You have a magic coin.
The first time you flip the coin it will land heads up.
The second time you flip the coin it will land tails up.
Every time you
flip the coin after that, the probability that the coin will land heads up is proportional to the number of "heads up" that "run".
So for example, the probability of the third toss landing heads up is 1/2.
If the third toss is heads, then the probability of the fourth toss landing heads is 2/3, otherwise the probability of landing heads is only 1/3.
And so on.
Find the probability that the coin will land "heads up" exactly 42 times
in the first 100 tosses (in a single run).
Source: NCH contest
In 10,000 trials of the scenario, it looks as if the probability of each of the 99 possible numbers of heads is equal. That would be 1/99. Certainly the probability of exactly 42 heads would be the same as the probability of 42 tails or 58 heads.
number times
of heads occurred
1 99
2 120
3 104
4 110
5 120
6 106
7 99
8 80
9 95
10 97
11 97
12 95
13 110
14 110
15 106
16 102
17 92
18 109
19 85
20 113
21 108
22 94
23 96
24 101
25 110
26 96
27 89
28 104
29 97
30 100
31 99
32 108
33 85
34 98
35 78
36 103
37 86
38 102
39 103
40 96
41 116
42 110
43 87
44 101
45 96
46 102
47 100
48 99
49 105
50 95
51 116
52 95
53 108
54 108
55 97
56 102
57 101
58 92
59 84
60 76
61 114
62 99
63 85
64 101
65 95
66 105
67 101
68 95
69 104
70 98
71 119
72 128
73 139
74 94
75 107
76 110
77 89
78 93
79 94
80 110
81 106
82 97
83 112
84 93
85 113
86 90
87 105
88 110
89 92
90 84
91 108
92 106
93 131
94 90
95 96
96 92
97 102
98 101
99 100
A simple case is the probability of that first heads being the only one:
(1/2)*(2/3)*(3/4)*...*(98/99) = 1/99
For two heads it gets more complicated:
case 1, the second heads is on the third toss:
H T T T T
(1/2)*(1/3)*(2/4)*...*(96/98)*(97/99) = 1/(98*99)
as the numerators go from 1 to 97 and the denominators from 2 to 99.
case 2, the second heads is on the fourth toss:
T H T T T
(1/2)*(1/3)*(2/4)*...*(96/98)*(97/99) = 1/(98*99)
case 3, the second heads is on the fifth toss:
T T H T T T
(1/2)*(2/3)*(1/4)*(3/5)...*(96/98)*(97/99) = 1/(98*99)
These all have the same 1/(98*99) value and there are 98 of them, making the total of all cases 1/99.
I'd hate to go into this for each possible number of heads, but I suspect the answer would be the same: 1/99.
The simulation was done by:
DefDbl A-Z
Dim crlf$
Private Sub Form_Load()
ChDir "C:\Program Files (x86)\DevStudio\VB\projects\flooble"
Text1.Text = ""
crlf$ = Chr(13) + Chr(10)
Form1.Visible = True
DoEvents
Dim score(100)
For trial = 1 To 10000
heads = 1: tosses = 2
For toss = 3 To 100
p = heads / tosses
tosses = toss
r = Rnd(1)
If r <= p Then
heads = heads + 1
End If
Next
score(heads) = score(heads) + 1
Text1.Text = Str(score(42)) & Str(trial)
DoEvents
Next trial
Text1.Text = ""
For i = 1 To 99
Text1.Text = Text1.Text & Str(i) & Str(score(i)) & crlf
Next
End Sub
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Posted by Charlie
on 2014-06-24 15:23:52 |
possible solution
