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        riddles >> easy >> Magic total with cards
        
(Message started by: Noke Lieu on Feb 14^th, 2005, 2:59pm)

Title: Magic total with cards
Post by Noke Lieu on Feb 14^th, 2005, 2:59pm

It came to while this morning dozing,
A question that I could be posing.
I hope in my haste to get it hosted
It's not already solved and posted

Take a ordinary deck of cards and deal nine of them in a square.
What's the probability that
(1) at least
(2) exactly
two lines of three would have the same total?

(edit)
The value of the cards is A=1, 2=2, 3=3 ... j=11, q=12, k=13

By lines of 3, I mean they can be vertical or horizontal lines of 3 cards where the total of the values of the card in the line of 3 (horizontal or vertical) is the same.
as in

1 3 2 this is fine, because the first two columns add up to
5 4 6 15. The middle row does as well. Thus, at least 2.
9 8 7

1 2 3 This has only 3+9+8 = 4+7+9. Thus exactly 2.
4 7 9
6 5 8

Hope this clears up my obfuscation...

Title: Re: Magic total with cards
Post by Noke Lieu on Feb 20^th, 2005, 8:32pm

in my haste to get it out there before I forgot, I think its a bit too tough... What say we change it to just one suit?

Title: Re: Magic total with cards
Post by Noke Lieu on Feb 27^th, 2005, 3:47pm

Well, I may have misjudged the difficulty, but its worse putting an Easy in Hard than the other way round.

Bearing that in mind, maybe my method needs peer reviewing?
I will share my thinking for finding the probability of dealing out an instant 3x3 magic square.

It seems to me that there are only a few magic sqaures using 1-13. 12 of them in fact- (1x15, 3x18, 4x21, 3x24, 1x27). Maybe that should have been a puzzle on it own.

Each of those squares, because of rotation and reflection could be laid out 8 different ways.
Thus, there are 96 different ways of dealing out a magic square.
Sadly, there is 715 different ways of chosing 9 cards at random from 13. And 362880 ways of setting each of those out. I would divide that by 8 to remove the reflection and rotation side of things, giving 45360.

So there are 32432400 possible ways of dealing out 9 cards from 13. Only 96 of those are magic squares. So, the probability is:

96/32432400 (= 4/675675) which is pretty small. Understandably.

I worked that out late last night, on a coach back from Sydney, just carbon based computing, so there could be big errors. But the technique still seems fine in the light of day.

Title: Re: Magic total with cards
Post by SWF on Feb 27^th, 2005, 4:55pm

This problem may have been ignored because it was unclear how much a jack, queen, or king is. Are they 11, 12, 13, or blackjack values of 10? Based on your last post, they are 11, 12, 13.

Title: Re: Magic total with cards
Post by Noke Lieu on Feb 27^th, 2005, 5:57pm

yeah, good point. Have a tendency to assume too much.

You're correct SWF: A is one, J is 11, Q is 12, K is 13. Never even thought about the face cards/royals all being 10.

Title: Re: Magic total with cards
Post by markr on Feb 27^th, 2005, 7:17pm

on 02/14/05 at 14:59:35, Noke Lieu wrote:

two lines of three would have the same total?

I'm confused by this. Does this mean 2 of 3 horizontally, vertically, or either?

Title: Re: Magic total with cards
Post by Noke Lieu on Feb 27^th, 2005, 9:14pm

two line of three cards, vertical or horzontal, will ignore the diagonals. So much like a magic square.
Sorry this has been so unclear- will go back and edit original question.

Title: Re: Magic total with cards
Post by markr on Feb 27^th, 2005, 9:43pm

Why are you multiplying 12 by 8 (reflections/rotations), AND dividing 13!/4! by 8 (reflections/rotations)?

Title: Re: Magic total with cards
Post by Noke Lieu on Feb 27^th, 2005, 9:51pm

Was questioning this myself... Seemed appropriate, that's why.

It seemed odd to do the reverse of each, but was also attacking it from either end. With the x8 had one instance of it being true, when in fact there are 8 using the same numbers in the same order, just reflected or rotated.

The other instance, I calculated the total number of combinations. By the same logic, rather than... and the bubble just burst. Thank you. Feeling much better now. :-[ :)

Title: Re: Magic total with cards
Post by markr on Feb 27^th, 2005, 10:26pm

Actually, since you don't care about the diagonals, the multiplier (or divider) is 72. There are 6 ways to arrange rows, 6 ways to arrange columns, and a factor of 2 for rotation.

So I would compare (13!/4!)/72 to 12.

Title: Re: Magic total with cards
Post by Noke Lieu on Feb 27^th, 2005, 11:08pm

fair enough, but the magic square example is a sepparate problem where I was still condisering the diagonals- else it isn't a magic square....

But am a little confused as to whether that's you answer for the original question, or a correction for the example. (there have been drinks at work, I skipped lunch...)

Title: Re: Magic total with cards
Post by markr on Feb 27^th, 2005, 11:49pm

Here are my results for a simpler version of the problem (I'm working my way up to the original problem). The deck is only 1-9 and diagonals don't matter.

There are 5040 ways to deal the cards (not counting row transpositions, column transpositions, or rotations).

I get:
4179 ways that no pairs of rows or columns have the same sum
755 ways that one pair of rows (or columns) have the same sum
35 ways that one pair of rows AND one pair of columns have the same sum
67 ways that three rows but no columns (or three columns but no rows) have the same sum
3 ways that three rows and two columns (or three columns and two rows) have the same sum
1 way that three rows and three columns have the same sum

Note that I never compare the row sums to the column sums.

Title: Re: Magic total with cards
Post by markr on Feb 28^th, 2005, 11:32pm

Here are my results for the posted problem. The deck is 1-13 and diagonals don't matter.

There are 3,603,600 ways to deal the cards (not counting row transpositions, column transpositions, or rotations).

I get:
3,035,431 ways that no pairs of rows have the same sum and no pairs of columns have the same sum
534,918 ways that one pair of rows but no columns have the same sum
22,387 ways that one pair of rows have the same sum and one pair of columns have the same sum
10,264 ways that three rows but no columns have the same sum
520 ways that three rows have the same sum and two columns have the same sum
80 ways that three rows have the same sum and three columns have the same sum (these aren't necessarily magic)

I'm giving precedence to rows, but "rows" and "columns" can be switched.

Note that I never compare the row sums to the column sums.