7 coins - probability over a dollar


by bloynoys
Tags: coins, dollar, probability
0xDEADBEEF
0xDEADBEEF is offline
#19
Jan4-13, 06:20 AM
P: 824
In this type of questions priors must always be stated otherwise you produce something like the "two envelopes problem" as an extreme case.
Bacle2
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#20
Jan6-13, 02:21 AM
Sci Advisor
P: 1,168
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 non-ways
1 quarter : 0 ways 3^6 non-ways.
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 non-quarter coins adding up to $0.25-or-more.
bpet
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#21
Jan6-13, 04:36 AM
P: 523
Quote Quote by Bacle2 View Post
...
So it comes down to having four non-quarter coins adding up to $0.25-or-more.
This step could be done using with generating function, ie expand [itex](x+x^5+x^{10})^4[/itex]
D H
D H is offline
#22
Jan6-13, 08:22 AM
Mentor
P: 14,456
Quote Quote by Bacle2 View Post
2 quarters: 1 way ( 5 10's)
That's not a solution because the total is exactly $1.00. The OP asks for "probability that a person has over a dollar in coins."

Quote Quote by bpet View Post
This step could be done using with generating function, ie expand [itex](x+x^5+x^{10})^4[/itex]
Even easier, don't break the problem down at all. Just use the generating function to form [itex](x+x^5+x^{10}+x^{25})^7[/itex] (expanded) and be done with it. Generating functions are magical.


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