The Math Thread

[Venexis]

In a two-child family, one child is a boy. What is the probability that the other child is a girl? What if the older child is a boy?

Hidden in case you don't want to ruin it: show

First part of the question, the probability should be 2/3, not 1/2. This is because in a family of two children, assuming there are only two sexes (sorry SJWs math doesn't give a ♥♥♥♥), there are only four unique combinations. Older son + younger son, older daughter + younger son, older son + younger daughter, and older daughter + younger daughter. The last case isn't valid, as the question is only concerned with families that have at least one son, so this leaves only three possible family variations... two thirds of which contain a girl.

For the second part of the question, we can exclude two of the above cases (the two with an older daughter, since it specifically asks for an older son). This leaves a 50% chance of the younger child being a boy.

[lordpat]

Ding, ding, ding, ding! Ven is right, congratulations!

[Supershroom]

Damn it, how could I forget that. (although I didn't explicitely say that the probability would be 1/2). The probability also isn't 1/2 if you say "Once child is a boy born at November 4th."

But in my case, I know the answer.

[Konradix]

Card answer attempt 2: show

One card has two red sides.
One card has two black sides.
One card has one of each.
You are saying that the card you show will have two sides that are of the same colour.

You therefore have a 2/3 chance of winning. Since you bet 10 bucks, 10/2=5 bucks.

Right now?

[Supershroom]

Nice one! Actually most people would really hold up 10 buckets saying that if two cards have the same probability, the two colours also have, which isn't true. Even great mathematicians sometimes fall for mistakes like these.

[GrandPiano]

The Collatz Conjecture: Take any natural number. If it's even, divide it by two, and if it's odd, multiply it by three and add one. Do the same for whatever number you arrive that. If you continue this process, you will eventually reach one. Is this conjecture true?

(Note: This conjecture is not known to have been proven or disproven by anyone yet.)

Example:
Start with 23.
23 is odd --> Multiply by 3 & add 1 --> 70
70 is even --> Divide by 2 --> 35
35 is odd --> Multiply by 3 & add 1 --> 106
106 is even --> Divide by 2 --> 53
53 is odd --> 53*3+1=160
160 is even --> 160/2=80
80 is even --> 80/2=40
40 is even --> 40/2=20
20 is even --> 20/2=10
10 is even --> 10/2=5
5 is odd --> 5*3+1=16
16 is even --> 16/2=8
8 is even --> 8/2=4
4 is even --> 4/2=2
2 is even --> 2/2=1
1 is the target number.

One thing I can say is that every number must reach a power of two to reduce to one, because you can only get to 1 from 2, as proven below:

3x+1=1
3x=0
x=0
0 is not a natural number (Even if it were allowed, it is an even number and therefore would take the /2 route, which would just cause it to repeat itself).

[Oranjui]

Okay I'm just going to try to play with this and see if anything helpful comes out of it, which it probably won't:

When you multiply an odd number by 3, because 3 is odd, the result will also be odd. When you add 1 to that odd result, it becomes even. You're always trying to get to even numbers. This is also a result of what you said; you can only reach one by using the procedure for even numbers.

For any odd number x, (3x+1)/2 is an integer.
For any even number y, y/2 is an integer.

wait forget all of that

You have to reach 2^x to be able to get to 1.

If a is an odd number, then, eventually,
3a+1 = 2^x
3a = 2^x - 1
a = (2^x - 1)/3

You know I don't know where I'm going with this at all so I'm just going to stop


Also, http://imgur.com/gallery/QAJXW
/r/mathpics is a good sub

[Supershroom]

You can reach and only reach the powers of 2 whose exponent itself is even, meaning: 3 divides 2^k - 1 ⇔ k is even.

This is true for k = 2, and for the step from k to k + 2 take

2^k+2 - 1 mod 3 = 0 ⇔ 2^k+2 mod 3 = 1 ⇔ 4 ∙ 2^k mod 3 = 1 ⇔ 2^k mod 3 = 1,
which is true by the induction hypothesis. Otherwise, if k is odd, take 2^k mod 3 = 2 ∙ 2^k-1 mod 3 = 2, since 2^k-1 mod 3 = 1, hence 2^k - 1 can't be divided by three.

[*Emelia K. Fletcher]

The Collatz conjecture has been proven to not have an easy generalization at all, so have fun

[Oranjui]

le solve

y = ( ln(x) + (sa - 1)ln(m) )/(r^2)

edit: Damn it guys it said Merry Christmas if you solved it
edit2: Yeah I knew that but I was disappointed that nobody solved it on Christmas