m00li
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Posts posted by m00li
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Suppose there are 500 pirates. You propose that all pirates will die until pirate #202 is reached.
If pirates #203 to #500 all decide they would rather live than die, and agree to take nothing while distributing coins to pirates up to #200, that would seem to fulfill their logical criteria and not lead to their deaths.
On the other hand, if pirate #499 makes a convincing argument that he would save pirates 203-499 while letting them all see #500 walk the plank...
No plasmid. What you are stating will work only if condition 3 is relaxed
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one such polygon exists where perimeter = area, then put an arbitrary point O on any one of its sides AB. introduce a infinitesimally small kink at O, such that AO and OB are almost almost almost colinear but are not.
this new polygon has the same area and same perimeter as the old one so it satisfies the original conditions and also has greater number of sides.
wouldn't the area be infinitesimally off
yup, sowwee
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Yes, bonanova. I think the answer depends on a (e.g. if a is obtuse the ray doesnt exit on left and never hits the top mirror). It coinicides with the icident ray if a is a factor of pi/2 radian
Everything in your post is correct. But remember what the question is.
Whatever value a has, how close does the ray come to the point O?
The phrase "before it exits to the left" is appropriate to a being acute, as in the figure.
If you want to include a being obtuse, so that the ray would not exit to the left,
that just gives you "forever" as the time frame to consider the ray's point of closest approach.
Although of course you wouldn't need "forever" to determine it.
In other words, the wording is not meant to be tricky: assume a to be acute if you like.
Hi Bonanova,
Sorry, I do not understand. Is my answer wrong? or incomplete?
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Yes, bonanova. I think the answer depends on a (e.g. if a is obtuse the ray doesnt exit on left and never hits the top mirror). It coinicides with the icident ray if a is a factor of pi/2 radian
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one such polygon exists where perimeter = area, then put an arbitrary point O on any one of its sides AB. introduce a infinitesimally small kink at O, such that AO and OB are almost almost almost colinear but are not.
this new polygon has the same area and same perimeter as the old one so it satisfies the original conditions and also has greater number of sides. -
hi bonanova,
i have given it for any arbitrary a (both when its sub multiple and when its not).
only case to be added is, if a is integer multiple of pi, then the ray doesnt return at all
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if we start backwards and think of what happens when only two pirates are left, and then extrapolate, its clear that, the captain should handover 1 coin each to the existing 2nd, 4th,6th,... mates.
In case of 5 pirates the captain keeps 98 and gives 1 each to 2nd and 4th mate.
for n <= 202, he retains 100-[(n-1)/2] where [3.5] = 3
for n>202, the captains keep dying till n reaches 202 -
Lets see what the light ray sees. The light ray, is travelling straight parallel to the horizontal mirror. It encounters a plane inclined at angle a. We can assume it keeps travelling straight through without getting reflected, and then it meets another surface inclined at 2a. it keeps travelling straight through, without getting reflected and meets the third surface at 3a, and so on.
It will be similar, if all these surfaces are assumed to be originating from O, inclined at a, 2a, 3a, .. angles in anticlockwise direction, and the light ray keeps going straight, cutting through all these surfaces. The light ray, thus, starts exiting from left, when it hits the first surface which is inclined at 90 degrees or more to the horizontal.
If 'a' is a factor of 90 degree, it will cut this surface exactly at d distance.
e.g, if a=30:
imagine three imaginary surfaces, inclined at 30, 60 and 90, all intersecting at O. Imagine the ray travelling horizontally, d distance below the horizontal mirror, cutting through these three surfaces and finally hitting the 90 degree mirror at d distance below O. Hence, it exits exactly at d distance from O IF a is a factor of 90.
but, if a is not a factor of 90 (but less than 90), then the distance is d/cos(ma mod 90) where ma>90>(m-1)a
In general, for any a, distance = d/cos(ma mod 90) where ma>=90>(m-1)a
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Had bmad doubled up her pace, you would have died waiting for her (note) call.
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Suppose the hr, min and sec hands behave like that of a normal watch at 00:00:00. Then, at 00:00:01 the sec hand is after the min hand which is after the hr hand. Note, that the hands remain in this configuration (i.e. h-m-s, hour hand before minute and minute before second hand in clockwise direction) but keep changing their speeds.
e.g. at 00:01:02 the configuration becomes (s-h-m) i.e. second hand is at hour's position, hr hand is at minute's
At 00:02:03 it becomes (m-s-h) i.e. minute hand is at hour's position and second hand is at minute's position
At 00:03:04 it returns to normal i.e. (h-m-s).
Note that all three configuration's above remain in this configuration (i.e. h-m-s, hour hand before minute and minute before second hand in clockwise direction) but keep changing their speeds. From the above 3 examples, Its clear that the hour hand position is taken up by second hand, then minute hand and then hour hand every time the second hand of a normal watch catches up with the hr hand of a normal watch i.e. after every 720/719 minutes. Similarly, the minute hand position is taken up by hr hand, then second hand and then minute hand, every time the second hand of a normal watch catches up with the minute hand of a normal watch i.e. after every 60/59 minutes. BUT, the configuration remains same, i.e. (h-m-s) in clockwise direction
UNTIL the hour hand and the minute hand get to swap their positions (i.e. every 720/11 minutes).
Thus, the clockwise order of hands, alternates between (h-m-s) and (m-h-s) every 720/11 minutes
The above example demonstrated that:
the clockwise order of hands, alternates between (h-m-s) and (m-h-s) every 720/11 minutes. Hence after 12 hours, the configuration reverses
the hour hand position is taken up by sec hand, then minute hand and then hr hand after every 720/719 mintes in h-m-s configuration
the minute hand position is taken up by hour, then second and then minute hand after every 60/59 minutes in h-m-s configuration
At 9:05:25 am the configuration is h-m-s. The difference between 9:05:25 am and 11:11:11 pm is 845.7667 minutes
Dividing this by 720/11 we get quotient as 12 which implies that the configuration remains as h-m-s
Dividing this by 720/719 we get quotient as 844 = 1 mod 3, hence hr position is occupied by second hand (OR second hand is moving the slowest)
Dividing this by 60/59 we get quotient as 831 = 0 mod 3, hence minute position is occupied by minute hand (OR minute hand is moving normally)
And second position is occupied by hr hand (OR hr hand is moving fastest)
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Isnt it just 1/n assuming the casino deck is made up of n distinct cards?
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Bonanova: using that formula with 3 balls being drawn 3 times...
probability of drawing 1 color
3! 13-1 / 33(3-1)! = 6*1 / 27*2 = 6/54
probability of drawing 2 colors
3! 23-2 / 33(3-2)! = 6*2 / 27*1 = 12/27
probability of drawing 3 colors
3! 33-3 / 33(3-3)! = 6*1 / 27*1 = 6/27
Sum of all three of those is less than 1.
I'm having trouble following m00li's logic. I got lost as soon as he started talking about selecting balls from an infinite choice of colors (which was pretty soon).
You are not alone Plasmid. I am having trouble too
I must have been high on something that dayMy solution is for cases where a 'selection' is being made out of n different colors. I should have calculated permutations and not selections.
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Has to be zero
At time t1, they are equally likely to be at distance d apart or distance d1 apart or distance d2 apart or distance 0 apart as their velocities are not given and can be assumed to be anything. Similar distribution holds true for time t2 as well. So, there are infinite equally likely possibilities, and running into each other again is but one of them. -
Awesome Rob!
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hi bmad, Is it purely mathematical or does it require some general knowledge too?
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If you mean unit area, the mean distance between two interior points of a unit regular n-gon converges to that of a disk.
If you mean unit side, then the mean distance diverges, since that type of unit regular n-gon becomes the entire plane.
how do we find the mean distance for a square? and for a regular unit pentagon? seems very very tough to me
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Two points are on a semicircular arc, even if they are opposite ends of a diameter. For larger n, the remaining n-2 points simply have to lie on the same side of the diameter, with equal probability. That gets us a (1/2)n-2 term, where DeGe stopped. We finally say that the argument applies to all of the n points. Thus p = n/2n-2.
I see now you had n-1, but I think you meant n-2.
This logic doesnt seem convincing. Suppose we have only three points to consider. Two issues:
1) "n-2 points simply have to lie on the same side of the diameter". Which diameter are we considering here? If we have fixed two points, and fixed 'the' diameter, the third can STILL lie on the other side of the diameter but be in the same semi circle. So the probability will not be 1/2 but more or less depending on the configuration of the first two points.
2) "We finally say that the argument applies to all of the n points". Since we did not chose 'the' two out of three points to begin with, this clause seems to 're-count' already counted cases. This clause seems to suggest that the ordering of selection is important.
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2 items from a group of 4 can be selected in 10 distinct ways if repetitions are allowed. e.g. {(1,1),(1,2),(1,3),(1,4),(2,2),(2,3),(2,4),(3,3),(3,4),(4,4)}
Are each of these selections equally likely?
Why or why not?
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Required probability = selecting T balls from an infinite choice of m colors where atleast 1 ball from each of m colors is present/ selecting T balls from an infinite choice of n colors (not all n colors need be present in the selection)
Calculating Denominator = Selecting T balls from an infinite selection of n color balls:
Assume a string of n-1 0s. These 0s divide the string into n sections. We can fill T 1s in these T sections in n+T-1CT ways. Now, assume each section represents one of the n distinct colors. The number of 1s in each section represents the number of balls of that color. So, this representation gives us the number of ways selecting T balls from a set of n color balls where T<=n
Hence, Denominator = n+T-1CT
Now, Numerator can be arrived at as
first selecting m colors = nCm ways
then selecting 1 ball from each of these m colors = 1 way
then selecting the remaining T-m balls from an infinite selection of these m color (not all m need be present in the selection) = m+(T-m)-1CT-m (refer denominator) = T-1CT-m
Hence required probability = nCm*T-1CT-m/n+T-1CT
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At 9:05:25 a regular wrist watch hands are at:
hr @ -87.291667 degrees
min @ 32.5 degrees
sec @ 150 degrees
therefore, a regular wristwatch cannot have its hands separated equally at 9:05:25. Hence, if a wristwatch has it hands separated equally at 9:05:25, it cannot be working properly. The question doesnt seem to be correct in stating BOTH "A strange wrist watch have it hands separated equally on 9:05:25 am." AND ".and seem to be working properly."
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(I) Lets look at a scenario where all blacks are finished first and only k whites are finished. No. of balls remaining are (w-k). The probability of this happening is
b-1+kCk(b*(b-1)*(b-2)..*1*w*(w-1)*(w-2)..(w-(k-1)) / (w+b)*(w+b-1)*...(w-(b+k-1))
= b-1+kCk(b!*k!wCk) / (b+k)!w+bCb+k
= (b/(b+k))*(wCk/w+bCb+k)
=(1/w+bCb)*(b/(b+k))*(b+kCb)
=(1/w+bCb)*b+k-1Cb-1
therefore, expected number of balls remaining = (1/w+bCb)*(S0->w-1[b+k-1Cb-1(w-k)] + S0->b-1[w+k-1Cw-1(b-k)]) .... where Sx->y denotes sum of series from x to y
(II) Alternative (I) could have also be interpreted as:
probability that w-k whites remain = (arranging b blacks and k whites such that last ball picked is a black)/(arranging b black and w white balls)
= (arranging b-1 blacks and k whites such that last ball picked is a black)/(arranging b black and w white balls)
= (b+k-1Cb-1/w+bCb)
therefore, expected number of balls remaining = (1/w+bCb)*(S0->w-1[b+k-1Cb-1(w-k)] + S0->b-1[w+k-1Cw-1(b-k)]) .... where Sx->y denotes sum of series from x to y
... gotta rush.. will continue later
continuing...
S0->w-1[b+k-1Cb-1(w-k)] = w*S0->w-1(b+k-1Ck) - S0->w-1(b+k-1Ck*k) ...(A)
Using nCr + n-1Cr-1 = n+1Cr : w*S0->w-1(b+k-1Cb-1) = w*b+w-1Cw-1
Using S0->ns+kCkk = (s+1)s+n+1Cn-1: S0->w-1(b+k-1Ck*k) = b*b+w-1Cw-2
Hence, (A) = w*b+w-1Cw-1 - b*b+w-1Cw-2 = b+wCw-1
Similarly S0->b-1[w+k-1Cw-1(b-k)] = b*w+b-1Cb-1 - w*w+b-1Cb-2 = b+wCb-1
Therefore, expected number of balls remaining = (b+wCw-1 + b+wCb-1)/b+wCb = w/(b+1) + b/(w+1)
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Yes, the solution is absolutely incorrect
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(I) Lets look at a scenario where all blacks are finished first and only k whites are finished. No. of balls remaining are (w-k). The probability of this happening is
b-1+kCk(b*(b-1)*(b-2)..*1*w*(w-1)*(w-2)..(w-(k-1)) / (w+b)*(w+b-1)*...(w-(b+k-1))
= b-1+kCk(b!*k!wCk) / (b+k)!w+bCb+k
= (b/(b+k))*(wCk/w+bCb+k)
=(1/w+bCb)*(b/(b+k))*(b+kCb)
=(1/w+bCb)*b+k-1Cb-1
therefore, expected number of balls remaining = (1/w+bCb)*(S0->w-1[b+k-1Cb-1(w-k)] + S0->b-1[w+k-1Cw-1(b-k)]) .... where Sx->y denotes sum of series from x to y
(II) Alternative (I) could have also be interpreted as:
probability that w-k whites remain = (arranging b blacks and k whites such that last ball picked is a black)/(arranging b black and w white balls)
= (arranging b-1 blacks and k whites such that last ball picked is a black)/(arranging b black and w white balls)
= (b+k-1Cb-1/w+bCb)
therefore, expected number of balls remaining = (1/w+bCb)*(S0->w-1[b+k-1Cb-1(w-k)] + S0->b-1[w+k-1Cw-1(b-k)]) .... where Sx->y denotes sum of series from x to y
... gotta rush.. will continue later
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Assume we have placed two points on the circle. the first at 0 radian and the second somewhere between x and x+dx radians. (the probability of doing this is dx/2pi). Here we are assuming a circle from -pi to pi, so that 0 is at the mid point of the interval (-pi,pi)
When x is in the interval (0,pi),
for any point to lie in the same semi circle as these two points, it can lie anywhere in the interval (x-pi, pi) with prob. (2pi-x)/2pi = 1-x/2pi
So, for n-2 points, prob. is (1-x/2pi)^(n-2)
Similarly, when x is in the interval (-pi,0),
prob. for n-2 points is (1-|x|/2pi)^(n-2)
Hence total probability is Int(-pi,pi)(1-|x|/2pi)n-2dx/2pi = (1/2pi)Int(-pi,pi)(1-|x|/2pi)n-2dx .... where Int(a,b) is definite integral from a to b
=2*(1-21-n)/(n-1) for n>2
The 'aha' solution you are looking for might not be applicable for a sphere. am i right?
Play on a classic: computers
in New Logic/Math Puzzles
Posted
this answer assumes that we know the number of bad computers, but we don't. what is the actual answer?