Best...Puzzle...Ever....

[orsal]

Do a web search for statistics and put in "mutually exclusive" with it. Basically, it's a rule that determines AHEAD OF TIME whether or not the result of the first trial influences the second trial.

No. The word you're thinking of is "independent". Two events are mutually exclusive if they can't both happen. Two events are independent if knowledge of one does not affect the probability of the other. Two variables are independent if any events associated with them are independent; it doesn't make sense to say that events are mutually exclusive.

And, on any given prob/stats exam, I guarantee at least 4-5 of my students will confuse those two concepts too. We learn them the same week, since we use them in parallel ways -- independence simplifies the calculation of the probability that event "A and B" happens, mutual exclusivity simplifies calculation of the probability that "A or B" happens.

 

[Tuzenbach]

Tuzenbach, did you post the answer to your "faces" riddle? If so, I missed it.

Due to my extreme paranoia about having a potentially "genius" riddle spoiled as a result of over-exposure, I decided that I wouldn't post the answer here. However, I also stated that if a person REALLY wanted to know, to email me: beloc_@hotmail.com

It must not have been enticing enough, as no one emailed me!

:-( ::. :. :.::. :.: ..:. :.. .:.:. :..:: ..: .:.. :.:..

 

[Abisashi]

Beaten to death, but...

Yet another explanation of the Monty Haul problem:
This has already been explained several times, but I was finally able to convince myself, thanks to the pea&cups puzzle.

As noted in the p&c puzzle, the fact that the guy reveals a cup with no pea adds no information to the puzzle... you already knew there was a cup with no pea under it that you didn't pick. Therefore, we can ignore this information.

For the Monty Haul puzzle:

You pick a door. You have a 1/3 chance of being right. You now have the choice to pick BOTH of the other doors; as long as the prize is behind one of them, you get it. Effectively, you temporarily forget what Monty revealed (because it adds no information), but once you choose to switch, you then remeber which door has no prize and will thus choose the other one. Reducing these two decisions to one, you choose between:
a) 1 door (1/3)
b) 2 doors (2/3)

Thus, you always switch.

 

[Tuzenbach]

And, on any given prob/stats exam, I guarantee at least 4-5 of my students will confuse those two concepts too. We learn them the same week, since we use them in parallel ways -- independence simplifies the calculation of the probability that event "A and B" happens, mutual exclusivity simplifies calculation of the probability that "A or B" happens.

Now you know why I failed it the first time around! You're a Stat teacher? That's pretty cool! So, can you state TECHNICALLY both sides of the Merak/Piel argument? I tried........and failed! I mean, from a certain standpoint, they can BOTH be right so long as the jargon is CORRECT in setting up the problem.

 

[Pielorinho]

Tuzenbach, read Piratecat's link several pages back; it gives a technical explanation of both sides of the problem. The solution that makes the most sense to me certainly does not assume that the gender of one chid affects the gender of the other child, and I don't see how you could read what I posted and conclude otherwise. It simply assumes that you get the information in exactly the order given, and that the information you receive is deliberate and specific -- that the problem could not have substituted "one of them is a girl" for the line "one of them is a boy".

On an aside, I think you should post your riddle solution here; that's how messageboards work.

Daniel

 

[orsal]

Now you know why I failed it the first time around! You're a Stat teacher? That's pretty cool!

Well, that's not my principal field, but I've taught now at two universities where statistics is part of the math department, and at both schools I taught introductory stats classes as part of my teaching load.

So, can you state TECHNICALLY both sides of the Merak/Piel argument?

OK. I won't bother with spoiler tags, since I don't think anybody is going to get anything from this explanation by casually glancing over it.

There are four possibilities for a two-child family: BB, BG, GB, GG. All of them are equally likely (probability 1/4). If you insist on not distinguishing BG from GB, just call them both "1 of each", that's OK, but you no longer have uniform probability -- now P(1 of each)=1/2, P(2 boys)=1/4, P(2 girls)=1/4.

Suppose you know there is at least one boy. Then you eliminate (2 girls) as a possibility. You're looking for conditional probabilities: P(2 boys|at least 1 boy), which is read "probability of two boys, given at least one boy", and P(1 of each| at least one boy). To calculate conditional probabilities, you divide by P(at least one boy), which is 1/4 + 1/2 = 3/4. This ensures that the probabilities still add up to 1. Thus we get P(2 boys | at least 1 boy)=(1/4)/(3/4)=1/3, and P(1 of each | at least one boy)=(1/2)/(3/4)=2/3.

This is the correct calculation on the assumption that if there is at least one boy in the family, we will learn that, but will never learn any more. Pielorinho, and I, and several other people, did this interpretation.

On the other hand, suppose that we were destined to learn the sex of one of the kids, but it could be either. It could be that we would be told the sex of whichever one was older, or whichever one we happened to meet first, or whichever one our friend randomly decided to tell us about. Then

P(we learn of a boy | 2 boys) =1
so P(there are 2 boys, and we learn of at least one boy)=(1/4)(1)=1/4

But P(we learn of a boy | one of each) =1/2
so P(there is one of each, and we learn of at least one boy)=(1/2)(1/2)=(1/4)

and P(we learn of a boy | 2 girls) =0
so P(there are two girls, and we learn of a boy)=(1/4)(0)=0

These calculations all come from the formula P(A and B)=P(A)P(B|A).

Altogether, P(we learn of a boy)=(1/4)+(1/4)=(1/2)
so we calculate the conditional probabilities as
P(2 boys | we learn of a boy) = (1/4)/(1/2) = (1/2)
P(one of each | we learn of a boy) = (1/4)/(1/2) = (1/2)
which justifies Merak Spielman's answer.

 

[MerakSpielman]

Gee, it's nice feeling justified. Sounds like you know your stuff, orsal.

 

[Tuzenbach]

BUMP!

How can such an amazing thread be pushed to page three?

 

[Tuzenbach]

Here's a riddle for all you weekend web-heads!

In the following statement, what does "it" refer to?



"A rope ends it."

 

[MerakSpielman]

reluctant to answer, since it's obviously not the rope, and it's probably not the obvious answer of "not enough information," or it wouldn't be a riddle.

So I'm going to go out on a limb and say "life." Perhaps somebody's being hanged?