All about flooble | fun stuff | Get a free chatterbox | Free JavaScript | Avatars
 perplexus dot info

 An enchanted coin (Posted on 2014-06-24)
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

 No Solution Yet Submitted by Ady TZIDON No Rating

Comments: ( Back to comment list | You must be logged in to post comments.)
 possible solution | Comment 1 of 3
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	     timesof 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\$

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
tosses = toss
r = Rnd(1)
If r <= p Then
End If
Next
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

 Posted by Charlie on 2014-06-24 15:23:52

 Search: Search body:
Forums (0)