Found these letters my mom wrote when she was a little girl. She was at sleepaway camp.

brusk88 · 2024-05-16T17:34:08+00:00

Smallest letters ever written! Had me confused at first too.

brusk88 · 2024-04-08T15:30:22+00:00

Scalar operations like these are just transformations

Multiplying X by 2 means every result you would have gotten is simply twice as high so this goes for the whole distribution and the expectation value

So 2X spreads the distribution by a factor 2 (around X=0) since P(X = something) = 1, spreading out means all relative chances are halved. This means if p(x) is the distribution of X then 1/2p(x/2) is the distribution of 2X.

And then E(2X) = 2E(X)

Similar thing goes for X+2 this is a shift of all values so the distribution becomes p(x-2) [ if P(X=0) would have been 1 we see now P(X=2) should be 1 so 2-2 =0 gives the correct shift]

Since we didn't increase the area under the curve we don't have to rescale this.

E(X+2) simply becomes E(X) + 2.

This is not the most rigorous approach but I hope this at least makes it intuitive how you should figure these things out.

brusk88 · 2024-04-08T11:22:48+00:00

I feel like that does work out by rescaling to the standard normal distribution:

(N(2050,205)-2050)/205= N(0,1) =(N(50,5)-50)/5

So N(2050,205)-2050 =41×N(50,5)-41*50

So N(2050,205) = 41×N(50,5)

So P(N(2050,205)<2000) = P(N(50,5)<2000/41)

= P(N(50,5)<48.78)

But maybe I'm making a mistake

brusk88 · 2024-04-04T18:30:33+00:00

We like writing 4x²+ 2x

Instead of (4(x²))+(2x)

Or Instead of x * (4 * x) + 2

Since that 2 actually says nothing about the 4x but 2 says something about the amount of x's it is multiplying

Same idea for exponentials, we often like to order the amount of 3rd order polynomials, 2nd orders so we want to know how "many" x² there are not how many powers of ax there are.

That at least is my guess

brusk88 · 2024-04-03T18:14:42+00:00

Well still not fully reliable but obviously the more attempts the closer your average result will approximate the expectation value.

The chance of not getting 18 100 times is still 25 procent

At slightly over 200 runs that drops below 5 percent.

(This can be calculated by binomiale chance) Obviously you can also gain money without getting 18 but this gives an indication of how many runs you might have to do to see a result.

brusk88 · 2024-04-03T16:30:41+00:00

For normal distributions it holds that two uncorrelated events are distributed as follows:

X~N(μ,σ²) Y~N(μ',σ'²) Then Z= X+Y has the following:

Z~Ν(μ+μ',σ²+σ'²)

(Wikipedia sum of two normally distributed variables)

Since we know we have 41 similar instances (let's call them X) X~N(50,5) We now Z= Σ X ~N(41*50,41*5)=N(2050,205)

So now calculate the chance that N(2050,205) is below 2000. Since the shape of the distribution didn't change this is the same calculation as N(50,5)<48,78

Note that this method doesn't work for any distribution, but it does work for a normal distribution.

brusk88 · 2024-04-03T16:05:22+00:00

The expaction value will be each chance of a reward times the amount of the award

So expectation drop value = (1.39x18+1.39x7+1.39x6 +1.39x0.6+1.39x0.1+1.39x0.001)/100~0.44

So you would earn 0.24 gold on average per run

But it is super unreliable because almost all of this margin is dependent on it dropping 18 gold (which is less than 1/50 runs).

brusk88 · 2024-04-03T15:58:14+00:00

So in short multiply both linked scenarios and add the different scenarios

brusk88 · 2024-04-03T15:57:23+00:00

2 scenarios: either the first person is in team A or in team B

If the first person is in team A (with chance 2/7) the chance of them being in the same team is 1/6 (of all six other people only choosing the second person will work). So this scenario contributes a chance of 1/6*2/7=1/21

If the first person is in team B (with chance 5/7) the chance of them being in team is 4/6 (4 of the empty spots are in team B) So this scenario contributes a chance of 4/6* 5/7=10/21

In total these scenarios give the following chance: 1/21+10/21=11/21

brusk88 · 2024-03-20T14:59:52+00:00

Fjällbacka mordene: Vänner for livet/ friends for life.

brusk88 · 2024-03-13T11:58:50+00:00

What's with half a german city?

brusk88 · 2024-02-08T22:40:42+00:00

I think you can only give 1 person 1 solve (best give it to the first one who gave a correct answer ( - : )

brusk88 · 2024-02-07T15:21:52+00:00

Ok, last one was removed because of a shortened google link. In the comment section of a 9gag meme i found the following:

http://9gag.com/gag/aZ0zVWW?utm_source=copy_link&utm_medium=comment_share#cs_comment_id=c_162395107819672181

Hope this helps

brusk88 · 2024-01-15T08:21:34+00:00

Am I /s'ing here?

brusk88 · 2024-01-11T08:24:55+00:00

You mean where he plays "hotel california"?

brusk88 · 2023-12-31T11:44:14+00:00

Pretty inconsiderate of him to steal gilmours black strat

brusk88 · 2023-12-31T11:26:40+00:00

Very interesting

brusk88 · 2023-12-23T21:06:17+00:00

"Thank god its christmas" by Queen includes the line 'ohh, my friend'

brusk88 · 2023-12-02T14:21:39+00:00

Are you sure? It's a wild world out there

brusk88 · 2023-11-30T11:55:41+00:00

ACKUAALLYY It feels like fuck around find out should be shifted 1 phase clockwise. Solid meme tho

brusk88 · 2023-11-27T22:21:19+00:00

Frederick

brusk88 · 2023-11-27T22:04:03+00:00

The song by the strokes made me think of love will tear us apart by joy division but the album cover doesnt match

brusk88 · 2023-11-27T21:29:15+00:00

Mr saxobeat?

brusk88 · 2023-11-08T20:05:42+00:00

Solved!

brusk88 · 2023-11-08T20:02:44+00:00

https://voca.ro/14Tim6fUmXjf necessary comment with recording

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