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7 coins  probability over a dollar 
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#19
Jan413, 06:20 AM

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In this type of questions priors must always be stated otherwise you produce something like the "two envelopes problem" as an extreme case.



#20
Jan613, 02:21 AM

Sci Advisor
P: 1,169

Maybe you can simplify your calculations a bit by using the fact that you need at least two quarters to have a total to add up to $1.00 . That leaves one option : 5 dimes.
Then you can have 3 quarters , and 3^4=81 cases, or 4 quarters , and certainty. So your winning cases of total larger than $1.00 are (using _exact_ number of quarters): 0 quarters: 0 ways , 3^7 nonways 1 quarter : 0 ways 3^6 nonways. 2 quarters: 1 way ( 5 10's) 3 quarters: only complicated cases 4 quarters: 3^3 ways 5 quarters: 3^2 ways, etc. So it comes down to having four nonquarter coins adding up to $0.25ormore. 


#21
Jan613, 04:36 AM

P: 523




#22
Jan613, 08:22 AM

Mentor
P: 15,058




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