I’ve just tried to calculate the expected duration, I make it 168 551/2310, so about 170 as you had it.

If you want to try this yourself don’t read on just yet.

An event which occurs with probability p will occur on average after 1/p occasions. For example if you repeatedly throw a die then the chance of a six is 1/6. You would expect to wait 1/(1/6) = 6 throws. This makes sense, if you kept throwing the die then one in six throws would be a six so you expect the average wait for the next six to be six throws.

In this puzzle there is one ‘counter’ and 11 ‘signallers’. We are looking for the following sequence:

first new signaller (probability 11/12 so expected wait is 12/11 visits)

counter (probability 1/12 so expected wait is 12)

second new signaller (probability 10/12 so expected wait is 12/10)

counter (probability 1/12 so expected wait is 12)

…..

eleventh new signaller (probability 1/12 so expected wait is 12/1)

counter (probability 1/12 so expected wait is 12)

The total expected wait is 12( 1 + 1/2 + 1/3 … 1/11) + 11×12 which I make 168 551/2310. (I did the sum by hand so apologies if it’s not quite right)

Cheers,

Steve

]]>Prisoner 2 is the accountant

Prisoner Claimed Accountant

Day Selected Switch Tallied

1 10 Y

2 6

3 7

4 3

5 10

6 8

7 2 Y

8 9 Y

9 3

10 6

11 5

12 6

13 5

14 8

15 11

16 8

17 1

18 9

19 10

20 7

21 0

22 2 Y

23 5 Y

24 1

25 3

26 9

27 1

28 10

29 3

30 1

31 5

32 5

33 7

34 0

35 8

36 10

37 0

38 3

39 7

40 4

41 9

42 5

43 10

44 6

45 1

46 1

47 2 Y

48 7 Y

49 11

50 5

51 2 Y

52 3 Y

53 7

54 11

55 8

56 10

57 8

58 9

59 1

60 11

61 2 Y

62 10

63 5

64 2

65 10

66 5

67 0 Y

68 11

69 0

70 11

71 7

72 10

73 4

74 9

75 8

76 10

77 10

78 11

79 5

80 9

81 8

82 11

83 5

84 3

85 10

86 1

87 5

88 10

89 3

90 10

91 1

92 5

93 8

94 10

95 11

96 7

97 4

98 11

99 10

100 4

101 3

102 5

103 6

104 7

105 6

106 3

107 9

108 4

109 6

110 2 Y

111 6 Y

112 2 Y

113 1 Y

114 11

115 9

116 3

117 4

118 2 Y

119 5

120 7

121 1

122 7

123 1

124 9

125 5

126 4 Y

127 8

128 1

129 4

130 6

131 10

132 11

133 3

134 4

135 6

136 9

137 11

138 4

139 6

140 5

141 10

142 4

143 11

144 4

145 3

146 8

147 11

148 7

149 11

150 5

151 7

152 4

153 0

154 8

155 5

156 5

157 4

158 2 Y

159 11 Y

160 0

161 0

162 9

163 11

164 8

165 5

166 6

167 9

168 5

169 2 Y

170 3

171 2

172 4

173 7

174 2

175 8 Y

176 2 Y

Firstly with 10^24 which is

The other is 10^-24 which is

Both of these extensions then lead back to one, making the sequence recur infinately with 0 1 8 10 19 90 10^24 10^-18 1 8 10 19 90 10^24 10^-18 1 … ]]>

I have another number to add to the sequence of 0 1 8 10 19 90 it’s 10^24, which is a Yotta (http://en.wikipedia.org/wiki/Yotta)

And I have to say I love the podcast, it’s awesome.

neil ]]>

You have to remember that the podcasts are re-broadcasted radio segments where the whole point is that you have a week (or more!) to work out the answer before the next programme.

Mr H ]]>

and, I definitely want to know the pattern to 0 1 8 10 19 90 and it can’t have another number.

Help! Thanks, love your podcast! margaret mmyatt01757@yahoo.com ]]>