Probability find color of iphone I have a question as following:
There are n customers forming a line inside Apple store, half of them have Rose iPhone and half have Gray iPhone, shuffled randomly. You are challenged to predict iPhone color of the next person exiting the store, where they will exit one by one.
At any point, if you correctly predict the iPhone color of exiting customer as Rose, you win. You can choose to skip any number of guesses and see their iPhone color until you decide to guess that the next iphone color is Rose. You will only win if predict the color to be Rose and people exiting the store carries a Rose iPhone.
I want to ask that is there a method that has better chances of winning than $\frac{1}{2}$, and if there is no such method, why is that?
Thank you
 A: Let the exiting customers be labeled by numbers $1$ through to $N$, where $N$ is even. It is given that exactly half of them have rose colored phones and the rest half have grey colored phones. Consider the proposition:
\begin{align}
R_i\equiv\textrm{The exiting customer $i$ has rose colored phone}
\end{align}
The negation of the above proposition means that customer $i$ has grey colored phone. Suppose $(n-1)$ customers have exited, $1\leq n\leq N$, and you guess that the $n$-th customer holds a rose colored phone. What is the probability that your guess is right? That depends on the color of the phones carried by the $(n-1)$ customers who have exited and which cannot be known in advance.
Let $m$ among the $(n-1)$ customers who have exited have rose colored phones. Since there are $N/2$ rose colored phones in all, there are $C_m^{N/2}$ ways of selecting $m$ rose colored phones. Rest of the customers who exited have grey colored phones, and there are $C_{n-1-m}^{N/2}$  ways of selecting $(n-1-m)$ grey colored phones. There are therefore $C_m^{N/2}\times C_{n-1-m}^{N/2}$ ways of having $m$ rose colored phones among $(n-1)$ customers who exited. Here we adopt the convention that $C_x^y=0$ if $x>y$.
Therefore the probability that exactly $m$ among $(n-1)$ customers who exited have rose colored phones is:
\begin{align}
P(m|n-1)=\frac{C_m^{N/2}\times C_{n-1-m}^{N/2}}{\sum_{k=0}^{n-1}\left( C_k^{N/2}\times C_{n-1-k}^{N/2}\right)},\quad 0\leq m\leq \textbf{min}(n-1,N/2)
\end{align}
and zero otherwise.
Given that $(n-1)$ customers exited with $m$ rose colored phones, the probability that your guess about the $n$-th customer will be correct is:
\begin{align}
P(R_n|(m,n-1))=\frac{N/2-m}{N-(n-1)},\quad 0\leq m\leq \textbf{min}(n-1,N/2)
\end{align}
and zero otherwise.
Therefore the unconditional probability that your guess for $n$-th customer will be correct is by Bayes theorem:
\begin{align}
P(R_n)&=\sum_{m=0}^{\textbf{min}(n-1,N/2)}P(m|n-1)P(R_n|(m,n-1))\\
&=\sum_{m=0}^{\textbf{min}(n-1,N/2)}\left( \frac{C_m^{N/2}\times C_{n-1-m}^{N/2}}{\sum_{k=0}^{n-1}\left( C_k^{N/2}\times C_{n-1-k}^{N/2}\right)}\times \frac{N/2-m}{N-(n-1)}\right)
\end{align}
I have tried a few values of even $N$ and $P(R_n)=0.5$ always, no matter which customer you guess (i.e. for any $n$ such that $1\leq n\leq N$). I am not sure if there is an intuitive explanation for this. Also there must be some simplification for the expression above that reduces it to $1/2$ although I couldn't figure it out.
