Riddle me this one. Completely OT!

Let's say a man and a woman (who are unrelated) each has two children (separately).  We know at least one of the woman's children is a boy and we also know the man's oldest child is a boy. 

What are the chances that the woman has two boys? 

What are the chances that the man has two boys?

Oh yah, and Badger DIver wins....a steaming pile.

Please try again.

they have the same last name?

easiest so far..

A) Grandpa, Son, Grandson.

or

B) They really went to a strip club and only had enough money left over after lap dances to buty 3 fish fom the marlet on the way home.

Easier way to look at the anser to the motorcylce question.

It would take 2 minutes for a car to travel around a 1 mile track at 30mph.

In order to average 60mph for 2 laps around a 1 mile track it would take 2 minutes.

You have already used up those 2 minutes on the first lap so there is no way you can go fast enough on the second lap to have it only take 2 minutes..you'd have to do the second lap in 0 seconds. 

I take upper level statistics, and that is the right answer.

If you know the car is not behind door 3, you know it is behind either door 1 or 2, those are the only two options, so each has a probability of 1/2 in succeeding.

badger, is there an answer to the two switch one?

I knew what I put was not 100% certain, but I figured the chances of that strategy failing were astronomical.

Smokey,

You always switch doors.  The Monty Hall Paradox. 

For those who don't believe you can wikipedia it.  The probability if you switch is 2/3.  That was one of the first questions given to me by my Greek History Professor...proving to us that we shouldnt jump to conclusions.

Congrats AC13!!!

If you decide to change your original pick you will win the car 2/3 of the time, if you decide to keep your original pick you will win the car 1/3 of the time.

Badger made the mistake of assuming that when the game show host decided to open a door that he knew didn't have a car behind it, it somehow improved his chances of winning.  It doesn't. 

If there were a hundred doors, the host could open 98 of the ones that he knew didn't have cars, but your still stuck with a 1/100 shot of winning....unless you switch doors, then it would be 99/100.

A_i = { There is a car behind door i }

B_i = { There is no car behind door i }

P(A_1) = P(A_2) = P(A_3) = 1/3

P(B_1) = P(B_2) = P(B_3) = 2/3

Using Baye's Formula,

P( A_1 | B_3 ) = P ( A_1, B_3 ) / P( B_3 )

P( A_1, B_3 ) = 1/3

P( A_1 | B_3 ) = (1/3) / (2/3) = 1/2