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[Jerk of All trades]

Reminds me of that old logic quiz/ problem.

You're on a game show, and Monty Hall shows you 3 doors.  2 of them are empty, and behind one of them is 1,000,000$.

You pick one door, but before you open it, Monty opens up one of the remaining doors and shows you that it is empty.  If you want to maximize your odds of winning the million, do you:
A: Switch your pick to the door that you did not chose, and Monty did not open,
B: Or do nothing, because it doesn't matter

The answer is A. If you switch, your expected return is 666k, while if you stay it's only 333K. I've known college educated people that refuse to believe this is true.

[Ozle]

Oh no....what have you done!!

[heron]

I like that one!
My favorite however, is this:
I am a boy with one sibling. What is the probability that my sibling is a girl?

[Tdog]

I like that one!
My favorite however, is this:
I am a boy with one sibling. What is the probability that my sibling is a girl?

50% right?

[Jerk of All trades]

Less than 50%, because of identical twins.

[WanderingWinder]

I like that one!
My favorite however, is this:
I am a boy with one sibling. What is the probability that my sibling is a girl?

50% right?

I have two children. (At least) one of them is a boy born on a tuesday. What is the probability (assuming all days of the week equally likely birthdays and equal probabilities of boys and girls and independence within a family, etc. etc.) that my other child is a boy?

[eHalcyon]

I like that one!
My favorite however, is this:
I am a boy with one sibling. What is the probability that my sibling is a girl?

I always have trouble sorting these out.  I think (using the regular assumptions) it is 50% with the way it was phrased here.

[Beyond Awesome]

Reminds me of that old logic quiz/ problem.

You're on a game show, and Monty Hall shows you 3 doors.  2 of them are empty, and behind one of them is 1,000,000$.

You pick one door, but before you open it, Monty opens up one of the remaining doors and shows you that it is empty.  If you want to maximize your odds of winning the million, do you:
A: Switch your pick to the door that you did not chose, and Monty did not open,
B: Or do nothing, because it doesn't matter

The answer is A. If you switch, your expected return is 666k, while if you stay it's only 333K. I've known college educated people that refuse to believe this is true.

I heard this one before, but to this day, I still don't understand. If anyone can offer me an explanation as to why this is, I would love to hear.

[SirPeebles]

I like that one!
My favorite however, is this:
I am a boy with one sibling. What is the probability that my sibling is a girl?

50% right?

I have two children. (At least) one of them is a boy born on a tuesday. What is the probability (assuming all days of the week equally likely birthdays and equal probabilities of boys and girls and independence within a family, etc. etc.) that my other child is a boy?

You have a child A and a child B.  Consider all assignments of birthdays and (binary) sexes.  There are 7 cases in which child A is a boy born on Tuesday and 7 cases in which child B is, but one of these cases is common, so there are 13 cases total in which at least one is a boy born of tuesday while the other is still a boy.  Analogously, there are 14 cases in which child A is a boy born on Tuesday if child B's gender isn't specified, and 14 for child B, with one shared resulting in 27 total cases.

So the conditional probability is 13/27

[eHalcyon]

Reminds me of that old logic quiz/ problem.

You're on a game show, and Monty Hall shows you 3 doors.  2 of them are empty, and behind one of them is 1,000,000$.

You pick one door, but before you open it, Monty opens up one of the remaining doors and shows you that it is empty.  If you want to maximize your odds of winning the million, do you:
A: Switch your pick to the door that you did not chose, and Monty did not open,
B: Or do nothing, because it doesn't matter

The answer is A. If you switch, your expected return is 666k, while if you stay it's only 333K. I've known college educated people that refuse to believe this is true.

I heard this one before, but to this day, I still don't understand. If anyone can offer me an explanation as to why this is, I would love to hear.

There are three doors, A, B, and C.  Two have Goats, the other has a car.  You pick one.  The chance you picked correctly is 1/3 (the car is correct; we don't want goats in this example).

The host opens one of the remaining doors to reveal a Goat.  Now you may switch doors if you wish.

The crux is the original choice and the guarantee that one incorrect option gets eliminated. 

What happens when you switch?  If you had picked correctly, your switch will cause you to lose.  This happens 1/3 of the time, because your chance of picking correctly to start was 1/3.  But if you picked wrong (2/3 chance), switching means you win.

[jonts26]

I like that one!
My favorite however, is this:
I am a boy with one sibling. What is the probability that my sibling is a girl?

50% right?

I have two children. (At least) one of them is a boy born on a tuesday. What is the probability (assuming all days of the week equally likely birthdays and equal probabilities of boys and girls and independence within a family, etc. etc.) that my other child is a boy?

Hint: the Tuesday part is relevant.

[theory]

Reminds me of that old logic quiz/ problem.

You're on a game show, and Monty Hall shows you 3 doors.  2 of them are empty, and behind one of them is 1,000,000$.

You pick one door, but before you open it, Monty opens up one of the remaining doors and shows you that it is empty.  If you want to maximize your odds of winning the million, do you:
A: Switch your pick to the door that you did not chose, and Monty did not open,
B: Or do nothing, because it doesn't matter

The answer is A. If you switch, your expected return is 666k, while if you stay it's only 333K. I've known college educated people that refuse to believe this is true.

I heard this one before, but to this day, I still don't understand. If anyone can offer me an explanation as to why this is, I would love to hear.

There are three doors, A, B, and C.  Two have Goats, the other has a car.  You pick one.  The chance you picked correctly is 1/3 (the car is correct; we don't want goats in this example).

The host opens one of the remaining doors to reveal a Goat.  Now you may switch doors if you wish.

The crux is the original choice and the guarantee that one incorrect option gets eliminated. 

What happens when you switch?  If you had picked correctly, your switch will cause you to lose.  This happens 1/3 of the time, because your chance of picking correctly to start was 1/3.  But if you picked wrong (2/3 chance), switching means you win.

The key is that the host picks the door to open (to reveal a goat) non-randomly.  If the host had randomly picked a door to reveal a goat, then it would indeed be 1/2. 

[Jerk of All trades]

Wow, I've never heard that tuesday boy question before. It gets even crazier if you say "A boy born in an even numbered year"  (assuming birth years are binary, like if they aren't siblings or something)

the probability drops to nearly 43%.  That is totally counter-intuitive.

[jonts26]

Reminds me of that old logic quiz/ problem.

You're on a game show, and Monty Hall shows you 3 doors.  2 of them are empty, and behind one of them is 1,000,000$.

You pick one door, but before you open it, Monty opens up one of the remaining doors and shows you that it is empty.  If you want to maximize your odds of winning the million, do you:
A: Switch your pick to the door that you did not chose, and Monty did not open,
B: Or do nothing, because it doesn't matter

The answer is A. If you switch, your expected return is 666k, while if you stay it's only 333K. I've known college educated people that refuse to believe this is true.

I heard this one before, but to this day, I still don't understand. If anyone can offer me an explanation as to why this is, I would love to hear.

There are three doors, A, B, and C.  Two have Goats, the other has a car.  You pick one.  The chance you picked correctly is 1/3 (the car is correct; we don't want goats in this example).

The host opens one of the remaining doors to reveal a Goat.  Now you may switch doors if you wish.

The crux is the original choice and the guarantee that one incorrect option gets eliminated. 

What happens when you switch?  If you had picked correctly, your switch will cause you to lose.  This happens 1/3 of the time, because your chance of picking correctly to start was 1/3.  But if you picked wrong (2/3 chance), switching means you win.

I find it becomes more intuitive when you increase the number of options. Let's say I spread out a deck of cards face down and ask you to pick the two of diamonds. So you pick one and you have a 1/52 chance of being right. I know which one is actually the two of diamonds, but you don't. Now lets say I go ahead and flip over 50 cards, none of which are the two of diamonds, leaving only the card you originally picked and another one. Now, it's really unlikely that you picked the right card at random, which means it's really likely (51/52) that the other card has to be the 2 of diamonds.

[heron]

The answer to mine was 66%, as of families with two children, 2/3 of the possibilities with at least one son also have a daughter.

[eHalcyon]

The answer to mine was 66%, as of families with two children, 2/3 of the possibilities with at least one son also have a daughter.

But it depends on how the question is presented.  I have trouble sorting out which wording leads to which math... refer to link above to the wiki page for the paradox.

Edit:

To expand on this -- I don't know your gender at first.  For all I know, you are a girl.  So the possibilities are (for you/sibling):

boy/boy
boy/girl
girl/boy
girl/girl

You reveal you are a boy.  Now the possibilities are

boy/boy
boy/girl

And the chance of either one is 50%.




Further edit:

I think, for 67% to be the answer (not 66% -- round up ) you would have to reveal it like this:

I have one sibling.  At least one of us is a boy.  What is the probability that the other is also a boy?

In this case, knowing that at least one of you two is a boy leaves three possibilities for you/sibling:

boy/boy
boy/girl
girl/boy

And now we have 2/3 female sibling.

(Edited: originally said 2/3 male sibling, which was the wrong conclusion.)

[theory]

Most of these probability problems do depend on extremely precise wording.  They make for poor cocktail party conversation, especially when you start having difficulty remembering the exact formulation.

[heron]

Actually, there is no reason to omit girl/boy I believe.
Lets say I flipped 2 coins, and stated that one was heads. This is equal information that you know about me and my family:
There are two children, at least one which is a boy.
You don't know which of those coins/children I am, since I don't specify if I am older/younger.

Now that I said something random I'm going to try explain better:

I'll start the same way as you: There are two siblings.
BB
BG
GB
GG

Now, I state that I am a boy. This means that there is at least one boy. However, since you don't know which column is me, you can't rule out either BG or GB.
So,
BB
BG
GB

Which leaves a 66.666666666666666667% probability that I am a boy.

Basically, I am arguing that both of your wordings mean the same thing.
Basically, this is a linguistics question, not math.
Basically, this is a matter of opinion.

[eHalcyon]

Actually, there is no reason to omit girl/boy I believe.
Lets say I flipped 2 coins, and stated that one was heads. This is equal information that you know about me and my family:
There are two children, at least one which is a boy.
You don't know which of those coins/children I am, since I don't specify if I am older/younger.

Now that I said something random I'm going to try explain better:

I'll start the same way as you: There are two siblings.
BB
BG
GB
GG

Now, I state that I am a boy. This means that there is at least one boy. However, since you don't know which column is me, you can't rule out either BG or GB.
So,
BB
BG
GB

Which leaves a 66.666666666666666667% probability that I am a boy.

Basically, I am arguing that both of your wordings mean the same thing.
Basically, this is a linguistics question, not math.
Basically, this is a matter of opinion.

No, see, the first column IS you.   I suggest that you read the wikipedia article I linked, which delves into it further:

http://en.wikipedia.org/wiki/Boy_or_Girl_paradox

In the article, their example differentiates the two siblings as older or younger.  If I know that the OLDER sibling is the one that is male, then the chance of the younger sibling being male is 50%.  In this case, the identifying factor is you vs. your sibling.

So the crux is whether you say "I am male" or "one of us is male".  In the former, you identify yourself specifically so the chance ends up at 50%.  In the latter, I don't know who it applies to, so the chance of the second sibling (which could be you) being male is 67%.

[Tables]

Reminds me of that old logic quiz/ problem.

You're on a game show, and Monty Hall shows you 3 doors.  2 of them are empty, and behind one of them is 1,000,000$.

You pick one door, but before you open it, Monty opens up one of the remaining doors and shows you that it is empty.  If you want to maximize your odds of winning the million, do you:
A: Switch your pick to the door that you did not chose, and Monty did not open,
B: Or do nothing, because it doesn't matter

The answer is A. If you switch, your expected return is 666k, while if you stay it's only 333K. I've known college educated people that refuse to believe this is true.

I heard this one before, but to this day, I still don't understand. If anyone can offer me an explanation as to why this is, I would love to hear.

I've found people understand it better if you increase the number of doors.

You pick one door out of 100, which has a 1% chance of having the prize behing it. Monty then opens 98 other doors, leaving just two doors - the one you picked, and the other one.

Should you switch or does it not matter.

As for an actual explanation (of the usual problem), there are essentially two cases: If you pick right or if you don't.
If you pick the right door to start with (prob 1/3), then Monty opens another door at random. If you switch to the final door then with probability 100% you will lose, and with probability 0% you will win.
If you pick a wrong door to start with (prob 2/3) then Monty opens the other incorrect door. If you switch to the final door then with probability 0% you will lose and with probability 100% you will win.

So what's the probability that you win?
If you stick:
(1/3)*100% + (2/3)*0% = 1/3
(Prob of picking right door to start * chance staying is right) + (Prob of picking wrong door to start * chance staying is right) = 1/3
If you switch:
(1/3)*0% + (2/3)*100% = 2/3
(Prob of picking right door to start * chance switching is right) + (Prob of picking wrong door to start * chance switching is right) = 2/3

[Qvist]

Which leaves a 66.666666666666666667% probability that I am a boy.

I think you forgot a 6 there. It's actually 66.6666666666666666667%
Sorry 

[HiveMindEmulator]

Now, I state that I am a boy. This means that there is at least one boy. However, since you don't know which column is me, you can't rule out either BG or GB.
So,
BB
BG
GB

Which leaves a 66.666666666666666667% probability that I am a boy.

[Synthesizer]

Another nice one that people tend to refuse to accept:
You are flipping a coin. Previously, you have established that the coin has equal odds of showing heads or tails; so no funny business there. You get heads ten times in a row. What are the odds of getting heads next coinflip?
And is this equal to the odds of getting eleven times heads in a row?

odds of the next coinflip are the same as any random coinflip, 50%. Interestingly, odds of eleven consecutive coinflips with equal result is 100%*((.5)^11) = ~0.05%. The difference lies in the fact that in the first case, the "history" has already taken place; it doesn't matter what result they gave; the information is irrelevant, I might just as well have stated (heads x3-tails x3 - heads) as history. Basically, I just kept flipping the coin, until variance gave me the desired history; if I try long enough it WILL happen eventually.  While in the second example, I have more uncertainty: I don't have the freedom to try, try, try again; I am calculating the odds of getting it right in one try.

Same problem works with throwing a die, or black vs. red in roulette but you have to adjust for the odds of each option (1/6 vs. 1/2 for the die, and for roulette the odds are slightly less than 50% due to the green fields.)

And for those of you who think this stuff is merely done for fun - I'm an engineer, and I have done probability calculations similar to this in that function in the past. Most difficult part is explaining any counterintuitive results to management...

[Jimmmmm]

Another nice one that people tend to refuse to accept:
You are flipping a coin. Previously, you have established that the coin has equal odds of showing heads or tails; so no funny business there. You get heads ten times in a row. What are the odds of getting heads next coinflip?
And is this equal to the odds of getting eleven times heads in a row?

odds of the next coinflip are the same as any random coinflip, 50%. Interestingly, odds of eleven consecutive coinflips with equal result is 100%*((.5)^11) = ~0.05%. The difference lies in the fact that in the first case, the "history" has already taken place; it doesn't matter what result they gave; the information is irrelevant, I might just as well have stated (heads x3-tails x3 - heads) as history. Basically, I just kept flipping the coin, until variance gave me the desired history; if I try long enough it WILL happen eventually.  While in the second example, I have more uncertainty: I don't have the freedom to try, try, try again; I am calculating the odds of getting it right in one try.

Same problem works with throwing a die, or black vs. red in roulette but you have to adjust for the odds of each option (1/6 vs. 1/2 for the die, and for roulette the odds are slightly less than 50% due to the green fields.)

And for those of you who think this stuff is merely done for fun - I'm an engineer, and I have done probability calculations similar to this in that function in the past. Most difficult part is explaining any counterintuitive results to management...

I have this debate in my head in Mafia all the time. "Eevee's been Town like 10 games in a row - he's GOTTA be scum this game!"

[Davio]

I think we humans sometime have a hard time accepting that probabilities are not universal truths.

In the Monty Hall problem, if a random stranger walked on the set and was asked to pick a door (without having picked a previous door), he would always have a 50/50 chance. Probabilities can thus be different for different people, based on the information they possess. And if they make choices based on their own probabilities, they will have an expected value in accordance with their percentage.

I find the birthday paradox (is it a paradox, though?) also interesting and many of you may know it.
The question is simple: What's the least amount of people that need to be in a room together for a better than 50% chance of at least two of them having the same birthday (date only of course, not the same year)? And how many for a 99% chance?