Simple Probability? Outcome of multiple people picking one of two options

With two choices, it's equivalent to tossing a coin (assuming those choices are random).

For 1 person, there are 2 outcomes: [H, T]
For 2 people, there are 4 outcomes: [HH, HT, TH, TT]
For 3 people, there are 8 outcomes: [HHH, HHT, HTH, HTT, THH, THT, TTH, TTT]

The number of outcomes doubles each time and is equivalent to 2^(#people). 2^15 is 32768. There is only 1 outcome where everyone has chosen the same so the probability of this happening is 1/32768 or approximately 0.0030517578%.

For the second question, we need to determine how many of those 32768 outcomes result in 10 people picking choice 1 and 5 picking choice 2. Excel has a function for this. Typing =COMBIN(15,10) into a cell (or COMBIN(15,5) - it's the same answer) gives us 3003 so the chance is 3003/32768 or approximately 9.1644%. In fact, the values for picking 0 to 15 of one choice are:

0.0031%, 0.0458%, 0.3204%, 1.3885%, 4.1656%, 9.1644%, 15.2740%, 19.6381%, 19.6381%, 15.2740%, 9.1644%, 4.1656%, 1.3885%, 0.3204%, 0.0458%, 0.0031%

Hopefully that makes it clear how to extend this to more people with a binary choice. Once you move up to providing more choices (Choice1, Choice2, Choice3) then the math is going to get much more complex.

WBD