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
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.
edited by AaaBattery sunday 24/11/2019, 13:15
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.
The prizes will be sent out within the next 2 hours
Didnt see i had to dm
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?
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.
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!!!
Is there anything left?
Tortuga, or anything :*
You can PM me anytime, I am more than happy to help you co-organize an event
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!