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I am probably missing something obvious but here goes: Lets say we have ten people, and over a period of five days, five of them die. One does each day The probability of any person dying on Day 1 is 1/10, Day 2 is 1/9, and so on. I started off with the (obvious) premise that the probability of dying was 1/2. I then calculated that there were 10P5 possible permutations of YES DYING and of NOT DYING. There are 9P4 permutations that just include NOT DYING. 9P5/10P5 = 1/2.

PROBLEM: Some of those 10P5 possibilities include YES DYING more than once, which is not really an valid possibility. So now I am left with the absurd situation where the probability of dying is less than 1/2! Where is my error? Thanks

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    $\begingroup$ What is the original question? What is the probability that there are 5 dead after 5 days? Each person has the same chance (1 in (11-day#)) to die on each particular day? $\endgroup$ – JMoravitz Feb 10 '15 at 1:46
  • $\begingroup$ No. That was not my question. Sorry. I edited it to clarify... $\endgroup$ – Rachel Feb 10 '15 at 2:13
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    $\begingroup$ So, one person is guaranteed to die each day. Still, what is the event that we are wanting to know the probability of? Are you one of the 10 people and your question is the probability that you live? $\endgroup$ – JMoravitz Feb 10 '15 at 2:35
  • $\begingroup$ Yes. Exactly what I meant. Or conversely, the P of dying... $\endgroup$ – Rachel Feb 10 '15 at 4:05
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Now with the full statement of the question (please try to include all details the first time you post), we are curious of the following:

You are one of 10 prisoners. On each day, one of the remaining prisoners is picked uniformly at random and killed. What is the probability that you have survived after five days (and five of the other prisoners have been killed).

Ignoring for the moment on what specific day people die, there will be five dead people out of the ten. Thus, after five days, there are $\binom{10}{5} = \frac{10!}{5!5!}$ different groups of five people to have died. Among those ways, in some of them you will have died. Count also the number of ways in which you will have survived. There are $\binom{9}{5} = \frac{9!}{5!4!}$ of these. By the definition of discrete probability then, if each outcome is equiprobable then the probability of an event occuring is the number of ways it occurs divided by the number of outcomes total.

$$P(\text{you survive}) = \frac{\binom{9}{5}}{\binom{10}{5}}=\frac{1}{2}$$

Similarly, you can consider this in the following way. On the first day the warden lines up the prisoners randomly and has them sit in numbered chairs. The first five spaces chairs correspond to the people who are killed. Now worrying about the order as well, we use permutations instead of combinations, but it won't make a difference in our calculations.

$$P(\text{you survive}) = \dfrac{\frac{9!}{4!}}{\frac{10!}{5!}}=\frac{1}{2}$$

Finally, a final way of seeing the problem even easier than the others. Imagine the scenario as though the warden asked you to sit in a chair at random first before any other prisoner was asked to sit. There are 10 chairs to sit in, and five of them will lead to your death, 5 of them will lead to your survival.

$$P(\text{you survive})=\frac{5}{10}=\frac{1}{2}$$


I think your misconception and confusion stems from not understanding what the permutations you used represent. For the first approach, a sequence $DDDLDLLDLL$, corresponds to the first, second, third, fifth, and eighth prisoners dying over the course of five days.

For the second approach, $5,2,8,7,1$ would correspond to the fifth prisoner dying on day one, the second prisoner dying on day 2, the eighth prisoner dying on three, the seventh prisoner dying on day 4, etc. Note that as we are dealing with permutations in the second scenario, you cannot repeat a number. I think you thought that if you were labeled as prisoner 1, that the sequence $2,1,4,1,5$ is possible, wherein it would seem that you are killed both on day two and killed a second time on day four (which is of course absurd). If you allow for repeated entries, there are $10^5$ such sequences. $~_{10}P_5 = \frac{10!}{5!}\neq 10^5$ however.

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Lets say we have ten people, and over a period of five days, five of them die. The probability of dying on Day 1 is 1/10, Day 2 is 1/9, and so on.

I'm interpreting your question as asking: What is the probability that 5 of 10 people die over a period of five days when the probability of any person dying on day $n$ is $\frac 1 {11-n}$, for $n\in \{1... 5\}$.

So the probability that a person will live through the five days is: $$\prod_{n=1}^5 \frac {10-n} {11-n} = \frac 9{10}\cdot\frac{8}{9}\cdot\frac{7}{8}\cdot\frac{6}{7}\cdot\frac{5}{6} = \frac{1}{2}$$

So that comes from the given information. (It's not an assumption.)

This leaves us with the count of the people who die during the five days having a binomial distribution with parameters, $N\sim \mathcal{Bin}(10, \tfrac 1 2)$

$$\begin{align} \mathsf P(N=5) & = {^{10}\mathrm C_5}(1-\tfrac 1 2)^5 (\tfrac 1 2)^5 \\[1ex] & = \frac{\binom{10}{5}}{2^{10}} \\[1ex] & = \frac {63}{256} \end{align}$$

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  • $\begingroup$ Oops. I forgot to add the clause that one person dies each day. That is very important. $\endgroup$ – Rachel Feb 10 '15 at 2:11

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