Talk:Jump Attack

Mathematical expectation

I'm the one who wrote the "Mathematical expectation" section. I'm not really good at math so I want you to check the program I used to calculate the data. It's written in Python 2:

one=1
five=2
x5=3
bowser=4

import itertools

l=[one,one,one,one,five,five,five,x5,x5,bowser,bowser]
p=itertools.permutations(l,5)                              # change to 7, 9

def coins(l):
    l=list(l)
    s=0
    while len(l):
        a=l[0]
        del l[0]
        if a==one:
            s+=1
        elif a==five:
            s+=5
        elif a==x5:
            s*=5
        elif a==bowser:
            s=0
            break
        else:
            raise NameError('wtf?')
    return s

ans=[coins(l) for l in p]
s=sum(ans)

print s/len(ans)
print s
print len(ans)

The result is:

11
661200
55440

11
18385920
1663200

3
71245440
19958400

Since the 9-Block Option has a bonus, the real number of coins earned is 6.

I daren't calculate the variance because my machine is slow. Another gossip-loving Toad (talk) 22:32, 20 October 2014 (EDT)

Hi, I was curious about the expected reward as well, and when I saw the wiki page didn't list them, I ended up writing uncannily similar code as yours (I only checked the "talk" page after writing it), and found the same approximate values.

I also saw that you'd added this to the page at first, but later removed them again since you didn't trust your numbers...I suppose my independently written code arriving at the same numbers would fully corroborate it, assuming the game does not secretly alter the game rules in the player's favor or something.

With that caveat, I'm adding it back into the article.

(Here's my code, python 3)

 import numpy as np
  
 bowser = 0
 one = 1
 five = 5
 xfive = 25
 blocks = [one, one, one, one, five, five, five, xfive, xfive, bowser, bowser]
  
 def evaluate(blocks):
     total = 0
     for block in blocks:
         if block == one or block == five:
             total += block
         elif block == xfive:
             total *= 5
         else:
             return 0
     return total
  
 count = 100_000 # Number of simulated games to run, increase for more accurate average
 mode = 5 # Chance to 7 or 9
 total = sum( evaluate(np.random.permutation(blocks)[:mode]) for _ in range(count) )
 print(total/count)

--Rezuaq (talk) 05:15, December 23, 2021 (EST)

Update: I realised you don't even need to simulate anything, it's actually faster to just compute every possible winning game and use that to get the exact expected values. I have amended the table.

Python 3 code:

   import itertools
   import math
    
   one = 1; five = 5; xfive = 25
   winningBlocks = [one, one, one, one, five, five, five, xfive, xfive]
    
   def evaluate(blocks):
       total = 0
       for block in blocks:
           if block == one or block == five:
               total += block
           elif block == xfive:
               total *= 5
       return total
    
   print()
   for mode in [5, 7, 9]:
       print('====== %d-Block Option =====' % mode)
       count = math.factorial(11)//math.factorial(11-mode)
       print('# of combos:', count)
       winCount = math.factorial(9)//math.factorial(9-mode)
       print('# of non-losing combos:', winCount)
    
       # Only evaluate the winning combos-- far less than the actual total number of combinations.
       total = sum(evaluate(perm) for perm in itertools.permutations(winningBlocks, mode))
    
       print()
       print('E[x | winning] =', total/winCount)
       print('E[x] =', total/count)
       if mode == 9: print('(9-block bonus multiplier: Multiply by 2 to get the in-game reward)')
       print()

--Rezuaq (talk) 06:03, December 24, 2021 (EST)