The combinatorics of the clumsy dishwasher Adapted from digital dice by Paul Nahin. I've been struggling with this for a few days.
Question: A restaurant employs 5 dishwashers. In a week they break 5 dishes, with 4 breakages due to the same individual. What is the probability of a single dishwasher smashing at least 4 of the 5 plates.
Attempt at solution:
There are 5^5 ways of assigning smashed plates to dishwashers. All we need to do now is work out how many ways one washer could have smashed at least 4 plates. This means adding up the case where they smash 4 and the case they smash 5. I think this means the numerator has to be 5 Permute 4(four smashed plates are assigned to a single washer) plus any one of the washers smashing all five, which would be another 5:
$$ \frac{p^5_4+5}{5^5} =  \frac{125}{5^5}$$
Another way to work this out would be to consider the probability of a single washer smashing all 5 plates :
$$({\frac{1}{5}})^5 $$
multiplying by 5 for each of the washers = $\frac{5}{5^5}$
OR a single washer breaking 4 and not breaking the last one:
$$ \frac{1}{5^4}*\frac{4}{5} $$
multiply by 5 for each position the breaking event could have come in and by 5 again for each of the washers =
$$ \frac{100}{5^5} $$
adding these all up:
$$ \frac{100}{5^5} + \frac{5}{5^5} = \frac{105}{5^5} $$
The answer in the book however is:
$$ \frac{4* {5 \choose 4}+{5 \choose 5}} {5^5} = \frac{21}{5^5} $$
I have a feeling the book answer is P(one washer smashes >4 plates) whereas I've worked out P(any of the 5 washers in the kitchen smashing >4 plates)
EDIT: The author provides a monte-carlo estimate which essentially amounts to:
total_clumsy_events = 0
for i in sample:
    for j in 1:5:
        breaks = 0
        break_prob = uniform[0,1]
        if break_prob < 0.2:
            breaks+1

    if breaks>3:
        total_clumsy_events+1

Which to my mind is only evaluating:
p(5 breaks by one washer) + p(4 breaks by one washer) =

$$ \frac{1}{5^5} + \frac{4*5}{5^5} $$
(multiplying by 5 for the different orders of events). This of course equals the book's answer of $\frac{21}{5^5}$.
I'd really appreciate someone helping me sort out my thinking; What am I doing wrong?

*

*If I am right and we're answering different questions, why doesn't my first attempt work? ($\frac{125}{5^5}$


*Is it ambiguous what is being asked?


*Why does $4* {5 \choose 4}+{5 \choose 5}$ give the correct result?

Far simpler approach: 1/5
Why?

with 4 breakages due to the same individual.

This means that it is not a random chance for each plate, and how many plates any dishwasher breaks is not distributed in any normal statistical distribution, since we are concerned with a posterior evaluation, aka everything has already happened. Someone already broke at least 4 of those 5 plates.
Any given dishwasher either has broken zero, one, four or five plates, with four or five having the same "value" for us (="true"), as well as zero and one being the same for us: that dishwasher has not broken "4 or more plates". ( a dishwasher breaking 2 or 3 plates is not possible since we know someone broke 4 or more out of the 5 total breaks)
Also, we know that there is a dishwasher who has broken 4 or more plates, so the only question is whether the one dishwasher we're inquiring about is also the one that broke multiple plates, which is a 1 in 5 chance assuming fair distribution.
[I know that this seems too simple, please tell me where I went wrong]
