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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?

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

  2. Is it ambiguous what is being asked?

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

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    $\begingroup$ What is the probability of a single dishwasher smashing at least 4 of the 5 plates Umm, given the previous sentence with 4 breakages due to the same individual, the probability is 1, right? What am I missing? $\endgroup$ Jun 28, 2022 at 1:15
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    $\begingroup$ @JohnGordon Right -- or on one interpretation I thought it might be 1/5, because the "same individual" could be any of the 5, not necessary the specific "single dishwasher". $\endgroup$
    – nanoman
    Jun 28, 2022 at 3:36
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    $\begingroup$ @nanoman exactly my thoughts, I read the answers and was so confused when the answer seems so obvious. Did I (we?) miss something?? $\endgroup$
    – Hobbamok
    Jun 28, 2022 at 8:32

3 Answers 3

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$\frac{21}{5^5}$ is the probability a pre-identified individual smashed $4$ or $5$ plates (assuming each plate was independently equally likely to be smashed by anybody), and is $P(X\ge 4)$ when $X \sim Bin(5,\frac15)$. As the books seems to say.

$\frac{21}{5^4} = \frac{105}{5^5}$ is five times that and is the probability somebody smashed $4$ or $5$ plates, since it is impossible that two people each smashed $4$ or $5$ of the $5$ smashed plates. As your second attempt seems to say. Note that $5(4{5 \choose 4}+{5 \choose 5})=105$

Your $\frac{125}{5^5}$ looks harder to justify, especially your $p^5_4 =120$ for $4$ plates smashed by somebody and $1$ by someone else. In that case there are $5$ people who could be the big smasher and $4$ other people the little smasher and $5$ possible plates being smashed by the little smasher, so $5\times 4 \times 5=100$ possibilities, to which you later add $5$ to get $105$.

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  • $\begingroup$ thanks @Henry. Could you explain why 4∗(5C4)+(5C5) gives the correct numerator? $\endgroup$
    – RNs_Ghost
    Jun 27, 2022 at 14:57
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    $\begingroup$ @RNs_Ghost The probability is $P(X \ge 4)$ $= {5 \choose 4}\left(\frac15\right)^4 \left(\frac45\right)^1 + {5 \choose 5}\left(\frac15\right)^5 \left(\frac45\right)^0 $ $=\frac{{5 \choose 4}4 + {5 \choose 5}}{5^5}$ $\endgroup$
    – Henry
    Jun 27, 2022 at 15:29
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I am going to generalise your problem to allow you to deal with any number of dishwashers and broken dishes. (Also, I interpret the problem as being a condtional probability given a fixed number of broken dishes; the answer will be different if you don't condition on this.) I'll show the relevant combinatorial analysis below, but first I'll introduce you to the general distribution you are dealing with and show you some facilities for computing values from this distribution in the general case (i.e., allowing for any number of dishes and dishwashers).


This question requires you to compute a probability from the maxcount distribution (see related question). In general, we can consider a set of observed values $U_1,...,U_n \sim \text{IID U} \{ 1,...,m \}$ and we are interested in the distribution of the maxcount statistic, defined by:

$$M_n \equiv \max_{k=1,...,m} \sum_{i=1}^n \mathbb{I}(U_i = k).$$

In the context of your problem the value $n$ is the number of dishwashers and the number $m$ is the number of broken dishes, and you are seeking the probability $\mathbb{P}(M_n \geqslant 4)$. This can be computed combinatorially (see below) or it can be computing using the general method for computing the probability mass function for the maxcount distribution.

An algorithm to compute the cumulative distribution function of the maxount distribution can be found in Bonetti, Cirillo and Ogay (2019) (pp. 6-7). The algorithm is quite complicated, but fortunately for you, all the relevant probability functions have already been programmed in the occupancy package in R. You can compute the distribution of interest and the specific probability you want using the commands below. The probability of interest is $\mathbb{P}(M_n \geqslant 4) = 0.0336$.

#Compute the maxcount probabilities
library(occupancy)
PROBS <- dmaxcount(0:5, size = 5, space = 5)
names(PROBS) <- 0:5

#Compute the probablity of interest
sum(PROBS[5:6])
[1] 0.0336

#Plot the maxcount probability mass function
barplot(PROBS, col = 'blue',
        xlab = 'Maximum number of broken dishes (for a single dishwasher)',
        ylab = 'Probability')

enter image description here


Combinatorial analysis: The above allows you to compute values from the maxcount distribution in general. However, in the present case you have $n=m=5$, and so the parameters of the distribution are small enough that it is feasible to compute the probability of interest using a simple combinatorial analysis. There are $5^5 = 3125$ possible allocations of $n=5$ dishwashers to $m=5$ broken dishes. Out of those allocations, we have:

$$\begin{align} \text{Number of allocations } (M_n = 4) &= 5 \times 4 \times {5 \choose (4,1,0,0,0) } = 100, \\[12pt] \text{Number of allocations } (M_n = 5) &= 5 \times {5 \choose (5,0,0,0,0) } = 5. \\[6pt] \end{align}$$

This gives the probability:

$$\mathbb{P}(M_n \geqslant 4) = \frac{100+5}{3125} = \frac{105}{3125} = 0.0336.$$

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    $\begingroup$ Thanks @Ben. That's a super interesting distribution. My Monte-carlo estimate is the same as yours :) $\endgroup$
    – RNs_Ghost
    Jun 27, 2022 at 14:57
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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]

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