A brain teasing probability puzzle

Is it a prop plane or a jet?

This problem also depends entirely on the host being REQUIRED to show you a $1/goat box. If he is motivated to avoid paying out and can do whatever he wants, you're playing rock/paper/scissors with him

The problem also doesn't seem to apply to Deal or No Deal. I'm not too confident on that assertion though

Entire thing reeks of "let's present some silly scenario with unstated imperatives and sound clever by misleading people"

Tea_sl said:

If you have a group of random people in a room how many do you need before there is a 50/50 chance that there is a shared birthday?

The answer may surprise you.

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Actually it wouldn't, I'd read up on that before. Still quite interesting though!

Iannis_sl said:

Is it a prop plane or a jet?

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Turboprop.

/mindblown

Dabamf_sl said:

The wording of the problem is stupid because it doesn't say "the host will always pick a goat." That's important and the reason I, and probably lots of others, misconceptualized it.



The second set of choices is not always the same.

Scenarios of switching, assuming you always pick A first (because the first choice is meaningless)
A is car, B/C are goats. You switch and lose (1/3 time)
A is goat, B is car, C is goat. Host shows C (he's forced to, it's the only goat option), you switch to B, you win (1/3 of the time)
A is goat, B is goat, C is car. Host shows B (he's forced to, it's the only goat option), you switch to C, you win (1/3 of the time)
Win 2/3, lose 1/3

Scenarios of not switching
A is car, you win
B is car, you lose
C is car, you lose
Win 1/3, lose 2/3

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I agree with your assessment that "the wording of the problem is bullshit." It's exceptionally hard to use written words to explain multiple quantities and their relationships with very much precision at all, which is exactly why math exists.

But I offer a somewhat more detailed, precise, and accurate explanation of the scenario:

Always switching =
A is car, B & C are goats. Host shows B. You switch and lose. (1/4 of the time)
A is car, B & C are goats. Host shows C. You switch and lose. (1/4 of the time)
A is goat, B is car, C is goat. Host shows C (he's forced to, it's the only goat option), you switch to B, you win. (1/4 of the time)
A is goat, B is goat, C is car. Host shows B (he's forced to, it's the only goat option), you switch to C, you win. (1/4 of the time)
Win 2/4, lose 2/4

Never switching =
A is car, B & C are goats. Host shows B. You win. (1/4 of the time)
A is car, B & C are goats. Host shows C. You win. (1/4 of the time)
A is goat, B is car, C is goat. Host shows C (he's forced to, it's the only goat option), you lose. (1/4 of the time)
A is goat, B is goat, C is car. Host shows B (he's forced to, it's the only goat option), you lose. (1/4 of the time)
Win 2/4, lose 2/4

What if I know the odds are better if I switch but I still don't want to switch? The prize is blood money and an innocent person will be killed if I win the 100,000 and I don't want to switch. But the poll says I can only answer no if I agree the odds don't change. This means I can't answer the poll!

Araysar_sl said:

you are all so fucking stupid, how the fuck do you even live with yourselves.

holy shit, i am gonna edit this post again and ask how is it possible that so many of you are so fucking stupid.

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I'm not even going to try to understand how some of these people remember to breathe. But it's pretty clear where they're being led astray about this problem. There's a false equivalence being made where both of the two different goats = you lose. The fallacy that "you lose" is only one single event leads to an incorrect enumeration of the possible outcomes.

If you replace the two goats in the original scenario with "the top half of a mtf transsexual" and "the bottom half of a ftm transsexual", and rephrase the rules so that players win whatever is behind the door they chose, then you'll start seeing people put enough effort in to correctly enumerate the possibilities.

I think we can dig deeper.

It's a continuous event, not two discrete ones. Your first choice informs your second. In itself the first choice is meaningless, in relation to the second choice the first choice is meaningful.

Pretty sure the problem is just the semantics of the original post, which most of us instantly recognized as a Monty variation and responded to as such. That would make it a hellish exam question -- a poorly stated monty hall problem. Khalid, don't do this shit to your poor hung-over half-retarded students.

If by faulty premise you mean we were responding to what he meant rather that what he wrote, then yes.

If by faulty premise you mean the logic of the 66% is bad, you're just dumb.

There's a 50/50 for you!

Big Dummy.

Iannis_sl said:

Pretty sure the problem is just the semantics of the original post, which most of us instantly recognized as a Monty variation and responded to as such. That would make it a hellish exam question -- a poorly stated monty hall problem. Khalid, don't do this shit to your poor hung-over half-retarded students.

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The crux of my argument (and Araysar's) is actually that the Monty Hall problem is specifically poorly worded in a way that allows lazy people to take an incorrect mental shortcut and end up correctly doing the incorrect math.

Melvin_sl said:

I'm not even going to try to understand how some of these people remember to breathe. But it's pretty clear where they're being led astray about this problem. There's a false equivalence being made where both of the two different goats = you lose. The fallacy that "you lose" is only one single event leads to an incorrect enumeration of the possible outcomes.

If you replace the two goats in the original scenario with "the top half of a mtf transsexual" and "the bottom half of a ftm transsexual", and rephrase the rules so that players win whatever is behind the door they chose, then you'll start seeing people put enough effort in to correctly enumerate the possibilities.

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It doesn't matter. If your going to count each goat as a different goat,(Goat1 and Goat2), then you'll need to account for that in your winning. So for wins you'll have to add:
A is car, B & C are goats. Host shows B. You switch and lose. (1/6 of the time)
A is car, B & C are goats. Host shows C. You switch and lose. (1/6 of the time)
A is goat1, B is car, C is goat2. Host shows C (he's forced to, it's the only goat option), you switch to B, you win. (1/6 of the time)
A is goat2, B is car, C is goat1. Host shows C (he's forced to, it's the only goat option), you switch to B, you win. (1/6 of the time)
A is goat1, B is goat2, C is car. Host shows B (he's forced to, it's the only goat option), you switch to C, you win. (1/6 of the time)
A is goat2, B is goat1, C is car. Host shows B (he's forced to, it's the only goat option), you switch to C, you win. (1/6 of the time)

Which is surprise surprise.... 4/6... or 2/3.

Think of the concept like this, without probability. You have 10 doors. You choose door 1. You either chose the car, or the car is in one of the other 9 doors. The host opens 8 of the other doors as he cannot open the prize (its implied.). So the car is either behind your 1 door, or it is *still* behind one of the other 9 doors. But you know 8 of them are empty. By switching, your choosing "the other 9 doors". So which is more likely, the car is behind 1 chosen door, or its somewhere behind the other 9 chosen doors?

Now instead of 10 doors, you have 3 doors. So its either behind 1 door, or its behind the other 2 doors.

Hope that cleared it up.

Edit: If your just arguing over the wording of the original problem, then I guess we can all agree it was worded poorly. I'll leave the above as an answer to the intended Monty Hall problem if anyone was still curious.

Close, but no cigar.

Always switching =
A is car, B & C are goats 1 and 2, respectively. Host shows B. You switch and win goat 2.
A is car, B & C are goats 1 and 2, respectively. Host shows C. You switch and win goat 1.
A is car, B & C are goats 2 and 1, respectively. Host shows B. You switch and win goat 1.
A is car, B & C are goats 2 and 1, respectively. Host shows C. You switch and win goat 2.
A is goat1, B is car, C is goat2. Host shows C (he's forced to, it's the only goat option), you switch to B, you win a car.
A is goat2, B is car, C is goat1. Host shows C (he's forced to, it's the only goat option), you switch to B, you win a car.
A is goat1, B is goat2, C is car. Host shows B (he's forced to, it's the only goat option), you switch to C, you win a car.
A is goat2, B is goat1, C is car. Host shows B (he's forced to, it's the only goat option), you switch to C, you win a car.

So to rephrase the problem: Choose one of 3 doors. Behind it is the truth about Araysar (a)trollinghardor (b)really being obtuse. Then one door is opened for us, in which it is revealed Araysar is (b)really being obtuse. So we switch - because we hoping to find he's just (a)trollinghard- but since he IS trolling REALhard, he instantly switches his stance (b)really being obtuse.

So the final door is opened and reveals... LUMIE. End of world ensues. The only thing that can now bring us salvation, is FAITH in the fact that Araysar is a rational person. And then it strikes us...

seatbelts.

/oblivion

Melvin_sl said:

Close, but no cigar.

Always switching =
A is car, B & C are goats 1 and 2, respectively. Host shows B. You switch and win goat 2.
A is car, B & C are goats 1 and 2, respectively. Host shows C. You switch and win goat 1.
A is car, B & C are goats 2 and 1, respectively. Host shows B. You switch and win goat 1.
A is car, B & C are goats 2 and 1, respectively. Host shows C. You switch and win goat 2.
A is goat1, B is car, C is goat2. Host shows C (he's forced to, it's the only goat option), you switch to B, you win a car.
A is goat2, B is car, C is goat1. Host shows C (he's forced to, it's the only goat option), you switch to B, you win a car.
A is goat1, B is goat2, C is car. Host shows B (he's forced to, it's the only goat option), you switch to C, you win a car.
A is goat2, B is goat1, C is car. Host shows B (he's forced to, it's the only goat option), you switch to C, you win a car.

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So what you arereallysaying is you are able to pick the car on your first choice in 4 of the 8 cases, instead of an expected 1 in 3?

DiddleySquat_sl said:

So what you arereallysaying is you are able to pick the car on your first choice in 4 of the 8 cases, instead of an expected 1 in 3?

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What I'mreally reallysaying is that I have an exceptionally well developed sense of smell. So well developed, in fact, that I can smell my own shit when someone points out that I've stepped in it. Looks like I completely misdiagnosed why the odds of the two different goats were halved.