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07-04-2011 , 03:59 PM
#51
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awesome thread, keep it up!
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07-04-2011 , 04:05 PM
#52
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hmm I'm sure others have good ones. You can post yours itt gabe, I'm sure most people haven't seen them.

Haven't used this one in a long time, hopefully I remember it right:

A man goes over to his neighbors house and ask him how old his 3 kids are. He tells the man that if you multiply their ages together, you get 36, if you add their ages together, you get his house number. The man thinks about it for a bit, and responds that he still needs more information. The neighbor then says, "my eldest son plays the violin". How old are the 3 kids?
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07-04-2011 , 04:10 PM
#53
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Quote:
Originally Posted by ibavly
vn, there are multiple solutions but they all follow the general concept of making sure you do not ask the randomizer Q#2.

An easy one (hopefully it will be clear!):

You have two identical glass balls, and a skyscraper with 100 floors. You're curious to know what is the highest floor you can drop the ball from without it breaking. Obviously if both balls break and you do not yet know the breaking point you lose. Your goal is to find the smallest number of drops you can use to guarantee you will find the breaking point.

Ex. one plan of action would be to start on floor 1 and keep moving up. that way when the ball eventually breaks you will know the answer. This could take up to 100 tries. Another possibility is to drop first from floor 50. If it doesn't break you start trying higher floors. If it does you are forced to start on floor 1 and go up one at a time since otherwise you would not be guaranteed finding the breaking point. This could take up to 50 turns.
Spoiler:
1+2+3+4+5+6+7+8+9+10=55+11=66+12=78+13=91+14=105.
I looked for the first number that made the sum >100. This number is 14. I think this is the minimal number of throws.
You start throwing from floor 14. If it breaks you need to go 1-13 = maximum 14 throws.

If the ball does not break from floor 14, go up 13 floors (to27). If it breaks, you need to go 15-26 = 12 throws + the throw from 27 + throw from 14=14 throws again.

Every time you go up one less floor and the maximum amount of throws remains at 14.
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07-04-2011 , 04:15 PM
#54
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Spoiler:
36 factorised -> 1 2 3 4 6 9 12 18 36

combination whose product = 36 with equal sum -> 9,2,2 ; 6,6,1

since the oldest son can play a violin, his age must be > pi

so the kids are 9,2 and 2.
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07-04-2011 , 04:20 PM
#55
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you are correct I'm not sure what you're talking about wrt pi
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07-04-2011 , 04:29 PM
#56
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Quote:
Originally Posted by ibavly
hmm I'm sure others have good ones. You can post yours itt gabe, I'm sure most people haven't seen them.

Haven't used this one in a long time, hopefully I remember it right:

A man goes over to his neighbors house and ask him how old his 3 kids are. He tells the man that if you multiply their ages together, you get 36, if you add their ages together, you get his house number. The man thinks about it for a bit, and responds that he still needs more information. The neighbor then says, "my eldest son plays the violin". How old are the 3 kids?
Spoiler:
The first two questions show the ages have to be either 6, 6 and 1 or 9, 2 and 2, as the only set of factors that add up to the same number. The violin statement implies the oldest child must be at least fairly old so the kids are 9, 2 and 2.

I'm not a huge fan of the problem since it's not impossible that a 6 year old would play the violin but in the context it is at least obvious what the solution is meant to be.
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07-04-2011 , 04:31 PM
#57
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Quote:
Originally Posted by ibavly
hmm I'm sure others have good ones. You can post yours itt gabe, I'm sure most people haven't seen them.

Haven't used this one in a long time, hopefully I remember it right:

A man goes over to his neighbors house and ask him how old his 3 kids are. He tells the man that if you multiply their ages together, you get 36, if you add their ages together, you get his house number. The man thinks about it for a bit, and responds that he still needs more information. The neighbor then says, "my eldest son plays the violin". How old are the 3 kids?
Spoiler:
2,2,9

If the house number is 13 that POE's it to 1,6,6 and 2,2,9 and distinguishing there being an eldest makes it 2,2,9
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07-04-2011 , 04:32 PM
#58
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Quote:
Originally Posted by Willd
Spoiler:
The first two questions show the ages have to be either 6, 6 and 1 or 9, 2 and 2, as the only set of factors that add up to the same number. The violin statement implies the oldest child must be at least fairly old so the kids are 9, 2 and 2.

I'm not a huge fan of the problem since it's not impossible that a 6 year old would play the violin but in the context it is at least obvious what the solution is meant to be.
Spoiler:
the violin is actually a red herring. The information he is giving is that he has an eldest son, they are not twins.
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07-04-2011 , 04:36 PM
#59
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Quote:
Originally Posted by Willd
Spoiler:
The first two questions show the ages have to be either 6, 6 and 1 or 9, 2 and 2, as the only set of factors that add up to the same number. The violin statement implies the oldest child must be at least fairly old so the kids are 9, 2 and 2.

I'm not a huge fan of the problem since it's not impossible that a 6 year old would play the violin but in the context it is at least obvious what the solution is meant to be.
Spoiler:
I take the wording to mean that "older" implies not of the same age, so the fact that there is an older child is the important part, not that he plays the violin. Yeah, if you wanna nitpick he could have two twins where one technically was 2 minutes older or two kids where one was 9 months older.
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07-04-2011 , 04:37 PM
#60
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Quote:
Originally Posted by ibavly
Spoiler:
the violin is actually a red herring. The information he is giving is that he has an eldest son, they are not twins.
Spoiler:
I considered it but rejected it as being even stupider since two children of the same age don't have to be twins and one could easily be elder.
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07-04-2011 , 04:39 PM
#61
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Quote:
Originally Posted by Willd
Spoiler:
I considered it but rejected it as being even stupider since two children of the same age don't have to be twins and one could easily be elder.
Spoiler:
The point is that the neighbor felt that giving away this information would help solve the problem. If the kids were 9 months apart or if he considered one of the twins to be specifically his eldest, why would he even make the statement?
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07-04-2011 , 04:42 PM
#62
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Quote:
Originally Posted by ibavly
Spoiler:
The point is that the neighbor felt that giving away this information would help solve the problem. If the kids were 9 months apart or if he considered one of the twins to be specifically his eldest, why would he even make the statement?
Spoiler:
The bolded is a good point actually, I hadn't considered it from that angle.
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07-04-2011 , 05:35 PM
#63
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Quote:
Originally Posted by TheDean1
Feels like there should be one of these where people can just drop random brainteasers and try to answer them and whatnot. I'll go first:

In celebration of Mets's 75k post, he and HiFi decide to host a party and invite 10 other couples (22 total ppl at the shin dig) . Throughout the course of the party, Mets decided to keep track of the handshaking going on and surveyed the entire party (i.e. everyone including HiFi but not himself). Note that although Mets isn't part of the data sample directly, somebody else shaking his hand does count toward their respective tally. He found 2 noteworthy facts

1) Everybody that he surveyed shook a different number of hands
2) Nobody shook hands with their spouse/significant other/etc

How many hands did HiFi shake?
Spoiler:
So I worked out intuitively that this pretty much had to be 10 but had no idea how to go about proving it. I've proved it to myself using a spreadsheet but putting it into words is tricky.

Basically though the number of hands shaken by each person is from 0,1...20 and the person who shook 0 hands must be the spouse of the person who shook 20 hands, the person who shook 1 hand must be the spouse of the person who shook 19 etc, leaving just the person who shook 10 hands left spouseless in the sample data and therefore must be HiFi.
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07-04-2011 , 05:39 PM
#64
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ibav, gabe, and willd all were correct with their solution to my problem fwiw
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07-05-2011 , 12:57 AM
#65
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How many times in a day do the 2 arms on a standard clock point at exactly the same direction?
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07-05-2011 , 01:15 AM
#66
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Spoiler:
So why isn't the eldest son 36?
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07-05-2011 , 01:19 AM
#67
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Quote:
Originally Posted by bojtun
Spoiler:
So why isn't the eldest son 36?
Spoiler:
Or 18?


Something is missing from the question.
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07-05-2011 , 01:26 AM
#68
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The ages add up to the house number, but the person needs more information. Which means there are >1 set of numbers whose sum is the house number.
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07-05-2011 , 04:02 AM
#69
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We play a game.

For starters I pay you X$.

After that you start throwing a coin.
If it comes heads the first time - you pay me $1
If it comes tails the first time and heads the second time - you pay me $2
If it comes tails the first two times and heads the third time - you pay me $4
If it comes tails the first three times and heads the fourth time - you pay me $8
etc.

The game ends when you throw heads.

How much do you want me to pay you to play this game? What is a fair amount for X?
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07-05-2011 , 05:56 AM
#70
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that game has infinite EV for you so i guess i want you to pay me wan exozilian dorrars
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07-05-2011 , 06:27 AM
#71
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Quote:
Originally Posted by Gabethebabe
We play a game.

For starters I pay you X$.

After that you start throwing a coin.
If it comes heads the first time - you pay me $1
If it comes tails the first time and heads the second time - you pay me $2
If it comes tails the first two times and heads the third time - you pay me $4
If it comes tails the first three times and heads the fourth time - you pay me $8
etc.

The game ends when you throw heads.

How much do you want me to pay you to play this game? What is a fair amount for X?
Spoiler:
The EV of the game is the sum of 1/2 * 1 + 1/4 * 2 + 1/8 * 4 etc, which amounts to just 1/2 + 1/2 + 1/2 + ... which is infinite. Therefore theoretically you could pay me any amount and still be +EV in the game.
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07-05-2011 , 06:57 AM
#72
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Good answers. Anti-intuitive if you see the setup, but in the looooong run my EV is inf.
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07-05-2011 , 07:38 AM
#73
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but in the short run your EV is inf.
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07-05-2011 , 07:45 AM
#74
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Quote:
Originally Posted by theHuntContinues
but in the short run your EV is inf.
You want to play 10 times, you pay me $250 every game?
Unless the sample size of the games you play is really big, you will end up losing (significant) money like 99% of the time.
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07-05-2011 , 08:11 AM
#75
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That means the variance is large (infinite in this case) but the EV is still the same.
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