This is kinda sad

[#686578]

Depending on the situation I add to subtract as well, but it only works for certain situations. It's a more algebraic form of thinking situationally.

[Conrad]

I do it a lot if I'm going to the next hundred/thousand/whatever; for instance, 1000-473 = 527 or something. Take the amounts to get to 9 for all places above the ones, then add the ones up to 0 and it should be your answer. It's kind of hard to explain, but it makes sense to me so meh.

[#686578]

Definitely for things that are easy to do like 1000 I would just add, or really any sort of "round" number, but it doesnt necessarily have to be round. Also common multipliers help with subtracting like 49-42=7 because they're only one multiplication factor of 7 away from eachother...

Shut up, it makes sense to me

[Conrad]

That makes sense to me actually, though I'd probably just do 2 > 9 = 7

[Darkslitter]

ok guys, I got a discrete math test in an hour. Someone tell me how to find the expected value of an infinite series, for example:
A slot machine wins 5% of the time, what is the expected number of times you will play before you win?

[Darkslitter]

Expected value = Sum from 1 to infinite of: n * .05 * .95^n-1

[Conrad]

You mean it's not 20?
Oh yeah I guess it's not

[Darkslitter]

i know you have to do something crazy like take the integral of it calculate than take the derivative

[Linliel]

http://www.seowoman.com/wp-content/u...hates-math.png

[Conrad]

Draw a picture of a giraffe

[Darkslitter]

Ok I got it in case anyone was wondering,

Sum from i to infinity(n * .05 * .95 ^n-1)
= .05 * Sum from i to infinity(n * .95^n-1)
= .05 * Derivative(Sum from i to infinity(Integral(n * .95^n-1)))
(call .95 = q from here on out and differentiate with respect to q)
= .05 * Derivative(sum from i to infinity(q^n))
= .05 * Derivative(1/(1-q) - 1)
=.05 * 1/(1-q)^2
=.05 * 1/(.05^2)
=20

bleh it ended up being 20 after all =x

[Conrad]

lol

[Darkslitter]

Draw a picture of a giraffe

did you see my hw where i did it? the TA gave me a point

[Charismatic]

That's a lot of math to come up with a common sense number

[Linliel]

Ok I got it in case anyone was wondering,


bleh it ended up being 20 after all =x

lol!!

[Darkslitter]

That's a lot of math to come up with a common sense number

well, I won't get credit on a test if I don't show the actual reason that it works

[Darkslitter]

FUCK FUCK FUCK FUCK FUCK FUCK FUCK

I FORGOT MY CALCULATOR

[#686578]

That makes sense to me actually, though I'd probably just do 2 > 9 = 7

It was a simplified example, 144-72=72
(12*12)-(6*12)=(6*12)

[Conrad]

For that I would do (7*2) and (2*2) and get 14 4 which is the answer

[#686578]

Ok I got it in case anyone was wondering,

Sum from i to infinity(n * .05 * .95 ^n-1)
= .05 * Sum from i to infinity(n * .95^n-1)
= .05 * Derivative(Sum from i to infinity(Integral(n * .95^n-1)))
(call .95 = q from here on out and differentiate with respect to q)
= .05 * Derivative(sum from i to infinity(q^n))
= .05 * Derivative(1/(1-q) - 1)
=.05 * 1/(1-q)^2
=.05 * 1/(.05^2)
=20

bleh it ended up being 20 after all =x

That has to be wrong, looking at any singular event you would obviously use the singular chance, 5%

Looking at it as a series of events you could never believe that it would take you 20 games before you can "expect" to win (what is the definition of a good enough chance to "expect", what is "expectable"?)

You have a 95% chance to lose once, a near 90% chance to lose twice, a near 85% chance to lose 3 times in a row and so on

Up to 10 times you get to about 60% which is a fairly good odd that you'll lose

At 15 times you get to about 46% which really starts to say what is a good standard to be expectable, what threshold do you need to reach?



I know someone will call me out on it if I dont say it. There's a difference between looking at a single event versus a single event. In each event every time you pull the lever you will have a 5% chance to win, but when you look at it as a series of events you have a 46% chance to lose 15 times in a row.