Saturday Night Contest - Guess Again

Mar 10, 2008
15
0
math.

Well, its (55*55)/2 to account for the doubles. Dividing by two will kill the unique issue. Or should.

-ThrallMind

not true.

the total number of events is just 55*54/2 + 55. think about it. make two sets. where we have two different cards and the same cards.

different cards:
55 choose 2.
55*54/2

same cards:
we have 55 pairs.

stop playing with cards and study math!
i'm in stat right now. :D
 
Dec 4, 2007
1,074
2
www.thrallmind.com
2x Ace of Spades

2x Ace of Hearts

2x Ace of Diamonds
2x 6 of Diamonds

2x Ace of Clubs
2x 2 of Clubs

2x Double Backers

2x Jokers

2x Any of the Jacks

And also... The ad cards have no back. They are the ad on both sides.

-ThrallMind
 
Mar 16, 2008
183
0
Melbz
Yo.

Guess 1: 6 of Spades/ 9 of Diamonds

Guess 2: King of Clubs/ 2 of Hearts

Guess 3: Ace of Spades/ Jack of Clubs

yeaaah
 
Dec 4, 2007
1,074
2
www.thrallmind.com
not true.

the total number of events is just 55*54/2 + 55. think about it. make two sets. where we have two different cards and the same cards.

different cards:
55 choose 2.
55*54/2

same cards:
we have 55 pairs.

i'm in stat right now. :D

Actually, (55*54) + 55 should do it.

(Uniques) + (Cases where we can have a double)

-ThrallMind
 

TKK

Mar 22, 2008
8
0
Melbourne, Australia
Omg! It seems that both my first and second guesses had been taken up by the others. I'm wondering if I should look through all of the posts, compile them into a list, and thoroughly analyse the guesses, and come up with a very educated guess using the process of elimination? By but the time I had done so, some others might just have had their wild guesses right! So I might just as well have a wild guess here... Lol!

Guess 3: Two Nines of Hearts

Don't know if anyone had the samething AGAIN before me. :p
 
Dec 4, 2007
1,074
2
www.thrallmind.com
Doubles
======

Spades
Ace
Jack

Hearts
Ace
9
Jack

Diamonds
Ace
6
7
Jack
Queen

Clubs
Ace
2
Jack

2x Double Backers

2x Jokers

And also... The ad cards have no back. They are the ad on both sides.

-ThrallMind
 
Nov 17, 2007
519
1
Doubles
======

Spades
Ace
Jack

Hearts
Ace
9
Jack

Diamonds
Ace
6
7
Jack
Queen

Clubs
Ace
2
Jack

2x Double Backers

2x Jokers

And also... The ad cards have no back. They are the ad on both sides.

-ThrallMind
Wow. ThrallMind. You really are putting a lot of thought into this very simple contest. =)
 
Mar 10, 2008
15
0
i know stat.

Actually, (55*54) + 55 should do it.

(Uniques) + (Cases where we can have a double)

-ThrallMind

i tell you sir, it does not. haha.

combinations (order does not matter a.k.a king of spades-ace of spades is equal to ace of spades-king of spades)-

nCx (n choose x)
n!/(x!*(n-x)!)
so (55*54*...)/((53*52*51..)(2*1))

permutations (order does matter a.k.a king of spades-ace of spades is NOT equal to ace of spades-king of spades)

nPx
simply. n!/(n-x)!


i stand by 55*54/2 + 55 = 1540 i think
 
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