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Math Puzzles (from FiveThirtyEight - new puzzle every Friday) (1 Viewer)

I'm getting an estimate of

88.4 wins, assuming that none of their games are against each other.

This is an instance of the type of question that asks "for n independent variables, each with the same distribution, what is the expected value of the largest of the n variables?" There is a nice answer to this question for uniform variables (it is n/(n+1), which would be 5/6 in this case). For normal variables it's messier - this post claims that the answer for n=5 is 1.16296 standard deviations above the mean. Here we are dealing with binomial variables, which are approximately normal for large sample sizes. Number of wins for a single team is approximately a normal variable with mean 81 and stdev 6.36396 (which is sqrt(np(1-p)) = .5 * sqrt(162)). For an individual team you have to correct for the fact that # of wins is an integer, but I think you don't need to worry about that for expected values so I'll just use the normal approximation without any corrections. 1.16296 times 6.36396 is 7.401032, so we should expect the team with the most wins to have about that many wins above average, which is about 88.401 wins.

The answer is different if the teams are sometimes playing each other. Focusing on the case with just two teams, for independent teams (who never play against each other) the expected value of the team with more wins is 1/sqrt(pi) standard deviations above the mean (according to the post linked above). For two teams that play all the games against each other, the mean of the team with more wins is sqrt(2)/sqrt(pi) standard deviations above the mean (since we're just dealing with a half-normal distribution, although it's possible that I've made a mistake by not correcting for continuity and therefore underweighting the case where they both finish 81-81). That gives an expected value of 84.59 wins for the better of two teams that never play each other, and 86.08 wins for the better of two teams that always play each other. My guess is that, with actual MLB schedules, this would bring the answer up to around 88.5 rather than 88.4 for the original question.
 
The last two weeks were a little beyond me with the math involved, but I think this week is easy:

6 0 1 0 1 0 1 0 1 0
 
Forgot all about it this week, sorry guys.  Here's the link.

Ten Perfectly Rational Pirate Logicians (PRPLs) find 10 (indivisible) gold pieces and wish to distribute the booty among themselves.

The pirates each have a unique rank, from the captain on down. The captain puts forth the first plan to divide up the gold, whereupon the pirates (including the captain) vote. If at least half the pirates vote for the plan, it is enacted, and the gold is distributed accordingly. If the plan gets fewer than half the votes, however, the captain is killed, the second-in-command is promoted, and the process starts over. (They’re mutinous, these PRPLs.)

Pirates always vote by the following rules, with the earliest rule taking precedence in a conflict:

Self-preservation: A pirate values his life above all else.

Greed: A pirate seeks as much gold as possible.

Bloodthirst: Failing a threat to his life or bounty, a pirate always votes to kill.

Under this system, how do the PRPLs divide up their gold?

Extra credit: Solve the generalized problem where there are P pirates and Ggold pieces.
Click to expand...

 
oldmanhawkins said:
The last two weeks were a little beyond me with the math involved, but I think this week is easy:

6 0 1 0 1 0 1 0 1 0
Click to expand...
I think I agree with this.  Here's the reasoning:

If it gets down to the final 2 pirates (pirates P9 and P10), P9 will propose giving himself all the gold.  He only needs his own vote for this to pass, so that's sufficient. 

If it gets down to the final 3 pirates (P8, P9 and P10), P10 will vote for any plan in which he gets more than 0 gold, since that's what he'll end up with if P8's proposal doesn't pass.  So P8 proposes [9 0 1] as the distribution.

If it gets down to the final 4 pirates, P9 will vote for any plan in which he gets any gold at all, since he'll get none if it proceeds to P8.  So P7 proposes [9 0 1 0]

It carries on this this down the line.  The proposals would be:

P6: [8 0 1 0 1]

P5: [8 0 1 0 1 0]

P4: [7 0 1 0 1 0 1]

P3: [7 0 1 0 1 0 1 0]

P2 [6 0 1 0 1 0 1 0 1]

So finally, P1 proposes P1 [6 0 1 0 1 0 1 0 1 0], and all the pirates with non-zero gold amounts vote for it to pass.

This also illustrates the answer to the generalized problem with P pirates and G gold.  The plan that will pass is as follows: The first pirate gets [G - floor ((P-1)/2)] pieces of gold and the rest get 0 and 1 pieces in alternating fashion.  
 
Ignoratio Elenchi said:
I think I agree with this.  Here's the reasoning:

If it gets down to the final 2 pirates (pirates P9 and P10), P9 will propose giving himself all the gold.  He only needs his own vote for this to pass, so that's sufficient. 

If it gets down to the final 3 pirates (P8, P9 and P10), P10 will vote for any plan in which he gets more than 0 gold, since that's what he'll end up with if P8's proposal doesn't pass.  So P8 proposes [9 0 1] as the distribution.

If it gets down to the final 4 pirates, P9 will vote for any plan in which he gets any gold at all, since he'll get none if it proceeds to P8.  So P7 proposes [9 0 1 0]

It carries on this this down the line.  The proposals would be:

P6: [8 0 1 0 1]

P5: [8 0 1 0 1 0]

P4: [7 0 1 0 1 0 1]

P3: [7 0 1 0 1 0 1 0]

P2 [6 0 1 0 1 0 1 0 1]

So finally, P1 proposes P1 [6 0 1 0 1 0 1 0 1 0], and all the pirates with non-zero gold amounts vote for it to pass.

This also illustrates the answer to the generalized problem with P pirates and G gold.  The plan that will pass is as follows: The first pirate gets [G - floor ((P-1)/2)] pieces of gold and the rest get 0 and 1 pieces in alternating fashion.  
Click to expand...
unless (p-1)/2 exceeds g 

 
New puzzle and solution to last week's. 

A university has 10 mathematicians, each one so proud that, if she learns that she made a mistake in a paper, no matter how long ago the mistake was made, she resigns the next Friday. To avoid resignations, when one of them detects a mistake in the work of another, she tells everyone else but doesn’t inform the mistake-maker. All of them have made mistakes, so each one thinks only she is perfect. One Wednesday, a super-mathematician, whom all respect and believe, comes to visit. She looks at all the papers and says: “Someone here has made a mistake.”

What happens then? Why?
Click to expand...

 
Nothing. Each of them knows that the other 9 are not perfect logicians (since they've all made mistakes in math papers) and thus are unable to draw inferences from each other's behavior.

BOOM!

 
Ignoratio Elenchi said:
Yeah, this one seems pretty straightforward.  Feel free to post your answer in spoiler tags and if you're interested there are links to all the previous puzzle discussions in the OP.  
Click to expand...
I agree that it sounds like a rewrite of the green eyed dragon problem.  The thing I'm not completely clear on is whether or not the mathematicians know their group behavior when someone makes mistakes.  I'm having trouble deciding if it even matters.  

 
So the lady comes on a Wednesday, and resignations happen only on Fridays.  The days of the week were mentioned for a reason.

I think that in theory, 10 groups are created - each of 9 women, and excluding the 1 woman those 9 will go over the previous work of.  Assuming they find the mistake, and fix it, the 10th woman (not in the group) won't resign. 
 
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I'm not familiar with the dragon problem but I remember something similar about redheads on an island or something.  Probably the same idea.  My solution:

Consider the trivial case where there’s only one mathematician (M1).  Obviously M1 must be the one who’s made a mistake so she resigns that Friday.

Now consider the case with two mathematicians.  M1 and M2 know that if only one of them has made a mistake, that person will realize it and resign on Friday.  When Friday passes with no resignation, they both realize they’ve each made mistakes and both resign on the second Friday. 

With three mathematicians it’s the same.  They all know that at least two of them have made mistakes, so they expect resignations on the second Friday.  When that day passes with no resignations, they realize they’ve all made mistakes and they all resign on the third Friday.

Etc. 

So in the case of ten mathematicians, once the ninth Friday passes with no resignations, they all realize they’ve made mistakes and they all resign on the tenth Friday. 
 
oldmanhawkins said:
The thing I'm not completely clear on is whether or not the mathematicians know their group behavior when someone makes mistakes.  I'm having trouble deciding if it even matters.  
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I think this:

"To avoid resignations, when one of them detects a mistake in the work of another, she tells everyone else but doesn’t inform the mistake-maker."

implies that they all know the behavior of the group.  

 
You also have to assume that the mathematicians will catch mistakes in each other's work 100% of the time. If the possibility exists that the super-mathematician found a mistake that no one in the group did, the whole thing breaks down and no one has to resign.

 
Ignoratio Elenchi said:
I'm not familiar with the dragon problem but I remember something similar about redheads on an island or something.  Probably the same idea.  My solution:

Consider the trivial case where there’s only one mathematician (M1).  Obviously M1 must be the one who’s made a mistake so she resigns that Friday.

Now consider the case with two mathematicians.  M1 and M2 know that if only one of them has made a mistake, that person will realize it and resign on Friday.  When Friday passes with no resignation, they both realize they’ve each made mistakes and both resign on the second Friday. 

With three mathematicians it’s the same.  They all know that at least two of them have made mistakes, so they expect resignations on the second Friday.  When that day passes with no resignations, they realize they’ve all made mistakes and they all resign on the third Friday.

Etc. 

So in the case of ten mathematicians, once the ninth Friday passes with no resignations, they all realize they’ve made mistakes and they all resign on the tenth Friday. 
Click to expand...
This is correct (same as dragon problem) - the concept on how to expose a 'secret de polichinelle' (open secret)

 
The difference between the puzzles seems like it might matter - each dragon not only sees 9 red eyed dragons, each also knows that every one of the others sees at least 8 red eyed dragons.  

Meanwhile each mathematician knows that the other 9 have made errors but cannot know for certain if the others know at least 8 made errors.  

Unless "informing others" is a rule they agreed to collectively follow, rather than individually follow.  

Or not.

 
As the problem is written, I would say that no one resigns.

Consider the case where there are two mathematicians, Alice and Bob, and Alice has not detected any mistakes in Bob's work. In Alice's mind, there are four possibilities: 1) neither Bob nor I have made any mistakes, 2) Bob has not made any mistakes but I have, 3) Bob has made mistakes (which I haven't detected) but I have not made any mistakes, or 4) Bob and I have both made mistakes. When the super-mathematician says "Someone here has made a mistake" that rules out possibility #1, but doesn't distinguish between the other three cases. So Alice does not resign on Friday.

Now, consider the case where Alice has detected a mistake in Bob's work. In this case, there are only two possibilities (#3 and #4 above), and that remains true after the super-mathematician speaks, and Alice still will not resign on Friday.

So, Alice will not resign on Friday regardless of what she has or has not seen. And, by similar reasoning, neither will Bob. And Bob will not learn anything new when he sees that Alice did not resign on Friday, because she doesn't resign then in any possible state of the world. So nothing will change the next week, or the week after that, etc.

The original variant of this problem which I heard is the "blue eyes" problem, which ends with everyone "resigning" at once. But in that variant, it is assumed that if Bob made a mistake then everyone knows that he made a mistake (and everyone knows that everyone knows, and everyone knows that, etc.; there is common knowledge). In that case, "Alice has not detected any mistakes in Bob's work" would mean "Alice is certain that there are no mistakes in Bob's work", which would rule out two of the four possibilities and make it so that Alice would resign on the first Friday after the super-mathematician came (in the case with two mathematicians where Bob hasn't made any mistakes). So, if Alice didn't resign on the first Friday, Bob would know that she hadn't seen a mistake and he would resign on the second Friday (in the case with two mathematicians). And we (and the mathematicians) could do induction up to higher numbers, so that if all 10 have made mistakes then all 10 would resign on the 10th Friday.
 
New puzzle and solution to last week's. 

Every morning, before heading to work, you make a sandwich for lunch using perfectly square bread. But you hate the crust. You hate the crust so much that you’ll only eat the portion of the sandwich that is closer to its center than to its edges so that you don’t run the risk of accidentally biting down on that charred, stiff perimeter. How much of the sandwich will you eat?

Extra credit: What if the bread were another shape — triangular, hexagonal, octagonal, etc.? What’s the most efficient bread shape for a crust-hater like you?
Click to expand...

 
The extra credit looks much easier than the main puzzle. Extra credit answer:

You'd be best off with circular bread, in which case you'd eat 1/4 of it. (And you'd be much worse off if you switched to 3D bread, like a roll.)
 
Haven't proved it yet, but my initial guess for the square is:

pi/16
ETA: that's wrong. :wall:

I know the shape that is created though!

 
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bostonfred said:
1/4

It doesn't matter
Click to expand...
Say you have a square with area = 1, and center O.  If you surround it with another square with area = 4 and center O, the smaller square doesn't satisfy the criteria in the problem.  To see this, consider any of the vertices of the smaller square.  The distance to O, measured radially, is sqrt(2)/2.  The distance to the edge of the larger square, measured perpendicularly, is 1/2.  Thus, the vertices are outside the necessary cut.  

 
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I think that the shape you get for any polygon-shaped bread is made of

intersecting parabolas. One definition of a parabola is the set of points which are equally distant from a particular point and a particular line, and that is basically what we're doing here (with each edge playing the role of the line, and the center as the point).
 
ZWK said:
I think that the shape you get for any polygon-shaped bread is made of

intersecting parabolas. One definition of a parabola is the set of points which are equally distant from a particular point and a particular line, and that is basically what we're doing here (with each edge playing the role of the line, and the center as the point).
Click to expand...
I believe that you are right.  My initial attempt was four intersecting circles.  It made a pretty picture, but I didn't have a logical reason for choosing circles.  It worked for the few points I tested so I went with it.  When you switch to parabolas it changes the answer to:

(1/3)[4sqrt2 - 5] or about 21.9% of the original 

Not sure I want to put the hours in needed to generalize it though...

 
I’ll take a swing at this and be as illustrative as I can for the solution I propose (quite difficult without any image).

Suppose a square of length 2 – so the distance from the center to any side (perpendicular) is 1.

As mentioned here, a circle of radius 1/2 yields a pi/16 eatable area (19.6%) which respect the criteria but is suboptimal. The intersect solution (with a diagonal of sqrt(2)/2) would yield a 25% result but does not respect the ‘closer to the center than the border’ criteria... so we know that our answer must be between 19.6% and 25%.

Also, as mentioned here – the parabola offers the optimal solution since by definition it is equally distant from a point and a line.

Let look at the ‘bottom triangle’ – from center to both corners of the square. On a xy graph – suppose that the center of the square is located at (0;1) thus the bottom line of the square extend from (-1;0) to (1;0).

The optimal parabola is represented by the y = (x^2 +1)/2 function – goes through (0;1/2), (1/2;5/8) and (1;1)... you can derive that formula by the y = 1/4f x^2 + f generic function with focus f=1/2 here.

Looking at the right side, we see that this parabola intersects the triangle diagonal at x=3/8... So, the ‘eatable’ area is found by subtracting from the rectangular of length 3/8 * 1, the above triangle of 3/8 * 3/8 and the area underneath the parabola – which can be found by integrating (x^2+1)/2 dx from 0 to 3/8.

Eatable area: 3/8 (rectangle) – 9/128 (upper triangle) – 201/1024 (area under parabola) = 111 / 1024.

There are 8 identical sections of that eatable area = 888 / 1024 and our overall square is of 4.

Overall eatable area = 111 / 512 = 21.7%

 
New puzzle and solution to last week's.

At RoboPizza™, pies are cut by robots. When making each cut, a robot will randomly (and independently) pick two points on a pizza’s circumference, and then cut along the chord connecting them. If you order a pizza and specify that you want the robot to make exactly three cuts, what is the expected number of pieces your pie will have?
Click to expand...

 
I am getting an answer of

5 by going through the cases. Start by picking 6 points at random, and then pick one of them to be the start of the first cut. There is a 2/5 chance that the cut will go to an adjacent point (which leaves all 4 remaining points on the same chunk of pizza; call this case 1), 2/5 it goes to a point which isolates one of the other points (case 2), and 1/5 it goes to the farthest point which leaves 2 points on each side (case 3).

In case 1, pick one of the 4 remaining points as the starting point of the next cut. There is a 2/3 chance the cut goes to an adjacent point (case 1a), which winds up dividing the pizza into 4 pieces once you add in the last cut, and a 1/3 chance it criss-crosses the third cut (case 1b) which winds up dividing the pizza into 5 pieces. Similarly, in case 2 there is a 2/3 chance of getting 5 pieces and 1/3 chance of getting 6 pieces, and in case 3 there is a 1/3 chance of ending up with 4 pieces, 1/3 6 pieces, and 1/3 7 pieces. Multiply out the numbers, and the average is 5.

I suspect there is a more elegant solution.
 
I get 5.5 (expected slices of pizza)

First chord will create 2 pieces

50% chance the second chord will intersect with the first leaving 3 or 4 pieces.

In the 3 piece scenario, 1/3 chance the last chord intersects 0, 1 or 2 lines leaving 4, 5 or 6 pieces.

In the 4 piece scenario, 1/3 chance the last chord intersects 0, 1 or 2 lines leaving 5, 6 or 7 pieces.

Average 5.5

and 39.369% of the sandwich... failed to divide by two so closer to 20% using an estimation

 
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You get to play Roger Federer at Wimbledon.

Bad news:  you only have a 1% chance of winning any given point.

Good news:  you get to pick the starting score of the match.

Q:  what score do you pick to maximize your chances of winning.

 
Been swamped with work and kinda forgot all about this, glad to see you guys kept it alive!  Just took a look at last Friday's puzzle, and am I missing something?  If you pick the score why wouldn't you just put yourself at match point and Federer at 0?  (i.e. up 40-0 in the 6th game of the 3rd set, with you up 5-0 and 2-0 in games and sets, respectively).  Figuring out your overall probability of victory from that point is still maybe a fun challenge but I don't know why the whole "pick whatever score you want" part is in there, other than as a total red herring.  

 
I don't know the rules of tennis very well, so it could also be that I'm missing some nuance. :shrug:

According to some tweets I saw apparently you can do better than being up 40-0, 5-0 in the 3rd set.  Maybe it's trickier than I thought?

 
Ignoratio Elenchi said:
I don't know the rules of tennis very well, so it could also be that I'm missing some nuance. :shrug:

According to some tweets I saw apparently you can do better than being up 40-0, 5-0 in the 3rd set.  Maybe it's trickier than I thought?
Click to expand...
wdcrob figured it out here.

 
Ignoratio Elenchi said:
I don't know the rules of tennis very well, so it could also be that I'm missing some nuance. :shrug:

According to some tweets I saw apparently you can do better than being up 40-0, 5-0 in the 3rd set.  Maybe it's trickier than I thought?
Click to expand...
As noted above, wcdrob had it.  To compare the two:

Up 40-0 means you need 1 of the next 3 points.  

Up 6-0 in a tiebreaker, you need 1 of the next 6 points.
 
heckmanm said:
As noted above, wcdrob had it.  To compare the two:

Up 40-0 means you need 1 of the next 3 points.  

Up 6-0 in a tiebreaker, you need 1 of the next 6 points.
Click to expand...
That makes sense, and I assume he's right.  It just seems like

if you don't get one of the next 6 points in the tiebreaker, you're in a much worse spot than if you don't get 1 of the next 3 points in the 6th game, no?

But I haven't done the math and I assume my gut feeling about that is wrong (it usually is :)  )

 
Isn't there NO tiebreaker in the final set at Winbledon? Remember that one match that went 120 games or whatever a few years ago?

I'm back to the 'Up two sets, 5-0, 40-love and serving' which seems obvious so what am I missing?

Maybe 5-0, 40-love, and him serving a 2nd serve?

 
No tie-breaker in the 5th set only.  So up 6-0 in the t-b in the 3rd set is as good as you can do.  Your odds of winning enough points to actually win a game is minuscule.

 
New puzzle and solution to last week's.

The archvillain Laser Larry threatens to imminently zap Riddler Headquarters (which, seen from above, is shaped like a regular pentagon with no courtyard or other funny business). He plans to do it with a high-powered, vertical planar ray that will slice the building exactly in half by area, as seen from above. The building is quickly evacuated, but not before in-house mathematicians move the most sensitive riddling equipment out of the places in the building that have an extra high risk of getting zapped.

Where are those places, and how much riskier are they than the safest spots? (It’s fine to describe those places qualitatively.)

Extra credit: Get quantitative! Seen from above, how many high-risk points are there? If there are infinitely many, what is their total area?
Click to expand...

 
As a simpler version, we can imagine that the shape

of the building is a circle. Then the laser has to cut along a diameter to cut it in half (I assume "planar ray" means that it cuts in a straight line). We can assume that all diameters are equally likely. So the mostly likely spots to get hit are near the center of the building (with the point at the center guaranteed to get hit), and the least likely spots to get hit are near the edge.

For example, you can look at a ring of the circle near the edge, between 99% of the way from the center to the edge and 100% of the way to the edge. Compare that to a ring near the center of the circle, between 1% of the way from the center to the edge and 2% of the way to the edge. The outer ring is 66x as big in area as the inner ring, but each of them will have the same amount of laser cut within them. So a given object is 1/66 as likely to get hit if you put it near the outer edge than if you put it in that inner ring.

The answer should be pretty similar given that the building is pentagon-shaped - you want to move the important objects close to the perimeter (and presumably close to the corners, which are farther from the center).
 
ZWK said:
As a simpler version, we can imagine that the shape

of the building is a circle. Then the laser has to cut along a diameter to cut it in half (I assume "planar ray" means that it cuts in a straight line). We can assume that all diameters are equally likely. So the mostly likely spots to get hit are near the center of the building (with the point at the center guaranteed to get hit), and the least likely spots to get hit are near the edge.

For example, you can look at a ring of the circle near the edge, between 99% of the way from the center to the edge and 100% of the way to the edge. Compare that to a ring near the center of the circle, between 1% of the way from the center to the edge and 2% of the way to the edge. The outer ring is 66x as big in area as the inner ring, but each of them will have the same amount of laser cut within them. So a given object is 1/66 as likely to get hit if you put it near the outer edge than if you put it in that inner ring.

The answer should be pretty similar given that the building is pentagon-shaped - you want to move the important objects close to the perimeter (and presumably close to the corners, which are farther from the center).
Click to expand...
I'm having some trouble understanding this (the problem itself and your explanation).  If you don't mind, could you explain how (with the exception of the center point) how any point on the circle was more likely to be hit than any other?  I agree that the "rings" on the outside have larger area than those on the inside, but that doesn't mean they have more points.  Or am I missing something?

I interpreted the problem to mean that given an arbitrary point on the edge of the pentagon, a straight line would pass through the shape in such a way that the shape was divided perfectly in half.  For instance, if you start on a vertex, the line would obviously go through the midpoint of the opposite side.  Of course, you need not start on the vertex.  So, if you consider the set of all lines with this property, they will sweep out some shape.  I think the shape is a 5 pointed star made by connecting the vertices of the pentagon.  I'll play around a bit more tonight.  

 
oldmanhawkins said:
I'm having some trouble understanding this (the problem itself and your explanation).  If you don't mind, could you explain how (with the exception of the center point) how any point on the circle was more likely to be hit than any other?  I agree that the "rings" on the outside have larger area than those on the inside, but that doesn't mean they have more points.  Or am I missing something?

I interpreted the problem to mean that given an arbitrary point on the edge of the pentagon, a straight line would pass through the shape in such a way that the shape was divided perfectly in half.  For instance, if you start on a vertex, the line would obviously go through the midpoint of the opposite side.  Of course, you need not start on the vertex.  So, if you consider the set of all lines with this property, they will sweep out some shape.  I think the shape is a 5 pointed star made by connecting the vertices of the pentagon.  I'll play around a bit more tonight.  
Click to expand...
Think in terms of object with some area, not dimensionless points. A given object takes up a smaller fraction of the area of its ring if it is more towards the outside, and there is the same amount of laser cuts in each ring, so the odds of getting cut must be dropping as you move outward.

For example, imagine that you divide the circle into "rings" which are each one foot wide. Say that a ring near the center is 50 feet around its circumference, and a ring near the outside is 5,000 feet around. Now, take an object that is one square foot in size. Whichever ring you put it in, it takes up the entire width and 1 foot of the circumference. If you put it in the ring near the center then there is a 1/50 chance that the laser will go through it. If you put it in the ring near the outside then there is a 1/5000 chance that the laser will go through it.

(In general, the chance that the laser goes through it is proportional to 1/r, where r is its distance from the center of the circle.)

 
ZWK said:
Think in terms of object with some area, not dimensionless points. A given object takes up a smaller fraction of the area of its ring if it is more towards the outside, and there is the same amount of laser cuts in each ring, so the odds of getting cut must be dropping as you move outward.

For example, imagine that you divide the circle into "rings" which are each one foot wide. Say that a ring near the center is 50 feet around its circumference, and a ring near the outside is 5,000 feet around. Now, take an object that is one square foot in size. Whichever ring you put it in, it takes up the entire width and 1 foot of the circumference. If you put it in the ring near the center then there is a 1/50 chance that the laser will go through it. If you put it in the ring near the outside then there is a 1/5000 chance that the laser will go through it.

(In general, the chance that the laser goes through it is proportional to 1/r, where r is its distance from the center of the circle.)
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Thanks, I understand what you are saying now!  It helped to (correctly) imagine the equipment as having some fixed area.  

Still, isn't there some shape in the center that every equal areas dividing line goes through?  Some collection of points with the property that, if you left the equipment there, there would be 100% chance of it getting cut?  

I tried to sketch it on geometer's sketch pad and came up with a tiny pentacle or 5 point star.  The radius of the circumscribed circle about the pentacle was almost 10% of the radius of the circumscribed circle of the pentagon.  I had to make a false assumption to make the sketch easier which might account for the difference.   

 
oldmanhawkins said:
Thanks, I understand what you are saying now!  It helped to (correctly) imagine the equipment as having some fixed area.  

Still, isn't there some shape in the center that every equal areas dividing line goes through?  Some collection of points with the property that, if you left the equipment there, there would be 100% chance of it getting cut?  

I tried to sketch it on geometer's sketch pad and came up with a tiny pentacle or 5 point star.  The radius of the circumscribed circle about the pentacle was almost 10% of the radius of the circumscribed circle of the pentagon.  I had to make a false assumption to make the sketch easier which might account for the difference.   
Click to expand...
With a circle, there are infinitely many possible cuts and they all go through the same point at the center. Every diameter is a possible cut, and those are the only possible cuts.

I think that is also true with a pentagon, although I'm more confident in the "infinitely many" part than the "same point at the center" part. There should be infinitely many possible cuts, because if you pick any one point on the perimeter as your starting point there must be some point across the pentagon which you can cut to in order to make it exactly a 50-50 split. I'm less sure about them going through the same point, but it does happen with circles and squares and my intuition is that it works with all regular polygons.

 
ZWK said:
With a circle, there are infinitely many possible cuts and they all go through the same point at the center. Every diameter is a possible cut, and those are the only possible cuts.

I think that is also true with a pentagon, although I'm more confident in the "infinitely many" part than the "same point at the center" part. There should be infinitely many possible cuts, because if you pick any one point on the perimeter as your starting point there must be some point across the pentagon which you can cut to in order to make it exactly a 50-50 split. I'm less sure about them going through the same point, but it does happen with circles and squares and my intuition is that it works with all regular polygons.
Click to expand...
I fiddled around enough to convince myself that these cuts (which are indeed infinite in number) don't all go through the same point.  But there is some star like shape that they all pass through.  So if your equipment were sitting on this shape it would definitely get destroyed.  

I will try to recreate the sketch on Desmos tonight so I can share.

 
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