CoCalc Public FilesProject / S21 / Final Project Draft / Place Final Project Draft here.txt
Author: Zhang Han
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1#. our project is no. 4 in the list .
24. Suppose that you roll four 6-sided dice. Estimate the probability that the die rolls are such that the four dice can be paired off so that one pair sums to 7 and the other sums to 9.
3becasuse the condition is paired dies' sum should be 7 or 9 ,so the choices for 2 chooses from 4 will be 4*3/2 =6 , that means there will be six different appearances that satisfy the requests(+1). besides these 6 phases, others are all no useful(+0).
4in order to make it more accurate, we did two ways  as follows and we have similar solution which is arround 0.075 which should be correct answer.
5
6solution 1 :
7
8times=[]                                              # int the trials that fit for the condition
9import numpy as np
10for e in range(0,1000000):                            # input the trials we want
11    res = np.random.randint(1, 7, size=1)             # die 1 : the random number on 6 sides die roll once
12    res2 =np.random.randint(1, 7, size=1)             # die 2 : the random number on 6 sides die roll once
13    res3= np.random.randint(1,7,size=1)               # die 3 : the random number on 6 sides die roll once
14    res4=np.random.randint(1,7,size=1)                # die 4 : the random number on 6 sides die roll once
15    if res +res2 == 7 and res3+res4 == 9:             # one posibility if the sum of no1 and no2 could be 7 and others two is 9, then add 1
16        i=1
17    elif res+res3==7 and res2+res4==9:                # one posibility if the sum of no1 and no3 could be 7 and others two is 9, then add 1
18        i=1
19    elif res+res4==7 and res2+res3==9:                # one posibility if the sum of no1 and no4 could be 7 and others two is 9, then add 1
20        i=1
21    elif res2+res3==7 and res+res4==9:                # one posibility if the sum of no2 and no3 could be 7 and others two is 9, then add 1
22        i=1
23    elif res2+res4==7 and res+res3==9:                # one posibility if the sum of no2 and no4 could be 7 and others two is 9, then add 1
24        i=1
25    elif res3+res4==7 and res+res2==9:                # one posibility if the sum of no3 and no4 could be 7 and others two is 9, then add 1
26        i=1
27    else:                                             # different from these 6 choices are unfit for the condition, then add 0
28        i=0
29    times.append(i)                                   # keep trying the trials untill 1000000 times
30pro=np.mean(times)                                    # def probability is mean of all "1"
31print(pro)
32
33the output :  0.074081
34
35
36solution 2 :
37def die_rolls():                     # defining die_roll for multiple runs
38    D1= np.random.randint(1,7)       # die 1 : random number on a six number die
39    D2=np.random.randint(1,7)        # die 2 : random number on a six number die
40    D3=np.random.randint(1,7)        # die 3 : random number on a six number die
41    D4=np.random.randint(1,7)        # die 4 : random number on a six number die
42    if (D1+D2==7 and D3+D4==9) or (D1+D2 ==9 and D3+D4==7): # This is one of the possible result we want where
43        return 1                                            # D1 & D2 is a pair and D3 and D4 is a pair. and one pair
44                                                            # adds up to 7 and the other adds up to 9. Which we return 1
45    elif (D2+D3==7 and D1+D4==9) or (D2+D3 ==9 and D1+D4==7):# This is one of the possible result we want where
46        return 1                                             # D2 & D3 is a pair and D1 and D4 is a pair. and one pair
47                                                             # adds up to 7 and the other adds up to 9. Which we return 1
48    elif (D2+D4==7 and D1+D3==9) or (D2+D4 ==9 and D1+D3==7):# This is one of the possible result we want where
49        return 1                                             # D2 & D4 is a pair and D1 and D3 is a pair. and one pair
50                                                             # adds up to 7 and the other adds up to 9. Which we return 1
51    else:                                                    # (Unwanted result) None of the statement above is true /
52        return 0                                             # Isn't a result where one pair adds up to 7 and the other
53                                                             # pair adds up to 9. So it returns 0
54  sum([die_rolls() for i in range(1000000)])/1000000             # adds up all the '1's (wanted results) and divide it by the
55                                                            # total number of trials. (in this case 1000000)
56the output :          0.073664
57
58