Probabilty Riddles

[closer2313]

I can see how the answers are the answers, but I dont see exactly why. Can you guys explain?

Imagine you just won a TV-quiz, and the quiz master gives you the chance to go home with a car that is hidden behind one of three closed doors. The quiz master asks you to select door A, B, or C, and you select door B. Then, the quiz master opens door C, behind which no car appears to be present, and he asks you whether you want to stick to door B or to change to door A. What should you do?

Answer
Go to Door A because it is 66% while B is only 33%.

I can see that if door B was only 33% with 3 doors, and one door is eliminated that the other door would double, but why wouldn't both doors increase to 50%?

You are at the park, pushing your daughter on a swing. You meet a woman and start a conversation. She says she has two children. You ask if she has any girls. She says yes. What is the chance that both children are girls?

Answer 1/3
Possibily combinations of children with one girl is
BG,GB,GG
so GG is 1/3 but aren't GB and BG the same thing? I dont see why the order matters in this problem.

Thanks for any help.

[Yads]

For question 1.
You either picked the right door off the bat or you didn't
Probablity that you picked the right door is 1/3 so the chances that door B is correct is 1/3. Conversly the probability that you picked the wrong door is 2/3 so the chances that the other door is correct is 2/3.

[closer2313]

wow, that makes a lot of sense. Thanks alot.

[TomCollins]

I think the Monty Hall problem is posted more often than a useless post by Vince!

[MtDon]

For #2:

[ QUOTE ]
I can see how the answers are the answers, but I dont see exactly why. Can you guys explain?

(2)

You are at the park, pushing your daughter on a swing. You meet a woman and start a conversation. She says she has two children. You ask if she has any girls. She says yes. What is the chance that both children are girls?

Answer 1/3
Possibily combinations of children with one girl is
BG,GB,GG
so GG is 1/3 but aren't GB and BG the same thing? I dont see why the order matters in this problem.

Thanks for any help.

[/ QUOTE ]

It is a matter of the probility distribution.

A woman with two children can have them in 4 ways: BG, BB, GB, GG. Each of those conditions are equaly probable. So it is twice as likely she has a girl and a boy as having two girls. So given that you know she has at least one girl, which is the same thing as saying she doesn't have BB, there is 1 chance in three that she has GG.

Here's another way of thinking about it.

Supose there are 100 women who give birth on the same day. Not considering randomness, 50 will have boys and 50 will have girls.

One year later, they all again give birth. Each still has a 50% chance of having either a boy or a girl. So, of the 50 women who had boys, 25 will have boys and 25 will have girls. Of the 50 women who had girls, 25 will have boys and 25 will have girls.

Therefore, 25 women will have two boys, 50 will have a boy and a girl, and 25 will have two girls.

75 will have at least one boy and 75 will have at least one girl.

The woman in the "probability riddle" is part of the 75 women who have at least one girl. 25 of the 75 women with at least one girl have two girls. So the probability she has two girls, given that she has at least one girl is: 25/75 = 1/3

-- Don

[gaming_mouse]

I kind of like this one (not too hard, though)....

Young Hans leaves each day from school and walks to the bus stop. Two different bus lines pass by this stop, one that goes to the airport and one that goes to the harbor. Hans likes to watch both the planes and the ships, so he just gets on the first bus that arrives.

Hans arrives at the bus stop at random time between 3 and 4 each day.
Each bus makes exactly 3 stops at the bus stop between 3 and 4 each day.
Yet, at the end of the year, Hans ended up going to the harbor 12 times as often as he want to the airport.

How can this be?

[uphigh_downlow]

easy. The time from the last bus to the current bus belongs to the current bus.

So just darw schedules such that sum of times for one kind of bus is 12 times that of the other bus.

Many ways of doing this.

Close answer would be that Harbor bus comes a little less than 5 mins after the aiport bus

[RocketManJames]

I think there are other ways that this can happen, but here's what I think will satisfy the conditions.

So, there is an hour window where the kid will go to the bus stop randomly. He happens to take the Harbor bus 12 times as often as the other.

If the bus schedule is weird, like:

3:50, 3:52, 3:55:37 - Harbor
3:55:38, 3:58, 4:00 - Airport

Then, wouldn't the Harbor bus cover 12/13 of the hour and the Airport bus cover the remaining 1/13?

Or am I off base?

-RMJ

[gaming_mouse]

No, that's it. I warned that it wasn't too hard, but I still kind of like it. People usually either get it right away, or just never consider the possibility that the buses aren't even spaced? Here's another old famous one that I like even though it's not terribly hard (or probability related).

There is a jar of oil and a jar of vinegar. One spoonful of vinegar is poured into the oil. A spoonful is then taken back out of the oil-vinegar mixture, and poured back into the vinegar?

Is there now more vinegar in the oil, more oil in the vinegar, an equal amount of each in the other, or is it impossible to tell?

[LetYouDown]

Vinegar in the oil, unless I'm missing something, and assuming that the vinegar distributes itself evenly. If there's 10 teaspoons in each jar, then after you put the vinegar in the oil, there's now 11 teaspoons total, with 10/11 oil, 1/11 vinegar. So a teaspoon taken from that jar will contain roughly 91% oil, and 9% vinegar. Which means that the vinegar jar gets some of its vinegar back and 91% of a teaspoon of oil.

I must be missing something, heh.