Puzzleing Question! (HARD) - Page 35

On November 22 2004 07:22 travis wrote:
i like this karlsberg guy

On November 22 2004 09:18 gg2w wrote:
I'll concede. The relief of not trying to rack my brain over this far outweighs the agony of being wrong.

On November 22 2004 10:33 gg2w wrote:
BigBalls, sadly I'm a math grad and was out to lunch pretty much the whole way. Was trying to rely on an old math article which I couldn't find and had hazy recollections of. Having said that, stats and probability is about as far away from my field of research as possible. I can honestly say that probability theory is about on the same scale as greek poetry and is only surpassed in difficulty by adding and multiplying (of which I am truly terrible).

On November 22 2004 10:39 Catyoul wrote:

heh :p
what are you working on ?

On November 22 2004 09:39 BigBalls wrote:
I have not read through the entire thread. But i will shed some light on the question posed by radar14 in around the 10th page.


There are a ton of applications to coding theory with problems like these. Let's suppose there are 3 people and each are wearing a blue or a red hat. You can't see your own hat, but you can see other people's hats.

To win the game, at least one person out of the 3 must guess the color of his hat correctly and nobody may guess wrong.

You can win this game 75% of the time. The strategy is, if you see 2 hats of the same color, guess the opposite. possible combinations of hats are as follows:

RRR
RRB
RBR
BRR
BBR
BRB
RBB
BBB

The only way you lose is if the combination is RRR or BBB. So, this game can be won 75% of the time (6 ouf of 8).

This same type of logic can be applied to radars problem.

As for the childbirth question, ill shed some light on that too.....


The answer is 50%. It is NOT ISOMORPHIC TO THE DOOR PROBLEM. Baal, you CANNOT use the same logic for each question.

Every math major ive seen in this thread has taken that stance as well.

On November 22 2004 11:10 gg_hertzz wrote:


This would mean that the original solution was wrong, which is what I've been championing last night in my half-zombie state of consciousness.

I think Rota's solution at least assumes that one of the OTHER inmate's hat is blue which is clearly not the case.

By the way, if anyone is keeping score, and if I was right about the blue card/red card riddle then I'm ahead by a score of 2.

On November 22 2004 11:10 gg_hertzz wrote:


This would mean that the original solution was wrong, which is what I've been championing last night in my half-zombie state of consciousness.

I think Rota's solution at least assumes that one of the OTHER inmate's hat is blue which is clearly not the case.

By the way, if anyone is keeping score, and if I was right about the blue card/red card riddle then I'm ahead by a score of 2.

On November 22 2004 11:16 radar14 wrote:
the card question is not about probability, it's about the beautiful melody of logic that will play if you listen hard enough

damn how corny

On November 22 2004 09:09 radar14 wrote:
haha famouzze thinks we spent 33 pages discussing his problem

On November 22 2004 11:16 radar14 wrote:
the card question is not about probability, it's about the beautiful melody of logic that will play if you listen hard enough

damn how corny

On November 22 2004 11:59 wAt-74 wrote:
Hey guys about the boy and girl puzzle the answer IS 2/3. I'm double majoring and one of my majors is probability & statistics and this is a problem that you get in a beginning class. Lemme try to break it down a bit.

you are given the fact that you know one of the childs genders, which is a girl. Now you want to find the probability that the other child is either a boy or a girl. You can break down this problem into a simple equation:

P(the other child is a boy | given that at least one child is a girl) = P(having a girl and boy)/ P(having at least one girl) < -- there is a long proof for this equation but i dont think you would want to hear it. Trust me its true, all statisticians now it, its a basic equation and you can find the proof in any basic statistics book.

so what is the probability of having a girl and boy? well your choices are... GG BB GB BG

so P(having a girl and boy) = 2/4 // BG GB
whats the probability of having at least one girl?
P(having at LEAST one girl) = 3/4 // GG BG GB

you plug it into the equation and you get (2/4)/(3/4) = 2/3 (if you cant do the division, use a calculator :p)

Some people here think BG GB are the same when they are not. Both come in different order and that is very important to accept it. Here is a visual:

EDIT: forget the visual i cant get it to line up properly...

new visual:1st you have G-->G
or G-->B
or B-->G
or B-->B

As you can see they are different, i hope tat helps
Oh btw haha GG is still the same as GG(same goes for BB)

If you still dont believe me ask a statistician or stats professor about this, its pretty simple he/she should confirm this. AND if you still dont believe me then go to college and spend thousands of dollars so you can solve this.

AND BTW: this is a great thread and keep them coming!