I thought about a probability problem and the solution I got doesn't make much sense

[pineapplewithmouse]

The problem:
There are x people in a room.
The people are separated into groups (any group has a random number of people ranging from 1 to x, groups can have the same number of people but they don't have to).
50% of the people in the room are going to be randomly selected, and they will die.
What are the chances, that if I choose a group, nobody in the group will die?

I tried to solve this by creating a simpler version:
There is a room with x people. The people are separated into 2 equal groups: red and blue. And then the "50% of the people..."
I thought about it and I came up with the formula: (((0.5x)!)^2)/(x!).
The problem is, when x is more than 40, the odds the formula gives are almost zero.
It's seems too small to me.
I did something wrong? The original problem has an answer?

 

[Steven G]

I would really have to think about this one.
If I were you, I would start off by using small numbers for x and hopefully you'll see a pattern.
Please try that and post back with all your work.

 

[Dr.Peterson]

The problem:
There are x people in a room.
The people are separated into groups (any group has a random number of people ranging from 1 to x, groups can have the same number of people but they don't have to).
50% of the people in the room are going to be randomly selected, and they will die.
What are the chances, that if I choose a group, nobody in the group will die?

I tried to solve this by creating a simpler version:
There is a room with x people. The people are separated into 2 equal groups: red and blue. And then the "50% of the people..."
I thought about it and I came up with the formula: (((0.5x)!)^2)/(x!).
The problem is, when x is more than 40, the odds the formula gives are almost zero.
It's seems too small to me.
I did something wrong? The original problem has an answer?

As I read it, the initial problem can be stated more simply:

Out of a total of x people, we choose y of them to be in a group. Then half of the people are independently selected and marked. What is the probability that none of the y people we chose are among the half that were marked?​

(It's possible to interpret the problem as saying you randomly choose one of an unknown number of preformed groups; I think that would require more information.)

In your work, you have taken y = x/2. Please show us how you got your formula, which suggests at least that you are thinking in a reasonable direction.

Then tell us why you think the answer shouldn't be very small when x = 40!

 

[Steven G]

As I read it, the initial problem can be stated more simply:

Out of a total of x people, we choose y of them to be in a group. Then half of the people are independently selected and marked. What is the probability that none of the y people we chose are among the half that were marked?​

I only wish I can see things as you do more often. I will put myself in the corner because I should have easily seen the problem as you did.

 

[JeffM]

Thinking about Dr. Peterson’s problem, it seems to me that we need two variables. For example, let the total population, 2x, be ten. (The terms of the problem require the population to be an even number.) Let the size of the relevant group, y, be nine. Then the probability that none in the group die is zero. Let the size of the relevant group, y, be two. Then the probability of none in the group dying is

[math]\dbinom{5}{2} \div \dbinom{10}{2} = \dfrac{5 * 4}{2!} * \dfrac{2!}{10 * 9} = \dfrac{2}{9}.[/math]
If that analysis is correct then the formula for none in the relevant group dying, P, is

[math]1 \le y \le x \implies P = \dbinom{x}{y} \div \dbinom{2x}{y}= \dfrac{x! * (2x - y)!}{(2x)! * (x - y)!};\\ x < y \le 2x \implies P = 0.[/math]
Suppose y = 1. Then

[math]P = \dfrac{x! * (2x - 1)!}{(2x)! * (x - 1)!} = \dfrac{x}{2x} = \dfrac{1}{2} = \dfrac{1}{2^y}.[/math]
If y = 2, then

[math]P = \dfrac{x! * (2x - 2)!}{(2x)! * (x - 2)!} = \dfrac{x (x - 1)}{2x(2x - 1)} = \dfrac{1}{2} * \dfrac{x - 1}{2x - 1} = \\ \dfrac{1}{2} * \left ( \dfrac{1}{2} - \dfrac{1}{2(2x - 1)} \right ) < \dfrac{1}{2^y}[/math].

I have not bothered to do an induction proof. I have left that for the OP.

 

[pineapplewithmouse]

In your work, you have taken y = x/2. Please show us how you got your formula, which suggests at least that you are thinking in a reasonable direction.

So:
I started small, x=2
The answer is obviously 0.5 chance that no one from group y will die.
Then I took x=4.
I thought the chances that group y won't die like that:
There are 4 people, and 2 of them will die. So just one combination out of every possible one will cause the y group to not die.
In x=6 saw the same pattern.
So the formula is 1/xCy
I defined y as 0.5x
And the combination formula is n!/r!(n-r)! where n is the number of the people and r is the number of people that will die.
So n=x and r=y=0.5x
A little bit of algebra and I got the formula (((0.5x)!)^2)/(x!).

Then tell us why you think the answer shouldn't be very small when x = 40!

No mathematical reason, it just doesn't seems right.
Just 40 people and the odds that all of the group that I chose will survive are almost 0?
When I say it now, it seems a little bit more logical because the less people in the other group, the more that the next one who will die is someone from my group, but still, feels wrong.

 

[Dr.Peterson]

So:
I started small, x=2
The answer is obviously 0.5 chance that no one from group y will die.
Then I took x=4.
I thought the chances that group y won't die like that:
There are 4 people, and 2 of them will die. So just one combination out of every possible one will cause the y group to not die.
In x=6 saw the same pattern.
So the formula is 1/xCy
I defined y as 0.5x
And the combination formula is n!/r!(n-r)! where n is the number of the people and r is the number of people that will die.
So n=x and r=y=0.5x
A little bit of algebra and I got the formula (((0.5x)!)^2)/(x!).

This isn't entirely clear, but as I read it, you see that in this special case where you choose the same number of people who are going to die, using combinations there are C(x,x/2) choices of those who will die, and only one of those consists only of your group, so you took the reciprocal. Good.

Figuring that you would have used either combinations or permutations, I did it an entirely different way (finding the probability that your first person lives, and you second person lives, and so on; and got the same result.

Now give it a try in the general case, if the other answers haven't already given it away. I do recommend the idea of calling the total number of people 2x rather than x, so that x are killed and it's clear that's a whole number.

No mathematical reason, it just doesn't seems right.
Just 40 people and the odds that all of the group that I chose will survive are almost 0?
When I say it now, it seems a little bit more logical because the less people in the other group, the more that the next one who will die is someone from my group, but still, feels wrong.

My own thinking is actually along the lines of your work: It seems very unlikely that you would happen to choose exactly the same people that were chosen to die. There are lot of other choices they could have made!

 

[JeffM]

So:
I started small, x=2
The answer is obviously 0.5 chance that no one from group y will die.
Then I took x=4.
I thought the chances that group y won't die like that:
There are 4 people, and 2 of them will die. So just one combination out of every possible one will cause the y group to not die.
In x=6 saw the same pattern.
So the formula is 1/xCy
I defined y as 0.5x
And the combination formula is n!/r!(n-r)! where n is the number of the people and r is the number of people that will die.
So n=x and r=y=0.5x
A little bit of algebra and I got the formula (((0.5x)!)^2)/(x!).

No mathematical reason, it just doesn't seems right.
Just 40 people and the odds that all of the group that I chose will survive are almost 0?
When I say it now, it seems a little bit more logical because the less people in the other group, the more that the next one who will die is someone from my group, but still, feels wrong.

Can you not see that

[math]\dfrac{(z!)^2}{(2z)!} = \dfrac{1}{2} \le \dfrac{1}{2^z} \text { if } z = 1.[/math]
So we can say that there exists at least one integer k such that

[math]k \ge 1 \text { and } \dfrac{(k!)^2}{(2k)!} \le \dfrac{1}{2^k}.\\ \therefore \dfrac{\{(k + 1)!\}^2}{\{2(k +1)\}!} = \dfrac {(k + 1)^2 * (k!)^2}{(2k + 2)(2k+ 1) * (2k)!} = \dfrac{k^2 + 2k + 1}{4k^2 + 6k + 2} * \dfrac{(k!)^2}{(2k)!}.\\ \text {But } k \ge 1 \implies 2k^2 > k^2, \ 3k > 2k \implies \\ k^2 + 2k + 1 < 2k^2 + 3k + 1 \implies \\ \dfrac{k^2 + 2k + 1}{4k^2 + 6k + 2} < \dfrac{2k^2 + 3k + 1}{4k^2 + 6k + 2} = \dfrac{1}{2}.\\ \therefore \dfrac{k^2 + 2k + 1}{4k^2+ 6k + 2} * \dfrac{(k!)^2}{(2k)!} \le \dfrac{1}{2} * \dfrac{(k)!}{(2k)!} \le \dfrac{1}{2} * \dfrac{1}{2^k} \le \dfrac{1}{2^{k+1}}.[/math]
This is really not hard to understand. On the first choice, your probability of choosing someone not marked for death is x/(2x). Given that, on the second choice, that probability is (x - 1)/(2x - 1) < 1/2. You are multiplying probabilities that are, after the first one, less than 1/2. The bigger the group, the more times you must be lucky. The answer of indistinguishably different from zero when you have to do better than flipping a fair coin for heads forty times out of forty tries should be highly intuitive.