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.

[vmacosta]

[ QUOTE ]
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.

[/ QUOTE ]
Huh? The new information tells you that your choice has 50% chance of being correct so switching does you no good. The answer quoted (switch doors) must be a silly play on words.

[Thythe]

[ QUOTE ]
[ QUOTE ]
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.

[/ QUOTE ]
Huh? The new information tells you that your choice has 50% chance of being correct so switching does you no good. The answer quoted (switch doors) must be a silly play on words.

[/ QUOTE ]

No, switching doors is indeed correct. I know I'm just restating the above, but think of it this way. If you pick the wrong door, the host shows you the other wrong door. You then switch and get the correct door. What is the probability of picking the wrong door originally? 2/3. Now if you pick the right door originally, the host shows you randomly one of the two wrong doors, and then you unfortunately switch to the other wrong one. What is the probability of picking the right door originally? 1/3. So it is indeed correct to switch and you will win with 2/3 probability.

[sirtimo]

I always found it easier to think of that question using 100 doors-

You pick one-
The host opens 98 empty ones-
Should you keep the one you originaly picked or the remaining unknown door?

Much easier to grasp the concept that way.

regards,
Tim

[OrangeKing]

[ QUOTE ]
[ QUOTE ]
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.

[/ QUOTE ]
Huh? The new information tells you that your choice has 50% chance of being correct so switching does you no good. The answer quoted (switch doors) must be a silly play on words.

[/ QUOTE ]

The problem is that the stated problem in this thread isn't EXACTLY like the Monty Hall problem. The missing bit of information is:

You know the host will always open a wrong door every time you play this game. So it's not a surprise when he shows you one of the losing doors.

If he just decides to show you a door for no reason, you've gained no information. But when you know ahead of time that he's going to show you a losing door no matter what you pick...then you've gained a lot of information.

Same goes for the 100 doors example. If you know he'll show you 98 of the losing doors after you pick your door, do you really think it makes no difference whether you stick with your original door, or switch to the one other door he didn't open? Of course not; there's a 99% chance that other door is the winner. On the other hand, if you and 99 other people each have a door, and you see 98 people open their doors and lose, you and the other remaining contestant each have a 50% chance of winning.

[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?