Are you still thinking about doing that lottery? Because if not I'd like my clintz back.
Also, you still owe me 100k from a previous math problem, I already sent you a message about it with the card I wanted and I know you saw it because it was the same message where I said I wanted to enter the lottery

100 join ups so far! Wow Thank you so much
Make sure you join up and state what packs you'd like opening so you don't miss out on the first draw of this event

Link: WAN'S ADVENT CALENDAR 2019

Here s smthing ez for everyone to enjoy..
So I planned for a trip in Seville on September 9th (so everything concerning travels and stay
is paid and I m left with some money to spend however I choose). That first week I decided to check
out the Royal Alcázar and the Isla Magica which was all withing my calculation. However I
end up overspending by the end of the week, as I attend the urban rivals IRL, "by 200$ more
than three fifths of my money". I m now left with "more than 400$ less than half of it left".
So if I start my vacation with a "whole number" of $, what would be the greatest amount I could ve had?

LOL okay. Only 3 ppl have gotten the question right. All of whom DMed, me.

Thus, I will be granting only 3 prizes.
1st Place: 100K+100K=200K card of choice.
2nd Place: 50K+50K=100K card of choice.
3rd Place 25K+25K=50K card of choice.

The winners are the following people:
1st Place: Pere-sonne
2nd Place: FBF_Luis
3rd Place: Manifold

Congratulations! Please dm for the character you want.

For this question, I will offer two solutions.

Solution 1: Let p be the probability that Suzy wins from the deuce position. Then consider two balls after the deuce. Either Suzy has won with probability 1/9, Doug has won with probability 4/9 or it is a deuce again with probability 4/9. Therefore, we can set up this equation: p=1/9+4/9p solving for p we get that Suzy wins with probability 1/5

Solution 2: We can reduce this to the sum of a geometric sequence. Namely the geometric sequence 1/9+4/9*1/9+4/9^2*1/9... The sum of this infinite geometric sequence is (1/9)/(1-4/9)=1/5.

If you have any questions let me know.

abcdefghik123

What the history?

The prizes will be sent out within the next 2 hours

Didnt see i had to dm

Up

Greetings! Thank you for trying the math question.

The answer is the following:
Let f(x) be the function x^3+5x^2+12x+32. Define a new function F(1/x)=f(x). Then F(1/r)=f(r)=0, and the same follows with r, s and t. So, F(1/r) has roots 1/r, 1/s and 1/t.

To find F, we plug in 1/x as r, this yields F(1/1/x)=F(x)=f(1/x)=1/x^3+5/x^2+12/x+32.

We multiply by x^3 to get a third-degree polynomial.
F(X)=32x^3+12x^2+5x+1

I accepted any constant multiplied by this polynomial. For example: x^3+3/8x^2+5/32x+1/32 was accepted!

For those of you who got it right, I've sent you your prize. for thoses that got it wrong, I sent 1 CHOKO anyway!!!

You can PM me anytime, I am more than happy to help you co-organize an event

Congrats infi

64/128 players
Rewards up to date.

Still waiting for Klouni and Darkdays VP.
The others should be good now

I don't even know what to do with so many Clintz

Insane event, looking up for it!

Damn..... Hope this wasn't a scam for all the people who enter.

No problem man, thanks for playing!