15 people flip coins, find the probability distribution

[Addez123]

Homework Statement:

15 people flip 2 coins each. Whats the probability distribution of X,
where X = amount of people with same results on both flips

Relevant Equations:

Binominal, poisson, normal, multinomial and hypergeometric distribution equations.

Say I make it so that the 2 coin flips count as a single number 1,2,3,4 representing head-head, head-tails, tails-head, tails-tails.
Then what do I do?

I'm just lost as to how I would even approach this problem.

 

[PeroK]

Homework Statement:: 15 people flip 2 coins each. Whats the probability distribution of X,
where X = amount of people with same results on both flips
Relevant Equations:: Binominal, poisson, normal and hypergeometric distribution equations.

Say I make it so that the 2 coin flips count as a single number 1,2,3,4 representing head-head, head-tails, tails-head, tails-tails.
Then what do I do?

I'm just lost as to how I would even approach this problem.

For a start you could find the range of $X$.

 

[Addez123]

Range would be 0-15. Nobody could have the number and all 15 could have had same combination.
I think this is suppose to be solved with multinomial equation, I'm testing it rn..

 

[PeroK]

Range would be 0-15. Nobody could have the number and all 15 could have had same combination.

The you need to calculate $p(X = n)$ for $n = 0 - 15$. And that's your distribution. All you need to do after that is identify the name of a distribution like this.

 

[Addez123]

Oh so the question ment, the amount of people who get head-head or tails-tails on their flip.
That probability is 50%. So the answer was X ~ Bin(15, 0.5)

 

[Master1022]

Range would be 0-15.

Perhaps I am misreading the question (and I apologize if I have done so), but how can $X = 0$? For two coins, there are only $2^2 = 4$ outcomes and therefore I am struggling to see how each of the fifteen individuals could have a different outcome.

Once again, apologies if I have misunderstood the question.

 

[Mark44]

Perhaps I am misreading the question (and I apologize if I have done so), but how can $X = 0$? For two coins, there are only $2^2 = 4$ outcomes and therefore I am struggling to see how each of the fifteen individuals could have a different outcome.

I agree. It seems to me that the Pigeonhole Principle is in effect here.

Suppose that only 5 people were involved. If the first 4 of these people all had different outcomes, the 5th person would necessarily have an outcome that was the same as one of the other 4. So X = 1 is the minimum value in this scenario, and the range of X would be 1 through 5. In my example, the pigeon holes are the number of possible outcomes (4), and the pigeons are the number of people (5). Therefore, at least one person must have the same outcome as one of the other four people.

What would the range of X be if there were 6 people involved?

 

[haruspex]

I agree. It seems to me that the Pigeonhole Principle is in effect here.

Suppose that only 5 people were involved. If the first 4 of these people all had different outcomes, the 5th person would necessarily have an outcome that was the same as one of the other 4. So X = 1 is the minimum value in this scenario, and the range of X would be 1 through 5. In my example, the pigeon holes are the number of possible outcomes (4), and the pigeons are the number of people (5). Therefore, at least one person must have the same outcome as one of the other four people.

What would the range of X be if there were 6 people involved?

The question cannot mean that. If three people get one pair of outcomes while another four all get some other pair, what would we mean by X?
The interpretation in post #5 looks right.

 

[Master1022]

The question cannot mean that. If three people get one pair of outcomes while another four all get some other pair, what would we mean by X?
The interpretation in post #5 looks right.

I think I understand what you mean (but I'm not sure how we should define X). Surely the binomial distribution above would mean that $P(X = 0) \neq 0$ which still doesn't make sense to me. I thought it is impossible to have completely different outcomes over 15 people (when only 4 outcomes are possible)?

 

I think I understand what you mean (but I'm not sure how we should define X). Surely the binomial distribution above would mean that $P(X = 0) \neq 0$ which still doesn't make sense to me. I thought it is impossible to have completely different outcomes over 15 people (when only 4 outcomes are possible)?

It's just saying that 15 people stand in a line and each of them flips 2 coins. Then we count how many people got either two heads or two tails, out of 15. If we call that number $X$, then$$X \sim B(15, \frac{1}{2})$$

 

[Master1022]

It's just saying that 15 people stand in a line and each of them flips 2 coins. Then we count how many people got either two heads or two tails, out of 15. If we call that number $X$, then$$X \sim B(15, \frac{1}{2})$$

Ah, thanks for the clarification. For some reason, post #5 just made sense now. Yes, then I agree. I thought the question was asking for the number of people with the same HT, HH, TT, or TH combination.

 

[jbriggs444]

I think I understand what you mean (but I'm not sure how we should define X). Surely the binomial distribution above would mean that $P(X = 0) \neq 0$ which still doesn't make sense to me. I thought it is impossible to have completely different outcomes over 15 people (when only 4 outcomes are possible)?

It seems to me that your interpretation of the question differs dramatically from that in #5.

Let us go back to the question and try to see if we can extract your interpretation...

15 people flip 2 coins each. Whats the probability distribution of X,
where X = amount of people with same results on both flips

Your interpretation seems to be that we are asked how many people get the same pair of results as at least one other person.

So if person 1 gets HH and while person 2 through person 14 get TT then that counts as 14 people with the "same results".

Under this interpretation it is indeed nonsensical for nobody to get the same result pair as anyone else. There are only four possible result pairs and fifteen people. Somebody has to share a result pair.

Edit: As you had already realized.

 

[PeroK]

Ah, thanks for the clarification. For some reason, post #5 just made sense now. Yes, then I agree. I thought the question was asking for the number of people with the same HT, HH, TT, or TH combination.

That would require four random variables $X(HT), X(HH), X(TH), X(TT)$, where each is randomly distibuted on $0 - 15$.