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07-17-2012 , 06:01 PM
#201
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A woman has two children. One is a boy born on a Sunday. What is the probability she has two boys?
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07-17-2012 , 06:04 PM
#202
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Your solution does not work.
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07-17-2012 , 06:06 PM
#203
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And your question is more based on a misunderstanding of Bayesian statistics.

There is no correct answer it is up for interpretation.
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07-17-2012 , 06:06 PM
#204
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Quote:
Originally Posted by iversonian
A woman has two children. One is a boy born on a Sunday. What is the probability she has two boys?
Spoiler:
possible gender pairings of children
FF
MF
FM
MM

knowing that one of them is male limits us to 3 scenarios (dropping FF), only 1 of which the boy has a brother as his sibling. So 1/3
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07-17-2012 , 06:08 PM
#205
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explain how

Spoiler:
weigh 2, then weigh 2 others.

if they're same, weight the 3rd to get your answer

if they're different, weigh one of the ones you've weighed before.

if exactly half of what it was when you weighed them before, the coin is in other pile. you know the weight of normal coins to be 1/2 of the weight of the coin you just measured. if the other pile is high or low by x amount, you know that the odd coin is high or low by x amount.

if it's not exactly half, then the problem is one of those two. if the weight of the coin is exactly half of double the other pair, then it's correct, and you can figure out what the weight of the odd coin is. if it's not half, then it's the odd coin.
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07-17-2012 , 06:09 PM
#206
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Quote:
Originally Posted by ibavly
And your question is more based on a misunderstanding of Bayesian statistics.

There is no correct answer it is up for interpretation.
her comment was not spontaneous. it was in reply to a specific prompt about sons born on sunday.
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07-17-2012 , 06:11 PM
#207
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oh wait, i'm not sure that solution works
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07-17-2012 , 06:14 PM
#208
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also dean,

lame solving riddles you've already heard of
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07-17-2012 , 06:16 PM
#209
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dean, you didn't address the sunday part.

possible i didn't ask it properly. lemme think it over again, as it's been a while.
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07-17-2012 , 06:20 PM
#210
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ah, i remember.

only ONE of them is a boy born on sunday. what are the odds that both are boys.
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07-17-2012 , 06:31 PM
#211
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Spoiler:
so there are 49*4 combos. remove 49 combos of FF, 43 combos of MM, 84 combos of FM, you are left with 6/20=30%
Last edited by ibavly; 07-17-2012 at 06:34 PM. Reason: fixed numbers
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07-17-2012 , 06:34 PM
#212
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Quote:
Originally Posted by ibavly
also dean,

lame solving riddles you've already heard of
well i didn't know if this solution applied to this riddle

also lmk if this is right for the red/blue coin flipping thing:

Spoiler:
if there are m red coins, move m coins to its own pile and flip all of them over


Quote:
Originally Posted by iversonian
ah, i remember.

only ONE of them is a boy born on sunday. what are the odds that both are boys.
Spoiler:
i'm guessing that this is 6/13 since there are 14 day/gender pairings, and we know that child #2 is not sunday/male
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07-17-2012 , 06:35 PM
#213
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correct
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07-17-2012 , 06:56 PM
#214
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We should also taken into account the % of the time the babies are fraternal twins and in 100% of these cases the other one MUST be a girl, thus raising the probability the other kid is a girl slightly. lol
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07-18-2012 , 01:33 AM
#215
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fraternal twins don't have to be boy/girl
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07-18-2012 , 02:33 AM
#216
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Quote:
Originally Posted by ibavly
A couple more

You have 5 coins, one of them weighs a different amount. You have a standard electronic scale, and you can use it 3 times. How do you find out a - which coin is different b - how much does it weigh

0, 2, 24, 252, 3120, ... what comes next

9, 10, 11, 12, 13, 14, 21, 100, ...

Prove that N = n(n+1)(n+2)(n+3)+1 is a perfect square for any whole number n
ok i think i got it for real now

Spoiler:
weigh coins a+b

weigh coins b+c+d

if a+b * 3/2 = b+c+d then coin e is odd, weigh that.

otherwise, weigh coins b+c.

if b+c=a+b, coin d is odd. its weight is b+c+d - (a+b).

cleared coins d,e so far.

since d = (b+c+d)-(b+c)

if d*2 = a+b, coin c is odd. its weight is b+c+d - (a+b)

cleared coins c,d,e

if d*2 != a+b, then

if (b+c)*3/2 = (b+c+d), then coin a is odd. its weight is a+b - (b+c)/2

cleared a,c,d,e

if (b+c)*3/2 != (b+c+d), then coin b is odd. its weight is (a+b) - ((a+b)-(b+c))/2
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07-18-2012 , 02:54 AM
#217
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Looks good. General idea is

Spoiler:
Make sure there is no coin there are no 2 coins that are in all the same equations. Bunch of different ways of doing that
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07-18-2012 , 03:04 AM
#218
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Quote:
Prove that N = n(n+1)(n+2)(n+3)+1 is a perfect square for any whole number n
Spoiler:

because n(n+1)(n+2)(n+3)+1 can be rewritten as:

(n^2 + 3n + 1)^2

which is obv a perfect square

i didnt actually factor it. i just looked at the pattern and worked backwards
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07-18-2012 , 07:57 AM
#219
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Quote:
Originally Posted by PyramidScheme
We should also taken into account the % of the time the babies are fraternal twins and in 100% of these cases the other one MUST be a girl, thus raising the probability the other kid is a girl slightly. lol
Wat
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07-31-2012 , 05:11 PM
#220
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Quote:
Originally Posted by Aksdal
fraternal twins don't have to be boy/girl
For example; Mary-Kate and Ashley are fraternal twins of the same gender.
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07-31-2012 , 05:32 PM
#221
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Quote:
Originally Posted by pwnsall
What's in my pocket?
not a fair quesssstion, precioussss!
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07-31-2012 , 05:34 PM
#222
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this question is from a math talent search thing from when I was in high school:

Prove there are no 2 integers, the sum of whose squares is 12,345,678
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07-31-2012 , 07:29 PM
#223
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Quote:
Originally Posted by Sun Tzu
this question is from a math talent search thing from when I was in high school:

Prove there are no 2 integers, the sum of whose squares is 12,345,678
I guess solving this algorithmically is against the spirit of the question.
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04-14-2015 , 06:45 PM
#224
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excellent find jh

there was a 2006 thread too but it's probably in the archives
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04-14-2015 , 07:36 PM
#225
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Spoiler:
call the numbers x and y

since 12,345,678 is an even number, they are both odd or both even and the difference between them is even

so y = x + 2n where n is a natural number

x2 + y2 = 12,345,678

2x2 + 4nx + 4n2 = 12,345,678

x2 + 2nx + 2n2 = 6,172,839

so x (and y) has to be odd

let x be 2k + 1 where k is a natural number

(2k+1)2 + 2n(2k+1) + 2n2 = 6,172,839

4k2 + 4k +1 + 4kn + 2n + 2n2 = 6,172,839

4k2 + 4k + 4kn + 2n + 2n2 = 6,172,838

2k2 + 2k + 2kn + n + n2 = 3,086,419

n(n +1) is odd, which can't be true for integers so our initial assumption has led to a contradiction and is therefore false
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