Maths problems (interesting ones)

[Alan Anderson]

Originally Posted by diobsidian

I hate to break it to you after all your work, but there are 12 coins, and lettering them A through I is only 9.

Whoops.

I still have a 3/4 chance of finding the coin and identifying it as heavy or light in three weighings if I simply ignore the last three, and it will only take one extra weighing in that 1/4 case, but that obviously doesn't meet the requirements of the problem, does it?

I know how to do the harder problem in theory, choosing appropriate groupings of coins to get the equivalent of three equations with three unknowns, but I lack the time to figure out the details today.

[Kris Verdeyen]

Originally Posted by Alan Anderson

You're missing the fact that there are two different ways in which the door may be answered by a two-girl family. Your analysis is looking at the probability of a girl answering, which doesn't matter for this problem, because a girl answering is given.

Actually, I'm the only one catching it - it's twice as likely that a girl will initally answer from a G/G family than from a B/G family, and there are twice as many B/G families. Therefore, the other sibling has a 50/50 shot of being girly.

[Alan Anderson]

Originally Posted by Kris Verdeyen

Actually, I'm the only one catching it - it's twice as likely that a girl will initally answer from a G/G family than from a B/G family, and there are twice as many B/G families. Therefore, the other sibling has a 50/50 shot of being girly.

No, you're counting families too soon. Only half of of the boy/girl families had a girl answer the door, so the others don't meet the criterion for this problem.

None of the boy/boy families are included. Half of the boy/girl families are included. All of the girl/girl families are included. But once a girl answers the door, you have to completely forget all the families that didn't get included. You can't count them, because they don't exist in the set of familes that did have a girl answer the door.

[FizMan]

Okay let's try this again...

Four possible combinations of children. Right? That's simple math too, you have two inputs, four outputs.

A girl answers the door... This is GIVEN.

We've now determined that it is impossible for the family to have a boy/boy combination.

Since a girl answering the door is GIVEN, it leaves us with three possibilities.

Two of those possibilites have boys. One possibility has a second girl.


I don't know how else to go aboot this, but I for one am going to trust my OAC finite before I trust logic that I feel is flawed.

[Joe Ross]

I think it comes down to, how are the kids selected to answer the door. I don't beleive the problem is clear enough, and that's why you have two different arguments for the answer.

When Kris reads the problem, he sees no method for picking which kid goes, and so assume random. That throws out the possibility of 2 boys, and also 1 of the 1 boy 1 girl possibilities. You get 1/2

When Denman and FizMan read the problem, they see that if a girl is in the family, she is sent out first. Thus, you can only eliminate the 2 boys possibility, and get 1/3.

Is one or the other the correct reading? Make your arguments to convince me.

[FizMan]

I think we can all agree on the four possible combinations of children, right?

So imagine four houses, each having one different combination of children.

The question specifically states that a girl answers the door. I read that as a given. Leaving three houses left; two with a boy, and one with a girl. If you randomnly chose either house than it'd be a 33% chance.


IF you guys were going by the probability that a girl answered the door in the first place... then consider this. We can naturally exclude the boy/boy combination from calculations. This leaves FOUR girls and TWO boys in our last three possibilities.

So what are the chances that a girl will answer the door? 66%

If you choose randomnly between the houses, House A (has girl/girl) will yield a 100% girl answering door chance. Whereas Houses B,C (girl/boy x 2) have a 50% chance each. But the probability of having either possible combination is 33% (three possibilities)

1/3 x 100% + 1/3 x 50% + 1/3 x 50%
33% + 16.5% + 16.5% = 66% (2/3)

So you have a 66% chance that a girl will open the door, this is where your calculations should start. I agree with the idea that at this point, it can be a 50/50 chance for the second child being a boy or a girl (just like flipping a coin really), but since we already have the 66% chance of a girl opening the door in the first place, we multiply it by the 50% and we end up with a 33% chance of a girl/girl possibility through this method.

[Denman]

If you know stats then think of it as permutations / combinations
i will write a full answer later, but its only 9am and i am tired lol

[Denman]

There is a five-digit number having digits 1-9, no two digits being identical. Two are prime numbers, two are square numbers, and one is neither. The third digit is twice the fifth, the fourth is six greater than the second and the last is three less than the first.
This is really hard actually, i havn't worked it out yet lol

[FizMan]

Originally Posted by Denman

There is a five-digit number having digits 1-9, no two digits being identical. Two are prime numbers, two are square numbers, and one is neither. The third digit is twice the fifth, the fourth is six greater than the second and the last is three less than the first.
This is really hard actually, i havn't worked it out yet lol

73894

Here's my notepad logic. Then I just fiddled around with numbers for five minutes at the end until I got something that worked.

abcde
c = 2e
d = b + 6
e = a - 3
c = 2a - 6
c == even
c < 10
2a - 6 < 10
2a < 16
a < 8

c < 10
2e < 10
e < 5

d < 10
b + 6 < 10
b < 4

e < 10
a - 3 < 10
a < 13

a is less than 8
e = (<8) - 3
e < 5

abcde
73894

[Denman]

its right!
the webpage i got it from had several riddles and this was the most interesting one. To getto thesolutions page you had to solve that one you just did lol, and i checked and it works!
Well done
i just didn't have the patience

[Mike Ciance]

here is my favorite math problem. it's pretty cool.

i tried uploading it again in response to the post below

[Steve Howland]

Originally Posted by afflictionblade

here is my favorite math problem. it's pretty cool.

The document had a figure and the question, but there was no shaded region...try uploading again maybe...

[Kris Verdeyen]

This page really explains where the two different answers come from:

http://mathforum.org/library/drmath/view/52186.html

It's interesting and worth a read if you're into this kind of thing.

[mtrawls]

Originally Posted by FizMan

You mean, for example, if I said, "Hey mtrawls, I'll pay you $20 to play <insert above mentioned game here> with me"?

Exactly. And for those interested in finding the answer without doing the math (shock! pure blasphemy, I daresay), try searching for the St. Petersburg Paradox from our good friend Bernoulli (ah, but which one?)

[Kris Verdeyen]

Originally Posted by afflictionblade

here is my favorite math problem. it's pretty cool.

Neat problem.


<SPOILER ALERT>
(use pythagoras and you get the answer to be pi)