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Topic: Here are the easy answers (spoilers inside) (Read 12630 times) |
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C. Clark
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Here are the easy answers (spoilers inside)
« on: Jul 25th, 2002, 8:04pm » |
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Here are all of the answers for the Easy Section:
Marble Jars:
The question is really what is the best way to distribute the 50 black and 50 white marbles to maximize the odds of white.The answer is 1 bottle with 1 white (100%) and the other jar with 49 white and 50 black (49.49%).
Arab Sheikh Camels:
The question is how to change a race to be the loser into a race to be a winner. The answer is to switch camels or take your brother's camel.
3 Hats:
This is a logical chain. If A saw white on B and white on C, then he knows that he has black. That means A saw at least one black hat. This means B and C could be black or white but not both white. So if B saw white on C then by the above he would know that he has black. But he doesn't know what color he has so he must have seen black on C.
Humming bird:
This is a bounded (above and below) oscillating linear function, bounded by linear functions. The secret is to make an x-y graph and to make some observations.
Say you call LA 0 and NY 5000 miles on the y-axis. Then you call the speed the slope of the lines that you will be inserting.
The function for the train from LA to NY is y=15x + 0.
The function for the train from NY to LA is y=-20x + 5000.
The you make the function for the bird. The bird is unique in that when it intercepts a bounding line it will invert the slope. So the equations for the bird are y = 25x + intercept
and y = -25x + intercept. This will yield a sort of zig zaging line between the two lines (until they meet and the zig zaging ends.) If you rearrange the line segments by the following, the answer becomes aparrent. Take the positive slope segments and move them onto a single line (move them up until they coincide with the line y =25x + 0. Then notice that if you reflect the negative slope segments across the x-axis they change to positive slope segments of the same length. Then these new segments will be able to fill the gaps on the line y = 25x + 0. This will make a complete line y = 25x + 0 until the impact of the 2 trains. So the length of this line is 25 * inpact time or 3571.43 miles.
(Draw the picture and you can figure it out much easier.) 
Foot Size Implies Spelling Ability:
I guess because anatomically, their body and therefore their brain is bigger. I don't know about this one.
Christopher Clark
University of San Francisco Student
cjsclark@aol.com
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C. Clark
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Re: Here are the easy answers (spoilers inside)
« Reply #1 on: Jul 25th, 2002, 8:06pm » |
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Nonhomogeneous Rope burning:
This is an interesting question. The answer is to light both sides of rope one and one side of rope 2 at the same time. Start your watch. Rope one will burn in 30 minutes. This will also burn 30 minutes for rope 2. After rope 1 is burnt (30 minutes elapsed) light the other end of rope 2. The half hour strand of rope 2 will burn in 15 minutes. The elapsed time is 30 minutes and 15 minutes, or 45 minutes.
Willy Wu Tang and the Burning Island:
I think that the secret is to burn a second fire around the 7th mile and stand in the middle of the two fires. This will burn the last 3 miles of the island and he can stand in the charred region when the fire approaches. The first fire should go out when it reaches the 7th mile.
Two Coin Flips:
This is a trick question. The chance of getting heads is independent of all previous outcomes. So it is 50% neglecting the fact that coins are not symmetrical.
Coin Weighing Machine:
I am working on this one, I will repost when I have the answer.
Hourglasses:
If you flip the 7 and 11 minute hourglasses, you will have 4 minutes left when the 7 minute hourglass is done. Start your watch with these four minutes. When the 11 minute hourglass is done (4 minutes in) flip it over for another 11 minutes. This makes 15 minutes total.
Logical Signs I:
If the python is in the silver chest then then the sign on the chest is true. If that sign is true, then the gold sign can't be true ( or both signs are true and the gold sign is false) but that means that the gold sign is false. This means that the silver sign is true and the gold sign is false. But that makes the gold sign true (read the sign). So there is a contradiction. So you open the silver chest because it must be false.
Logical Signs II:
The best way to look at this is to choose a chest and look at the results of the signs. If you choose Silver then all the signs are true. If you choose the Gold chest then all the signs are false. If you choose the Bronze chest then the signs are false, false, true. So you choose bronze.
Chess Puzzle I:
From top to bottom and left to right the pieces are
Row 1:Knight, Queen
Row 5:Rook
Row 6: King
Row 8: Bishop
Chess Puzzle II:
Row 1: King
Row 3: Queen
Row 5: Bishop
Row 8: Knight, Rook
Cork, Bottle, Coin:
Push the cork in.
Family Relations:
This man's father is my Father's Son. That is the key line in all this. Taken piece by piece "my father's son" is the speaker. Substituting yields "This man's father" is the speaker. It is the speaker's son.
Analog clock I:
The angle is the angle between the minute hand and the hour hand (which has moved past 3). So the answer is 15 minutes / 60 minutes * 360 degrees / 12 hours = 7.5 degrees.
Analog clock II:
The clock hands meet 24 times in the 24 hour period.
The times are calculated as follows: 12:00 is a given and occurs 2 times. Then for hours = 1 to 11 the following must be calculated. minutes = 5 * hour + 5 * minutes / 60. This will yield all the times.
Apples and Oranges:
The secret is to see that the jar labeled apples and oranges is either all apples or all oranges. You choose one from there. If it is an apple, then you label that jar apples. Now you know that the orange jar is labeled wrong so it must be apples and oranges ( the apple jar has been found). This leaves the orange jar as the last one. If you get an orange in the beginning, the process is the same switching apples and oranges.
Stupid:
The letter "E" is the answer. Kinda funny huh.
Climbing snail:
The only thing that you need to do here is take your time. The snail is not moving at 1 meter a day but moving up 5 then falling 4. This means that on day 1 he is up to 5 then down to 1. Day 2 he is up to 6 then down to 2. On the 16th day he will go up to 20 and not go down any more (he is out of there!!)
8 Way Cake Slicing:
You cut the cake in half, then in fourths (just like a pizza). Then the secret is to cut it half way up from the table (cut it through the side.)
Chessboard Square Count
There are many types of squares in the chessboard. 1x1 , 2x2, ... 8x8.
So there are 8 squared 1x1 squares and there are 7 squared 2x2 squares.... so the answer is the sum of n squared from 1 to 8 which is 204.
Monty Hall Show:
There is still a 50% chance that either door will be the prize so switching or staying on the door that you have, you still have 50 percent either way.
Glass Half Full:
If you tilt the glass, until the water is touching the lip. If the water is it just touching the bottom of the opposite side, then the glass is half full. If the water is up the side of the opposite side, it is more than half full. If there is no water on the opposite side bottom, then the glass is less than half full.
Hanging Chain:
This is a geometry question. I did it with a line intergral and I am looking for a more elegant solution.
Two Condoms, Three Woman:
The Way to do it is to put on 2 condoms for the first woman. For the second woman take one of the condoms off, leaving one on ( don't forget to wash your hands ) Then for the third woman take the condom that you took off and turn it inside out and put it on over the other condom. Three woman and 2 condoms. You should always be prepared.
Christopher Clark
University of San Francisco Student
cjsclark@aol.com
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C. Clark
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Re: Here are the easy answers (spoilers inside)
« Reply #2 on: Jul 25th, 2002, 8:13pm » |
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8 Queens:
Here are the positions of the Queens for one possible solution:
A5, B8, C4, D1, E3, F6, G2, H7
Christoper Clark
University of San Francisco
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D. Spigarelli
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Re: Here are the easy answers (spoilers inside)
« Reply #3 on: Jul 25th, 2002, 8:25pm » |
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Foot Size Implies Spelling Ability:
<snip>
I guess because anatomically, their body and therefore their brain is bigger. I don't know about this one.
</snip>
Correlation doesn't mean causation. The foot size doesn't cause someone to do better or worse on a spelling bee or test. The study referred in the question is more than likely on children. A child with a size 4 shoe will do worse on a spelling test than a child with a size 9 shoe (usually). As children grow (feet size), there brain power also improves. Their shoe size is more indicative of their age than their spelling ability.
This correlation probably can't be shown beyond a certain range (say 0-15 years) but holds in that range. Elementary statistical question.
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C. Clark
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Re: Here are the easy answers (spoilers inside)
« Reply #4 on: Jul 25th, 2002, 8:29pm » |
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Coin Weighing Machine:
You label each coin machine 1-20. From coin machine 1 you put one coin. From machine 2 you put 2 coins ... etc. Since you know the weight of the coins then the weight should be (1+20)*20/2 multiplied by the weight. You take this number and subtract what you get on the scale and the answer is the machine that is wrong.
Christopher Clark
USF
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Frothingslosh
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Re: Here are the easy answers (spoilers inside)
« Reply #5 on: Jul 25th, 2002, 8:38pm » |
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For the chain problem the answr must clearly be that the chain goes down 3 and up three, it's full length, so the two top pegs must be together.
For the Monty Hall problem, sorry but you're wrong. This is a clasic problem and you're not the only one to make major fals assumptions. Monty is actually giving you important information and you should always change your choice after he gives it to you. I know that doesn't sound lgical, since he can and always will show you where a goat is and he can always find a goat if you have selected the cash or not, but it's true. Look at it this way: If you decide to always stick with your first choice and never take the alternate choice after being given the information and the chance to change, then you have a 1 in 3 chance of making the right choice. If you play this game many times you will win 1/3 of the time (not 1/2 of the time, since you made your choice before knowing where a goat was). Three choices, pick one and stick to it no matter what, and you win one time out of three. Now consider what happens if you always switch after Monty shows you a goat: Clearly every time you would have won if you had stayed with your first choice you will loose. And every time you would have lost, you now win. So now by changing your selection after Monty gives you more information, you win 2/3 of the time! Monty really is giving you useful information after all!
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klbarrus
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Posts: 29
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Re: Here are the easy answers (spoilers inside)
« Reply #6 on: Jul 25th, 2002, 10:28pm » |
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I'm not convinced FrothingSlosh's explanation of Monty Hall is correct. On the other hand, I'm not convinced it is wrong, I gotta think about it more 
Basically I think the explanation itself shows the following counting error: how many ways can 2 coins land? The naive count is 3: both heads, both tails, one of each... without regarding that "one of each" is actually 2 cases. And we all know that 1 head 1 tail is 50% likely to happen, not 33% likely if you count the naive way.
So, back to Monty Hall. When you choose the money the first time, Monty can show you either goat 1 or goat 2. Assume the layout is as follows:
Door 1: $
Door 2: goat 1
Door 3: goat 2
The possibilities are:
you choose door 1, monty shows door 2, you switch -> lose
you choose door 1, monty shows door 3, you switch -> lose
you choose door 2, monty shows door 3, you switch -> win
you choose door 3, monty shows door 2, you switch -> win
That looks like 50% chance of winning after switching, so no advantage.
I think it is a probability fallacy to collapse both door 1 cases into the same case.
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| « Last Edit: Jul 25th, 2002, 10:30pm by klbarrus » |
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Frothingslosh
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Re: Here are the easy answers (spoilers inside)
« Reply #7 on: Jul 25th, 2002, 11:02pm » |
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Sorry klbarrus, but 2 coins and 2 goats are different isues. Two coins can land 4 ways: HH, TT, HT, TH.
The Goats and cash can ony be distributed 3 ways: GG$, G$G, $GG.
OK, six ways if you want to swap around the 2 goats in the above,
but the math will work out exatly the same either way:
Lets assume you always guess door 1 and the goats are placed randomly:
1/3 of the time you get the cash, 2/3 you get Monty's goat.
Same if you always choose door 2, or always choose door 3.
And same if you randomize your choices, 2/3 of the time you get the goat.
(or 4/6 if you want to consider goat 1 and goat 2 different, but 4/6=2/3).
You can't, on the average, win more than 1/3 of the time if you stick
with your original pick.
Here's another way to look at it: If you believe that Monty's revealing
where a goat is does not change the odds in any way, and so you never
change your choice after making it, then he can skip showing where a
goat is (since you say it will have no effect on the game). With 2 goats
hiding behind the curtains and only one pile of cash, your chances of
getting the cash must be less than 50 percent, so you clearly want him
to reveal the information. But if you never use the information then it
can't change your odds for the original chance of 1 out of three to be right.
This is actually something you can run a simple simulation against to
see it happen. Just be sure to simulate everything: a random set up of
1 in 3 to win, a selection of 1 of those 3 (random or not, that
doesn't matter!), reveal a goat behind one of the unselected
positions (there must always be a goat behind AT LEAST one), and then
an option to swap. Do this 3 million times and you'll see that, on the
average, if you don't trade you'll win about 1,000,000 times, and if you
do trade you'll win about 2,000,000 times. Hint: you don't need six
million goats to simulate this, you can reuse the same two ;-)
As Penn Gelette said in Penn and Teller's promotion for Las Vegas:
Gambling - the excitement of bad math!
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klbarrus
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Posts: 29
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Re: Here are the easy answers (spoilers inside)
« Reply #8 on: Jul 25th, 2002, 11:17pm » |
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As I said in the first sentence of my post, I'm not convinced it is correct... or wrong. I gotta think about it more.
Actually, after searching around via google and finding several Monty Hall puzzle explanations, the correct answer is indeed switch.
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Kozo Morimoto
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Re: Here are the easy answers (spoilers inside)
« Reply #9 on: Jul 25th, 2002, 11:19pm » |
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Yeah, I did the Monte Carlo simulation on the Monty Hall prob years back and its 1/3 to stay and 2/3 if you switch.
The key to the problem is that the door that Monty picks is not random and thus biased. If the door Monty picks is random (ie sometimes he opens a door with the cash) then yes, switching won't alter your odds. However we know that Monty NEVER picks the door with the cash so his decision is non-random.
If you are given a choice of picking 1 door or 2 doors at once, what would you do? You know that if you choose 2 doors, one of them would NOT be the cash, and this it the door that Monty will open. But your odds would still be 2/3.
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aelvin
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Re: Here are the easy answers (spoilers inside)
« Reply #10 on: Jul 26th, 2002, 1:35am » |
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Heh! Actually, there's a much easier solution to the hummingbird riddle. The trains are heading toward each other at a closing speed of 35 mph (20 + 15). So the time it will take to cover 5000 miles (i.e., to collide) is about 143 hours. If the hummingbird flies 25 miles per hour for 143 hours, it'll go about 3575 miles.
An interesting thing about this class of riddle is that it doesn't actually matter if the hummingbird flies back and forth between the trains. If it flies that fast for that long, it doesn't matter where it flies, it'll always go the same distance. Most of the time you're told to ignore headwind, the time it takes the bird to turn around between the trains, and all that. Cutting out the inessential will reduce a lot of these riddles to really simple questions (e.g., "how far can a car going 25 mph go in the time it takes a car going 35 mph to go 5000 miles?").
[Note also that the figures given in the question are not stated to be exact, so a more correct answer for a Real Scientist would only incorporate one (!) significant figure (due to the "about 5000 miles" and "20 mph" numbers). The valid answer given the inputs is that the hummingbird would fly approximately 4000 miles. Pointing this out during your interview will give you a chance to belittle your interviewer for either lack of precision, lack of understanding, or poor communication. You will either be hired and immediately made Director of Engineering, or escorted from the building. In either case, they aren't worthy of you, so don't let it bother you.]
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aelvin
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Re: Here are the easy answers (spoilers inside)
« Reply #11 on: Jul 26th, 2002, 1:42am » |
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Read the shoe size one really carefully:
scientific studies have shown that there is a direct, positive correlation between foot size and performance in spelling bees / spelling tests.
It doesn't say a thing about children, or about anything else being equal. Don't compare children of the same age, compare a 2-year-old to a 20-year-old.
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tot
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Re: Here are the easy answers (spoilers inside)
« Reply #12 on: Jul 26th, 2002, 4:29am » |
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One way to understand Monty Hall problem is to think 100 doors. You pick one, and 98 others are opened with goats. You have 1% chance that your pick was the right one, so remaining 99% must be on the remaining closed door, so switch.
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C. Clark
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Re: Here are the easy answers (spoilers inside)
« Reply #13 on: Jul 26th, 2002, 7:55am » |
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Monty Hall:
Here is the way that I see the Monty Hall Problem:
When you initially choose a door, you have a one in three chance. Since there are 2 goats and Monty never opens your door, that means that there is always a goat behind one of the other two doors that he will choose. So when he opens one of the other doors (which he will always be able to do by the above) there are still 2 doors closed. One is the one that you picked and the other door. Now if you stop and look at the situation, you have the door that you chose and the other door. There are 2 items behind them, a goat and the money. So there is a 50 % chance that either door could contain the money. There is no better decision. Stay or change, Monty didn't do anything special. It is something that you expect.
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Dean0
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Re: Here are the easy answers (spoilers inside)
« Reply #14 on: Jul 26th, 2002, 8:48am » |
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Monty Hall:
there are three equally occuring combinations:
1 2 3
A: P G G
B: G P G
C: G G P
so 2/3 of the time you choose the wrong door. this is a fact, look at column 1.
now let's see what happens after monty removes a door, we assume you chose door 1 for simplicity.
(remember, he shows a goat, taking that door out of play)
1 2 3
A: P G x
B: G P x
C: G x P
now if you were in situation A, where you had originally chosen the correct door, you shouldn't switch. But if you are in either situation B or situation C, switching would get you the prize.
in 2/3 of the situations you should switch, so in general switching is the best thing to do.
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Superman
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Re: Here are the easy answers (spoilers inside)
« Reply #15 on: Jul 26th, 2002, 9:38am » |
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I am not convinced the answer to 3 hats is correct (he never actually answers it)
I also think that this riddle is impossible. The only way for A to know if he has black is if he sees 2 white. B has no way of knowing unlesshe hears A say yes I know. If C hears A and B say they know then he knows too.
Someone wanna show me what I am missnig.
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C. Clark
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Re: Here are the easy answers (spoilers inside)
« Reply #16 on: Jul 26th, 2002, 9:48am » |
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3 Hats:
This is a logical chain. If A saw white on B and white on C, then he knows that he has black. That means A saw at least one black hat. This means B and C could be black or white but not both white. So if B saw white on C then by the above he would know that he has black. But he doesn't know what color he has so he must have seen black on C.
You need to assume that they can hear each other.
The answer is C has on black.
The reason is this:
1. A did not see White on B and C or A knows that it has black on.
2. This means that B and C are one of the following combinations: BW WB or BB
3. If B saw white on C then only the first combination is possible. This would make B have a black hat.
4. Since B doesn't know what he has then C must not have white on his head. (By the above.)
5. Therefore C has a black hat.
QED
Christopher Clark
USF
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AFamiliarFace
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Re: Here are the easy answers (spoilers inside)
« Reply #17 on: Jul 26th, 2002, 1:00pm » |
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Willy Wu Tang and the Burning Island:
I don't agree on this one. If you set another fire further down then it will burn to the far end of the island and leave a charred spot that would be safe to stand in. Unfortunately, the fire would also burn back towards you even with a 2 mph wind, and you'd be stuck in the middle.
I don't have an answer for this one. All I could think of were lame things, like hang out on the beach (assuming there is one). Or get to chopping down trees and hope to make a fire break in time.
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jrray
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Re: Here are the easy answers (spoilers inside)
« Reply #18 on: Jul 26th, 2002, 1:22pm » |
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The Monty Hall problem has been a topic of endless debate in my circle of friends, it's funny how some people will decide that you should switch or shouldn't switch and cannot be convinced otherwise.
I take a different approach when trying to explain why you should switch doors.
Monty is going to show you where one of the goats is, so it is in your best interest to guess a door that contains a goat. Assuming you picked a door with a goat, and Monty shows you where the other goat is, the 3rd door must contain the prize. Your strategy should be to pick a door with a goat, then switch doors.
So if your goal is to initially pick a goat, you have a 2 in 3 chance of picking a goat. Therefore, by switching doors, you have a 2 in 3 chance of winning the prize.
If you don't switch doors, you have a 1 in 3 chance of winning the prize.
You never have a 50% chance of doing anything.
- Robert
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Frothingslosh
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Re: Here are the easy answers (spoilers inside)
« Reply #19 on: Jul 26th, 2002, 1:24pm » |
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to C. Clark
Monty Hall:
Here is the way that I see the Monty Hall Problem:
When you initially choose a door, you have a one in three chance. ....
Absolutely correct so far ......
....... Now if you stop and look at the situation, you have the
door that you chose and the other door. There are 2 items behind them, a goat and the
money. So there is a 50 % chance that either door could contain the money. There is no
better decision. Stay or change, Monty didn't do anything special. It is something that you
expect.
Now that's incorrect. There is never a 50% chance, even though there
are 2 doors. Look at what you said, you sarted out saying that there was
a 1/3 chance that you picked the cash, but then you say that Monty didn't
do anything special, and yet you also say that there is now a 50% chance
for what was before a 1/3 chance, even though you think what Monty did
doesn't matter! It does matter, and it matters even more than you realise.
He eliminated a choice (a known bad choice, not just a random elimination
that sometimes would take the cash out of the game). That leaves your
original choice with the exact 1 out of 3 chance it always had to be the
winner (Monty will never reveal what is behind your curtain) and effectively
transfers (adds) the 1/3 chance to be the winner from the curtain he
revealed to the still unselected one. Yea, that sounds like a load of crap,
but it's what happens! Your choice still has a 1/3 chance of being right,
same as it always did, but the other curtain has the ramaining 2/3 chance.
Want to see it in action? Lets do a simulation: Lets further assume that your
lucky number is 2 and that you always start with a pick of 2 (players can play
their lucky number, this still works as long as Monty's distribution is random
and even. We can prove the more generalized case where you select randomly
later.
Lets play 3 times:
Monty gives you a choice. You select 2. Monty shows you a goat behind 3.
Monty offers you a swap, but you refuse. The cash was behind 1. You loose.
Monty gives you a choice. You select 2. Monty shows you a goat behind 3.
Monty offers you a swap, but you refuse. The cash was behind 2. You win.
Monty gives you a choice. You select 2. Monty shows you a goat behind 1.
Monty offers you a swap, but you refuse. The cash was behind 3. You loose.
These are basically the 3 cases that can happen. In 3 cases with your
approach you win 1 out of 3. Play these 300 times rather than 3 times and you
win 100 times out of 300. You can run the same test where you select curtain 1
300 times and you will win 100 times, and you can do the same for curtain 3. You
can combine them all and you still win 1 time out of 3 (on the average).
Now do the same test again, but see if swapping helps:
Monty gives you a choice. You select 2. Monty shows you a goat behind 3.
Monty offers you a swap, and you do. The cash was behind 1. You win.
Monty gives you a choice. You select 2. Monty shows you a goat behind 3.
Monty offers you a swap, and you do. The cash was behind 2. You loose.
Monty gives you a choice. You select 2. Monty shows you a goat behind 1.
Monty offers you a swap, and you do. The cash was behind 3. You win.
See any difference in the results?
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Aimeric
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Re: Here are the easy answers (spoilers inside)
« Reply #20 on: Jul 26th, 2002, 5:26pm » |
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The answer given for Analog Clock II is incorrect.
It depends on how your analog clock works. If the hour hand is only moved
at discreet intervals, i.e. once per minute, then the above answer is (almost)
correct. *However*, if the hour hand is moved with constant angular velocity
(as is, ideally, the minute hand), then the answer is not correct.
Let m(0) be time it takes for minute hand to reach the point where the hour
hand was at the start of the hour.
Let m(k) be the time is takes for the minute hand to go from its point at m(k-1)
to reach the point where the hour hand was at time m(k-1).
[Note that the answer above stops with k = 1. For example, for 3 o'clock,
it notes the minute hand will take 15 minutes to get to m(0), then 1.25 minutes
to get to where the hour hand was at 3:15, which is m(1). But in 1.25 minutes,
the hour hand has already had 1.25 minutes to move *beyond* this point. So we
need an infinite series. Note: To get the correct answer when the clock moves
the hour hand once a minute, stop when k = a number such that m(k) < 1. Note
that, k is not always 1. In the example above, it should be k = 2, since
m(1) = 1.25]
We will find that the sum of m(n), n from 0 to infinity, converges.
Let h be the current hour.
m(0) = 5h
The hour hand takes 60 minutes to move the same space that it takes 5 minutes
for the minute hand to move. Therefore:
m(k) = 5*m(k-1)/60 = m(k-1)/12.
The infinite series is
5h + 5h/12 + 5h/12^2 + 5h/12^3.....
A geometric series whose solution is 5h*(1/(1 - 11/12)) = 5h(12/11) = 60h/11.
So, at h hours and 60h/11 minutes, they overlap.
Note what happens when h == 11. The number of minutes is *60*. This means
that they will *never* overlap between [11,12). So the number of overlaps
in a 24 hour period (if you decide to count both midnights) is only *23*.
Some intuition that may make this easier to accept:
Since the angular velocities are constant, the faster will lap the slower one
at a constant rate. If the hands met within [11,12), they would meet very close
to the 12...the last time they met would be a little over an hour ago (at
10-something), and then they'd meet *only a few minutes later*, at 12, which
contradicts the constant angular velocities.
You may wonder, if this is true, then how come one can (I believe) see the
hands meet at 11-something on a regular clock? The answer is that most analog
clocks move the hour hand once a minute, not at constant angular velocity.
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BigBadBert
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Re: Here are the easy answers (spoilers inside)
« Reply #21 on: Jul 29th, 2002, 12:15pm » |
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Monty Hall Show:
Pure probabilty assumes that when you can choose between two objects, one "correct" and one "incorrect", you have a 50% chance of getting the right answer. However, this problem introduces a human element which completely disrupts the pure. The answer is "switch to the other door".
See here for a more elegent description and arguement:
http://www.straightdope.com/classics/a3_189.html
Foot size and spelling:
As people get older they have better spelling skills.
Hourglass:
Start both hourglasses. At the end of seven minutes, flip over the 7-minute-hourglass. In another 4 minutes, the 11-minute hourglass will be empty and four minutes will have elapsed from the 7-minute-hourglass. Flip the 7-minute-hourglass over and watch the remaining 4 minutes fall away. 7+4+4=15. Doing it this way is 7 minutes faster than C Clark's solution.
Two Coin Flips:
0% or 50% depending on how you read it. Look at the difference:
"(Only) one of the coins came up heads"
"(At least) one of the coins came up heads"
Assuming the first way, it's a trick question.
Assuming the second way, the second coin flip is completely independant.
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william wu
wu::riddles Administrator
Gender: 
Posts: 1291
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Monty Hall Problem and U.S. "Innumeracy"
« Reply #22 on: Jul 29th, 2002, 9:48pm » |
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Here is an amusing article on how Americans responded to the Monty Hall Problem after it was presented by Marilyn vos Savant, the lovely holder of the world's highest IQ ( 228 or 218 ), in her weekly Parade magazine column. If you didn't solve the Monty Hall Problem correctly, don't feel too bad ... hundreds of math professors have humiliated themselves before you
Nation's Mathematicians Guilty of Innumeracy
http://members.aol.com/garypos/vos_Savant.html
Contrast with the following website, sent to me by a visitor who was trying to convince me that the problem is impossible -- apparently he said it depends on the motivations of the game show host, whatever that means.
Marilyn is Wrong
http://www.wiskit.com/marilyn/gameshow.html
This latter website is quite amusing. The webmaster, Herb Weiner, is some kind of creep that tries to compile every possible error that Marilyn has (or has not) committed. The main page is here: http://www.wiskit.com/marilyn.html
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[ wu ] : http://wuriddles.com / http://forums.wuriddles.com
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Brion
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Re: Here are the easy answers (spoilers inside)
« Reply #23 on: Jul 30th, 2002, 3:02pm » |
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I must disagree with the marble question.
The problem does not state that the marbles will be (no pun intended) jarred or shaken after you have arranged them, so the best arrangement is to split black and white into equal piles of 25 marbles. Then take the two black piles and place them in the jars respectively. Finally take the two white piles of 25 and place them on top of the black marbles in each jar. In this case you end up with nearly a 100% chance of getting a white marble regardless of which jar you're handed - which is a much better chance than 49.9%!!
Cheers,
Brion
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Brion
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Re: Here are the easy answers (spoilers inside)
« Reply #24 on: Jul 30th, 2002, 3:10pm » |
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-- Willy Wu Tang --
What if Willy were to tear out a 10-foot square patch of grass and brush (or dig a 10 foot circle 2 feet wide - shallow but wide) and light a fire on the perimeter of the circle? then the fire would spread out away from the circle and in a few hours leave a charred area he can stay safely within to escape the approaching fire?
Cheers,
Brion
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