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Probability Question
sicnarf
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#2
2009-10-25, 05:36 PM
Probability of 1st candy: 12/12
2nd: 11/12 (everything but the first candy found)
...
12th: 1/12

So the expected amount for the first would be 1, second would be 12/11, etc.

Total expected amount should be: 37.239ish

I think that's right. D:
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Messages In This Thread
Probability Question - by Hazzy - 2009-10-25, 04:01 PM
Probability Question - by sicnarf - 2009-10-25, 05:36 PM
Probability Question - by Hazzy - 2009-10-25, 05:53 PM
Probability Question - by y0y0y0y0shi0 - 2009-10-25, 06:06 PM
Probability Question - by sicnarf - 2009-10-25, 06:51 PM
Probability Question - by Hazzy - 2009-10-25, 06:55 PM
Probability Question - by Noah - 2009-10-25, 07:15 PM

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