Bobbie shoots free throws on a basketball court. She hits the first and misses the second, and thereafter the probability that she hits the next shot is equal to the proportion of shots she has hit so far. What is the probability she hits exactly 50 of her first 100
shots?
(In reply to
Solution: P&P, computer by Larry)
My simulation led me to believe all the possibilities (1 - 100) have equal probability, consistent with 1/99 for each of the non-zero cases:
clearvars
ct=zeros(100);
for trial=1:20000000
hits=1; misses=1;
for shots=3:100
r=rand;
if r<hits/(hits+misses)
hits=hits+1;
else
misses=misses+1;
end
end
ct(hits)=ct(hits)+1;
end
ct(50)/trial
for i=1:100
fprintf('%3d %6.4f \n',i,100*ct(i)/trial)
end
fprintf('\n')
1 1.0195
2 1.0168
3 1.0037
4 1.0150
5 1.0268
6 1.0128
7 1.0121
8 1.0133
9 1.0350
10 0.9921
11 1.0204
12 1.0005
13 1.0087
14 1.0054
15 0.9949
16 1.0195
17 1.0058
18 1.0157
19 1.0077
20 1.0035
21 1.0190
22 0.9956
23 1.0078
24 1.0144
25 1.0126
26 0.9963
27 0.9987
28 1.0049
29 1.0013
30 1.0344
31 1.0038
32 1.0090
33 1.0168
34 1.0150
35 1.0090
36 1.0060
37 0.9947
38 1.0027
39 1.0103
40 1.0174
41 1.0141
42 1.0007
43 1.0166
44 1.0141
45 1.0204
46 1.0122
47 1.0050
48 1.0091
49 1.0133
50 1.0163
51 1.0108
52 1.0265
53 1.0245
54 1.0294
55 1.0085
56 1.0041
57 0.9969
58 0.9991
59 0.9989
60 1.0086
61 1.0126
62 1.0067
63 1.0167
64 1.0057
65 1.0258
66 1.0114
67 1.0108
68 1.0065
69 1.0142
70 1.0138
71 0.9890
72 1.0056
73 1.0112
74 0.9976
75 1.0068
76 1.0010
77 0.9938
78 0.9979
79 1.0066
80 1.0071
81 1.0105
82 0.9940
83 1.0112
84 1.0131
85 1.0174
86 1.0056
87 1.0097
88 1.0084
89 1.0312
90 1.0078
91 1.0222
92 1.0128
93 1.0161
94 1.0064
95 1.0270
96 1.0269
97 0.9852
98 1.0159
99 1.0098
100 0.0000
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Posted by Charlie
on 2025-04-28 12:51:52 |
re: Solution: P&P, computer
