header image

tuesday 03/12/2019

Ok thats the last

deleted

I'm really hoping to get this one off the ground. My goal is to start it during New Years, as I have time to run it around then. The game will have at the very least, 1 million clintz from my own pocket involved, not including the cards that I already own.

How to enter:
Just send me at least 100k worth in cards to my private sales for 50 clintz. I will not buy them until the game actually starts. If you have any questions, please message me.

monday 02/12/2019

Hi everybody !!
December 5th it's my birthday in urban rivals !!

I just made a lottery

Loterie Urbannif 10 ans

come shawty !!! you to my friends you are all welcome come on !!!

sunday 01/12/2019

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 smileysmiley Thank you so much smiley
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 smiley

Link: WAN'S ADVENT CALENDAR 2019

sunday 24/11/2019

Please can you remake the event so the event has a 20k entrance fee. Lotteries where the creator keeps some of the entrance fee are not allowed and the way this event is set up stops us from being able to check. You can say the winner gets the entrence fee rounded up to 1000000 but you would have to post if people enter more than once to keep track of it.

https://www.urban-rivals.com/en/community/forum/?mode=viewsubject&id_subject=1762832

edited by UM_AaaBattery sunday 24/11/2019, 13:15

saturday 23/11/2019

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?

wednesday 20/11/2019

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

friday 15/11/2019

11 messages

What the history?

monday 11/11/2019

The prizes will be sent out within the next 2 hours

sunday 10/11/2019

wednesday 06/11/2019

THIS JUST IN:

In a twist, an bonus prize Oon Cr (Worth: 350K) will be awarded to the third person to get the answer right!

The third-place prize is greater than the first-place prize and the second-place prize.

Can you get third place? smiley

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!!!

monday 04/11/2019

Is there anything left?

Tortuga, or anything :*

monday 28/10/2019

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

Congrats infi

friday 25/10/2019

sunday 13/10/2019

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

Create a subject