[quote]dollarbill44 wrote:
What if, when you open the envelope, instead of money, there is a baby rattlesnake?
DB[/quote]
If it’s the first envelope, I think naturally you would choose to switch.
[quote]JLu wrote:
dollarbill44 wrote:
What if, when you open the envelope, instead of money, there is a baby rattlesnake?
DB
If it’s the first envelope, I think naturally you would choose to switch.[/quote]
That sounds like a good idea, until you wind up with two baby rattlesnakes.
[quote]dollarbill44 wrote:
What if, when you open the envelope, instead of money, there is a baby rattlesnake?
DB[/quote]
Hilarious.
Luckily, they don’t bite as hard as grown ass rattlers.
[quote]wushu_1984 wrote:
Maths geek to the rescue:
Let X denote your monetary return in the game.
X = 10 if you don’t switch
X = 5 or 20 if you do switch
i.e. either the envelopes are {5,10} or {10,20}.
Now if we don’t switch we always get 10.
If we switch our expected returns (i.e. average returns in repeated experiments) are:
E = 5Prob({5,10} occurs)+20Prob({10,20} occurs)
Now without better information we will assume that {5,10} and {10,20} are equally likely for the envelopes, i.e.
Prob({5,10} occurs)=Prob({10,20} occurs)=1/2
So,
E = 51/2+201/2 = 12.5 > 10
So as a strategy for maximising the expected returns over repeat occurences one should always switch.
Intuition: Half the time when you switch you’ll get 5 dollars the other half the time you get 20 dollars so on average you get:
5/2+20/2 = 12.5[/quote]
Well, I was just going to repeat the last paragraph, but yeah; switching wins.
[quote]jtrinsey wrote:
So do you switch or not?[/quote]
Always switch
Where x is what’s in your hand:
If you switch and lose you lose only 1/2 x to end with 1/2x
If you switch and win you win another x to end with 2x
Saying the same gets you x
So:
50% you get 1/2 x
and 50% you get 2x
Then the average amount is 1.25x with always switching
Always staying the same gets you 1x
Assuming an even distribution of x, we can average x to be the same number for all tries with enough attempts
So always switch
[quote]sen say wrote:
LankyMofo wrote:
Nevermind, I get it.
It’s good to know you can admit when you’re wrong. Now head over to Funniest Guy on T-Nation and mop up the mess you created there.[/quote]
Man, I thought we got over that!
Here’s a much harder one I’ve come across.
There are a thousand switches numbered from 1 to 1000 in a long corridor. There are a thousand people wearing t-shirts numbered from 1 to 1000 at one end of the corridor.
All the switches start out in the “off” position.
Each person walks down the corridor looking at each of the 1000 switches along the way. The person hits a switch (i.e. changes it from “off” to “on” if it is “off” or changes it from “on” to “off” if it is “on”) if the number on their t-shirt divides evenly into the number on the switch. For example the person with 1000 on their t-shirt would only hit switch 1000.
After all 1000 people have walked down the corridor hitting all relevant switches along the way how many switches are in the “on” position?
PS I actually got asked this question in a job interview!
Is there a reason this fucking thread says (No Goats)?
so i have 10 dollars
if i switch and lose i will have 5 dollars, difference of 5
if i switch and gain i will have 20 dollars, difference of 10
i would switch because its a lower risk/greater reward bet
[quote]wushu_1984 wrote:
Here’s a much harder one I’ve come across.
There are a thousand switches numbered from 1 to 1000 in a long corridor. There are a thousand people wearing t-shirts numbered from 1 to 1000 at one end of the corridor.
All the switches start out in the “off” position.
Each person walks down the corridor looking at each of the 1000 switches along the way. The person hits a switch (i.e. changes it from “off” to “on” if it is “off” or changes it from “on” to “off” if it is “on”) if the number on their t-shirt divides evenly into the number on the switch. For example the person with 1000 on their t-shirt would only hit switch 1000.
After all 1000 people have walked down the corridor hitting all relevant switches along the way how many switches are in the “on” position?
PS I actually got asked this question in a job interview![/quote]
lol
did they at least give you a calculator?
i can tell you off the bat the only numbers that will go in are number thats can be factored down to 2 and 5.
i guess my answer didnt go up
but id switch
if you have 10 dollars and one envelope is 2x the amount than the other envelope is either 5 or 20 dollars (2x5=10 and 2x10 = 20)
so if you do have the envelope of greater value your risk is a 5 dollar loss
however if you do not have the envelope and switch you have a 10 dollar gain
so it isnt about a better chance, but its a smarter move to make the switch because worst that could happen you lose 5 dollars but since the gain is greater than the risk it is worth it.
it doesnt take a fancy equation to solve this, its common sense.
i take a math course right now that is pretty much all questions like this.
[quote]LiveFromThe781 wrote:
lol
did they at least give you a calculator?
i can tell you off the bat the only numbers that will go in are number thats can be factored down to 2 and 5.[/quote]
Nope no calculator just pen and paper. The interviewer had asked me about 3 or 4 easy enough maths questions, kinda like the one the OP asked. I got all of them right. So then she got this smug look on her face and asked me this question. At first I was like WTF! Then I figured it out and she was like WTF!
Didn’t get the job in the end. Apparently too technically minded! I guess I wasn’t supposed to figure out the question. Oh well…
[quote]wushu_1984 wrote:
Here’s a much harder one I’ve come across.
There are a thousand switches numbered from 1 to 1000 in a long corridor. There are a thousand people wearing t-shirts numbered from 1 to 1000 at one end of the corridor.
All the switches start out in the “off” position.
Each person walks down the corridor looking at each of the 1000 switches along the way. The person hits a switch (i.e. changes it from “off” to “on” if it is “off” or changes it from “on” to “off” if it is “on”) if the number on their t-shirt divides evenly into the number on the switch. For example the person with 1000 on their t-shirt would only hit switch 1000.
After all 1000 people have walked down the corridor hitting all relevant switches along the way how many switches are in the “on” position?
PS I actually got asked this question in a job interview![/quote]
Also just to explain it better. For example:
- The person with 23 on their t-shirt would hit switches 23,46,69,92,115, … , 989.
- The person with 87 on their t-shirt would hit switches 87,174,261, … , 957.
So the person only hits the switches from 1 to 1000 that the number on his t-shirt divides evenly.
Then we are interest what happens after everyone with t-shirts 1,2,3,4, … , 1000 has already walked down the corridor and hit all the relevant switches. Then the question is after all this which switches are in the “on” position?
the only ones that would go in would be like
2, 5, 10, 25, 50, 100, 125, 150, 200, 250, 500 and 1000
i think
so itd be 12 on and 988 off
it doesnt really seem hard if you know the rules of divisibility.
[quote]wushu_1984 wrote:
wushu_1984 wrote:
Here’s a much harder one I’ve come across.
There are a thousand switches numbered from 1 to 1000 in a long corridor. There are a thousand people wearing t-shirts numbered from 1 to 1000 at one end of the corridor.
All the switches start out in the “off” position.
Each person walks down the corridor looking at each of the 1000 switches along the way. The person hits a switch (i.e. changes it from “off” to “on” if it is “off” or changes it from “on” to “off” if it is “on”) if the number on their t-shirt divides evenly into the number on the switch. For example the person with 1000 on their t-shirt would only hit switch 1000.
After all 1000 people have walked down the corridor hitting all relevant switches along the way how many switches are in the “on” position?
PS I actually got asked this question in a job interview!
Also just to explain it better. For example:
- The person with 23 on their t-shirt would hit switches 23,46,69,92,115, … , 989.
- The person with 87 on their t-shirt would hit switches 87,174,261, … , 957.
So the person only hits the switches from 1 to 1000 that the number on his t-shirt divides evenly.
Then we are interest what happens after everyone with t-shirts 1,2,3,4, … , 1000 has already walked down the corridor and hit all the relevant switches. Then the question is after all this which switches are in the “on” position?[/quote]
ooo wtf i thought it was what numbers divide 1000 evenly.
she really expexted you to sit there and list every number, 1 through 1,000 and then the numbers that it would divide… and you did it? how long did that take like 45 minutes?
wait wouldnt all the switches be on? if you have numbers 1-1,000 and people with shirts 1-1,000 then everyone could just stand by their number
i must be interperting the question wrong b/c every answer seems way too easy.
hmmm
Is it, the only ones that are on are squares?
So 4, 9, 16, … 1000 etc?
So the number of squares <= 1000
edit: the reason being numbers that are squares of others have an odd number of factors, while every other number has an even number of factors. At least I think so 
I just went through a few under 20 and realized everything had an even number of factors except certain numbers, and they happened to be squares.
[quote]chrisb71 wrote:
hmmm
Is it, the only ones that are on are squares?
So 4, 9, 16, … 1000 etc?
So the number of squares <= 1000
edit: the reason being numbers that are squares of others have an odd number of factors, while every other number has an even number of factors. At least I think so I just went through a few under 20 and realized everything had an even number of factors except certain numbers, and they happened to be squares.[/quote]
its numbers you can factor to 2 and 5 or exponets of 2 and 5 like 2^2 5^2 etc
this is because 2 and 5 are the only things that can divide 10 evenly.
so like 25 once u break it down u get 5^2
50 you get 5^2 * 2
100 would be 5^2 * 2^2 and it keeps going
you can also use that information tell if a fraction is going to be terminating or irrational.
[quote]chrisb71 wrote:
hmmm
Is it, the only ones that are on are squares?
So 4, 9, 16, … 1000 etc?
So the number of squares <= 1000
edit: the reason being numbers that are squares of others have an odd number of factors, while every other number has an even number of factors. At least I think so I just went through a few under 20 and realized everything had an even number of factors except certain numbers, and they happened to be squares.[/quote]
Yes, the only ones turned on after all is said and done will be the squares (don’t forget “1”). It’s because every number has an even number of factors – for every factor, there has to be another number to multiply it by – except the squares because one of their factors is multiplied by itself.
So, for every flip “on”, the next factor will flip it back “off”.
Cool puzzle.
[quote]malonetd wrote:
chrisb71 wrote:
hmmm
Is it, the only ones that are on are squares?
So 4, 9, 16, … 1000 etc?
So the number of squares <= 1000
edit: the reason being numbers that are squares of others have an odd number of factors, while every other number has an even number of factors. At least I think so I just went through a few under 20 and realized everything had an even number of factors except certain numbers, and they happened to be squares.
Yes, the only ones turned on after all is said and done will be the squares (don’t forget “1”). It’s because every number has an even number of factors – for every factor, there has to be another number to multiply it by – except the squares because one of their factors is multiplied by itself.
So, for every flip “on”, the next factor will flip it back “off”.
Cool puzzle.[/quote]
Yep you’re both absolutely right. Just to finish it; the number of squares less than 1000 is the integer part of sqrt(1000) which is 31, since 31^2 < 1000 < 32^2.
So there will be 31 switches in the “on” position when all is said and done.