I'm trying to do the following problem, and want to know why my approach does not work.
A “gender reveal” party is held to announce the gender of an expected newborn. 15 cups are filled in advance with colored beads and covered: if the baby is a boy, 8 are filled with blue beads and 7 with pink. If the baby is a girl, 7 are filled with blue and 8 with pink. When the audience arrives the cups are knocked over (revealing bead colors) in a uniformly random order until the audience has seen 8 cups of the same color (and thereby knows the gender). Let N be the number of cups turned over before the gender is known. Compute the probability that N = k for k ∈ {8, 9, 10, . . . , 15}. Is the probability that N = 15 (and one has to wait to the very end) more or less than 1/2?
Let $E_i$ be the event that the gender of the baby is known after the $i^{\text{th}}$ cup is turned over, for $i \in \{ 8,9,10,...,15 \}$. Let Boy be the event that the baby is a boy, and Girl the event that it is a girl. Then the desired probability is: $$P(E_i) = P(E_i| \text{Boy})P(\text{Boy}) + P(E_i| \text{Girl})P(\text{Girl}) $$
For a particular $i$, we try to compute $P(E_i| \text{Boy})$. Given that the baby is a boy, there would be 8 cups filled with blue beads and 7 cups filled with pink beads. If we place the cups in a line, then the outcomes we are interested in, i.e., the ones where we have to turn over the $i^{\text{th}}$ cup (from left to right) before the gender of the baby is known, are the cases where the $i^{\text{th}}$ cup is a cup filed with blue beads (denoted $B$), and all the cups to the right of it are filled with pink beads (denoted $P$), i.e., the situation in the following diagram:
$${\underbrace{\_ \phantom{.} \_ \phantom{.} \_ \dots \underline{B} }_\text{i}} \phantom{.} {\underbrace{\underline{P} \phantom{.} \underline{P} \phantom{.} \underline{P} \dots \underline{P} } _\text{15-i}}$$
In the above diagram, there are $7 \choose {15-i}$ ways to choose the rightmost pink-beads cups, with $(15-i)!$ ways to permute each selection, 8 ways to choose the blue-beads cup at position $i$, and $(i-1)!$ ways to arrange the remaining cups. So we'd have:
$$P(E_i| \text{Boy}) = \frac{{7 \choose {15-i}} \cdot (15-i)! \cdot 8 \cdot (i-1)!}{15!}$$
Given that the situation is similar when the baby is a girl, and that $P(\text{Boy}) = P(\text{Girl}) = \frac{1}{2}$, we'd have: $$P(E_i) = \frac{{7 \choose {15-i}} \cdot (15-i)! \cdot 8 \cdot (i-1)!}{15!}$$
But this is wrong, since, with $i = 15$, for example, we'd have $P(E_{15}) \approx 3.73$. Where did I go wrong with my analysis?
I don't see an issue with your analysis, and when I calculate $P(E_{15})$ according to your formula, I get $$ \frac{\binom 70 \cdot 0!\cdot 8\cdot 14!}{15!} = \frac{1\cdot 1\cdot 8\cdot 14!}{14!\cdot 15} = \frac 8{15} $$
However, there is a much faster way to do this.
The gender is known before the last cup if the last cup contains a bead of the color there are 7 of, and it will still be unknown after flipping 14 cups if the final cup contains a bead of the color there are 8 of. The probability that it contains an 8-bead is $\frac{8}{15}>\frac12$.