Discussion about math, puzzles, games and fun. Useful symbols: ÷ × ½ √ ∞ ≠ ≤ ≥ ≈ ⇒ ± ∈ Δ θ ∴ ∑ ∫ π -¹ ² ³ °

You are not logged in.

- Topics: Active | Unanswered

Pages: **1**

gAr:

The way I looked at the problem, I would stop looking once I hit a pair.

Thank you all for thinking about this one.

**MathPro101**- Replies: 5

Hello all,

I am just passing time playing with a deck of cards and thought of a probability question. It is a very straight forward problem, but for some reason I cannot figure out the solution.

If I have a deck of 52 cards, shuffled, and I turn over one card at a time, what is the probability that I will see 2 of the same number in a row? (i.e., flipping over a 9, followed by another 9).

Let me know what you all think! Thank you.

Oh I like the You-Substitution one. Very nice.

**MathPro101**- Replies: 447

Just two math pickup lines I thought were pretty humorous.

"You and I add up better than a Reimman sum"

"My love for you is like dividing by 0; its so undefined"

Gotcha, I have a lot of work to do

WOOOO!!!! What's next?

Hi gAr,

Thank you very much.

**MathPro101**- Replies: 6

Hey, just a quick question about the forum. What do I have to do to get rid of my "Novice" title? Thank you.

Are you referring to uv transformations in calculus?

Slope intercept form is the form y=mx+b.

This means that in each equation, we must solve for y in terms of x and a constant, b.

To solve for y, we need to have it alone on one side. In order to do that, we must move x to the other side. In all three equations, x is being added to y on the left side, so in order to move x over, we must subtract x from the left side. But whatever we do to the left side we must do to the right side. So after one step, you should have

x + 2y - x= -4 - x

x + y - x= 3 -x

x + 3y - x= -6 - x

which is the same as

2y= -4 - x

y= 3 -x

3y= -6 - x

Now, the second equation is in slope-intercept form, but the first and third are not. We must apply the same algebraic strategy that we did in the first step to get y alone. For example, in the first equation, y is being multiplied by 2. So in order to get rid of the two, we must divide the y by 2. But we must also divide the WHOLE right side by two, not just the -4. Once you do that for the first and third equation, you should end up with

y= -2 - x/2

y= 3 - x

y= -2 -x/3

Does that help?

**MathPro101**- Replies: 2

Hi Everyone,

My name is David from Colorado. I am an undergraduate studying Applied Mathematics at an engineering school. I joined the forum to talk to people who love math as much as me. I look forward to chatting about math. Thanks!

Very Good! One quick addition to that would be to say that θ = π/2 ± 2πn, where n=0,1,2,.. This is the same as rotating on the unit circle and getting back to the same place each time. Therefore the true answer should be e^(-π/2 ± 2πn), which is still a real number!

There is another way of solving this problem that uses a different technique. Can anyone figure that one out? It simplifies things greatly.

**MathPro101**- Replies: 2

I found this video on youtube of an MIT Physics lecture. If you jump ahead to around 45 minutes, the professor offers a math problem that I found interesting.

http://www.youtube.com/watch?v=OvDePn41fPY

It took me a long time to work through it, but if you know the trick, it can be solved in less than 10 seconds.

i^i, where i = sqrt(-1)

the clue is that i=e^i(pi/2 plusminus 2*pi*n), n=0,1,2,...

Have fun with it, let me know what you think.

Pages: **1**