math awareness video contest

[cris4884]

Allow me to untangle this web of confusion:

You are correct in your calculation of the number of ways of choosing 50 mil from 100 mil. Your calculator got an overflow which basically means you got a really huge number (infinity). Now you just need to take that answer and then divide it by the total number of outcomes possible in the election (see below). Btw, you don't need to use 100 mil as you are doing a fraction anyway so you might as well use 100 or 10.

[philelly]

i tried 100 million because i think 2^x increases with x faster than x choose x/2 increases with x; therefore the probability that your vote counts is less with increased voter number x. i think. i'm working on a sequel video to further analyze these issues.

[cris4884]

Continuing, now the answer is the number of ways of choosing 50 mil from 100 mil times the probability that a person will vote for one candidate(raised to the 50 mil) times the probability that a person will vote for the other candidate (raised to the 50 mil) Which, if we drop the mils (it's a proportion anyway), is: 100!/((50!)(50!)) * ((1/2)^(50)) * ((1/2)^50)

which reduces to:

100!/(50!^2) * (1/2)^100

which is approximately according to my calculator:

1*10^29 * 8*10^-31

equals = .08

[cris4884]

Continuing, the way to calculate the total number of outcomes is really very close to your second idea, which is flipping a coin 100 mil times. The answer there is quite easy though, and that is, 2^(100 mil). As you probably know, this number will not show up in your calculator. However, there is a very easy way to find the answer you are looking for using the binomial distribution as you say. (see below)

[stisaac]

this is really funny.

[imanonlychild]

not sure i'd want you as a doctor, but glad to have you as a non-voter.

[chuckdickenstfb]

bless you, holy truth speaker

[0hhai]

incisive, cogent.

[yinxzon]

i dont understand why you have stethoscope when you are doing this. i think you have told the wow jokes numerous times. booyacacha

[philelly]

best wishes.