Probability of matching set of cards

Question

I understand everything apart from 1 aspect: The numbers multiplying the combinations. I assume these are the possibilities that the remaining cards do not have any matches, but how can I calculates these numbers without writing out all the combinations. Thanks

The Question and Answers

Answer 1

For part (a), you can describe this very simply using derangements. The probability that Sophie wins is the probability that Ella's arrangement of her cards is a derangement with respect to Sophie's arrangement. In this case, $\dfrac{!5}{5!}=\dfrac{44}{120}=\dfrac{11}{30}$, no brute force listing of arrangements required.

As for how to calculate $!5=44$, the linked wikipedia article has plenty of information on this. Your pictured solution went through the trouble of finding exactly one match, exactly two matches, etc... which is tedious and has strong chance of making a mistake. Further, it glosses over how to have found the $9,2,1,0,1$ that appear in the products which leads exactly to your question here. These numbers happen to be $!4, !3, !2, !1,$ and $!0$ respectively, but the solution makes no mention how these were found and one can only guess that the author of the solution found them by brute force (which isn't so bad for $!3, !2, !1, !0$ but for $!4$ this really should have been explained in greater detail)

A better way to do this directly with little to no knowledge of derangements would have been with inclusion-exclusion over the events where the first card matched, the second card matched, etc... At least one card matching would have been given by $\sum\limits_{k=1}^5 \left(-1\right)^{k+1}\binom{5}{k}(5-k)!$ which amounts to $120-60+20-5+1=76$ and subtracting this away from $120$ to give the number of arrangements with no matches would have been $120-76=44$, as found before. This inclusion-exclusion argument leads to one of the common methods to calculate the number of derangements which is outlined further in the linked wikipedia article. There are faster ways to do this which require a bit more theory to understand, also outlined in the article.

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Now, as for part (b), the only thing to recognize is that this will follow a binomial distribution and you can use the well known results for that.