Probability of 5 Heads in Binomial Distribution

[aaaa202]

Suppose you have a coin with 4 fair sides, flip it 5 times, and want to know the probability of 5 heads. This is
K(10,5) * (0.25)⁵ * (1-0.25)⁵ = K(10,5)*0.25⁵*0.75⁵
Or more generally for any binomially distributed outcome:

1) p(x=r) = p^r*(1-p)^n-r*K(n,r)

But also we must have that:

2) p(x=r) = K(n,r)/total combinations = K(n,r)/4ⁿ
How do you show that 1) and 2) are equivalent?

 

[alan2]

They're not equivalent. Your concept in 2) is not applicable in this case because the events are not equally likely.

Take for example a single coin flip of a fair coin where p=1-p=1/2. The probability of heads is the number of events resulting in heads divided by the total number of possible outcomes or 1/2. All events are equally likely. Now consider a weighted coin where p=3/4=p(heads). The number of ways to get heads is still 1 and the total number of possible outcomes is still 2 but p(heads) does not equal 1/2.

 

[aaaa202]

hmm...

you were to suppose that each side was equally probable :)

 

[alan2]

Ooops, I'm really sorry, you have a 4 side coin. My bad.

Anyway, your expressions are incorrect. I'm assuming only one side of the coin is heads. In that case, the probability of 5 heads is (0.25)^5=(1/4)^5.

Alternatively, the number of ways to get 5 heads is 1 and the number of possible outcomes is 4^5, so P(5 heads)=1/(4^5). Same result. Where did the 10 come from?

 

[blue_raver22]

To show that 1) and 2) are equivalent, we can start by expanding the first equation using the formula for K(n,r):

1) p(x=r) = pr*(1-p)n-r*K(n,r)
= pr*(1-p)n-r * n!/(r!(n-r)!)
= pr*(1-p)n-r * n*(n-1)*(n-2)*...*(n-r+1)/(r*(r-1)*(r-2)*...*1)
= (n^r * p^r * (1-p)^n-r) / (r!)

Next, we can expand the second equation using the formula for K(n,r) and the fact that there are 4n total combinations:

2) p(x=r) = K(n,r)/total combinations
= n!/(r!(n-r)!) / 4n
= (n*(n-1)*(n-2)*...*(n-r+1))/(r*(r-1)*(r-2)*...*1) / 4n
= (n^r * (n-1)^r * (n-2)^r * ... * (n-r+1)^r)/(r! * (n-r)! * 4n)

We can see that both equations have the terms n^r and r!, but the rest of the terms are different. However, we can simplify the second equation by cancelling out some terms:

= (n^r * (n-1)^r * (n-2)^r * ... * (n-r+1)^r)/(r! * (n-r)! * 4n)
= (n^r * (n-1)^r * (n-2)^r * ... * (n-r+1)^r)/(r! * (n-r)! * 4^r * n^r)
= (1 * (1-1/n)^r * (1-2/n)^r * ... * (1-(r-1)/n)^r)/(r!)

We can then use the binomial theorem to expand the terms in the numerator:

= (1 - r/n + (r*(r-1))/(2!*n^2) - (r*(r-1)*(r-2))/(3!*n^3) + ...) / (r!)
= 1 - r/n + (r*(r-1))/(2!*n