[notanewbie]

Who thinks my $25 contest is worth more money?

I asked if I should split the prize and pay out for 1, 2 or 3 correct answers (after a restart) should the contest not be won in the next few days.

It has also been suggested that I leave things as they are (guess the 3 numbers) but increase the prize money.

Can you please let me know how you want this to go?

Thanks.

[onizuka]

~Raise my hand~

I vote for leave things as they are, because it was reach more than 200 posts, and let it be 2000 maybe? Something like Talkgold's mega jackpot with $25 prize and increasing until $50 maybe?

Let's say a long run contest like this held once every 6 months, that's fun..

Just an idea..

[omega007]

[quote=notanewbie]LOL - I read every single post, I have to. If you guys were consistent in how you post your answers my browser would do the job for me. When I get your numbers, some are separated by commas, others by dashes or whatever, it's easier for me to check every post. QUOTE]any suggestion for consistency?

these are some of the formats...

xx, yy, zz (comma and space)
xx,yy,zz (comma)
xx yy zz (space)
xx-yy-zz (dash)

sorted? add 0 for single digit?

[satria]

Originally Posted by notanewbie

I asked if I should split the prize and pay out for 1, 2 or 3 correct answers (after a restart) should the contest not be won in the next few days.

Agree with that, More winners, more fun.

[mjomha]

but maybe guessing 2 number 3 number there are over 100k possible combination its gonna take a long time to guess the three right numbers

maybe increasing the prize every week 10$ if no one gets the right answer..

[satria]

Originally Posted by mjomha

but maybe guessing 2 number 3 number there are over 100k possible combination its gonna take a long time to guess the three right numbers

maybe increasing the prize every week 10$ if no one gets the right answer..

100k possible combination ?? NOT exactly

Guessing 1 number from 1-50 there will be 50 possibilites

Guessing 2 number from 1-50 there will be 1225 possibilites
C(50, 2) => 50! / 2!(50-2)! => 50 x 49 x 48! / 2! 48! = 50 x 49 / 2 x 1 = 1225

Guessing 3 number from 1-50 there will be 19.600 possibilites
C(50, 3) => 50! / 3!(50-3)! => 50 x 49 x 48 x 47! / 3! 47! = 50 x 49 x 48 / (1 x 2 x 3) = 19.600

[LameCore]

Split the prize and let me be one of the winners

[mjomha]

Originally Posted by satria

100k possible combination ?? NOT exactly

Guessing 1 number from 1-50 there will be 50 possibilites

Guessing 2 number from 1-50 there will be 1225 possibilites
C(50, 2) => 50! / 2!(50-2)! => 50 x 49 x 48! / 2! 48! = 50 x 49 / 2 x 1 = 1225

Guessing 3 number from 1-50 there will be 19.600 possibilites
C(50, 3) => 50! / 3!(50-3)! => 50 x 49 x 48 x 47! / 3! 47! = 50 x 49 x 48 / (1 x 2 x 3) = 19.600

oops forgot that rule....well i meant the samething lol alot of combinations!!!!

[retep]

Originally Posted by satria

100k possible combination ?? NOT exactly

Guessing 1 number from 1-50 there will be 50 possibilites

Guessing 2 number from 1-50 there will be 1225 possibilites
C(50, 2) => 50! / 2!(50-2)! => 50 x 49 x 48! / 2! 48! = 50 x 49 / 2 x 1 = 1225

Guessing 3 number from 1-50 there will be 19.600 possibilites
C(50, 3) => 50! / 3!(50-3)! => 50 x 49 x 48 x 47! / 3! 47! = 50 x 49 x 48 / (1 x 2 x 3) = 19.600

hmm strange calulations.

I would say

2 nummer 1-50

first numer change of good 2 in 50 so (1 in 25)

second nummer 1 in 49

49 * 25 = 1225 (is the same)

3 in 30 good

first 3 in 50 (1 in 16.666666)
second 2 in 49 ( 1 in 24.5)
tird 1 in 48

= 1 in 19600

ansser is the same

[satria]

Originally Posted by retep

hmm strange calulations.

I would say

2 nummer 1-50

first numer change of good 2 in 50 so (1 in 25)

second nummer 1 in 49

49 * 25 = 1225 (is the same)

3 in 30 good

first 3 in 50 (1 in 16.666666)
second 2 in 49 ( 1 in 24.5)
tird 1 in 48

= 1 in 19600

ansser is the same

That was not strange calculation, do you learn math ?
If you know the term Permutation then you should also know Combination, a two different way how to arrange a set of items in a group.