Easy math puzzles

Quote: Ace2

Suppose you roll a fair die until some face has appeared six times. For instance, your rolls could be 42132253261242 (for six 2's).On average, how many rolls would it take?

Quote: Ace2

Suppose you roll a fair die until some face has appeared six times. For instance, your rolls could be 42132253261242 (for six 2's).

On average, how many rolls would it take?

Quote: EdCollins

Show Spoiler

19.74

Quote: Ace2

Suppose you roll a fair die until some face has appeared six times. For instance, your rolls could be 42132253261242 (for six 2's).

On average, how many rolls would it take?

Quote: Ace2

Please show your calculation

Quote: Wizard

Answer

Integrate the following from 0 to infinity:

x*(exp(-x/6)*(1+x/6+x^2/72+x^3/1296+x^4/31104+x^5/933120))^5*exp(-x/6)*(x^5/933120)

The answer is 2597868106693535971 / 131621703842267136 = Approximation: 19.73738396371749

Quote: EdCollins

Sorry. No "calculation." My answer was based upon a quick & dirty simulation I wrote via Excel.

Quote: Wizard

Here is my solution in more detail (PDF).

Recommended integral calculator.

Quote: Ace2

Recommended integral calculator.

Quote: Gialmere

A white rook and a black bishop are randomly placed on an empty chessboard.

What is the probability that one of the pieces is attacking the other?

Quote: Gialmere

A white rook and a black bishop are randomly placed on an empty chessboard.

What is the probability that one of the pieces is attacking the other?

Quote: chevy

Show Spoiler

I get 1456/4032 = 36.111%

for each possible position of rook, it can attack 7 vertical + 7 horizontal = 14

for each possible position of rook, it can be attacked by bishop.....depending on location of rook
-outer "ring" of squares (8x8 border only = 28 squares) attacked by bishop from 7 positions
-next "ring" of squares (6x6 border only = 20 squares) attacked by bishop from 9 positions
-next "ring" of squares (4x4 border only = 12 squares) attacked by bishop from 11 positions
-inside block of squares (2x2 = 4 squares) attacked by bishop from 13 positions

[ (64*14)+(28*7+20*9+12*11+4*13) ] / [64*63] = 1456/4032

Quote: ThatDonGuy

Answer

Select the bishop's space first
Each of the 28 spaces on the board edge has 7 spaces that can attack the rook
Each of the 20 spaces one space away from the board edge spaces has 9 spaces that can attack the rook
Each of the 12 spaces two spaces away has 11 spaces
The 4 spaces in the center each has 13 spaces
No matter where the bishop is placed, there are 14 spaces from where the rook can attack it, none of which are on spaces where the bishop can attack the rook.

There are 64 squares on which the bishop can be placed; for each one, there are 63 on which the rook can be placed.
The probability is (28 x (7 + 14) + 20 x (9 + 14) + 12 x (11 + 14) + 4 x (13 + 14)) / (64 x 63) = 13/36

Quote: charliepatrick

Show Spoiler

Assume for simplicity the Rook is placed in one of the quarters, say NW. Then the other quarters would create identical chances.
Wherever the rook is, it can take the bishop on 7 squares on the same vertical column and 7 on the same row.
If the rook is in on the edge, i.e. A, then there are 7 places (on the diagonal) the bishop can be where it could take the rook.
B means 9 places, C means 11, D means 13.

There are 7 A's, 5 B's, 3 C's and 1 D. Thus the average number of squares where the bishop could be taken is
(7*21+5*23+3*21+1*23)/16 = 22.75.
So the chances of being taken is 22.75/63 = 91/252 (x4) = 13/36 (/7).

A A A A
A B B B
A B C C
A B C D

joke

Quote: Ace2

Any chance of a problem for people who don't play chess?

Quote: Gialmere

A single die is rolled until a run of six different faces appears. For example, one might roll the sequence
535463261536435344151612534 with only the last six rolls all distinct.

What is the expected number of rolls?

Quote: EdCollins

My answer is...

Show Spoiler

64.8 rolls

Quote: Gialmere

A single die is rolled until a run of six different faces appears. For example, one might roll the sequence
535463261536435344151612534 with only the last six rolls all distinct.

What is the expected number of rolls?

Quote: ThatDonGuy

I'm not entirely sure, as I am having a little problem simulating it to confirm the answer...

Answer

Let E(N) be the expected number of rolls needed if the N most recent are all different
Suppose the last 5 rolls are 1, 2, 3, 4, 5, in order.
If the next roll is 1, then you have 2, 3, 4, 5, 1; if it is 2, you have 3, 4, 5, 2; if it is 3, you have 4, 5, 3; if it is 4, you have 5, 4; if it is 5, you have 5; if it is 6, you are done.
E(5) = 1 + 1/6 E(1) + 1/6 E(2)+ 1/6 E(3) + 1/6 E(4) + 1/6 E(5) + 1/6 x 0
Similarly, if the last 4 are 1, 2, 3, 4, then a 5 or 6 results in a "run" of 5:
E(4) = 1 + 1/6 E(1) + 1/6 E(2)+ 1/6 E(3) + 1/6 E(4) + 1/3 E(5)
E(3) = 1 + 1/6 E(1) + 1/6 E(2)+ 1/6 E(3) + 1/2 E(4)
E(2) = 1 + 1/6 E(1) + 1/6 E(2)+ 2/3 E(3)
E(1) = 1 + 1/6 E(1) + 5/6 E(2)
E(0) = 1 + E(1)
The first five can be written as five equations in five unknowns:
1/6 E(1) + 1/6 E(2) + 1/6 E(3) + 1/6 E(4) - 5/6 E(5) = -1
1/6 E(1) + 1/6 E(2) + 1/6 E(3) - 5/6 E(4) + 1/3 E(5) = -1
1/6 E(1) + 1/6 E(2) - 5/6 E(3) + 1/2 E(4) = -1
1/6 E(1) - 5/6 E(2) + 2/3 E(3) = -1
-5/6 E(1) + 5/6 E(2) = -1
Solving results in E(1) = 52.8, so the expected number = E(0) = E(1) + 1 = 53.8

Quote: Gialmere

Also incorrect, but also hot on the trail.

Quote: ThatDonGuy

I think there's a bug in my equation solver - when I do it by hand, I get:

Answer

416/5 = 83.2

Found the problem in my equation solver: it panicked if, when trying to normalize row N, column N was already zero.

Official Solve

Shirt for Dice Lovers

Quote: teliot

For this question, watching this video is absolutely encouraged!

So here's the question: Is there a value x=t so that if Gabriel's Horn goes from x=1 to x=t, then the volume and surface area of Gabriel's Horn are equal? If so, can you solve for this value?

Quote: unJon

Cool video. Though it left my questioning why you can get away with using dx to integrate the volume instead of needing to use ds.

Quote: teliot

So here's the question: Is there a value x=t so that if Gabriel's Horn goes from x=1 to x=t, then the volume and surface area of Gabriel's Horn are equal? If so, can you solve for this value?

Quote: ThatDonGuy

Click Here for Math

The only value where they are equal is t = 1 (i.e. both volume and surface area are zero), and even this assumes that the disc of radius one at t = 1 is not included in the area.

The surface area = INTEGRAL[1, t] (2 PI / t) dt = 2 PI INTEGRAL[1, t] (1/t) dt = 2 PI (ln t - ln 1) = 2 PI ln t.
The volume = INTEGRAL[1, t] (PI / t^2) dt = PI INTEGRAL[1, t] (1/t^2) dt = PI (-1/t + 1/1) = PI (1 - 1/t).
These are equal when 2 ln t = 1 - 1/t.
2 ln 1 = 1 - 1/1, so they are equal when t = 1.
If t > 1, then they are equal when 2 ln t + 1/t - 1 = 0.
Let f(t) = 2 ln t + 1/t - 1
df/dt = 2/t + 1/t^2; since this is positive for all t > 1, f(t) is strictly increasing for all t > 1, which means f(t) > f(1).

If the disc is included, then the surface area = 2 ln t, and they are equal when ln t = -1/t, but for t > 1, ln t > 0 and -1/t < 0.

Quote: chevy

Is it a paradox?

I guess I have always disagreed with this as a "paradox". The problem I always had was with equating surface area to volume. Gabriel's Horn hides this in a calculation where we concentrate on one integral being finite but another is infinite. But consider the x-y plane. The surface area of that is infinite, but the volume of paint required to "paint" it is 0. In the case of Gabriel's Horn, you go out and buy your can of paint with volume, pi "units", you paint the inside (or outside) using up 0 "units", and you still have pi "units" of paint left to fill the horn with.

As to Teliot's original question, I agree with the calculations Teliot and ThatDonGuy discussed...but if I go back to comparing surface area to volume, does equating them have some physical or geometric meaning I am missing?

Quote: teliot

Quote: ThatDonGuy

Click Here for Math

The only value where they are equal is t = 1 (i.e. both volume and surface area are zero), and even this assumes that the disc of radius one at t = 1 is not included in the area.

The surface area = INTEGRAL[1, t] (2 PI / t) dt = 2 PI INTEGRAL[1, t] (1/t) dt = 2 PI (ln t - ln 1) = 2 PI ln t.
The volume = INTEGRAL[1, t] (PI / t^2) dt = PI INTEGRAL[1, t] (1/t^2) dt = PI (-1/t + 1/1) = PI (1 - 1/t).
These are equal when 2 ln t = 1 - 1/t.
2 ln 1 = 1 - 1/1, so they are equal when t = 1.
If t > 1, then they are equal when 2 ln t + 1/t - 1 = 0.
Let f(t) = 2 ln t + 1/t - 1
df/dt = 2/t + 1/t^2; since this is positive for all t > 1, f(t) is strictly increasing for all t > 1, which means f(t) > f(1).

If the disc is included, then the surface area = 2 ln t, and they are equal when ln t = -1/t, but for t > 1, ln t > 0 and -1/t < 0.

Your solution was my original thought on this. But now, I disagree about the surface area calculation you've given ... thinking out loud, if you replace your formula with "The Surface Area > ..." and the rest of your argument works.

Quote: ThatDonGuy

Quote: teliot

Quote: ThatDonGuy

Click Here for Math

The only value where they are equal is t = 1 (i.e. both volume and surface area are zero), and even this assumes that the disc of radius one at t = 1 is not included in the area.

The surface area = INTEGRAL[1, t] (2 PI / t) dt = 2 PI INTEGRAL[1, t] (1/t) dt = 2 PI (ln t - ln 1) = 2 PI ln t.
The volume = INTEGRAL[1, t] (PI / t^2) dt = PI INTEGRAL[1, t] (1/t^2) dt = PI (-1/t + 1/1) = PI (1 - 1/t).
These are equal when 2 ln t = 1 - 1/t.
2 ln 1 = 1 - 1/1, so they are equal when t = 1.
If t > 1, then they are equal when 2 ln t + 1/t - 1 = 0.
Let f(t) = 2 ln t + 1/t - 1
df/dt = 2/t + 1/t^2; since this is positive for all t > 1, f(t) is strictly increasing for all t > 1, which means f(t) > f(1).

If the disc is included, then the surface area = 2 ln t, and they are equal when ln t = -1/t, but for t > 1, ln t > 0 and -1/t < 0.

Your solution was my original thought on this. But now, I disagree about the surface area calculation you've given ... thinking out loud, if you replace your formula with "The Surface Area > ..." and the rest of your argument works.

I can see where the surface area in "if the disc is included" is wrong; it should be PI (2 ln t + 1). Is that what you were referring to?

Quote: chevy

My objection was 2D vs 3D, and not really an infinity issue. My objection still holds for a 1x1x1 cube....takes zero volume of paint to coat the inside(surface area), but 1 volume unit of paint to fill it. We wouldn't compare 6 "units" to coat it with 1 "unit" to fill it.

Quote: chevy

unJon:

Interesting comparison of using infinite paper to create a finite volume Gabriel's Horn, but not a sphere. To that end I concede, there are intrinsic relationships between surface area and volume.

I think I like your implied style "Can you make an object with infinite surface area and finite (non-zero) volume?" as a better wording than the video's "paint it" paradox.

My objection was 2D vs 3D, and not really an infinity issue. My objection still holds for a 1x1x1 cube....takes zero volume of paint to coat the inside(surface area), but 1 volume unit of paint to fill it. We wouldn't compare 6 "units" to coat it with 1 "unit" to fill it.

Quote: teliot

if you want an exact value for the surface area over a finite interval, you cannot disregard the term in the surface area integral that is disregarded in the video in the infinite upper limit case.

For example, find the limit x = t that gives a surface area equal to twice the volume. I have no idea how to do this.

Quote: ThatDonGuy

Maybe I am calculating it wrong, but I thought the surface area = the integral that sums the circumferences of the circles from x = 1 to x = t.
The circumference of the circle at x is 2 PI (1/x), whose integral with respect to x is 2 PI ln (x), so the integral over x = 1 to t is 2 PI ln t - 2 PI ln 1 = 2 PI (ln t - 0) = 2 PI ln t.
Then again, that method doesn't seem to work when I try it with calculating the surface area of a sphere...

Quote: teliot

watch the video @13:30 and you will see what's going wrong with your calculation method.

Quote: teliot

Then space filling curves should be equally objectionable. A one dimensional line filling two-dimensional space.

Quote: gordonm888

How can you fill the horn with paint without painting the interior surface?

Well, there is a parameter that we are missing and assuming is everywhere equal - the thickness of the coat of paint.

As the flare of the horn stretches out to infinity, the finite thickness of a coat of paint will eventually place some of the coat outside of the "defined volume" of the interior of the horn. Even if the thickness of the coat is one molecule in width, there is an infinite length of horn's flare that is closer to the "length" of the horn than one molecular width.

The problem is really in saying that we fill the horn with paint -because the paint volume becomes infinitely thin as the horn flares out to infinity. Thus, the paint itself must progressively become infinitely thin as the horn flares to infinity. It is this requirement for infinite thinning of the paint layer on the interior of horn that allows a finite volume of paint to cover an infinite surface area.

Quote: Gialmere

Hmm. I'm not sure what the status of this thread is but, here is a classic math puzzle (which I hope hasn't been posted before)...
There is a building of 100 floors
-If an egg drops from the Nth floor or above it will break.
-If it's dropped from any floor below, it will not break.
You're given 2 eggs.

How do you find N in the minimum number of drops?

Quote: Gialmere

How do you find N in the minimum number of drops?

Quote: charliepatrick

How do you find N in the minimum number of drops?

Here are various plans - starting with the worst!

Plan A: It seems reasonable that with one egg you could drop from floor 1, floor 2... and just work your way up until you find N. Average number of drops about 50.

With two eggs Plan B, not optimal, is to narrow that range to either 0-50 or 51-100 by dropping one from floor 50, and repeating as above. Average number of drops about 25 + 1. A similar plan is to go up in even numbers (dropping from floor 2,4,6 etc and then when the egg breaks see if it would have dropped from one floor lower. Similarly 25+1.

Plan C, not optimal, is to drop the egg from floor 33. If it breaks go from 0-32. If it doesn't try floor 66. This has a lower average.

Plan D, not optimal, is to drop egg from 10,20,30 etc then hone in on the last digit with the second egg. While this is quicker for low numbers, it takes quite a while with the higher ones.

Plan E, is the best I have found so far (although I don't know how to prove whether it's a minimum).

Suppose you try to ensure the worst case scenario is you needed 14 drops, then that might be better. So for instance if your first drop was on floor 14, then it could take a worst case of 14 to find the floor (14,1,2,3...13). However If you had more drops before the first egg broke, then you might want the gaps smaller. So one plan is drop an egg at 14,14+13=27,27+12=39 etc. and the work your way up. 14 27 39 50 60 69 77 84 90 95 98 seems a good plan and ensures the worst case is 14, I think this gives an average of about 10.35. (I've also ignored that once 99 doesn't break you know it's 100, so the average would be 10.34.)

Quote: Wizard

Are you asking for the minimum mean N or the minimum maximum N?

If it's the maximum N, then I get the same answer as Don. Looks like the maximum number of drops needed is ...

Show Spoiler

14

It's probably the same answer as for the mean N as well.