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Can anyone get this tricky math problem?
- Thread starter Frisco
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Ok, fun question:
Suppose you have two unmarked buckets: one has a capacity of 3 L and the other has a capacity of 5 L. Let's say you're in a room with just a hose - no measuring instruments, no scale, etc, nothing, just a hose with running water, and these two empty, unmarked buckets. How could you obtain EXACTLY 4 L using the two buckets?
Don't look this one up on the net because I'm sure it's a popular one - try and figure it out!
Suppose you have two unmarked buckets: one has a capacity of 3 L and the other has a capacity of 5 L. Let's say you're in a room with just a hose - no measuring instruments, no scale, etc, nothing, just a hose with running water, and these two empty, unmarked buckets. How could you obtain EXACTLY 4 L using the two buckets?
Don't look this one up on the net because I'm sure it's a popular one - try and figure it out!
Someone's been watching Die Hard with a Vengence,
Ok, fill the 5 L, then pour 3 L from the 5L to the 3L (so there is now 3L in the 3L and 2L in the 5L.
Now, empty the 3L, and pour the 2L remaining in the 5L into the 3L (so there is now 2L in the 3L and 0 in the 5L)
Now, fill the 5L, and pour 1L into the 3L, leaving 4L in the 5L
Someone's been watching Die Hard with a Vengence,
Ok, fill the 5 L, then pour 3 L from the 5L to the 3L (so there is now 3L in the 3L and 2L in the 5L.
Now, empty the 3L, and pour the 2L remaining in the 5L into the 3L (so there is now 2L in the 3L and 0 in the 5L)
Now, fill the 5L, and pour 1L into the 3L, leaving 4L in the 5L
Yes and yes!
That one's kind of old.
I still like the 12 unmarked coins with 1 counterfeit and only a basic "heavy" "light" scale problem. That one required good old fashioned thinking power.
That one's kind of old.
I still like the 12 unmarked coins with 1 counterfeit and only a basic "heavy" "light" scale problem. That one required good old fashioned thinking power.
Yeah, I like that one too, of course, the hard part is that you are restricted to a small number of weighings
Yeah, but you knew which one I was talking about.
It took me a while to get that one the very first time I had heard it.
This problem comes up regularly on sci.math. Unfortunately, we don't have enough information about the rules to be able to answer. I'll assume that your choices as the player are to either take what is behind door #2, or to take what is behind door #3.
The first question is: What possible choices did the guard have when he opened the door? Did he have the choice to open a door or not open a door? Did he have the choice to open a door that showed wealth (and you are stuck with death
), did he have the choice to show you door #3 or door #1? Someone posted that you should always switch. Now assume the guard really doesn't like you. Here is what he does: If your door holds wealth, he opens a door showing death. If your door holds death, he opens a door showing wealth. If he is allowed to do this, and you follow the advice you were given if possible, you will certainly die.
Another possibility: If your door holds wealth, he opens a door showing death. If your door holds death, he doesn't open a second door. So if he opens a door and you follow the advice, you die. But if you know that this is the strategy he follows, then you can beat him: If he opens a door sowing death, you stay with your choice. Otherwise you switch. Now your chances are one in three to win.
That's a tricky math problem? Wow... I feel like i'm losing IQ points as i'm writing this out but: 14 girls legs, 1372 big cat legs, and 9604 little cat legs. Total = 10990 legs.
edit: Nevermind. I looked at the original post again and it says it's a 5th grade math problem. Now i'm not so confused anymore.
edit: Nevermind. I looked at the original post again and it says it's a 5th grade math problem. Now i'm not so confused anymore.
If it's eating through the 1,000th page it's not eating through the last book cover (atleast it's not stated in the question). So it would be 1/16" less distance.
How would 1+1 equal a window?
Only one I know
Impossible. You need an equal sign at the end to do that! That's an old one. The ones are the left and right sides of the window and the equal sign is the top and bottom. The plus is that middle thing that some people have in their windows. I don't have em so I dunno the name.
I'm not sure about that solution. For some reason it seems to assume that all children are girls.
Ok, since you asked.
There are 3 doors. Behind 1 door lies immense wealth and riches. Behind the other two lie certain death.
A guard asks you to choose a door and you pick #2. The guard then reveals to you what's behind Door #1 and you see that behind it was one of two chances of certain death.
The guard now asks you if you would like to change your selection from Door #2 to Door #3.
Mathematically is there a probability increase that benefits you if you change doors?
BTW, I don't know the answer to this, but I've seen it thrown around a few times
Yes - there is a probability increase if you switch. The odds of picking the right door initially is 1 in 3. You are more likely to pick certain death than you are the correct answer. If another door is opened and it is certain death, switching your bet would give you odds in 1 in 2 rather than 1 in 3.
I disagree.
The odds of picking the right door initially is 1 in 3. You are more likely to pick certain death than you are the correct answer. If another door is opened and it is certain death, switching your bet would give you odds in 1 in 2 rather than 1 in 3. But... keeping door #2 will also give you odds of 1 in 2 rather than 1 in 3. So your chances improved when you were told what's behind door #1, but your chances won't improve further by switching from door #2 to door #3.
You can also see this by enumerating the cases, based on assumption that the three arrangements are equally likely:
Case 1: {wealth and riches, certain death, certain death}
Case 2: {certain death, wealth and riches, certain death}
Case 3: {certain death, certain death, wealth and riches}
Case 2: {certain death, wealth and riches, certain death}
Case 3: {certain death, certain death, wealth and riches}
Case 1 is ruled out in the problem statement. So it's Case 2 or Case 3 with equal probability. You pick door #2. Door #1 is shown to be certain death. If you change to door #3, you win half the time (Case 3) and lose half the time (Case 2). If you stick with door #2, you win half the time (Case 2) and lose half the time (Case 3).
It could be that there are two types of certain death, making 6 equally likely cases:
Case 1: {wealth and riches, certain death by tickling, certain death by accordion music}
Case 2: {wealth and riches, certain death by accordion music, certain death by tickling}
Case 3: {certain death by tickling, wealth and riches, certain death by accordion music}
Case 4: {certain death by accordion music, wealth and riches, certain death by tickling}
Case 5: {certain death by tickling, certain death by accordion music, wealth and riches}
Case 6: {certain death by accordion music, certain death by tickling, wealth and riches}
Case 2: {wealth and riches, certain death by accordion music, certain death by tickling}
Case 3: {certain death by tickling, wealth and riches, certain death by accordion music}
Case 4: {certain death by accordion music, wealth and riches, certain death by tickling}
Case 5: {certain death by tickling, certain death by accordion music, wealth and riches}
Case 6: {certain death by accordion music, certain death by tickling, wealth and riches}
But you get the same answer. Cases 1 and 2 are ruled out in the problem statement. So it's Case 3, 4, 5, or 6 with equal probability. You pick door #2. Door #1 is shown to be certain death by either tickling or accordion music. If you change to door #3, you win half the time (Cases 5 and 6) and lose half the time (Cases 3 and 4). If you stick with door #2, you win half the time (Cases 3 and 4) and lose half the time (Cases 5 and 6).
Sorry, I mixed up my odds slightly, but the fact remains your chances double when you switch.
http://www.youtube.com/watch?v=P9WFKmLK0dc
Sorry, still not true. The odds are the same whether you switch doors or not, in the problem stated above. The odds might change if the other party gave you new information, e.g., by picking which door to show you based on already knowing what's behind them. That's not the case here, where what's revealed to you is predetermined to be door #1.
As additional confirmation, I wrote a test program based on random numbers and ran one million trials eight times, four of them keeping door #2 and four of them switching to door #3.
Here are my results:
If I kept the same door, I won this percentage of the time in the four trial runs: 49.9995%, 50.0665%, 50.0933%, 50.0150%
If I changed doors, I won this percentage of the time in the four trial runs: 50.0103%, 50.0423%, 49.8875%, 50.0214%
In other words, 50% either way.If I changed doors, I won this percentage of the time in the four trial runs: 50.0103%, 50.0423%, 49.8875%, 50.0214%