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Help with Weird Probability/Math Question?

  
 
Perfect Tommy
11:11 / 21.10.01
I'm not quite sure how to properly phrase my questions so that Barbelithian math-headz are intrigued before they are offended at how dumb this all is...

Eh, screw it. I play geeky collectible card games -- Magic: The Gathering is the most well known (though it's not one that I've played in several years). When I was recently learning a new one, I realized that the reason I enjoy building decks to play with is the process is that it's sort of like building a machine out of salad: you take various elements (the cards) in various proportions, mix it thoroughly (shuffling), and hope that you get the right elements at the right stages of the game. I have been trying to think of how to apply this idea to programming projects... f'rinstance, I thought that one might be able to represent various things to do when improvising music as "cards" and see if you can build a "deck" in a computer which spits out jazz after you shuffle it. That kind of thing.

However, I haven't taken a math class for a long time, and when I did, probability was a subject we tended to skip over as quickly as possible.

First question: If you draw 6 cards from a 60-card deck, I seem to remember that the chance of getting a particular card is 1/60 + 1/59 + 1/58 + 1/57 + 1/56 + 1/55. Intuitively, one might think it's 6/60. Is 6/60 a close approximation or a horrible lie in all cases?

The second question is the bit that I get embarrassed at, because while Magic and Shadowfist are games I play that seem pretty easy, the World Wrestling Federation card game (so help me) is the hardest for me to figure out, mathematically.

Cards in the WWF game have a Fortitude (F) requirement, and a Damage (D) value. To play, say, a 10F card, you must have 10 Fortitude on the table, which you get by playing cards with a combined D value of 10 or greater. For example, something like a Punch does only 3D, but it takes 0F to play. Something fancy like the Undertaker's Tombstone Piledriver might do 25D (ouch!), but it takes 30F to play. So now we know why wrestlers take so long to get to their fancy maneuvers -- they haven't built up enough Fortitude.

All WWF decks are exactly 60 cards, but different wrestlers draw a different number of cards at the beginning of the game, just to complicate matters further. I'm trying to figure out the answers to such questions as, "How probable is it that The Dudley Boyz, who draw 9 cards, will have 13F at the end of their first turn, if they have 3 0F/6D cards, 6 0F/5D cards, 6 0F/4D cards, 3 6F/5D cards, 1 5F/8D card, 3 10F/6D cards, considering the remaining 38 cards blank for the moment?"

Not that I want someone to do my math for me, but I don't even know where to begin... anyone have a hint? I promise that it's not just because my friend and I got 2nd in the Tag Team tournament, it's for Mad Science!

[ 21-10-2001: Message edited by: doubting thomas ]
 
 
Magic Mutley
11:41 / 21.10.01
I'll take at shot at the first question - you're right about the answer, but it might help to see why.

You're looking to draw a certain card from a 60 card pack, in a deal of 6 cards.

1st draw - there are 60 cards in the pack. Chances of drawing a particular card is 1/60

2nd draw - there are now fifty nine cards in the pack (you've taken one). Chances of drawing a particular card is 1/59

3rd draw - there are now fifty eight cards in the pack. Chances of drawing a particular card is 1/58

3rd draw - 1/57

4th draw - 1/56

5th draw - 1/55

6th draw - 1/54

The overall chances of drawing a certain card in the deal is the sum of these
= 1/60 + 1/59 + 1/58 + 1/57 + 1/56 + 1/55 + 1/54

I'm not quite clear about the notation in the second question -

"if they have 3 0F/6D cards, 6 0F/5D cards, 6 0F/4D cards, 3 6F/5D cards, 1 5F/8D card, 3 10F/6D cards"

Say for 3 0F/6D - am I right that 0F/6D is a card valued "zero Fortitude & six Damage". Do you mean there are 3 of these cards in the pack?
 
 
Perfect Tommy
13:13 / 21.10.01
Yes, that's right -- 3 out of the 60 cards are 0 Fortitude for 6 Damage, 6 out of the remaining 57 are 0 Fortitude/5 Damage, etc.
 
 
Magic Mutley
15:43 / 21.10.01
Another question: what do you get for Fortitude if you pick one of the other 38 'blank' cards? Do you get 0, in which case we can count the number of cards which add zero to Fortitude as 3 + 6 + 6 + 38 = 53 - or am I misunderstanding something?

[ 21-10-2001: Message edited by: Wheaty Goodness ]
 
 
Perfect Tommy
23:21 / 21.10.01
I was assuming those cards add nothing to fortitude; in fact, that they do nothing at all for the purpose of the question. Keep in mind that when I say something is "0F/5D", it takes zero Fortitude to play it, but it adds 5 to the amount of Fortitude you have available. I should have changed the terminology, because when I'm re-reading all this it sounds more confusing than I'd realized.

Perhaps it would be more clear to say "Fortitude Required" (r) and "Fortitude Built" (b):
quote:How probable is it that The Dudley Boyz, who draw 9 cards, will have 13b at the end of their first turn, if they have 3 0r/6b cards, 6 0r/5b cards, 6 0r/4b cards, 3 6r/5b cards, 1 5r/8b card, 3 10r/6b cards, considering the remaining 38 cards as 0r/0b for the moment?
In the real deck, they're either bigger maneuvers that require more fortitude and seem to complicate matters, or various non-damaging wrestler actions like "Throw Opponent into the Corner Turnbuckle", or "Offer Handshake", or "Study for Your G.E.D." because education is very important to pro wrestlers. (I am not making that card up.) The reason I chose 13 Fortitude as a target value was that most of the remaining 38 cards are playable once you have that first 13F -- it's a kind of hurdle to clear for most of the deck to work.

In addition, the reason I'm interested in how probable that first 13 Fortitude is, one doesn't generally want to fill the deck with 0r maneuvers, because the cards with high Fortitude requirements are more helpful to actually winning -- you don't want to draw a bunch of Backhand Slaps and Hair Pulls in the endgame, because they're wimpy for actual damage-dealing.

Quick show of hands: Who is surprised and shocked that I dropped out of electrical engineering as fast as I could find the administrative office? Anyone? Anyone?
 
 
Magic Mutley
17:17 / 22.10.01
I reckon the best way to do this generally is to write a computer simulation of the deck. Put all the cards in an array & pick out nine - be careful to mark the cards so you can't deal them twice. Take the total, and register if it's above 13. If you repeat this thousands of times, you should get a good estimate of the probability.

Anyway, since I'm unemployed &...erm...didn't have much to do today, I gave it a shot working this out using maths. It's a long time since I tried anything like this, so it's pretty likely I've got it very wrong. If anyone can point out mistakes - or a quicker way of doing it - I'd be grateful.

I'll try and answer the question -

For a pack of 60 cards made up of the following cards -
6 of 4b
9 of 5b
6 of 6b
1 of 8b
38 of 0b

What's the probability of dealing a hand of 9 cards where the total of the b values comes to at least 13b?

As far as I can see, the 'r' values don't come into this - if this isn't the question you asked, then damn, I'm already wrong...

I think it's easier to start by working out the prob of getting less than 13b.

Ok, dealing 9 cards, these are all the ways I can think of of getting under 13b

0 is a zero scoring card
s is a scoring card as specified

i) 4b 4b 4b 0 0 0 0 0 0

ii) s s 0 0 0 0 0 0 0 where s s =4b 4b, 4b 5b, 4b 6b, 4b 8b, 5b 5b, 5b 6b, 6b 6b, or 8b 4b

iii) s 0 0 0 0 0 0 0 0 where s =4b, 5b, 6b, or 8b

iv) 0 0 0 0 0 0 0 0 0

I've shown deals with the 's' cards at the begining, but the deal is valid in any order.
i.e. 4b 4b 0 4b 0 0 is a valid deal for i)

So i) means six zero scoring cards and three 4b cards in any order
ii) means seven zero scoring cards and any of the pairs listed, in any order
iii) means eight zero scoring cards and any of the single cards listed, in any order
iv) is all zero scoring cards.

[ 22-10-2001: Message edited by: Wheaty Goodness ]
 
 
Magic Mutley
17:22 / 22.10.01
CASE i)
=======

a) 4b 4b 4b 0 0 0

Card_____ 4b___4b___4b___0b___0b___0b___0b___0b___0b
Probability 6/60 5/59 4/58 38/57 37/56 36/55 35/54 34/53 33/52

(I had to use underscores 'cos UBB strips spaces - they're supposed to be in columns, & the probabilities refer to the cards in the row above them)

The prob of this deal is 6/60 x 5/59 x 4/58 x 38/57 x 37/56 x 36/55 x 35/54 x 34/53 x 33/52

To get the probability of dealing these in any order, you have to multiply this by the number of arrangements of these cards.
4b 4b 0b 4b 0b 0b 0b 0b 0b
4b 4b 0b 0b 4b 0b 0b 0b 0b
4b 4b 0b 0b 0b 0b 0b 0b 0b
...
etc

For 9 cards made up of three indistinguishable 4bs and six indistinguishable 0bs, there are possible 84 arrangements - 9!/(3!x6!)

! means 'factorial'
1! = 1
2! = 2x1 = 2
3! = 3x2x1 = 6
4! = 4x3x2x1 = 24
etc.

Total = (6*5*4*38*37*36*35*34*33/60*59*58*57*56*55*54*53*52) * 9!/(3!*6!)

= (238522838400 / 5364506808460800) * 84

TOTAL for Case i) = 0.0037349

[ 22-10-2001: Message edited by: Wheaty Goodness ]
 
 
Magic Mutley
17:42 / 22.10.01
Case ii)
=========
s s 0 0 0 0 where s s =4b 4b, 4b 5b, 4b 6b, 4b 8b, 5b 5b, 5b 6b, 6b 6b, or 8b 4b

b) s s=4b 4b

4b 4b 0b 0b 0b 0b 0b 0b 0b
6/60 5/59 38/58 37/57 36/56 35/55 34/54 33/53 32/52

This is 9 cards made up of two indistinguishable 4bs and seven indistinguishable 0bs
Number of possible arrangements = 9!/(2!x7!)


Total = (6*5*38*37*36*35*34*33*32 / 60*59*58*57*56*55*54*53*52) * 9!/(2!*7!)

= 1908182707200 / 515817962352000 * 36

c) s s=4b 5b
4b 5b 0b 0b 0b 0b 0b 0b 0b
6/60 9/59 38/58 37/57 36/56 35/55 34/54 33/53 32/52

Arrangements = 9!/7!

Total = 6*9*38*37*36*35*34*33*32 / 60*59*58*57*56*55*54*53*52) * 9!/7!

= 3434728872960 / 5364506808460800 * 72

d) s s=4b 6b
4b 6b 0b 0b 0b 0b 0b 0b 0b
6/60 6/59 38/58 37/57 36/56 35/55 34/54 33/53 32/52

Arrangements = 9!/7!

Total = (6*6*38*37*36*35*34*33*32 / 60*59*58*57*56*55*54*53*52) * 9!/7!

= 2289819248640 / 5364506808460800 * 72

e) s s=4b 8b
4b 8b 0b 0b 0b 0b 0b 0b 0b
6/60 1/59 38/58 37/57 36/56 35/55 34/54 33/53 32/52

Arrangements = 9!/7!

Total = (6*1*38*37*36*35*34*33*32 / 60*59*58*57*56*55*54*53*52) * 9!/7!

= 381636541440 / 5364506808460800 * 72

f) s s=5b 5b
5b 5b 0b 0b 0b 0b 0b 0b 0b
9/60 8/59 38/58 37/57 36/56 35/55 34/54 33/53 32/52

Arrangements = 9!/2!7!

Total = (9*8*38*37*36*35*34*33*32 / 60*59*58*57*56*55*54*53*52) * 9!/(2!*7!)

= 4579638497280 / 5364506808460800 * 36

g) s s=5b 6b
5b 6b 0b 0b 0b 0b 0b 0b 0b
9/60 6/59 38/58 37/57 36/56 35/55 34/54 33/53 32/52

Arrangements = 9!/7!

Total = (9*6*38*37*36*35*34*33*32 / 60*59*58*57*56*55*54*53*52) * 9!/7!

= 3434728872960 / 5364506808460800 * 72

h) s s=6b 6b
6b 6b 0b 0b 0b 0b 0b 0b 0b
6/60 5/59 38/58 37/57 36/56 35/55 34/54 33/53 32/52

Arrangements = 9!/2!7!

Total = (6*5*38*37*36*35*34*33*32 / 60*59*58*57*56*55*54*53*52) * 9!/(2!*7!)

= 1908182707200 * 5364506808460800 * 36

i) s s= 8b 4b
8b 4b 0b 0b 0b 0b 0b 0b 0b
1/60 6/59 38/58 37/57 36/56 35/55 34/54 33/53 32/52

Arrangements = 9!/7!

Total = (1*6*38*37*36*35*34*33*32 / 60*59*58*57*56*55*54*53*52) * 9!/7!

= 381636541440 / 5364506808460800 * 72


TOTAL for case ii) = 0.3098903

[ 22-10-2001: Message edited by: Wheaty Goodness ]
 
 
Magic Mutley
17:43 / 22.10.01
Case iii)
=========
s 0 0 0 0 0 where s=4b, 5b, 6b, or 8b

j) s=4b
4b 0b 0b 0b 0b 0b 0b 0b 0b
6/60 38/59 37/58 36/57 35/56 34/55 33/54 32/53 31/52

Arrangements = 9!/8! = 9

Total = (6*38*37*36*35*34*33*32*31 / 60*59*58*57*56*55*54*53*52) * 9

= 11830732784640 / 5364506808460800 * 9

k) s=5b
5b 0b 0b 0b 0b 0b 0b 0b 0b
9/60 38/59 37/58 36/57 35/56 34/55 33/54 32/53 31/52

Arrangements = 9!/8! = 9

Total = (9*38*37*36*35*34*33*32*31 / 60*59*58*57*56*55*54*53*52) * 9

= 17746099176960 / 515817962352000 * 9

l) s=6b
6b 0b 0b 0b 0b 0b 0b 0b 0b
6/60 38/59 37/58 36/57 35/56 34/55 33/54 32/53 31/52

Arrangements = 9!/8! = 9

Total = (6*38*37*36*35*34*33*32*31 / 60*59*58*57*56*55*54*53*52) * 9

= 11830732784640 / 5364506808460800 * 9

m) s=8b
8b 0b 0b 0b 0b 0b 0b 0b 0b
1/60 38/59 37/58 36/57 35/56 34/55 33/54 32/53 31/52

Arrangements = 9!/8! = 9

Total = (1*38*37*36*35*34*33*32*31 / 60*59*58*57*56*55*54*53*52) * 9

= 1971788797440 / 5364506808460800 * 9

TOTAL for Case iii) = 0.3526390

[ 22-10-2001: Message edited by: Wheaty Goodness ]
 
 
Magic Mutley
17:44 / 22.10.01
Case iv)
=========
0b 0b 0b 0b 0b 0b

n)

0b 0b 0b 0b 0b 0b 0b 0b 0b
38/60 37/59 36/58 35/57 34/56 33/55 32/54 31/53 30/52

Arrangements = 1

Total = 38*37*36*35*34*33*32*31*30 / 60*59*58*57*56*55*54*53*52) * 1

= 59153663923200 / 5364506808460800

TOTAL for Case iv) = 0.0110269
 
 
Magic Mutley
17:46 / 22.10.01
Total for all cases = 0.6772911

Probability of getting less than 13b is 0.6773 (rounded to 4 decimal places)

So for the probability of getting 13b or more I get 1-0.6773 = 0.3227 or 32.27%

Dunno if it sounds right...

[ 22-10-2001: Message edited by: Wheaty Goodness ]
 
 
Perfect Tommy
00:17 / 23.10.01
!!!!!

Thank you, Wheaty G!
 
 
Magic Mutley
19:25 / 24.10.01
No probs - though it could well be completely wrong...

As I said, I reckon you're best writing a computer simulation.
 
 
toadchrist
03:05 / 03.11.01
On the subject of improvised music using cards, John Zorn used table-top role playing games to give directions to his musicians on how to improvise on a given section. The group was called Cobra, I think....
 
 
Perfect Tommy
06:23 / 03.11.01
*slaps forehead*

I can't believe I forgot to start hunting down stuff on Cobra. I saw him play with Mike Van Patten, but missed the night that he did Cobra.

Thanks!
 
  
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