[Question]
  [
In Magic: the Gathering, mages (known as "planeswalkers") battle each other by casting spells. Spells cost mana. Five colors of mana exist: White, Blue, Black, Red, and Green, represented as {W}, {U}, {B}, {R}, and {G}, respectively.
A spell's cost is slightly more complex. The cost can be any combination of the following:
* One or more colors
* One or more colorless, represented as {X}, where X is a positive integer
* One or more hybrids, represented as {Y/Z}, where Y and Z are either a color (represented by one of the five letters) or colorless, represented by a positive integer
The following rules apply when attempting to cast a spell:
* A color in a cost must be satisfied by one mana of that color
* A colorless cost {X} may be satisfied by X mana of any color
* A hybrid cost {Y/Z} may be satisfied by satisfying either Y or Z
	+ Note that braces are not nested
	+ Y and Z are not hybrid
Write a program or function that, given a pool of mana and a cost, prints or returns true (or some truthy value) if and only if the mana in that pool can satisfy the cost, else false (or some falsy value).
A mana pool is a non-empty string of the format:
```
Color1,Color2,Color3,...,Colorn-1,Colorn
```
A cost is a non-empty string of the format:
```
Cost1,Cost2,Cost3,...,Costn-1,Costn
```
**Examples**
In the format `Pool Cost -> ExpectedOutput` (with a space between Pool and Cost):
```
{R},{R},{G},{B},{R} {4},{R} -> True
{G},{G},{G},{G},{W},{W},{W} {2/W},{2/U},{2/B},{2/R},{2/G} -> False
{G},{G},{R} {R/G},{G/B},{B/R} -> True
{R},{R},{R},{G} {1},{G},{2/G}-> True
{R} {R},{R},{R},{R},{R} -> False
{W},{R},{R} {2/W},{W/B} -> True
{U},{U} {1} -> True
{W},{R},{G} {1},{2} -> True
```
  
[Answer]
# Pyth, ~~55~~ ~~53~~ ~~52~~ 50 bytes
```
FN*Fmsm?k}kG^Gvkcd\/ceKc-rz0`Hd\,#=sN)I!.-NhK1B)E0
```
Try it online: [Demonstration](https://pyth.herokuapp.com/?code=FN*Fmsm%3Fk%7DkG%5EGvkcd%5C%2FceKc-rz0%60Hd%5C%2C%23%3DsN)I!.-NhK1B)E0&input=%7BW%7D%2C%7BR%7D%2C%7BR%7D+%7B2%2FW%7D%2C%7BW%2FB%7D&debug=0) or [Test harness](https://pyth.herokuapp.com/?code=zFz.zp%22%20%3D%3E%20%22zFN*Fmsm%3Fk%7DkG%5EGvkcd%5C%2FceKc-rz0%60Hd%5C%2C%23%3DsN)I!.-NhK1B)E0&input=Test-Cases%3A%0A%7BR%7D%2C%7BR%7D%2C%7BG%7D%2C%7BB%7D%2C%7BR%7D%20%7B4%7D%2C%7BR%7D%0A%7BG%7D%2C%7BG%7D%2C%7BR%7D%20%7BR%2FG%7D%2C%7BG%2FB%7D%2C%7BB%2FR%7D%0A%7BR%7D%2C%7BR%7D%2C%7BR%7D%2C%7BG%7D%20%7B1%7D%2C%7BG%7D%2C%7B2%2FG%7D%0A%7BR%7D%20%7BR%7D%2C%7BR%7D%2C%7BR%7D%2C%7BR%7D%2C%7BR%7D%0A%7BW%7D%2C%7BR%7D%2C%7BR%7D%20%7B2%2FW%7D%2C%7BW%2FB%7D%0A%7BU%7D%2C%7BU%7D%20%7B1%7D%0A%7BW%7D%2C%7BR%7D%2C%7BG%7D%20%7B1%7D%2C%7B2%7D&debug=0)
Notice that the time and memory complexity is really bad. So the second example doesn't work. I allocates about 1.6 GB of Ram before it crashes on my machine. 
### Explanation
The explanation is for the 53 solution. The only difference is, that the initial parsing happens in the middle instead of the beginning. 
```
Kc-rz0"{}"dFN*Fmsm?k}kG^Gvkcd\/ceKc-rz0`H\,#=sN)I!.-NhK1B)E0
```
So here's the initial parsing. 
```
Kc-rz0`Hd
   rz0     convert input() to lowercase
  -   `H   remove all curly brackets (`H = "{}")
 c      d  split at the space
K          assign to K
```
So the input `"{W},{R},{R} {2/W},{W/B}"` gets converted into `['w,r,r', '2/w,w/b']`. 
```
m               ceK\,    map each cost d of the costs split by "," to:
 s                         the sum of
  m         cd\/           map each value k of cost split by "/" to:
    k                        k
   ? }kG                     if k in "abcdef...xyz" else
        ^Gvk                 Cartesian product with "abc...yz" of int(k) repeats
```
So what does this do? The cost input `'2/w,w/b'` gets converted into:
```
[['aa', 'ab', 'ac', ..., 'zx', 'zy', 'zz', 'w'], 'wb']
```
Every string in `['aa', 'ab', 'ac', ..., 'zx', 'zy', 'zz', 'w']` satisfies `{2/W}` and every char in `'wb'` satisfies `{w/b}`. 
Now we generate the Cartesian product of these lists (or strings) and see, if any combination can be produced with the mana-pool. 
```
FN*F...              )      for N in Cartesian product of ...:
       #   )                   while 1:
        =sN                      N = sum(N)
                               this flattens N
            I!.-NhK            if not (subtract mana pool from N):
                   1             print 1 (True)
                    B            break
                      E      else:
                       0       print 0 (False)
```
[Answer]
# Python 2.7, 412 characters
```
import re,collections as C
r,C=re.findall,C.Counter
def g(m,h,c,v):
 try:return t(m,h,c+int(v))
 except:
  if m[v]:return t(m-C({v:1}),h,c)
def t(m,h,c):return any(g(m,h[1:],c,v)for v in h[0].split('/'))if h else sum(m.values())>=c
def f(m,c):m=C(r(r'\w',m));c=[filter(None, x)for x in zip(*r(r'(\w+/\w+)|(\d+)|(\w)',c))];m.subtract(C(c[2]));print all(x>=0 for x in m.values())*t(m,c[0],sum(int(x)for x in c[1]))
```
The function `f` is the one that does the check. It takes the mana pool and cost as string arguments, and prints `1` when the mana satisfies the cost and `0` otherwise. For example, `f('{R},{R},{G},{B},{R}', '{4},{R}')` prints `1`.
Ungolfed, it basically looks like this
```
import re
from collections import Counter
def helper(mana, hybrids, colorless, option):
  try:
    option = int(option) # See if option is an integer
    # For colorless hybrid, just add the value to the colorless amount
    # to check at the end.
    return check_hybrids(mana, hybrids, colorless + option)
  except ValueError: # Option is a mana letter
    # For colored hybrid costs, check if any of that color is
    # available, then try to pay the rest of the cost with 1 less
    # of that color.
    if mana[option] >= 0:
      return check_hybrids(mana - Counter({option: 1}), hybrids, colorless)
    else:
      return False
def check_hybrids(mana, hybrids, colorless):
  '''Check whether the given mana pool can pay the given hybrid costs and colorless costs'''
  if hybrids:
    # For each option in the first hybrid cost, check whether the
    # rest of the cost can be paid after paying that cost
    return any(helper(mana, hybrids[1:], colorless, option) for option in hybrids[0].split('/'))
  else:
    # When there are no remaining hybrid costs, if there is enough
    # remaining mana to pay the colorless costs, we have success
    return sum(m.values()) > colorless
def can_cast(mana_str, cost_str):
  mana = Counter(re.findall(r'\w', mana_str))
  # transpose to get separate lists of hybrid, colorless, and colored symbols
  cost = zip(*re.findall(r'(\w+/\w+)|(\d+)|(\w)',cost_str))
  cost = [filter(None, sublist) for sublist in cost] # Remove unfound symbols
  mana.subtract(Counter(cost[2]))
  # After subtracting the single-colored cost from the mana pool, if
  # anything in the mana pool is negative, we didn't have enough to
  # pay for that color.
  if any(x <=0 for x in mana.values()):
    return False
  return check_hybrids(mana, cost[0], sum(int(x)for x in cost[1]))
```
]