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I am trying to solve a probability question.

5 people are competing in a cooking competition and there are 3 raters. Each person will cook 4 dishes and the dishes will be rated from 0 (bad) to 10 (good). I would like to determine the probability that two raters always rate within 5 points of each other for the 4 dishes for the five people.

Please could we not get into the advanced topics such as reliability of the raters and so on. My initial thought was to apply principle of inclusion/exclusion. So first compute the probability for each individual question and then multiple all these probabilities to the the final answer. I thought I did something like this in high school or college, but cannot really comprehend the answer. Can anyone solve this? Some steps of explanation are highly appreciated.

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  • $\begingroup$ So, basically there are 5x4 = 20 decisions that raters made and the answer would be the same if 20 people were making 1 dish only? And, you probably assume that a rater can rate each of 0-10 by 1/11 probability (i.e. uniformity) $\endgroup$ – gunes Sep 9 at 14:34
  • $\begingroup$ This seems right. Is it possible that you could write out the formula of computation? $\endgroup$ – don Sep 9 at 20:40
  • $\begingroup$ Is the question 'only two' or 'at least two' raters will always be within 5 points of each other? Or that any two raters will be within 5 points of each other for each dish? $\endgroup$ – Elenchus Sep 10 at 5:15
  • $\begingroup$ Also how is 'within 5 points' working? Is it inclusive? E.g. are 0 and 10 within 5 points of 5, or not? (0,1,2,3,4 are 5 numbers, so it could be argued that 5 is outside that) $\endgroup$ – Elenchus Sep 10 at 5:24
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$\bf{Problem:}$ There are three judges $X$, $Y$ and $Z$. There are 5 cooks, each receiving a random score in [0, 1, ..., 10] from each of the judges. So there are 15 scores, all independent from each other. What is the probability that there are at least two judges, whose scores are within 5 points on each of the 5 cooks?

$\bf{Notations:}$ Let $x_i, y_i, z_i$ be the scores received from the 3 judges for cook #$i$, for $i = 1, 2, ..., 5$. Define $X:Y$ as $|x_1 - y_1| \le 5 \cap ... \cap |x_5 - y_5| \le 5$, and likewise for $Y:Z$ and $Z:X$.

$\bf{Answer:}$ From Principle of Inclusion and Exclusion:

$P(X:Y \cup Y:Z \cup Z:X) = 3P(X:Y) - 3P(X:Y \cap Y:Z) + P(X:Y \cap Y:Z \cap Z:X)$.

where

$P(X:Y) = P(|x_1-y_1| < 5)^5 = (\frac{91}{121})^5$ $P(X:Y \cap Y:Z) = P(|x_1-y_1| < 5 \cap |y_1-z_1| < 5)^5 = (\frac{781}{1331})^5$ $P(X:Y \cap Y:Z \cap Z:X) = P(|x_1-y_1| < 5 \cap |y_1-z_1| < 5 \cap |z_1-x_1| < 5)^5 = (\frac{671}{1331})^5$.

Further details: https://drive.google.com/file/d/1BiFzJhpHxlGJQRK488zTFi2sgQPTMQSM/view?usp=sharing

Comments on "Probability thru Geometry": This method can easily handle the case when scores are uniformly (rather than discretely) distributed. Just need to compute the volume of various polyhedrons. But this method cannot handle the case of 4 or more judges.

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  • $\begingroup$ If the problem is for each judge to produce 20 scores, one on each dish, then just change all the powers of 5 into powers of 20. $\endgroup$ – Shang Zhang Sep 10 at 17:02

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