice, not once, in the possible outcomes because you're not telling me which dice you've looked at. Without position (whether it's dice or raptors) the odds are different than with position.

Edited 2011-08-17 18:48 (UTC)

Then you will understand that the revised explanation below I have illustrated marking the female are F* holds true.

You've actually got the answer right there:

Of the 36 equal possibilities of two dice, you eliminate the 25 that involve no sixes, leaving you with the 11 still-equally-likely where at least one die is a six.

Of those 11, one is 6-6, and the rest are not.

So your odds of the second die also being a 6 are 1/11.

Another one, since we're pulling out the dice:

Sue and Bob take turns rolling a 6-sided die. Once either person rolls a 6, the game is over. Sue rolls first. If she doesn’t roll a 6, Bob rolls the die; if he doesn’t roll a 6, Sue rolls again. They continue taking turns until one of them rolls a 6.

Bob rolls a 6 before Sue.

What is the probability Bob rolled the 6 on his second turn?

Hints: The answer is not 1/6. It is not 5/36. It is not 125/1296.

The answer is 5/6 x 5/6 x 5/6 x 1/6

Nope! That's the odds of Bob winning on the second turn of any game.

What you want is, given that Bob won, what is the probability that he did so on turn 2?

Badly phrased, but given this then the answer is 1 x 5/6 x 1 x 1/6

Which is ofcourse 5/36 Which you have already thought was wrong. I'm afraid you clearly don't understand statistics.

You seem to be taking this personally, and I don't know why.

But no - "5/36" is still not the right answer. That's the odds of Bob winning in turn 2 if he's playing alone.

Er no. You just said to assume he wins. That makes Sue the loser. I don't have to take any of her rolls into account at all.

What you are saying is "Sue's FIRST roll is not a 6". Not the same thing at all.

You could calculate it that way, but it's harder, and now you have to filter out the games Bob doesn't win.

The fact that Sue can win in turn 2 changes what percentage of Bob's wins will come in turn 2.

Imagine Bob is playing alone. 1/6 of his wins will happens in turn 2, 5/36 of his wins will happen in turn 2, etc. Right? That's what you just calculated.

Now remember that, 1/6 of the time when you hit turn 2, Bob loses outright and you start over, and Bob loses that game, even if, had he been playing alone, he would have won it.

Sue's presence affects *when* Bob wins, even if we only consider the games where Bob wins. Sue is a spoiler.

Like I said in the other comment: Imagine that Bob is the 1000th player. For him to win at all, he needs to beat the odds massively, but we can run the trial for long enough that he gets a few wins.

In order for him to win on the second turn, though, he has to beat the odds *again*, just to get a chance. Most of the time, even IF Bob got a shot in turn 1, he'll never get a shot in turn 2 - so Bob's wins that come in turn 2 are going to be much rarer than his wins in turn 1, because of all the things that can go wrong with a game before Bob has a chance to win, right?

I already have filtered out all the other possibilities. That is why I multiplied the odds of rolling a 6 on the second time (1/6) by the odds of NOT rolling a 6 on the first time (5/6).

You clearly just don't understand this at all.

I already have filtered out all the other possibilities.

And in doing so, you have calculated the odds of Bob's win coming in turn 2 if nobody else is playing.

And that is why this is *the wrong answer*.

You already told me he IS going to win. Therefore it is EXACTLY THE SAME as if he was playing by himself because every single throw the other person makes WILL NOT be a 6.

Christ alive, how many time do I need to explain this?!?

Except it's not the same, because he doesn't win every game.

He won *this* game, and the question is, what are the odds that he won on turn 2, instead of turn 1 or turn 3 or turn 11?

Another way of expressing this same question is "what percentage of his wins happen on turn 2?"

I'm not repeating myself further. Come back when you have found something online on the problem, along with an apology when you realise you are wrong.

You probably should have googled it yourself. It's an old, old problem, and I didn't even change the words on it the way I did with the raptor babies.

The raptor babies? That's the Boy Or Girl Paradox.

Have you actually READ that link? Because it confirms what I said.

It even explains where you are wrong due to the ambiguity.

"From all families with two children, at least one of whom is a boy, a family is chosen at random. This would yield the answer of 1/3."

"From all families with two children, one child is selected at random, and the sex of that child is specified. This would yield an answer of 1/2."

You stated clearly that "I've checked the first one, and it's female!" in the question which is the same as the 2nd statement.

You see the difference between one of them is female and AT LEAST one of them is female?

"One is female, one is unspecified, and they're not ordered" is the same as "at least one is female".

And you probably should have checked on the problem clarification, or listened to any of the explanations - or, y'know, simply not claimed that "It's not 1" meant "4 is twice as likely as before" so many times.

Edited 2011-08-17 20:57 (UTC)

You clearly state that the head scientist only looked at the first one and it was female. Don't come back trying to throw the blame because you got it wrong. Most of the article you linked talks about the ambiguity in detail and yet you clearly didn't take it in.

You are wrong. It IS 50%. You own article explains why.

You clearly state that the head scientist only looked at the first one and it was female.

More, that the first one the scientist checked was female. Clarified, again, repeatedly, to everyone's satisfaction, three hours ago.

I don't understand why you seem to be taking this so personally. Did Bob and Sue kick your dog or something?

Because you are being so patronising and yet you are wrong. That you have proven yourself wrong with your very own link only serves to make you seem even more of an arse on the subject.

Even the Bob and Sue link is filled with people pointing out the problems with the wording. It is ambiguous as to whether we assume that sue always rolls a 1-5 or whether we take all of the games possible and only look at the percentage of those where Bob wins.

Your problem seems not to be maths, but english.

I'm afraid that's not correct. The odds of Bob winning ("BW") satisfy the equation

x = 1/6 + (5/6)(1-x)

so it's 6/11.

The odds of Bob winning on the second turn ("BW2") are as you calculated 5/6 x 5/6 x 5/6 x 1/6 = 125/1296.

So P(BW2|BW) = P(BW2 & BW)/P(BW) = P(BW2)/P(BW) = (125/1296)/(6/11) = 1375/7776.

This is wrong; 6/11 is SUE's chance of winning. Bob's is 5/11. So that's 275/1296. Sorry.

I realise this thread is all over the shop, so I will paste a bit from later on here.

"Also with regards to Sue and Bob, I think a better way to phrase it is "Of the times that Bob wins, what proportion will be on his 2nd turn?", which would rule out any ambiguity."

The way I read it was that there was a 100% certainty that Sue would lose rather than working out how often Bob would win on his 2nd turn of the times that he wins. One rules out Sue's dice rolls and the other is dependant upon them but only when the outcome is in favour of Bob.

The thread got derailed because it was about English rather than Maths; a different kind of expression :)

Wrong again, because that fails to account for the possibility that, in any given game, Sue can win in turn 2.

You're correct that Sue's first turn is actually irrelevant: In the set of all games that Bob wins, Sue *will always* lose on the first turn. However, the fact that Sue can end the game in turn 2 will change the percentage of Bob's wins that will come in turn 2.

A more intuitive way of looking at Sue's influence on Bob's results is to imagine a thousand players, with Bob going last. In order for Bob to win *at all*, the first 999 players must fail, so we can discard those on the first turn - but on the second turn, they all must lose *again* in order for Bob to have a chance to win on turn 2.

Can you see how that will affect the percentage of Bob's wins that come on turn 2, versus turn 1 or turn 3+?

"Bob rolls a 6 before Sue."

This is definitive statement. If you want me to assume this then I never need to worry about what Sue rolls.

Nope. Still wrong, for the same reasons. The fact that Sue can win changes *when* Bob wins, even if we only care about games where Bob wins.

It's an unintuitive problem. Simulate it if you don't believe me - which will, by the way, also converge on the correct result of 275/1296.

It it only unintuative to you. I suggest if you want me to take you seriously from this point on you start linking me to some sort of reference online which actually backs up your way of thinking. Something that isn;t just a friend of yours on LJ but an actual statistical reference. Mainly as at this stage I think you are probably trolling rather than just being stupid.

If you don't believe me, simulate it. Or take a look in the comments of xkcd, where I originally got it.

One of the correct answers is here.

And the short version is Bayes' Theorem and the sum of an infinite series.

(rather than sum the infinite series, a more elegant solution: Consider the two die rolls as simultaneous. If the first is a 6 (6/36 of the time), Bob loses. If the first is not a 6 and the second is a 6 (5/36) of the time, Bob wins. If neither is a 6, the test repeats.

Therefore, you can view this as an infinite loop with 11 possible exits, 6 of which are Sue wins and 5 are Bob wins. Given that SOMEONE wins, we see that Bob wins 5/11 of all games.)

If you don't like that, then yank out your geometric progression rules and sum the infinite series the hard way. I'll wait, you'll get 5/11.

Now: You've already shown that, for any game, the odds of Bob winning in turn 2 are 5/6*5/6*5/6*1/6: not-6, not-6, not-6, 6. That's 125/1296.

So, the odds of him winning in ANY game in turn 2, divided by the percentage of games he wins, gives you (125/1296) / (5/11).

Which is 275/1296.

275/1296 of Bob's wins will come in turn 2, which means that, for any given game where Bob wins, the odds of his win happening in turn 2 are about 21%.

Mainly as at this stage I think you are probably trolling rather than just being stupid.

You're demonstrably, aggressively wrong about the dino babies, the pub coins, and Bob and Sue, and you're calling *me* stupid because of it. You have no leg to stand on with accusations of "trolling"

Edited 2011-08-17 20:40 (UTC)

See other comment save me repeating myself.

Without a position, i.e. the, "I've checked one and it's female" as discussed elsewhere, surely that's:

MM 0%
MF 33%
FM 33%
FF 33%

Okay if we are going to stick to this modal, Lets assume it is a FF scenario. How do you know the first female she saw is the one in the 1st slot and not in the 2nd? If you want to keep this option then the true options are:

M F*
F* M
F* F
F F*

Where the F* is the one that we know about. Now given them all 25% and you have a true modal.

So you think the FF probability really does double once you learn that either of them is F?

But that's not the (revised) wording. In your scenario you're saying, "I've checked -that- one and she's female." You've added position to it.

If you are applying it to a position, why do you have MF and FM? You can't have it both ways!

If we were to check both our raptors and assign positions then there would be four equally possible combinations:

1: MM
2: MF
3: FM
4: FF

At this point we have all the information, the raptors have position and we can state exactly which of the equally likely scenarios we're in.

However, when we have partial information - in this case the knowledge that one, and we don't know which, of our raptors is female then we can only specify that we're not in scenario one because we have a female somewhere in the offspring. We haven't physically changed the raptors so the remaining three scenarios are still equally likely.

In two of those three scenarios the other raptor is male, therefore there's only a 1/3 chance we have two female raptors.

If we were to specify position in our partial information by saying, for example, "Raptor 1 is female", then we can rule out scenarios one and two and now the odds of two females is 50/50 as scenarios three and four are still equally likely.

I tell you what here's an idea when we are in the pub. You toss two coins and look at one of them. You will tell me what it is (let's just assume it is a Head this time for an example). You will then shuffle them about so we don't know the position.

I will then allow choose H-T T-H or H-H. If I am right, as my odds are 33% or 2-1 you will give me £2. If I am wrong, I will give you £1.

I will be happy to repeat this gambling game for as long as you want until you are satified you are wrong or I have all your money.

(Hint - I will ALWAYS choose H-H)

But that's not a correct description of the problem, here.

Instead, let's go to a pub, and I will flip two coins. I will examine them, and if both are tails, we discard that round (two MM babies, which we know didn't happen) and no money changes hands, and we flip again.

If one or the other is heads, I reveal them. If BOTH are heads, I give you £1. If one is heads and one is tails, you will give me £1.

We will continue this until one of us is out of money, at which point I will buy the last round and we will go home.

Fair?

Why should I accept only £1? You told me that the odds of HH were the same as HT and TH. That would mean you believe the odds of HH are 33% which is 2-1.

Because *you* say it's 50/50.

And since you insist it's a fair bet, I will happily take your money as if it was a fair bet. Giving you 2-1 means we'll break even on average, and that doesn't make the point nearly as well as you going broke betting the farm that a 2/3 chance is 50/50.

And because the scenario you described is not the same as the one in the question you gave. In the one you just offered you get to look at BOTH coins. That doesn't happen in the raptor question. What happens in that one is that a single raptor is looked at and its sex is determined. What you are doing is looking at both of the coins and letting me know that AT LEAST one of them is a Head. That is not the same thing at all and infact in THOSE circumstances the odds ARE 33% of it being 2 Heads.

You see the major difference now?!?

I think, after a long evening of posting, I see the problem.

It was clarified elsewhere in the threads that the original raptor question was worded sloppily and "the first one's female" was meant to imply, "the first one I looked at is female but it doesn't matter which one of the two raptors it was" - or more simply, "one of them is female".

I suspect, if you missed this clarification early on, Si (and I'm not sure if it was in a thread of the comments you were in or not and there's waaay too many to check now) that it would explain all the arguing.

Everyone step away from the interwebs and take a deep breath?

Just had a look and found the "clarification" The below is taken from his conversion with Andy in a different branch of the thread.

"She's checked one of them.

In the set of all two-raptor pairs, you don't actually know which one she's checked.

My wording "the first one" really should have been the more clear "I have checked one". The traditional Science Announcement is "Yes, at least one is female".

All you know is that they're not both male."

I'm afraid even this is contradictory. If she checked only one then it would still be 50% (she can't be certain at least one is female without knowing that the only one she saw is female). If she saw both of them, that would change matters entirely (She could see the first was male and still see the second and use the same phrase were it a female). This was the difference between the coin scenarios above too.

This started as a thread on Stats, but sadly has come to be one on English and wording of questions. I'd be less annoyed if I hadn't wasted time on trying to explain myself to someone who had access to a whole webpage describing the differences between the two and the pitfalls of being ambiguous.

*sigh*

I actually do see your argument now, yes.

I still think you're *wrong*, but now I understand where you're coming from

The problem here is that you're not supposed to know which coin you looked at - only that examination of the coins has shown that one is heads.

You, instead, are aiming to exclude all cases where the first coin examined is tails, which *would* result in 50/50 - except we're not discussing that. Not even in the original question, where I phrased it poorly and implied a possible ordering to the raptors.

Instead, the original phrasing, corrected and clarified three hours ago, was meant to imply that examination of a random raptor from the pair had resulted in a female, and the second raptor had not yet been examined.

Which, given our original four equally likely possibilities of
MM
MF
FM
FF
Means we're SOMEWHERE in the bottom three, but we don't know which.
We have either FX or XF, and in 2/3 of the possible cases X is M.

Edited 2011-08-17 21:13 (UTC)

ARGH!

Thats STILL 50%

The point is the 33% comes from being able to look at BOTH of them and say that at least one is female!

If you look at the first one and it is female you can say with certainty "at least one is a female" whether or not you look at the second, which happens 50% of the time. The point is that if you see the first one is male then you can STILL look at the second one and if that is female you can STILL say "at least one is a female", which happens 25% of the time.

In THAT scenario it is 33% that both are female. Only identifying ONE of them without looking at the other means that you CANNOT assume ANYTHING OTHER THAN the one she looked at IS female. That means the other one is the ONLY unknown and therefore it is a 50-50.

The key to phrasing this is that a female can be either or both rather than a specific one.

Please PLEASE try and understand this. Read your own link again and then maybe we can agree and move on.

The problem here is the same one you had with Bob and Sue: We're talking about a single specific instance of a larger generalisable problem.

Just because we know Bob won this one game, doesn't mean Bob always wins games, so we can't discount Sue when determining *how often* Bob wins. Just because Sue *can* win games doesn't mean that she might have won this specific game, so we need to take into account that we know she didn't win this one.

Back to the babies: We haven't addressed what might have happened has the first baby been male, because in our specific, that didn't happen.

But we didn't generate raptor pairs guaranteeing one is female.

We generated a random pair of random raptors, and then learned that one, or the other, or both, was female.

The fact that Sue didn't win the game in question doesn't mean that Sue can't win, and thus that Sue's existence doesn't alter the distribution of Bob's wins.

The fact that the Head Scientist determined that one or both of the raptors was definitely female did not change the fact that they *could* have both been male.

You appear to be arguing that by examining only one of them, the head scientist has introduced an ordering to the raptors - the examined and the unexamined......

....and, I think I actually just got what you're aiming at.

Rephrase as "This one is female! The other one escapes and eats you. Was it male or female?" and suddenly the question becomes much clearer.

So you're right, and I do see your point. In the situation where you know which one she examined, it is not the same as "at least one is female".

From the perspective of

andrewducker, who owns Jurassic Park 3.0, though, he only knows that the scientist knows one is female without knowing which. If she was to tell him "one is female, argh they've got my neck blergh I am teh dedzorz" all Lovecraft-character-blogging-his-execution-style, though, wouldn't the question of "were both the raptors who ate the scientist female" sit on 33%?

Deleted my comment as I went off on one without reading the second half of your comment, which I apologise for.

Yes. You got it. By rephrasing it as you have, it is now 33%. Agreed.

Also with regards to Sue and Bob, I think a better way to phrase it is "Of the times that Bob wins, what proportion will be on his 2nd turn?", which would rule out any ambiguity. I appreciate that this was not YOUR fault as the question was taken as a whole from another forum, but it is misleading none the less.

If you give me 10p for every time it comes up T-T and you always choose H-H you're on. It'll take a while but I'll bankrupt you eventually with the 10p's, otherwise we'll stalemate.

Wee bit of code to illustrate. Source code and exe are in the zip file.

Raptors.zip

Nupe, I would give you nothing for T T.

As mr Weasel King has provided a link that happens to completely disprove his own theory, I will happily link it further:

http://en.wikipedia.org/wiki/Boy_or_Girl_paradox

You'll note the difference between stating that one of the raptors is female and AT LEAST one of the raptors is female. He didn't but hopefully you will and finally agree with me the odds are 50%. If the Head Scientist had come back and said "I've seen BOTH of the raptors and AT LEAST one is a female" then yes the odds would be 33%. He didn't say that though.

I prefer my solution.

chuma insists that it's a fair bet, so let's treat it like one.

Flip two coins. If both are tails, no money changes hands. If both are heads, you give

chuma £1. If only one is heads, he gives you £1.

Your solution is yet again not what was discussed or offered, nor is it an accurate description of the raptor puzzle. At this point when your own evidence proves you wrong, you seem to be doing the equivilent of sticking your fingers in your ears and repeating "I can't hear you".

I genuinely do not know what you think I've done to deserve your infantile behaviour.

You've called me stupid and demanded an apology for being correct, and then stopped responding entirely in that thread when I proved that yes, my answer was correct and yours was not.

Here, we have a question that was acknowledged to be ambiguous almost immediately after it was posted, and the ambiguity was corrected, and you're still, hours later, stamping your feet and launching personal attacks because we keep telling you you're answering the wrong question.

Seriously, what makes you think this is even slightly appropriate behaviour?

What has caused it? Your attitude and the way you come across. You still think you are correct and answered the same question even though in your "clarification" (which was never given to me, but to Andy in a different section) you still contradicted yourself within a few lines. You're still incapable of acknowledging you didn't phrase keys parts of either question right, and instead prefer to bang on answering what you thought you had said, even when there is overwhelming evidence to the contrary. What makes you think THAT behaviour is appropriate?

In the interest of supporting people who appear to be right on the internet, I've read the wikipedia page and thought about it, and you appear to be right.

The crux is clearly

From all families with two children, at least one of whom is a boy, a family is chosen at random. This would yield the answer of 1/3.

From all families with two children, one child is selected at random, and the sex of that child is specified. This would yield an answer of 1/2.

You (or nature, or hot velociraptor sex, or the scientist breeding velociraptors, or people who don't understand stats on the internet) have selected a family of two velociraptors at random from the set of all possible families of two velociraptors, and have then acquired the additional information that one velociraptor is a girl. It's clearly the second scenario, not the first. Unless the scientist was producing embryos in some strange probability space where they had to have at least one girl embryo. But if they'd done that they wouldn't need to check them, because they'd know they had at least one girl ;-)

Bless you :)

I knew someone would put it more succinctly than me. :D

[In the interests of transparency I think I have to add that I am still managing to confuse myself every time I think about this.]

I have made an LJ poll to confuse myself further, and friended you so you can read it if you want to and are not entirely sick of this debate already ;-)

I think theres a section of the thread that already deals with everything that was said here more neatly than it was here, with the outcome that everyone agrees on the appropriate wording required to get the two different percentages (In the section started by Woodpijn). Infact his conclusion on being convinced it was 50% is "I think the boy-girl paradox is as much a linguistic problem as a mathematical one." I entirely agree with him.

My reading of "the first one" means the options are:

MM 0%
MF 0%
FM 50%
FF 50%

Doesn't "the first" imply a position?

If she'd said, "I've checked one, and it's female," rather than "I've checked the first one, and it's female," then I'd agree with you. I can see your point, but the implied position in the quote means I'm with Andy on this one... for now :)

You're right, it's a bad phrasing. "the first one" was meant to mean "I have checked one, and will soon check the other".

Either way, there's no position.

Yeah, I saw your reply to Andy below. With the revised phrasing and no position, I agree that it's a 2/3 chance the other is male.

If she'd checked raptor two then that would leave universes 1 and 3. Which still gives me a 50% chance. Which one she checked doesn't matter.

She's checked one of them.

In the set of all two-raptor pairs, you don't actually know which one she's checked.

My wording "the first one" really should have been the more clear "I have checked one". The traditional Science Announcement is "Yes, at least one is female".

All you know is that they're not both male.

That's not true.

I know either that I'm in universes 1/3 or in universe 3/4. In either case I have a 50% chance of the other one being male and the other one being female.

Only if you accept "I've checked one and it's female" to mean "I have numbered them, and determined that #1 is female, thus eliminating TWO of the possible results", instead of the intended (and sloppily worded) "I have determined that one of them is female, without caring which"

It depends.

Let's assume that there are two doors, each with a raptor behind it.

If they say
"I went behind door 2 and checked, and that one is a female." then we've eliminated two of the results.

If they say "I went behind both doors and checked, and I remember that at least one of them was a female, but I can't remember which one." then we've only eliminated one of the results.

I think this question requires more careful phrasing than the Monty Hall one :->

"If they say "I went behind both doors and checked, and I remember that at least one of them was a female, but I can't remember which one." then we've only eliminated one of the results."

This example would indeed be a 33% chance of the 3 remaining options.

It does.

But I love this kind of problem because of the *breadth* of the wrong answers you get, and the paths people use to get there.

(I just, here at work, got two more smart educated geeky people to fall for both the Monty Hall and the puppies.)

I'm sorry but the only wrong answers here are your own. If I could drag others into this argument, I know a statistician who I am sure could explain it more succinctly than I could.

Au contraire, there are at least three wrong answers in the original poll, and at least two wrong answers to the beagle puppies problem. Not counting the two in my office just now!

This is not to say that my answers to this kind of thing are always correct - the Bob And Sue are rolling dice one had me bouncing off wrong answers for hours before I found the trick - but I really do love this kind of problem and this kind of thread because I love seeing *how* people come up with the answers they reach.

Yes. Seeing how people's intuitions are broken is always fascinating. Especially when they're mine!

And now you have another example, of mine getting all twisted up!

I understand this now! You're wrong, because 'I have checked one and it is female' is not the same as 'at least one is female'. Some cases of the latter will not be picked up by the test.

Ie, if the raptor family is a boy and a girl, it would always be included in 'at least one is female'. But half the time it would not be included in 'I checked 1 and it was female' as you might have checked the boy.

Or as a friend of mine said so much better than I could '"First-born" and "first one I looked at" are equivalent as far as probabilities are concerned, and the probability is 1/2.

If instead I took a sample of both velociraptor's DNA, mixed them together and said "Hey I've found some Y chromosomes, at least one velociraptor is male" THEN the probability that both are male is 1/3.'

If instead I took a sample of both velociraptor's DNA, mixed them together and said "Hey I've found some Y chromosomes, at least one velociraptor is male" THEN the probability that both are male is 1/3.'

This is brilliant.

"I have checked one and it is female" obviously is not the same as "at least one is female". The latter statement may imply having checked both.

The intended interpretation was that the owner of the park wouldn't know which was which - similar to "One is female, the other is OH GOD IT HAS MY NECK ARGH WHY DID WE BREED RAPTORS WHYYYYYYY"

The owner knows she checked one and stopped. The stopping is critical, because it rules out one of the two ways to find only one female (the way ruled out is to find a male, say "that sucks", and then find a female).

The 2:1 odds (1/3 probability) solution requires there to be two ways to get one female, against only one way to get two females. Here, there's one way to get one female, and one way to get two females, and that's all. So the solution is 1:1 odds (1/2 probability).

By killing the scientist, the deadly raptor stops her playing word games with the park owner; otherwise she would have mischievously turned the uniquely distinguishable states "MF" and "FM" into the two indistinguishable states "one F". But these aren't just word games. Our measurements of chemical entropy confirm that the loose verbal description "this is a gas: the molecules are all over the room" really does describe many more possibilities than "this is a solid: the molecules are in this dish" does.

Consider the Ambiguous Monty Hall problem.

Somehow, you end up being one of the first contestants on this new game show. The host, who you have learned is called Monty Hall, presents you three doors, tells you there's a prize behind one, and nothing behind the other two, and asks you to pick a door. You pick a door, expecting him to open it. But instead, he opens another door, and asks you if you want to change your mind. You did not expect this. You know no more than what you have been told.

Unanswerable questions:

a) Does he always do this?
b) Does he know where the prize is?
c) Is he on your side or not?

Calculating the probabilities requires an answer to these questions.

Good point. I believe this was the case in the original, real-world Monty Hall quiz show. Sometimes he would open a door, sometimes he would take your first pick. At this point it is probably rational to go:

(a) The man likes to save his show's budget.
(b) He does this sort of thing for a living.
(c) He is probably better at mind games than you.
(d) I should stick with my first choice, because that way I guarantee my one-third chance of being right.

Exception: if you were an utterly cute thirteen year old, swap if he suggests it. Because if he's nasty to you, his sponsors will be cross with him.

I don't believe switching is important because you can still choose the original box and hence have 50% chance.

Sadly not.

You had a 1 in 3 chance when you picked the original door. Opening doors on the other side doesn't change that chance.

If we could take you, once the other door has been opened, knock you out, remove all of your memories, and then ask you to choose again with no bias, _then_ you'd have a 50:50 chance of getting it right. But as you already know things, this isn't the case.

Monty Hall

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