SQ2.1 Which bet?
If you are willing to assume that birthdays in the class are independent (not a terrible assumption) and that birthdays are distributed randomly throughout the year (a terrible assumption, as it turns out!), you should take Bet #2. Here’s a way to explain why:
-
Consider that 25 people can be combined into pairs \({25 \choose 2}\) ways (which you can read “as 25 choose 2”), which is equal to \(\frac{25 \times 24}{2} = 300\) (and that little calculation is where those binomial coefficients I mentioned in my hint come in handy).
-
Generalizing the logic from part b of the textbook exercise last week problem, I have assumed that each of these possible pairings are independent and thus each one has a probability \(P = (\frac{364}{365})\) of producing an unmatched set of birthdays.
+
Consider that 25 people can be combined into pairs \({25 \choose 2}\) ways (which you can read as “25 choose 2”), which is equal to \(\frac{25 \times 24}{2} = 300\) (and that little calculation is where those binomial coefficients I mentioned in my hint come in handy).
+
Generalizing the logic from part b of the textbook exercise last week, I have assumed that each of these possible pairings are independent and thus each one has a probability \(P = (\frac{364}{365})\) of producing an unmatched set of birthdays.
Putting everything together, I can employ the multiplication rule from OpenIntro Ch. 3 and get the following: \[P(any~match) = 1 - P(no~matches)\]
\[P(no~matches) = (\frac{364}{365})^{300}\]
And I can let R do the arithmetic:
diff --git a/psets/pset2-worked_solution.rmd b/psets/pset2-worked_solution.rmd
index be15c57..501021a 100644
--- a/psets/pset2-worked_solution.rmd
+++ b/psets/pset2-worked_solution.rmd
@@ -245,9 +245,9 @@ The plot reveals a very strong positive relationship between average daily dista
If you are willing to assume that birthdays in the class are independent (not a terrible assumption) and that birthdays are distributed randomly throughout the year (a terrible assumption, as it turns out!), you should take Bet #2. Here's a way to explain why:
-Consider that 25 people can be combined into pairs ${25 \choose 2}$ ways (which you can read "as 25 choose 2"), which is equal to $\frac{25 \times 24}{2} = 300$ (and that little calculation is where those [binomial coefficients](https://en.wikipedia.org/wiki/Binomial_coefficient) I mentioned in my hint come in handy).
+Consider that 25 people can be combined into pairs ${25 \choose 2}$ ways (which you can read as "25 choose 2"), which is equal to $\frac{25 \times 24}{2} = 300$ (and that little calculation is where those [binomial coefficients](https://en.wikipedia.org/wiki/Binomial_coefficient) I mentioned in my hint come in handy).
-Generalizing the logic from part b of the textbook exercise last week problem, I have assumed that each of these possible pairings are independent and thus each one has a probability $P = (\frac{364}{365})$ of producing an *unmatched* set of birthdays.
+Generalizing the logic from part b of the textbook exercise last week, I have assumed that each of these possible pairings are independent and thus each one has a probability $P = (\frac{364}{365})$ of producing an *unmatched* set of birthdays.
Putting everything together, I can employ the multiplication rule from *OpenIntro* Ch. 3 and get the following:
$$P(any~match) = 1 - P(no~matches)$$