Mathematics marathon

[cmhardw]

Spoiler

Ah yeah, you're given a random face, not a random card! I should've thought "what's the probability that the card has the same colour on both sides", which is effectively what it's asking! Awesome problem more!

Yes exactly! If you like this style problem, and now that you have the basic idea, try some more.

There are 6 cards in a deck. 1 card is red on both sides, 2 cards are black on both sides, and 3 cards have red on one side and black on the other side. If this deck is shuffled randomly, and one card is produced that shows red on top, what's the probability that when you flip this card over it shows black on the other side?

And now for the fun question, what's the probability that if you shuffle this deck randomly and produce a card, that flipping this card over shows the other color?

 

[Rinfiyks]

Yes exactly! If you like this style problem, and now that you have the basic idea, try some more.

There are 6 cards in a deck. 1 card is red on both sides, 2 cards are black on both sides, and 3 cards have red on one side and black on the other side. If this deck is shuffled randomly, and one card is produced that shows red on top, what's the probability that when you flip this card over it shows black on the other side?

And now for the fun question, what's the probability that if you shuffle this deck randomly and produce a card, that flipping this card over shows the other color?

Spoiler

First question: 3/5?
Second question: 1/2. Part of me wants it not to be 1/2 so that my mind is blown again like before!

 

[cmhardw]

Spoiler

First question: 3/5?

Yep

Second question:

Spoiler

Second question: 1/2. Part of me wants it not to be 1/2 so that my mind is blown again like before!

Ok I can't let this one slide, you have to be confident in your answer Is it 1/2 or is it not 1/2?

 

[Rinfiyks]

Yep

Second question:

Spoiler

Ok I can't let this one slide, you have to be confident in your answer Is it 1/2 or is it not 1/2?

Spoiler

Haha, yes I am confident

 

[cmhardw]

Spoiler

Haha, yes I am confident

Spoiler

Yep, 1/2 is right here. There are two common ways to prove this. I like the method that says that only 3 of the 6 cards are multi-colored, so 1/2 chance you picked one of those cards from the deck.

The other method is to evaluate the probabilities in the following formula:
P(red on top side) * P(black on other side | red on top side) + P(black on top side) * P(red on other side | black on top side)

Both methods yield 1/2 as the answer, so I agree

 

[cmhardw]

Another probability question that I always loved:

What's the probability that at least two people out of a group of 23 people have the same birth month and birth day, assuming that no one in the group has a birthday on Feb. 29th?

 

[Rinfiyks]

Spoiler

Yep, 1/2 is right here. There are two common ways to prove this. I like the method that says that only 3 of the 6 cards are multi-colored, so 1/2 chance you picked one of those cards from the deck.

The other method is to evaluate the probabilities in the following formula:
P(red on top side) * P(black on other side | red on top side) + P(black on top side) * P(red on other side | black on top side)

Both methods yield 1/2 as the answer, so I agree

Spoiler

I prefer the former too it's the way I did it.
Reminds me of another probability problem:
You have two boxes. You know that one of the boxes contains a certain amount of money while the other one contains twice as much. But you do not know which is which.
You pick a box and assume it contains a value of money we shall call x. (You don't open it yet)
You then reason that the other box contains half as much or twice as much: x/2 or 2x.
You then work out the "average" value of the other box.
(x/2)/2 + 2x/2
(1/4)x + x
(5/4)x.
So statistically the other box contains, on average, more money. So you should switch!
You could then apply the same reasoning and switch again and again...

 

[keemy]

Spoiler

\( 1- \frac {23! \times \binom{365}{23}}{365^{23}} \)

since we are on combo already

lets say you have a 3 by n binary matrix what is the minimum n required to ensure that some row has has either 3 1s or 3 0s

 

[Rinfiyks]

Another probability question that I always loved:

What's the probability that at least two people out of a group of 23 people have the same birth month and birth day, assuming that no one in the group has a birthday on Feb. 29th?

I've heard this one before! I don't know how to calculate it exactly but

Spoiler

It's >50%

Edit:

Spoiler

\( 1- \frac {23! \times \binom{365}{23}}{365^{23}} \)

since we are on combo already

lets say you have a 3 by n binary matrix what is the minimum n required to ensure that some row has has either 3 1s or 3 0s

Spoiler

5?

 

[keemy]

that's correct (though you should try to prove it) also in the extend the question further find a function \( f \) such that \( f(k)= \) minimum n to ensure there are either k 1s or k 0s in some row.

 

[Rinfiyks]

that's correct (though you should try to prove it) also in the extend the question further find a function \( f \) such that \( f(k)= \) minimum n to ensure there are either k 1s or k 0s in some row.

Spoiler

You can get up to k - 1 of both 0s and 1s in a row safely. This makes 2(k - 1). Then if you add another column, that row must have either k 0s or k 1s. So it's 2(k - 1) + 1 = 2k - 1.

 

[keemy]

true but not a proof.

 

[qqwref]

Here's a cool one:

Find a polynomial p(x,y) so that, when x and y are integers, p gives a unique value for each possible value of x and y. So, if you know x and y are integers, and you are given a value of p, you should be able to figure out the single possible (x,y) pair that gives that value.

 

[cmhardw]

Here's a cool one:

Find a polynomial p(x,y) so that, when x and y are integers, p gives a unique value for each possible value of x and y. So, if you know x and y are integers, and you are given a value of p, you should be able to figure out the single possible (x,y) pair that gives that value.

If p is irrational for some x and some y, then must it be written in decimal form?

 

[keemy]

A polynomial over what field? I mean I can easily choose

Spoiler

\( p(x,y)= x+iy \)

 

[qqwref]

Pshh, alright. All the coefficients in p(x,y) must be integers.

 

[cmhardw]

Pshh, alright. All the coefficients in p(x,y) must be integers.

Ok, I like this. I was thinking of:

Spoiler

\( p(x,y)=x\sqrt{2} + y\sqrt{3} \)
personally.

With integer coefficients... *scratches head*

 

[Keroma12]

I always liked this problem.

There is a deck of cards that contains only 3 cards. One card is red on both sides. One card is black on both sides. One card is red on one side and black on the other. The deck is shuffled randomly and one card is produced from the deck and laid on the table. This card shows red on top. What is the probability that when you flip this card over that it will be black on the the other side?

I love these problems! Got any more? (Yes, I got this one right.)

 

[cmhardw]

I'm still working on Michael's, but I have another good one I just remembered (and Michael you may like this one).

Prove that for any 5 points placed randomly on the surface of a sphere, that at least four of those points are always contained on the same hemisphere.

 

[keemy]

ok I think I have one (and can probably make some other with the same idea)

Spoiler

\( P(x,y)=y^2+2yx+y+x^2+3x = (x+y)(x+y+1) +2x= \) double the (x+y)th triangular number + 2x uhh I am not positive if this will work for all integers but it should be good for positive ones