Even I got a probability of 0.85....

The key to the solution is to figure out how many cases you can construct wherein a sum total of 50 coins is $1....

As Andy pointed out, there are only 2 such cases, which I think is the most important aspect to solving the problem... I came up for the number of ways by using brute force method....

Once we agree that there are only 2 cases possible, we assume that each of them is equally likely... Also we assume that in any given scenario any coin can get lost...

P[success| case 1] = 0.8 and P[success|case 2] =0.9... since each case is equally likely we get P[success]=0.85..

Here is how I worked out the problem...

In some senses we need to find all non negative integer solutions to :

x+y+z+w=50 (there are 50 coins)

x+5y+10z+25w=100 (they add up to $1)

x,y,z,w >= 0 and integer

I am sure there must be neat and clean ways (some more linear algebra ehhh

) to solve the above system but unfortunately i needed to get down to case by case analysis .....

First consider at max how many quarters can you have in the solution set ----

Obviously 4 or anything more is not possible

Can we have 3 quarters :-- no because then we have to distribute 25 cents within 47 coins... Not possible

Suppose we have 2 quarters.... In that case we will have to distribute 50 cents over 48 coins.... That is not possible becasue we must have a coin other than a penny (with all pennies you get 48 cents, 2 cents too short), and even if its the next smallest dimension a nickel ( key assumption nickel = 5 cents, am never sure of that) then we have 45 cents and 47 coins ---- No solution exists...

Hence we can have only 2 cases :--

Case 1 0 quarters

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if we start out with 0 quarters we can probably have at max 5 dimes -- with 6 dimes, you are left with 40 cents and 44 more coins

So we need to find out if non negative integer solutions exist for any of the following system of equations

a) 5 dimes

x+y=45 (45 coins left)

x+5y= 50 ( 50 cents to be filled in)

no solution

b)4 dimes

x+y=46 (46 coins left)

x+5y= 60 ( 60 cents to be filled in)

no solution

c) 3 dimes

x+y=47 (47 coins left)

x+5y= 57 ( 70 cents to be filled in)

no solution

d) 2 dimes

x+y=48 (48 coins left)

x+5y= 80 (80 cents to be filled in)

voila solution

40 pennies and 8 nickels

e) 1 dime

x+y=49 (49 coins left)

x+5y= 90 (90 cents to be filled in)

no solution

f) 0 dimes

x+y=50 (45 coins left)

x+5y= 100 (100 cents to be filled in)

no solution

Case 2 1 Quarter

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Well given the size of this posting, I think I can spare the details.... There is one solution within this category....

And hence we end up with only 2 cases, in with 40 pennies, and one with 45 ...