| Author |
 Replies: 678 / Views: 16,137 |
|
|
|
Pillar of the Community
United States
2669 Posts |
Was actually doing this instead of A1 A2
J 1 A 1 6 1
J 1 2 1 6 1
J 1 3 1 6 1
J 1 4 1 6 1
J 1 5 1 6 1
J 1 6 1 6 1
J 1 7 1 6 1
J 1 8 1 6 1
J 1 9 1 6 1
J 1 10 1 6 1
J 1 J 1 6 1
J 1 Q 1 6 1
J 1 K 1 6 1
then
J 1 A 2 6 1
For the first group, 2 are skipped because J1 and 61 are already in use, iterations 2, 3, 4 are fine so that'd be 50 each. I don't remember where the other 4 went.. I blame Scooby for tying my brain in knots.
|
|
Pillar of the Community
 United States
4000 Posts |
Quote:
I blame Scooby for tying my brain in knots.  You know that's what I strive for! That, and a few of these ---> |
|
Valued Member
Canada
287 Posts |
Quote:
J 1 A 2 6 1 Curious, did you count J1-62-61 or J1-J2-61. |
|
Valued Member
Canada
287 Posts |
Ok, one last thing before I go to bed. A thought about defining the numerator/denominator.
Numerator:
The number of possible combinations where a Jack and six are within two cards of each other multiplied by the four Jacks in the deck.
Denominator:
The total number of possible combination of cards including a Jack.
|
|
Valued Member
Canada
287 Posts |
Or would you multiply the quotient by four?
G'nite all.
|
|
Pillar of the Community
United States
2669 Posts |
That caused one of the re-do's of the calculations - I wasn't allowing them first, but needed to... so yes, those were being counted.
|
|
Pillar of the Community
United States
2669 Posts |
Quote:
Or would you multiply the quotient by four? I'll get back to you tomorrow on that one.. lol |
|
Pillar of the Community
United States
8904 Posts |
|
|
Pillar of the Community
United States
2669 Posts |
Help us Moe! My last brain cell reminded me I had calc'ed this when we were working on the 2xJx2 (5 slot) theory.. 132600 3-number combinations 2652 2-number combinations There are 2598960 combinations of 5-block numbers. 5-number blocks / all potential combinations = 2598960 / 2734212 = 95% Funny how you can work the math.. Quote:
Numerator:
The number of possible combinations where a Jack and six are within two cards of each other multiplied by the four Jacks in the deck.
I think this would be 800 (50 each J1 61, 62, 63, 64 = 200 * 4 = 800) Quote:
Denominator:
The total number of possible combination of cards including a Jack.
Total permutations * 4/52? (which would be 10200 3-card only) |
|
Pillar of the Community
United States
602 Posts |
|
|
Pillar of the Community
United States
4000 Posts |
So, what's the answer? Have we solved this?
Or do you want to know what I've been babbling?
|
|
Pillar of the Community
United States
4000 Posts |
Wolf, we know you like to mush. Now mush!
|
|
Valued Member
Canada
287 Posts |
Quote:
I think this would be 800 (50 each J1 61, 62, 63, 64 = 200 * 4 = 800) I think this accounts for J#-x-6# but does not account for 6#-x-J# nor J#-6#-x, 6#-J#-x. |
|
Valued Member
Canada
287 Posts |
J#-x-6# where x is any other card = 50
Do this for all 4 Jacks =200
Do this again with all 4 6's = 200
Total for right handed matches is 400.
Repeat same process using 6#-x-J# =400
Total for left handed matches = 400
Total all matches = 800.
There would be no additional counts for J#-6# as these would have been seen during previous exerises.
eg.
...
J(1)-10(3)-6(1)
J(1)-10(4)-6(1)
J(1)-J(2)-6(1)
J(1)-J(3)-6(1)
...
Therefore xshift=correct
|
|
Valued Member
Canada
287 Posts |
Total combinations possible of any three cards would be (52*51*50) which covers all possible 3 card combinations, both left and right.
52*51*50=132600
Therefore xshift=does not appear as correct as previously stated.
-----------------------------
Final Answer for all the marbles = (50*4)* 4 / 52*51*50 = 800/132600 = 0.006
Edited by rikcando
10/26/2010 07:32 am
|
| |
Replies: 678 / Views: 16,137 |
First Page
Previous Page
Posted 10/25/2010 9:57 pm
