Easy math puzzles

Quote: ThatDonGuy

Quote: Wizard

What is the expected number of spins in double-zero roulette to see five reds or five blacks in a row? The previous spin landed on 0.
link to original post

Answer

Let E(n) be the expected number after having n consecutive of a color; E(5) = 0
Let p be the probability of getting a particular color, and r = 1 - 2p = the probability that it comes up green

E(4) = 1 + p E(1) + r E(0)

E(3) = 1 + p E(1) + r E(0) + p E(4)
= 1 + p E(1) + r E(0) + p (1 + p E(1) + r E(0))
= 1 + p + (p + p^2) E(1) + (r + pr) E(0)

E(2) = 1 + p E(1) + r E(0) + p E(3)
= 1 + p E(1) + r E(0) + p (1 + p + (p + p^2) E(1) + (r + pr) E(0))
= (1 + p + p^2) + p (1 + p + p^2) E(1) + r (1 + p + p^2) E(0)

E(1) = 1 + p E(1) + r E(0) + p E(3)
= (1 + p + p^2 + p^3) + p (1 + p + p^2 + p^3) E(1) + r (1 + p + p^2 + p^3) E(0)
(1 - p - p^2 - p^3 - p^4) E(1) = (1 + p + p^2 + p^3) + (1 - 2p) (1 + p + p^2 + p^3) E(0)
E(1) = (1 + p + p^2 + p^3) / (1 - p - p^2 - p^3 - p^4) + ((1 - 2p) (1 + p + p^2 + p^3) / (1 - p - p^2 - p^3 - p^4)) E(0)

E(0) = 1 + (1 - 2p) E(0) + 2p E(1)
E(0) = 1 / 2p + E(1)
E(0) = 1 / 2p + (1 + p + p^2 + p^3) / (1 - p - p^2 - p^3 - p^4) + ((1 - 2p) (1 + p + p^2 + p^3) / (1 - p - p^2 - p^3 - p^4)) E(0)
(1 - (1 - 2p) (1 + p + p^2 + p^3) / (1 - p - p^2 - p^3 - p^4)) E(0) = 1 / 2p + (1 + p + p^2 + p^3) / (1 - p - p^2 - p^3 - p^4)
E(0) = (1 / 2p + (1 + p + p^2 + p^3) / (1 - p - p^2 - p^3 - p^4)) x (1 - p - p^2 - p^3 - p^4) / p^4
E(0) = (1 - p - p^2 - p^3 - p^4) / (2 p^5) + (1 + p + p^2 + p^3) / p^4
E(0) = (1 - p - p^2 - p^3 - p^4) / (2 p^5) + (2p + 2p^2 + 2p^3 + 2p^4) / (2p^5)
= (1 + p + p^2 + p^3 + p^4) / (2p^5)
= (1 - p^5) / (2 p^5 (1 - p))

For p = 9/19, this is 4,592,395 / 118,098, or about 38.8863

link to original post

Quote: teliot

What I finds odd about this problem is this related problem: what are the chances that starting with the present spin you get a sequence of exactly 5 reds or 5 blacks?

The answer is, of course, (36/38)*(18/38)^4*(20/38) = 0.025103 or 1-in-39.836. I find it odd that this is even close to the other question Mike asked (the expected number of spins to get a sequence Red or Black of length 5). I was almost going to go with 39.836 as my answer.
link to original post

Quote: charliepatrick

no calculators used, but a glass of wine was!

Person N gets N/100 of what's left. In which case person (N+1) gets (N+1)/100*(100-N)/100. This means 100N=100N+100-N²-N which means N²+N-100=0. Using N=-b2+/-SQRT(etc) gives -1+-SQRT(1+400)/2 which is about 10.
When N=10 the 10th person gets 10% and the 11th person get 11% of 90% = 9.9%. Similarly the 9th person gets 9% and the 10th person gets 10% of 91% = 9.1%. Confirming the 10th person gets the most.

link to original post

Quote: Wizard

It takes Alice and Bill 3 hours to paint a house.
It takes Alice and Cindy 4 hours to paint a house.
It takes Bill and Cindy 5 hours to paint a house.

How long does it take if they all paint?
link to original post

Quote: Wizard

It takes Alice and Bill 3 hours to paint a house.
It takes Alice and Cindy 4 hours to paint a house.
It takes Bill and Cindy 5 hours to paint a house.

How long does it take if they all paint?
link to original post

Quote: ThatDonGuy

Easy one today

Let A, B, and C be the amount of the house Alice, Bill, and Cindy can paint in an hour
A + B = 1/3
A + C = 1/4
B + C = 1/5
2 (A + B + C) = 1/3 + 1/4 + 1/5 = 47/60
A + B + C = 47/120
It takes all three 120/47, or 2 26/47, hours

link to original post

Quote: MichaelBluejay

How about a variant?

It takes Alice 3 hours to paint a house.
It takes Bill 4 hours to paint a house.
It takes Cindy 5 hours to paint a house.

How long does it take if they all paint?

Also, I hope I can hire any of them at a normal hourly rate for a painter, because their throughput is phenomenal.
link to original post

Quote: MichaelBluejay

This is from my local weekly newspaper's math column. I didn't get this one, because I'm not good with spatial concepts.

Flip the triangle upside-down by moving only 3 balls.

⬤⬤⬤⬤
 ⬤⬤⬤
  ⬤⬤
   ⬤

Solution

Move the top-left and top-right balls down to the row of 2, making it a row of 4.
Then move the single ball at the bottom to the very top.

link to original post

Quote: aceside

This solution is a little confusing. To be clearer, I would say, move the top-left and top-right balls 2 rows down making it the bottom row.

Quote: MichaelBluejay

This is from my local weekly newspaper's math column. I didn't get this one, because I'm not good with spatial concepts.

Flip the triangle upside-down by moving only 3 balls.

⬤⬤⬤⬤
 ⬤⬤⬤
  ⬤⬤
   ⬤

link to original post

Show Spoiler

Quote: Wizard

Here is another way of putting the solution.

Quote: charliepatrick

Seats in a Cafe

I was sitting in a local cafe earlier this week and noticed that all the chairs were either red, black or white. Also each table had four seats. It got me thinking if all the tables had been circular (i.e. a seating arrangement RRBW is the same as RBWR etc.) and every table had four chairs with a different arrangement from any other table, what would be the largest possibe cafe size.

In other words how many possible ways are there to arrange four chairs, each being either Red, Black or White, around a circular table.

I think it's fair to say that the answer is bigger than the number of tables my local cafe actually has!!
link to original post

Quote: MichaelBluejay

Diabetes

Um, 2/3? Is it as simple as taking the average of 7 and 9, and dividing by 12?

link to original post

Quote: chevy

Is wine healthier than syrup?

My guess assuming the bottom 7 inches are all cylindrical....

Turn it upside down and the first 5" is what was above the previous 7" of wine.
The next 4" are what would have been the top 4" of the previous 7" of wine.

So the first 5" (when inverted) must be the same volume as the bottom 3" when right side up.

So if just a cylinder, it would be 3" + 4" (original 7") + 3" (the equivalent of the 5" inverted)

So original is 7/10 = 70%

But then again I don't like wine, so what do I know.

link to original post

Quote: Wizard

Suppose you have a partially full bottle of pancake syrup in an unusually shaped bottle.

The bottle is 12" high.

When the cap it at the top, the syrup makes it to 7" point.

When the bottle is turned upside down, the syrup makes it to the 9" point.

What ratio of the bottle is full?
link to original post

Quote: Wizard

A rubber band is wrapped around two circles. The large circle has radius 5. The small circle has radius 2.

What is the distance of the rubber band?

For purposes of discussion, feel free to use...

this diagram

θ is an angle.

link to original post

Quote: chevy

quick try

mine based on your original drawing and labels.....


d = b = 2
a+b = 5+2 = 7
d+e = 2+e = 5 .... so e = 3

So c^2 = 7^2 - 3^2 = 40.....c=sqrt(40)=2sqrt(10)

There are 2 "c" legs for rubber band

Angle t = arccos(3/7) (in radians)
So angle of rubber stretched on big circle = (2pi - 2t)
So length of rubber band around big circle = (2pi-2t)*5 = 10pi - 10arccos(3/7)

small angle in Wiz's triangle is arcsin(3/7) (in radians)
So angle of rubber stretched on small circle = 2pi - 2*(pi/2 + arcsin(3/7)) = pi-2arcsin(3/7)
So length of rubber band around small circle = (pi-2arcsin(3/7))*2

total = 4sqrt(10) + (10pi-10arccos(3/7)) + (2pi - 4arcsin(3/7))

= 4 sqrt(10) + 12pi - 10arccos(3/7) - 4arcsin(3/7)

(+/- typos....and if you need something without arches or arcsin....I wave the red flag of submission)

link to original post

• If rock beats scissors, scissors pays rock $1
• If scissors beats paper, paper pays scissors $2
• If paper beats rock, rock pays paper $3
• No money changes hands on a tie

Quote: Wizard

This one is probably not "easy," so all due apologies for putting it here.

Consider a game of rock-paper-scissors where:

• If rock beats scissors, scissors pays rock $1
• If scissors beats paper, paper pays scissors $2
• If paper beats rock, rock pays paper $3
• No money changes hands on a tie

Assume two logicians are playing. What is the optimal strategy for each?
link to original post

Quote: ThatDonGuy

My answer

Let p be the probability of playing paper, and s the probability of playing scissors; (1 - p - s) is the probability of playing rock

Expected values of the strategy, if the opponent plays:
Rock: 3p - s
Paper: 2s - 3(1 - p - s) = 5s + 3p - 3
Scissors: (1 - p - s) - 2p = 1 - s - 3p

Assuming an equilibrium (i.e. all EVs = 0):
5s + 3p - 3 = 3p - s -> s = 1/2
1 - s - 3p = 3p - s -> p = 1/6
Check:
Rock = 3p - s = 0
Paper = 5s + 3p - 3 = 0
Scissors = 1 - s - 3p = 0

Each player rolls a 6-sided die, and plays scissors on 1-3, rock on 4-5, and paper on 6

link to original post

Quote: aceside

I recently played a 6-deck automatic-shuffle blackjack game. Everything was normal except one thing, that is, 3 of the 6 decks of cards had a red back and the remaining 3 decks of cards had a blue back. Can a player gain some EV edge from this game by taking into account the card back color information?
link to original post

Quote: aceside

I recently played a 6-deck automatic-shuffle blackjack game. Everything was normal except one thing, that is, 3 of the 6 decks of cards had a red back and the remaining 3 decks of cards had a blue back. Can a player gain some EV edge from this game by taking into account the card back color information?
link to original post

Quote: Wizard

I apologize if I've asked this before.

A prisoner is presented with three closed doors, labeled A, B, and C. The warden explains that behind two doors is freedom and behind one is immediate death. The prisoner may point to one door and ask one yes/no question of the warden.

If the prisoner points towards a door leading to freedom, the warden will answer truthfully. If the prisoner points towards a door leading to death, the warden will answer yes or no randomly.

What should the prisoner do?
link to original post

Quote: avianrandy

I thought this question sounded familiar so I used the search function here. You asked it March of last year. It is on page 262 of this thread. It is still a good question though
link to original post