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12-03-2016 , 08:03 PM
#176
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Quote:
Originally Posted by xander biscuits
Isn't part of your circle in contradiction of X and Y not being the longest sides?
No, since the circle lies within the unit square, both X and Y are <= 1, which is the longest side length. The possibility of one of them also being 1 contributes nothing to the areas involved. If I could have put in a useful diagram, I could have explained my idea better.
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12-03-2016 , 08:07 PM
#177
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Quote:
Originally Posted by iraisetoomuch
#11

Spoiler:
Mix them together as the dissipation of heat is related to surface area, so having them together would reduce the total surface area exposed to the air, slowing the change in temperature?
Don't you want the cream to change temperature more rapidly and the coffee to change slowly?
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12-03-2016 , 09:23 PM
#178
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Would it have anything to do with the temperature differential between the coffee/cream and the room? Mixing first brings the temp differential down, and lessens the cooling rate?



I kind of just want to try this as an experiment.
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12-03-2016 , 09:24 PM
#179
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It lessens the cooling rate but also lessens the temperature

When coffee that's not mixed with cream's temperature lessens then so does the cooling rate of said coffee...
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12-03-2016 , 09:33 PM
#180
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Quote:
Originally Posted by xander biscuits
Puzzle #15
Spoiler:
1st set is a diamond inscribed in the garden. 2nd set is a square inscribed in the diamond.
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12-03-2016 , 09:42 PM
#181
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12-03-2016 , 11:39 PM
#182
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The triangle question doesn't really work unless you specify the distribution of how the points are selected. Having a uniform probability over all points from -infinity to infinity is not a well-defined distribution (see here).
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12-03-2016 , 11:42 PM
#183
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Puzzle 16:

A country implements a law where families have to keep having babies until they have a boy, then they have to stop. What would the ratio of boys to girls in the new baby population be?

Assume chance of boy vs girl is 50%
Last edited by housenuts; 12-03-2016 at 11:48 PM.
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12-03-2016 , 11:46 PM
#184
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#16

Spoiler:
Would be a ratio of 1:1 wouldn't it? (assuming that there is an equally likely chance to have a boy or a girl)
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12-03-2016 , 11:48 PM
#185
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I'm not smart enough to adjust the triangle question to overcome the infinity obstacle it seems
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12-04-2016 , 12:02 AM
#186
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Quote:
Originally Posted by xander biscuits
#16

Spoiler:
Would be a ratio of 1:1 wouldn't it? (assuming that there is an equally likely chance to have a boy or a girl)
Correct I believe.

Spoiler:
# of boys will always be 1
# of girls will be (1/4)*1 + (1/8)*2 + (1/16)*3 + (1/32)*4... and that series converges to 1
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12-04-2016 , 12:06 AM
#187
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Spoiler:
Or you could think of it as: Every child born has a 0.5 chance of being a boy and therefore half of the children will be boys.
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12-04-2016 , 12:08 AM
#188
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Quote:
Originally Posted by xander biscuits
I'm not smart enough to adjust the triangle question to overcome the infinity obstacle it seems
You can just make it a finite plane instead of infinite (i.e. x=0..1 and y=0..1). Then it just becomes a math exercise
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12-04-2016 , 12:12 AM
#189
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Quote:
Originally Posted by xander biscuits
Spoiler:
Or you could think of it as: Every child born has a 0.5 chance of being a boy and therefore half of the children will be boys.
Spoiler:
In theory, if all births stay 50:50. The problem gets a little more interesting if you consider that natural selection may favor families that have more girls :P
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12-04-2016 , 12:22 AM
#190
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Quote:
Originally Posted by DarkMagus
You can just make it a finite plane instead of infinite (i.e. x=0..1 and y=0..1). Then it just becomes a math exercise
Make it a finite plane instead then...
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12-04-2016 , 12:23 AM
#191
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Quote:
Originally Posted by xander biscuits
Spoiler:
Or you could think of it as: Every child born has a 0.5 chance of being a boy and therefore half of the children will be boys.
Spoiler:
but this type of thinking and the fact that families have to keep having babies until they have a boy and then stop, could lead someone to say "well 50% of the time they have to stop after the first baby, so 100% boys. If the first one is a girl, they have another and 50% that is a boy, so 75/25.
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12-04-2016 , 12:23 AM
#192
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Puzzle #17

I have 2 children. One day you meet one of my children who is a boy.

What is the probability that my other child is a girl?
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12-04-2016 , 12:38 AM
#193
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Quote:
Originally Posted by xander biscuits
Puzzle #17

I have 2 children. One day you meet one of my children who is a boy.

What is the probability that my other child is a girl?
Don't go there, man! Many years ago this puzzle generated so much virtual fighting; almost as bad as Monty Hall.
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12-04-2016 , 12:40 AM
#194
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Quote:
Originally Posted by xander biscuits
Puzzle #17

I have 2 children. One day you meet one of my children who is a boy.

What is the probability that my other child is a girl?
Spoiler:
2/3
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12-04-2016 , 12:44 AM
#195
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12-04-2016 , 01:13 AM
#196
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Quote:
Originally Posted by housenuts
Spoiler:
2/3
Quote:
Originally Posted by xander biscuits
Spoiler:
But, after I meet little dweezil biscuits, isn't he as likely to have a brother as he is to have a sister. Or you could have met either of two brothers, but only one boy with a sister, so that's 1/3 for a sister.


I'm not really trying to stir up trouble, but this puzzle has pitfalls, and any answer has to be carefully explained, including all explicit and implicit assumptions.
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12-08-2016 , 09:26 PM
#197
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Puzzle #18

You have 2 blank six-sided dice and you get to choose your own numbers to go on them.

Restrictions:

-Each face must display a unique number
-The sum total whenever you roll the 2 dice must be a prime number

Some research is allowed for this puzzle. You can search for lists of prime numbers etc but you obviously are not allowed to google the puzzle itself
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12-08-2016 , 09:56 PM
#198
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Lol. I got a few pages into the wiki on the Goldbach conjecture and I decided
'Ill just wait for the answer.'
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12-08-2016 , 10:20 PM
#199
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Quote:
Originally Posted by xander biscuits
Puzzle #18

You have 2 blank six-sided dice and you get to choose your own numbers to go on them.

Restrictions:

-Each face must display a unique number
-The sum total whenever you roll the 2 dice must be a prime number

Some research is allowed for this puzzle. You can search for lists of prime numbers etc but you obviously are not allowed to google the puzzle itself
I'm tired and don't think I can finish this off tonight, but I have it done for five-sided dice if someone wants to take this and find a 6th face for each:

Spoiler:
0, 2, 8, 12, 18
5, 11, 29, 71, 269
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12-08-2016 , 11:00 PM
#200
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Quote:
Originally Posted by Lattimer
I'm tired and don't think I can finish this off tonight, but I have it done for five-sided dice if someone wants to take this and find a 6th face for each:

Spoiler:
0, 2, 8, 12, 18
5, 11, 29, 71, 269
Spoiler:
32
431
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