I wish I'd read this when you first put it up

"First you point towards a door" he says "then I'll open one of the other doors to reveal a goat"....

This advance guarantee that the host will "take out" a goat which appears in the Mueser and Grenberg version of the puzzle but not the Parade Magazine summary makes the whole thing make sense.

If (as on my previous reading of the puzzle) the host removed a goat through discretion rather than as a fixed rule of the game, I think my original logic holds (although I'm not as sure of that now as I was!)

Oh yeah. Well then you can get into the whole "does the host want the contestant to win?" thing. But yeah, if the host always reveals a goat, it's better to switch.

Ahem - no need to mention this thread in "You know you're addicted to forums when......" is there?

I think I'll be counting goats in my sleep

Well I'm not at school, I need something to fill up the day with

Oops, didn't realise this topic would run so fast while my back was turned...to try to clarify the goat thing for those that are still confused about the missing fraction, think of it like this:
In total there are 6 possible combinations of contestant/compere actions:
1 Contestant goat1, compere goat2, door3 car
2 Contestant goat1, compere car, door3 goat2
3 Contestant goat2, compere goat1, door3 car
4 Contestant goat2, compere car, door3 goat1
5 Contestant car, compere goat1, door3 goat2
6 Contestant car, compere goat2, door3 goat2

Assuming these are all equally likely, note that at this point (before any doors have been opened), contestant had a 1/3 chance with the car. However, we know from the text that options 2 and 4 can't happen (compere did not reveal the car), leaving the 4 equally likely outcomes, 2 of which involve the car being behind door 3.

As for the boys one, YAP is right, in the end it boils down to a philosophical debate about whether you have a particular child in mind when you specify the problem.
In total there are 4 combinations of 2-child families:
Older boy, younger boy
Older boy, younger girl
Older girl, younger boy
Older girl, younger girl

If you specify only that one of the children is a boy (and you don't say which), you are knocking out just one of the 4 combinations from your total sample (Older girl, younger girl), leaving 3 equally likely possibilities, only one of which involves 2 boys.
But if you specify that (say) the oldest child is a boy, then you knock out the last 2 possibilities, putting the probs back to 1/2.
I have to say though, that this still hasn't been universally agreed, even amongst statisticians. It all boils down to whether you can ever be truly non-specific about children (it seems much clearer if you reduce it to bags of apples and oranges).

Which only goes to show that:
1 in stats, specifying the population and sample space is usually the trickiest bit of the problem; and
2 if any of you ever have a stats probability exam question, read it 6 times before you start the answer!! Usually the actual maths in these things is quite straightforward, so they make up for it using weasel words to try to catch people out!

Thouston
(professional research statistician)

Exactly why I hate stats! Can't anyone think up a nice, straightforward algebra problem?

Yes, I've just bought her a set of 11 plus maths paper recently. To test her since she would like to try and get a scholarship to City of London's Girls. This school has a fantastic music department! She's only 9 but we were told that this school requires exceptionally HIGH maths std so we got her the 11yr olds std paper. Well, we WERE told that she's a gifted child and three yrs ahead so we let her attempt them. Sure enough, she did them easily and quite quickly within 20 mins (not the given 30mins) BUT she gets tripped up by the same deceptively straightforward-looking algebraic sequence each time. I think they have never taught such stuff at school so she has never learnt the rationale behind them? For example, what would be the 16th or 32th letters of xxyzzxxy? Looks straightforward but not to her. By the way, the timing constraints do not allow you to list/count forward so I will have to find the right formula for her to apply.

Tess

EDIT
By the way, I'm not asking the question!!! Just illustrating! Hubby will teach her when he comes home tonight.

Thats good because I havn't a clue how to go about that! So much for being G+T with a predicted A* in math.

Theoretically then, both Trebor and YAP were right. In the first case, where it is not specified that the first child is a boy, just simply there is one, then the chances of the other child being a boy is a 1/3, thus, in this case scenario, Trebor was right.

In this case scenario, which in actual fact was how the original question was phrased, you could actually try to 'choose' the families using doors like the other problem. You are on a gameshow (incidentally, not a very good one). You have 4 lots of 2-child families to reflect families around the world, assuming that the chances of the 4 combinations were equal. Thus, behind each of the doors is a family of boy-boy, a family of boy-girl, a family of girl-boy, and a family of girl-girl. The gameshow host now opens one of the doors, revealing the family with two girls, thus leaving the other families with at least one boy. He now asks you to choose the door with a family of two boys. In this scenario, your chances of winning is a 1/3, providing you are not dumb enough to choose the door that the host just told you was wrong.

However, with rephrasing of the question, saying the boy is the eldest child, and what is the chance of the second child being a boy, YAP is right - the chance of this is 1/2, assuming the non-existance of twins. The gender of the second child is irrespective of the gender of the first child, so is still half. The reason why in this scenario, the answer is 1/2, and in the former is a 1/3 is because of the dispute of whether you allow the merging of the boy-girl probability and the girl-boy probability. In the former, this is allowed, but in the latter, this is not allowed, as you have already specified that the first child must be boy, so you eliminate two instead of just one probability, i.e. the girl-girl and the girl-boy possibility. However, in the former scenario, just because the first may be a girl does not mean there is no probability of there being a boy at all, and so you only need eliminate the girl-girl probability.

OK, maybe I could have phrased this better, but basically, yep, I agree, it all depends on the wording of the question, and that in one case, Trebor is right, and in the other, YAP is right.






Oh yeah, on to the case of the door thing, wasn't there a similar programme where it comes to the end where there are two contestants 'Shaft' or 'Share' (I can't remember if those were the exact terms) and the prize money. If both contestants 'Shafted', they both don't get the money. If one 'Shafts' and the other 'Shares', then the 'Shafter' gets all the money. If they both 'Share', there is a split pot. In the end, you should always 'Shaft'... If you were to draw a Game Theory style grid, although I'm no expert, so the numerical values are probably wrong, the Sharer, at best, in totalling all possibilities will get 1, but the Shafter will get 2. In other words, if you Share, the best you can get is half the money, and the worst you can get is none of the money. But if you Shaft, the best you can get is all the money, and the worst you can get is none of the money. So, by probability, you should always Shaft.

However, to throw in an extra factor, let's say you watched TV and you realised now that every contestant that went on Shafted. In which case, forever, no-one gets the money. So, Shafting or Sharing would make no difference to you: if the other guy is always going to Shaft, you still get no money, regardless. So you could Share, and at least have the satisfaction of knowing that you were kind enough to let the other contestant have the money. But, let's say after so many trials, there was somebody else that thought just like you! Then, you would both Share, and you would both get split pot and get some money: more so than any other pair of contestants would that would both Shaft, thinking this is the best way of winning by probability!

But, let's say this happens numerous times, then everyone may be set on Sharing, and so one crafty person may decide to Shaft, and win the lot. Thus starting a trend of Shafting, letting the cycle go on! By the way, for those interested, this is the 'Prisoner's Dilemma'. But anyway, I guess that's not much use anymore, as the gameshow no longer exists!

OK, enough Stats for one day...

~Wobby~

Rosemary,
Being great in maths has nothing to do with exams! You can be predicted as one who could get distinctions and yet that means nothing much. I had distinctions all along in maths throughout my school career but I must have been wrongly taught! Today I believe I can pass no further than GCSE maths even though O Levels maths in those days were supposed to be a LOT harder.

Tess

Tess, I know that being good at exams and good at a subject are totally unrelated. But I get hold of mathematical concepts almost as soon as they are explained to me, causing a lot of frustration as I wait for everyone else to catch up!

I have 'the perfect sausage and other formulas' one and it has equasions useful for GCSE&A-levels