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.

[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.

[vmacosta]

[ QUOTE ]
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.

[/ QUOTE ]
Actually, by your logic, they both get the same amount of impurity (10-1+1/11=100/11 for vinegar and 10-10/11=100/11 for oil). Is the answer based on lower density of oil or do we assume uniform mixing?

[uphigh_downlow]

equal amounts in each, if you start with equal volumes in the jars

Xo and Xv to start X is the # spoonfullss in each

Xo + 1v and (X-1)v after the first exchange

(Xo + 1v){1- (1/X+1)} and (X-1)v + (Xo +1v)/X+1
after second exchange

Turns out that there is X/(X+1) of one in the other, both ways

[LetYouDown]

Yeah, I got the concept but didn't even think about it. I vote for same amount.

[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

[gaming_mouse]

[ QUOTE ]
equal amounts in each, if you start with equal volumes in the jars

Xo and Xv to start X is the # spoonfullss in each

Xo + 1v and (X-1)v after the first exchange

(Xo + 1v){1- (1/X+1)} and (X-1)v + (Xo +1v)/X+1
after second exchange

Turns out that there is X/(X+1) of one in the other, both ways

[/ QUOTE ]

Your answer is correct, but I believe this calculation is assuming that they are getting evenly mixed, which you cannot assume. That's why the question uses oil and vinegar. In fact, you cannot know how much of each will be in the other -- there might be 0, and there might be a full spoonful. But even so, you CAN always know that the amount of contamination will be equal. There's a simple logical argument, which requires no algebra, and is the reason I like the puzzle.

gm

[RocketManJames]

When I first heard this problem a couple years ago, I ended up reasoning it out using jars of marbles. One jar had N red marbles, and another had N blue marbles.

Using this approach, it was pretty easy to see why the contamination was equal.

-RMJ

[gaming_mouse]

[ QUOTE ]
When I first heard this problem a couple years ago, I ended up reasoning it out using jars of marbles. One jar had N red marbles, and another had N blue marbles.

Using this approach, it was pretty easy to see why the contamination was equal.

-RMJ

[/ QUOTE ]

Yes, that works. Or just notice that the amount of oil being transferred over to the vinegar side is exactly the same as the amount of vinegar which remains in the oil side. To put it mathematcially:

(vinegar being returned to vinegar) + (oil moving to vinegar) = 1 spoonful
(vinegar being returned to vinegar) + (vinegar left over in oil) = 1 spoonful

The result follows.

[chaosuk]

It is perhaps simpler and key just to acknowledge that both jars finished with the same volume of liquid they started with.Since the liquids are conserved between the jars, whatever one jar lost the other jar gained. Whatever they lost must have been replaced by an equal volume from the other jar - the jars simply exchange a volume of eachothers liquid which muist be the same since, they both end up with the the same volume they started with.

EG If the vinegar jar has 75% vinegar left then it must have 25% oil. So the oil jar must have 75oil, 25% vinegar.

[uphigh_downlow]

I feel really stupid having even attempted the math. This is such an elegant and simple reasoning.

Sometimes the easiest things are right infront of your eyes, and you cant see them.

Thanks for openingg my eyes /images/graemlins/smile.gif

[Thythe]

Yeah I didn't see that either. Very clever.

[Siegmund]

[ QUOTE ]
It is perhaps simpler and key just to acknowledge that both jars finished with the same volume of liquid they started with.Since the liquids are conserved between the jars, whatever one jar lost the other jar gained. Whatever they lost must have been replaced by an equal volume from the other jar - the jars simply exchange a volume of eachothers liquid which muist be the same since, they both end up with the the same volume they started with.


[/ QUOTE ]

Yes.

[ QUOTE ]

EG If yhe vinegar jar has 75% vinegar left then it must have 25% oil. So the oil jar must have 75oil, 25% vinegar.

[/ QUOTE ]

And no.

The result applies to absolute quantity in the wrong jar, NOT to percentage. The percentages will only be equal if both jars have the same volume (which was not specified in the original post.)

[gaming_mouse]

nice.

[chaosuk]

'And no.'

A fair cop, but it was 4am. Though I thought the problem was a mixture problem , which, as you said, required equal sizes and which is, of course, a special case of the problem amount.

I only just caught end of the thread and hence the answer. No doubt had this been an exam question, I'd have gone down the same maths route, but since I had the answer (& the maths had been done) I could look for a more simpler explantion. It is particularly disturbing, though, how we've trained our minds to approach these problems - a solution that involves simple addition and subtraction and one that a small child could understand is not immediately apparent.

[stankphish]

"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?"

I think im missing something but. When you put a teaspoon of vinegar into the oil and then you put a teaspoon of the oil and vinegar back into the oil.

Dosnt that mean that you are putting more vinegar into the oil b/c you are putting oil and vinegar back into the vinegar meaning that there is more vinegar in the oil?

I understand the logic that they both have the same amount of liquid that they started with but thats not what the problem asks. It asks which jar has a larger amount of the other substance in it. Hypothetically if you spooned out of jar with oil and vinegar in it and the spoon contained 3/4 oil and 1/4 vinegar and you put that back in the vinegar jar there would be 3/4 spoonfull of vinegar in the oil and 3/4 oil in the vinegar... ok i just figured it out

[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.

[PairTheBoard]

[ QUOTE ]
[ QUOTE ]
equal amounts in each, if you start with equal volumes in the jars

Xo and Xv to start X is the # spoonfullss in each

Xo + 1v and (X-1)v after the first exchange

(Xo + 1v){1- (1/X+1)} and (X-1)v + (Xo +1v)/X+1
after second exchange

Turns out that there is X/(X+1) of one in the other, both ways

[/ QUOTE ]

Your answer is correct, but I believe this calculation is assuming that they are getting evenly mixed, which you cannot assume. That's why the question uses oil and vinegar. In fact, you cannot know how much of each will be in the other -- there might be 0, and there might be a full spoonful. But even so, you CAN always know that the amount of contamination will be equal. There's a simple logical argument, which requires no algebra, and is the reason I like the puzzle.

gm

[/ QUOTE ]

Right. All you have to notice is that after exchanging teaspoons the volumes are as before so whatever oil is in the vineger is missing from the oil and must therefore have been replaced by an equal amount of vineger.

PairTheBoard

[theTourne]

[ QUOTE ]
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

[/ QUOTE ]

A guy and I were trying to explain this concept to a friend of mine. We used your approach, and said "pretend there are 100 doors, he's going to show you 98 empty doors," etc. Well my friend was still confused, and after a few more agonizing explanation attempts the other guys said, "OK, well then imagine that you have a million doors..."

It was hilarious at the time.

[DonkeyChip]

Hi all, first post here.

[ QUOTE ]
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.

[/ QUOTE ]

OK, NOW I get it. I thought you guys were crazy. That's VERY important!

Think of it another way (under the original premise that the second door opened is RANDOMLY chosen): Suppose you originally picked the correct door; in that case then EVERY time Monty opens another door (randomly), it will be empty. But if you originally chose incorrectly, then only 50% of the time Monty RANDOMLY opens another door it will be empty. So once Monty opens that second door and it's empty, then it's TWICE as likely that your original choice was correct. So using that logic, you should STICK with your original choice. But this logic is just as flawed as the original logic (provided we think that the second door is RANDOMLY chosen).

If that's a bunch of horse-hockey, lemme know. /images/graemlins/smirk.gif

Regarding the vinegar and oil: Vinegar and oil for the most part don't mix and vinegar is more dense than oil, so the teaspoon of vinegar just sinks to the bottom of the oil. So you've put 1 teaspoon of vinegar in the oil and then put 1 teaspoon of oil into the vinegar. There are equal amounts of each in the other. They could've chosen a lot of different liquids/substances but I'm assuming they chose oil and vinegar for a reason.

I like these kinds of riddles, gives me something to think about while driving to/from work.

[kyro]

[ QUOTE ]
[ QUOTE ]
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

[/ QUOTE ]

A guy and I were trying to explain this concept to a friend of mine. We used your approach, and said "pretend there are 100 doors, he's going to show you 98 empty doors," etc. Well my friend was still confused, and after a few more agonizing explanation attempts the other guys said, "OK, well then imagine that you have a million doors..."

It was hilarious at the time.

[/ QUOTE ]

Sickeningly enough, this happened to me when my teacher explained the problem. He did the 100 door example and I was still adamant he was wrong. About 5 minutes later, it finally hit me, and I'm usually pretty quick with this kinda stuff.