A plate of pasta contains 100 strands of spaghetti.
Tie two loose ends.
Keep doing this until there are no more loose ends.
What is the expected number of loops at the end?
A plate of pasta contains 100 strands of spaghetti.
Tie two loose ends.
Keep doing this until there are no more loose ends.
What is the expected number of loops at the end?
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IF E(1)=1, then E(n+1)=E(n)+1/(2n+1), So E(100)=1/199+1/197+1/19
-jinjing-
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08/10/2012 postreply
13:51:59
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E(100)=1/199+1/197+1/197+.....+1/3+1=3.27869....
-jinjing-
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08/10/2012 postreply
14:15:05
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谢谢!忘了说请教啦!我再想想 E(n+1)=E(n)+1/(2n+1)
-绿袖儿-
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08/10/2012 postreply
15:32:18
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谢题!We can get it when n=1,2,3 .if you are not mathematician.
-jinjing-
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08/10/2012 postreply
16:05:58
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E(1) = 1, E(n+1) = E(n) + 1. E(100) = 100.
-wxcfan123-
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08/11/2012 postreply
14:31:56
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请注意,loops,100根有200个头,自成,loop概率1/199,所以E(100)=E(99)+1/199)*1
-jinjing-
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08/11/2012 postreply
15:44:23
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抱欠。将loops理解成程序中的多少次循环操作。职业性思维。
-wxcfan123-
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08/11/2012 postreply
17:12:27
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你是不是只考虑了一根的loops?但是几根也可以成loop
-绿袖儿-
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08/12/2012 postreply
18:48:26
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细说一下递推公式,算是将功补过吧。
-wxcfan123-
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08/12/2012 postreply
19:36:29
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谢!我从来没有概率思维,搞不清各种情况。
-绿袖儿-
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08/13/2012 postreply
07:26:16
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