Brainteaser/Puzzle Corner (Part 5)

Oh I see. And I assume _________________________________________________ is that line. I didn't know that.

so it is ;D

 

As promised I'm putting this riddle up with no loose ends.

1. There is a lamp in a room with no windows on the seventh floor of a building.  On the first floor of the building there are three switches, one of which is connected to the lamp.  The other two switches are not connected to anything.  The question is how may you determine which switch turns on the light if you may only go upstairs to check on the lamp once.

Other conditions:  You may not have assistance from any other person or creature.  You have no tools or equipment.  The swithces and the wires connecting one of them to the lamp are isolated from you.  You may not tamper with the functionality of the switches.  A switch can either be on or off; there is no in between state.

And for kicks I'll put up a second riddle/brainteaser

2. You are on a game show. There are three curtains. Behind each curtain is a prize. Two of the prizes are nearly worthless. One of the prizes is very valuable. The game show host asks you to pick one of the curtains. After you make your decision, the hostess opens one of the curtains, it is never curtain you picked and it is never the curtain with the good prize behind it. This is possible since the hostess knows where all the prizes are and also knows which curtain you picked.

At this point the host asks you if you wish to stick with the curtain you picked or change your guess and pick the other unopened curtain.

If you desire to win the valuable prize, what choice should you make, stay with the curtain you picked, switch to the other curtain, or does it matter which choice you make?

You must also give an explaination as to how you caim to your conclusion.

I will tell you if your answer is right or wrong, I will not say if your guess is correct and reasoning incorrect or if your guess is wrong.

 

turn the first switch on for half an hour or so
turn it off and turn second on
quickly go upstairs and check, if the lamps on it's nr 2
if it's off touch it, if it's warm then nr 1, if not nr3

 

ahhhhhhh d@m it meee posted before me.

 

very quick, Mee wins and there is also a second riddle, hopefully this one will take longer lol.

 

How can there be only one unopened curtain after choosing 1 of 3?

 

'the other unopened curtain'
the non-other is the one you chose before and it's still unopened too

 

I'll have to choose the same curtain twice, methinks.
I choose 1, I'll get 2 (3 being the valuable).
I choose 1 again, I must get 3 because 2 is already open.

 

hmm, maybe I need to be a bit more clear.

I'll give an example, lets say you choose curtain 1
Now the hostess will open either curtain 2 or 3 depending on which one is a bad prize, she'll pick one at random if they are both bad.
The host asks you to keep the curtain you choose or switch to the other unopened curtain
Whatever is behind the curtain decide up the final time is the curtain you get to keep.

I hope that clears things up

 

The switches are A B and C

You light up switch A, and wait a long time. Then you light up B and run to the seventh floor. If the light is turned on, it is lamp B, if you feel the light bulb and it is hot (for being turned on for a long time), it is A, and if none of the above happen, it is C.

You should switch. The original choice has 1/3 chance of being right. The woman removes a curtain which had a third of being right. After that there is still 1/3 chance of the one you picked being right, and 2/3 of the other one being right. By switching you double your chances.

 

What decides which curtain gets the extra 1/3 chance?

 

Z is right owerup:

I'll spell it out.  If you pick a curtain it has a 1/3 probability of being the valuable prize and thus there is a 2/3 probability that one of the others has the valuable prize.  After the hostess reveals a bad prize, it does not change the fact that your curtain has a 1/3 probability of being valuable and the other two had a probability of being valuable.  What it does change is that the curtain that has been proven to be not valuable now has a 0 probability of being valuable.  Which means for that curtain and the other curtain to still have a combined 2/3 probability of being valuable the other curtain must have a 2/3 probability of being valuable.

Therefore, you should switch your answer to the other curtain to get the best odds of winning the valuable prize.

 

Okay I have been thinking of a riddle. This one is going to be really hard and the prize will be really big, I'm going to make a new Brainteaser corner.

 

Wait, what? How? Either I'm bad at math, or that doesn't make any sense. Why would the unchosen curtain suddenly have a 2/3 chance if the chosen one doesn't?

If you select curtain 1, and curtain 3 is eliminated, then staying with your current curtain would be the equivalent of saying "It's either in curtain 1 or curtain 3."

If you switch, it's the equivalent of saying "It's either in curtain 2 or curtain 3."

The first choice, you obviously have a 1/3 chance of hitting the correct curtain. The second choice, you would have a 1/2 chance of hitting the correct curtain, because one is gone. Why would one option be less likely just because you pointed at it the first time and a completely different curtain was removed?

 

Paragon, say you had 10000 curtains, with only 1 correct answer, and I ask you to choose one out of those 10000. Are your odds high?

I then remove 9998 empty curtains, leaving only one curtain unknown. There are only two curtains, so according to your logic, each one has half chance of being right.

Think again

 

But... it does. If there are 500 vases, and I say,
"One of these vases has water in it, but it's not the 498 vases on the left."

That's the equivalent of saying:

"One of these vases has water in it, and it's one of the two on the far right."

One of the two. 1/2. 50% chance. You KNOW it's not the 498 on the left. As such, they're completely irrelevant. You only care about the two on the right. Why would you include them in your calculations?

I mean, I get what you're saying about me happening to pick the right curtain out of 10,000. But from a math standpoint, I don't see how that adds up. Logic agrees with you, I would probably say it's most likely the one I didn't choose. But since when does basic math take probability into account?

 

if you use common sense on this one it will fail. You choosing, THEN the person removing is DIFFERENT then the person removing, then you choosing.

The probability doesn't change. If you have 10 vazes and make a choice, only ONE out of TEN will be right. Reveal an empty vase. Your odds are STILL 1/10, they don't change. Keep removing empty vases your odds won't change.

Say that out of 3 million tickets only one will win the lottery. Now say that all the wrong tickets were burnt expect for one (leaving one wrong and one right one). And say you have a choice between the two. That way the odds are going to be the same, because one is right while the other is not, and you haven't made your choice yet.

Now say that out of those 3 million tickets, you pick one. Are your odds high? No. They are extremely extremely low. Now say we do the same thing, we burn all of the wrong tickets and leave one, which will also be wrong, if you have the right ticket , or right, if you have the wrong one. But wait, weren't the odds that you had the right one so slim? What does burning the other tickets change in your odds, it is still the same ticket, it still has the same chances of being right than being wrong. So, from those two tickets, you should switch!

Do you understand the difference between these two scenarios?

Hope that helps.

 

I am not sure how to say this but Bizzaro is correct in a way. lets take for example of a hat with 5 numbered balls in it. At the beggining you have 1/5 chance of grabbing the correct ball and that will stay true until you are given another choice or trial at the point you have to revaluate the probabilities taking into account what has changed. Now for an example given the circumstance you set up. If I start removing balls besides a specific 2 that cant be removed then odds of both of those balls beign correct increase. So if I removed 2 balls and evaluated the probabilities you would have a 1/3 chance grabbing the correct ball. Now if you remove the last ball that can be romoved from the equation you would still have a 1/3 chance of grabbing the correct ball, until you evaluated the probabilities which you would have to because the problem has changed. Now after you revaluate the probabilities you would have a 1/2 chance of grabbing the correct ball.


Now quanta's answer is still valid if you dont revaluate the probabilities.

 

you're logic seems flawed to me. after she opens a curtain, there will be 2 prizes left. whether you change or stick with your old choice, there is still a 50% chance that you picked the big prize. that it had a 33% chance before doesn't matter, it has a 50% chance this time.


but i already heard this riddle. when i heard it, it would be favorable to change, because the game host doesn't want you to win and would try and get you to change. if you picked the bad one, she would know, and would hardly give you the option of changing, I.E. definitely getting the good prize

 

I'll try to explain it gain.

You have a standard deck of 52 cards. You win a prize if you draw the ace of spades. Pick a card at random and do no look at its value. Then the other person picks up the remainder of the deck, looks throught it and picks out one of the cards. They then discard the rest of the deck. They tell you truthfully that the ace of spades was not among the discarded cards. They then ask you if you wish to keep the card you chose are switch to the card they chose.

Now you had a 1/52 chance of drawing the ace of spades when you first drew a card. The other person got to look through the deck and as long as you weren't lucky enough to hit that 1/52 chance they pick out the ace of spades. In other words there is only a 1/52 chance that they don't draw the ace of spades which means there is a 51/52 chance that they do. Now do you think you need to re-evaluate the probabilities?

The exact same logic applies to the 3 curtains it just isn't as obvious.