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Let there be a circular board that is divided into $2^k$ sectors (similar to a dartboard). Each sector can be red with probability $p$ and green with probability $1- p$. If we rotate the board, when it comes to rest, I am interested in an event that a red sector covers the 6 o'clock direction.

Now let $X$ denote the reward that one gets when the above event occurs. The random variable $X$ takes a value according to the following rule: $X = 1/t$, when $1 \leq t\leq 2^k$ sectors are red and one of them covers the 6 o'clock direction, i.e., the reward of $1$ is shared among those red sectors. I am interested in the probability mass function of $X$. When none of the sectors that are red covers the 6 o'clock direction, the reward is $0$, i.e., $X = 0$.

My attempt: There are $2^k + 1$ combinations depending on how many of the $2^k$ sectors are red, e.g., $0$, $1$, $2$, $\dotsc$, $2^k$. Let $p_n$ denote the probability that $n$ sectors are red before rotating the board. Then we have $$p_n = {2^k \choose n}p^{n}(1-p)^{2^k-n}, \,\,\,\, 1 \leq n \leq 2^k.$$ Now, for the case when there are $1 \leq t \leq 2^k$ sectors are red, the probability that a red sector covers the 6 o'clock direction is $$q_t = \frac{t}{2^k}.$$

So, one has $$ X = \begin{cases} 1, \quad \text{w.p.}~~ p_1q_1 \\ \frac{1}{2}, \quad \text{w.p.}~~ p_2q_2\\ \vdots \\ \frac{1}{2^k}, \quad \text{w.p.}~~ p_{2^k}q_{2^k} \\ 0, \quad \text{w.p.}~~(1-p)^{2^k}. \end{cases} $$

Then the probability that a red sector covers the 6 o'clock direction is $$P_{c} = \sum_{n = 1}^{2^k} p_n q_n.$$ The probability that no red sector covers the 6 o'clock direction is $$P_{n,c} = (1-p)^{2^k}.$$

Then the following equation should hold: $$P_{c} + P_{n,c} = 1.$$ But it does not hold true. To check this, let us consider the case of $k = 1$. In this case,

$$P_{c} = \frac{1}{2}2p(1-p) + \frac{2}{2}p^2$$ and $$P_{n,c} = (1-p)^2.$$ Here, $P_c + P_{n,c} \neq 1$.

What am I missing here?

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  • $\begingroup$ +1 to the query. Debit is that you didn't proofread before. For me, that is much more than offset by excellent presentation and work shown, for a complex math problem. $\endgroup$ – user2661923 Mar 3 at 16:10
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    $\begingroup$ @Oscar, did you mean to say The event that a green (not red) sector covers the required direction? If so, your comment is quite correct. But all that means is that $P_{n,c}\gt(1-p)^{2^k}$, since there are plenty of ways in which the required direction can be covered by green even if some of the sectors are red. $\endgroup$ – Barry Cipra Mar 3 at 16:13
  • $\begingroup$ Ah, I just realized that. I think the error is in the calculation of $P_{n,c}$. I am considering only one event that leads to not covering the required direction; there are other ways also to not cover the direction. $\endgroup$ – Oscar Mar 3 at 16:14
  • $\begingroup$ You have to assume the colors of the sectors have independent probability in order to come up with these answers. Interestingly, the expected value of $X$ is insensitive to that assumption. You can skip all of these calculations when evaluating $E(X).$ $\endgroup$ – David K Mar 4 at 13:10
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Where your attempt goes awry is in equating the probability, $P_{n,c}$, that no red sector covers the 6 o'clock position with $(1-p)^{2^k}$. The latter is the probability that there are no red sectors at all. But it's entirely possible to have a mix of red and green sectors with the 6 o'clock position covered by green. So $P_{n,c}$ lies somewhere between $(1-p)^{2^k}$ and $1-p^{2^k}$ (since $p^{2^k}$ is the probability that all sectors are red).

In fact, the probability $P_c$ and $P_{n,c}$ are simply $p$ and $1-p$ respectively. That's because the 6 o'clock position is necessarily covered by some sector, and whichever sector that is will be red with probability $p$ and green with probability $1-p$.

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  • $\begingroup$ Sorry for the ill-posed problem. I have improved it. $\endgroup$ – Oscar Mar 3 at 15:56
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    $\begingroup$ @Oscar, your original problem was not ill-posed. It may just not have been exactly what you actually wanted to ask. I've added a note to my answer to indicate it no longer quite applies. $\endgroup$ – Barry Cipra Mar 3 at 16:05
  • $\begingroup$ I've revised my answer to address the OP's essential question, which concerns the error in their attempt. $\endgroup$ – Barry Cipra Mar 3 at 16:39
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I haven't read your calculation to look for an error.

The probability that a random position on the wheel is red is just $p$, the probability that each sector is red.

Edit in response to the edited question, that introduces the reward for red.

The reward information is independent of the error. The argument still relies on a calculation of

the probability that a red sector covers the 6 o'clock direction

that relies on the sum over the possible numbers of red sections. But that probability is still just $p$. So I suspect an error somewhere in that calculation.

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  • $\begingroup$ Sorry for the ill-posed problem. I have improved it. $\endgroup$ – Oscar Mar 3 at 15:56
  • $\begingroup$ @Oscar See my edit. $\endgroup$ – Ethan Bolker Mar 3 at 16:08
  • $\begingroup$ Yes, I agree. Summing over relevant events of one or more red sectors covering the direction, the answer is $p$. The mistake is in the calculation of the complementary event, i.e., in the calculation of $P_{n,c}$. Thanks. $\endgroup$ – Oscar Mar 3 at 16:23

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