199 réponses à ce sujet

[DinoCrisisFan]

The answer is 2. Anyone who believes otherwise is a ****ing moron who needs to go back and learn elementary school math.

[Creature 1]

DinoCrisisFan wrote...

The answer is 2. Anyone who believes otherwise is a ****ing moron who needs to go back and learn elementary school math.

So what's your answer, do you stick with your original pick or do you switch to the other box, and why?

[Wittand25]

Tree friends go shopping. Together they spend 30€ . After they left the shop the manager notices that he charged them 5€ too much and sends his assistant to give back the money. Since 5 cannot be divided by three the assistant gives each of the friends 1 € back and keeps the remaining 2€ for herself.
So now the three firends spend each 9€ and the assistant has 2€. Making in sum 29€ instead of the 30€. Who has the missing money?

Modifié par Wittand25, 08 avril 2011 - 05:56 .

[Nattfare]

Creature 1 wrote...

Ok, now assume I present you with three boxes. I tell you one contains $1,000,000 and if you pick that box, you get to keep the money. After you pick, I open one of the other two boxes. It is empty. I then give you a chance to change your first pick. Should you stick with your initial guess, or switch to the other box?

I'd shout look behind you, a hedgehog! You'll look behind you and I'll grab both boxes and run.

[RainyDayLover]

Alright, this thread is getting waaay too nerdy for my taste. <_<

[BeljoraDien]

Rose of Mars wrote...

Creature 1 wrote...

Ok, now assume I present you with three boxes. I tell you one contains $1,000,000 and if you pick that box, you get to keep the money. After you pick, I open one of the other two boxes. It is empty. I then give you a chance to change your first pick. Should you stick with your initial guess, or switch to the other box?

When I first picked the box, I had 3 choices, ergo 33% chance of getting the prize.

Since now I know your box is empty, I have a 50% chance of getting the prize. Because there's only 2 choices left.


Hmmmm... It's a 50/50 chance. Sooo (since I'm stubborn), I would keep my box 

Yeah, but you had a 66% chance of picking the wrong box at first. You had a 33% of picking the right box. You're twice as likely to have picked the wrong box in the beginning. So it is much more logical to switch.

[Creature 1]

Strangely Brown wrote...

Okay here is something really strange.
I have tried 2 different calculators. The one I have in my hand is a Sharp EL-520W scientific calculator.
When I enter the equation exactly as it is written in the OP : 48÷2(9+3) = With this ÷ division symbol the answer I get is 2.

However when I use a different calculator and I use the / as the division symbol so that the equation reads 48/2(9+3) =

I get the answer 288.

I can tell you which calculator I wouldn't use when the right answer really matters. 

Does it make more sense when you change to 48*0.5*(9+3)?  The answer is 288, I would bet my beloved mother's life on it. 

[ejoslin]

Wittand25 wrote...

Tree frinds go shopping. Together they spend 30€ . After they left the shop the manager notices that he charged them 5€ too much and sends his assistant to give back the money. Since 5 cannot be divided by three the assistant gives each of the friends 1 € back and keeps the remaining 2€ for herself.
So now the three firends spend each 9€ and the assistant has 2€. Making in sum 29€ instead of the 30€. Who has the missing money?

Hah, this one is funny -- it aims to confuse.  I'm going to use $ so I don't have to pull up symbols.  But what you need to do is subtract $2 from the $27, not add $2 to it.

The change = $5 (each friend has $1 and the assistant has $2).  There is no missing dollar.

Edit: Oh, and with which box?  Though I know statistically I should have a better chance if I switch boxes, I'm sticking with my first choice because it seems lucky to me!

Modifié par ejoslin, 08 avril 2011 - 05:57 .

[Dominus]

 The answer is 42.

[Rose of Mars]

BeljoraDien wrote...

Rose of Mars wrote...

Creature 1 wrote...

Ok, now assume I present you with three boxes. I tell you one contains $1,000,000 and if you pick that box, you get to keep the money. After you pick, I open one of the other two boxes. It is empty. I then give you a chance to change your first pick. Should you stick with your initial guess, or switch to the other box?

When I first picked the box, I had 3 choices, ergo 33% chance of getting the prize.

Since now I know your box is empty, I have a 50% chance of getting the prize. Because there's only 2 choices left.


Hmmmm... It's a 50/50 chance. Sooo (since I'm stubborn), I would keep my box 

Yeah, but you had a 66% chance of picking the wrong box at first. You had a 33% of picking the right box. You're twice as likely to have picked the wrong box in the beginning. So it is much more logical to switch.

Ooooo makes sense. I think I would have a 50% chance if I picked my box AFTER the other person picked theirs. 

Math can be so fun 

[Ponce de Leon]

It's 288. I was wrong, I apologize. My father explained me why... it is also true however that it's not entirely clear as the OP wrote it BUT!!!! because there is no parenthesis in the final segment, the answer is in fact 288, as if I wrote :

48
--- *(9+3)
2

In which case it's 24*12 = 288.

[BeljoraDien]

Creature 1 wrote...

Ok, now assume I present you with three boxes. I tell you one contains $1,000,000 and if you pick that box, you get to keep the money. After you pick, I open one of the other two boxes. It is empty. I then give you a chance to change your first pick. Should you stick with your initial guess, or switch to the other box?

Based on this, I just thought of something... Imagine a scenario where you're on a gameshow with 2 players and 3 boxes. Player 1 gets to pick a box first, but then must wait as player 2 lifts up his/her box from the two remaining choices. If player 2 doesn't lift up the box with cash, player 1 is then allowed to switch boxes. Would you want to be player 1 or player 2? And, if player 1, would you still want to switch boxes?

[Creature 1]

BeljoraDien wrote...

Creature 1 wrote...

Ok, now assume I present you with three boxes. I tell you one contains $1,000,000 and if you pick that box, you get to keep the money. After you pick, I open one of the other two boxes. It is empty. I then give you a chance to change your first pick. Should you stick with your initial guess, or switch to the other box?

Based on this, I just thought of something... Imagine a scenario where you're on a gameshow with 2 players and 3 boxes. Player 1 gets to pick a box first, but then must wait as player 2 lifts up his/her box from the two remaining choices. If player 2 doesn't lift up the box with cash, player 1 is then allowed to switch boxes. Would you want to be player 1 or player 2? And, if player 1, would you still want to switch boxes?

I guess I should have specified that in the original example I know which box has the money, and will always open an empty box.  This produces the 2/3 odds for switching.  The second variation is when the host does not know what is in the box and happens to pick an empty box by luck.  In this case the odds are 1/2 for switching.  

With the caveat that the Monty Hall problem makes no intuitive sense to me (a little more sense if there are far more boxes than 3, and the host opens many many empty boxes), I think the odds for switching in your example would be 1/2, so it is still best to switch.  Initially both people would have a 1/3 chance, if the second person fails I think the first person would increase their odds to 1/2 by switching, so it's better to go first, wait, and take the option to switch if your opponent misguesses.  But I'm satisfied to be proven wrong.  

[Shirosaki17]

wonder if math is taught differently around the world, because people can't be this dumb.

[Godak]

Yup. 288. I had just woken up earlier, and my brain was malfunctioning.

PEMDAS is left to right when it comes to M/D or A/S. It's not in order. I'm sure Bio test nerves didn't help.

[Darthnemesis2]

I can't believe I read this whole thing :/

[BeljoraDien]

Looks like no one else wants to play....

You're right in your choice to be player 1 and to switch the boxes:

Whether 'player 2' (in my example) knows what's in the box beforehand or not, he still picked the wrong box, leading to the 66% vs 33% chance for player 1. The only real difference is that player 2 now has a chance of picking the right box, a 33% chance as it were.

There are a lot of different ways to think about it... but after some mind-puzzling I think the best way to think of it is that, in the event that player 2 picks the wrong box, you merely revert to the scenario before. There's a 2/3 chance that player 2 will pick the wrong box, so

if player 1 choose to switch we have 2/3*66%=44%.
If player 1 chooses not to switch, we have 2/3*33%=22%.
vs. player 2 and his 33%.

So, choosing to be player 1 and switching boxes would give you an advantage over choosing to be player 2 (not switching would give you a disadvantage). Either way, you have less than a 50% chance of winning.

There would be a 23% chance that no one wins if player 1 chooses to switch.
vs. a 45% chance that no one wins if player 1 chooses not to switch.

Edit: Actually... this CAN'T be true. In a scenario where player 1 is dead-set on switching no matter what, picking the 'right' box in the first turn would always end in no one winning... So a situation where no one wins would have to occur atleast 33% of the time when player 1 chooses not to switch... Wow, I'm stumped now.

Edit 2: I figured out the problem. The actual chances of no one winning is:
If player 1 chooses to switch 2/3*(100-44)=37%
If player 1 chooses not to switch 2/3*(100-22)=52%

Modifié par BeljoraDien, 08 avril 2011 - 08:33 .

[OBakaSama]

*Ahem* Everyone has been wrong so far. The answer is.....219 apparently.

*Chucks calculator out of the proverbial window*

[Bjelseth]

Poll added

http://social.biowar...79/polls/18184/

[Wicked 702]

I got 288. Then I read the thread. Then I got confused. Now I just want to hide in a corner and cry.

Have to admit that the follow-up statistical and mathematical brain teasers have been fun.

[Timon44]

LOL great discussion. Ok I'll join
The way it is written the result is clearly 288. Just type it in Google and you will see.

END OF DISCUSSION!

[Timon44]

AwesomeName wrote...

Bjelseth wrote...

Poll added

http://social.biowar...79/polls/18184/

So, which is it?  (48÷2)(9+3) or 48÷(2(9+3))?

I think that's the main issue really...

It is the former.

48/2(9+3) = (48/2)(9+3)          The extra bracket isn't really needed
48/2(9+3) != 48/(2(9+3))