Brain Teasers

[Pauldohert]

Ok, next one:

Two guards are guarding safes. One of the safes contains $1,000,000.00 and the other will explode on opening, killing everyone around. You are only allowed to ask one guard one question, which must be answered either yes or no. The problem is that one guard always tells the truth and the other always lies, and you have no idea which is which. What one question would you ask and why?

Without giving it away to spoil the fun

1 * -1 = - 1 as is -1 * 1

so - 1 * -1 = 1

 

[ChipperT]

Without giving it away to spoil the fun

1 * -1 = - 1 as is -1 * 1

so - 1 * -1 = 1

Yes, it quite a boolean problem, isn't it?

 

[Brianwarnock]

Yes, it quite a boolean problem, isn't it?

You guys seem to like to complicate matters.

This question is always cropping up and is easily answered by people who have never heard of boolean but can apply simple logic.

Brian

 

[Pauldohert]

Another classic - (I and probably others have posted it here before) , even after knowing the answer.


Suppose you're on a game show, and you're given the choice of three doors: Behind one door is a car; behind the others, goats.

You pick a door, say No. 1.

The host then under the rules of the game must open one of the two remaining doors to show you a goat.

The host opens door No. 3, to reveal one of the goats.

He then says to you as you final option , "Do you want to switch from door No 1 and pick door No. 2?" Is it to your advantage to switch your choice?

(For those who know it - I've rewritten this a little - so if there an error in there please tell us)

 

[Vassago]

Another classic - (I and probably others have posted it here before) , even after knowing the answer.


Suppose you're on a game show, and you're given the choice of three doors: Behind one door is a car; behind the others, goats.

You pick a door, say No. 1.

The host then under the rules of the game must open one of the two remaining doors to show you a goat.

The host opens door No. 3, to reveal one of the goats.

He then says to you as you final option , "Do you want to switch from door No 1 and pick door No. 2?" Is it to your advantage to switch your choice?

(For those who know it - I've rewritten this a little - so if there an error in there please tell us)

It'd naturally be better to switch doors. Just because he opens a door, it doesn't remove it from the mathematical probability of 33.3% of winning, but if you switch doors, you'd have a 50% chance of winning. It's not much, but it's enough to make it worth it.

 

[Brianwarnock]

It'd naturally be better to switch doors. Just because he opens a door, it doesn't remove it from the mathematical probability of 33.3% of winning, but if you switch doors, you'd have a 50% chance of winning. It's not much, but it's enough to make it worth it.

That can't be right, surely you now have, as you stated a 50% chance, but that applies whichever door is chosen, ie sticking or moving.

Brian

 

[Vassago]

If my math is correct:

They have the same amount (which isn't what I thought it would be when I first read it). Using bigger numbers for ease of math, if both cups have 5 units and I take 1 from A to B and then back. After the first move, cup B has 5 units of tea one 1 coffee. When I take out 1 unit, it is 1 of 6 units or 16.6667%. That means I take out .833333 of tea and .166667 of coffee. 5 minus .833333 is 4.166667 of tea, and 4 plus .166667 is also 4.166667 of coffee. There's probably a better way of explaining that, but I'm hurrying to make dinner so my excuse is hunger-induced brain cramps.

You are correct. They are exactly the same (assuming all environmental conditions prevent any affection of said liquids as defined by the people here who over-analyze the problem. )

A simpler way to explain it is, you can't possibly have more total volume of coffee or tea than you started with. Since you started with equal amounts of each, you must end with equal amounts of each. You used exactly 1 teaspoon in each transaction, making both cups still contain exactly the same amount of liquid. Mathematically, one cannot contain more tea than the other does coffee or vice-versa.

To make it even easier, find 10 of two small objects, say paper clips and thumbtacks, but any objects around your desk will do. Separate them into different piles. Now, grab any number of paper clips and add them to the thumbtack pile. Next, grab the exact same number of items from the original thumbtack pile, and place them into the original paper clip pile. It doesn't matter what you grab, as long as you grab the same amount, they will be equal of each other's opposite original items.

 

[Pauldohert]

Brian -

Its even better after a few pints.

Another classic - , even after knowing the answer.

 

[Vassago]

That can't be right, surely you now have, as you stated a 50% chance, but that applies whichever door is chosen, ie sticking or moving.

Brian

That's not true. It's a mathematical phenomenon, but staying where you are still keeps a 33% chance. He didn't really tell you anything about the door that you are on, to raise the chance of winning. But he did tell you that by switching, since there are two mystery doors left, you'll have a 50% chance. By switching, the third door is taken out of the equation.

I remember this riddle from AP Physics, we had to prove it mathematically. I don't remember the formula though.

 

[Pauldohert]

It'd naturally be better to switch doors. Just because he opens a door, it doesn't remove it from the mathematical probability of 33.3% of winning, but if you switch doors, you'd have a 50% chance of winning. It's not much, but it's enough to make it worth it.

Almost - but I think you may have a % wrong. Its a good one - even knowing the answer, its still makes you think.

 

[Brianwarnock]

That's not true. It's a mathematical phenomenon, but staying where you are still keeps a 33% chance. He didn't really tell you anything about the door that you are on, to raise the chance of winning. But he did tell you that by switching, since there are two mystery doors left, you'll have a 50% chance. By switching, the third door is taken out of the equation.

I remember this riddle from AP Physics, we had to prove it mathematically. I don't remember the formula though.

I would like to see the proof, tho' I doubt I'd understand it.

I think that you have now got a new choice, door1 or door2, it doesn't matter what you chose before its a new choice. So 50 -50 stick or move.

Brian

 

[ChipperT]

That's not true. It's a mathematical phenomenon, but staying where you are still keeps a 33% chance. He didn't really tell you anything about the door that you are on, to raise the chance of winning. But he did tell you that by switching, since there are two mystery doors left, you'll have a 50% chance. By switching, the third door is taken out of the equation.

I remember this riddle from AP Physics, we had to prove it mathematically. I don't remember the formula though.

I have heard this one for years and seen the mathematical "proof" but still have a very difficult time following the logic, which seems to say that if you switch and they remove your original door and then he gives you a final chance to choose the door you did not choose the second iteration, and you switch yet again, you have a 100% chance of winning? That cannot be true. The prize could have been behind the door you chose originally or the second door. Or am I just really dense on this one?

 

[pbaldy]

By switching, the third door is taken out of the equation.

I guess I'm not following this one, but I never took Physics. The third door is already out of the equation at this point. I may have had a 33% chance when I made my original choice, but with door 3 being revealed, that choice now has a 50% chance of being correct. Switching does not increase that percentage. Either door I choose now has a 50% chance of being correct.

 

[Vassago]

Almost - but I think you may have a % wrong. Its a good one - even knowing the answer, its still makes you think.

I went back and read it again. You are right, I do have the match wrong. You actually have a 66.7% chance of winning upon switching. Let me explain better if I can.

If you stay, you still only have a 1/3 chance of winning. Logically, since there is only one door left, that door must have a 2/3 chance of winning. The third door is taken out of the equation. My original explanation didn't even sound right to me.

Does that make more sense Brian?

 

[Vassago]

Let me see if I can break it down... let's assume Door #2 has the prize...

3 scenarios -

You choose Door Number 1, stay, lose
switch, you win...

You choose Door Number 2, stay, win
switch, you lose...

You choose Door Number 3, stay, lose
switch, you win...

These are the only scenarios possible. If you switch, you win 2/3 times. If you stay, you win 1/3 times. Get it?

 

[Brianwarnock]

I went back and read it again. You are right, I do have the match wrong. You actually have a 66.7% chance of winning upon switching. Let me explain better if I can.

If you stay, you still only have a 1/3 chance of winning. Logically, since there is only one door left, that door must have a 2/3 chance of winning. The third door is taken out of the equation. My original explanation didn't even sound right to me.

Does that make more sense Brian?

Sorry , no.
Its like tossing a coin, each toss is a new chance what went before doesn't matter. Here the new chance is between 2 doors the fact that you chose 1 before doesn't matter, sticking is just as much a chose as moving and the odds are the same for either door, just as it was before, but changing from 33.3% to 50%.

Brian

 

[Brianwarnock]

Let me see if I can break it down... let's assume Door #2 has the prize...

3 scenarios -

You choose Door Number 1, stay, lose
switch, you win...

You choose Door Number 2, stay, win
switch, you lose...

You choose Door Number 3, stay, lose
switch, you win...

These are the only scenarios possible. If you switch, you win 2/3 times. If you stay, you win 1/3 times. Get it?

But door3 has gone for the final choice. you cannot include it in the 2nd choice scenario.

Brian

 

[Brianwarnock]

Vass

It looks like we are going to have to agree to disagree on this

Bri

 

[ChipperT]

I would like to see the proof, tho' I doubt I'd understand it.

I think that you have now got a new choice, door1 or door2, it doesn't matter what you chose before its a new choice. So 50 -50 stick or move.

Brian

Where this logic fails is that now BOTH chances could be wrong since Door 1.may have had the prize.

So, you switch to door 2, here are the chances:

Door 2 has the prize, Door 3 does not
Door 2 has no prize, Door 3 has the prize
Door 2 has no prize, Door 3 has no prize

So, your odds of winning are still one in three or 33%

 

[Brianwarnock]

Where this logic fails is that now BOTH chances could be wrong since Door 1.may have had the prize.

So, you switch to door 2, here are the chances:

Door 2 has the prize, Door 3 does not
Door 2 has no prize, Door 3 has the prize
Door 2 has no prize, Door 3 has no prize

So, your odds of winning are still one in three or 33%

You have 2 doors left, 1 has the prize, that's the scenario, if none have it you have already lost.

Brian