# Understanding general formula for probability

I have two questions about probability in terms of formulas.

Going through a couple of videos on probability, I have seen a general function for calculating the probability an object with 2 outcomes but with more than one object. Let's use a coin for an example. The formula would be the following:

$$P(k\text{ heads}, n-k \text{ tails}) = \dbinom n k\dfrac{1 } {2^n}$$

where:

$$n$$ = # of objects: in this case coins

$$k$$ = # of a certain outcome: in this case heads

$$\displaystyle {n \choose k} = \frac{n!}{k!(n-k)!}$$

My questions:

• Can still formula remain true for other object with unique outcomes (6 six sided die, 3 side die, etc) with we replace the 2 with a 6 or 3? Can you explain why this is or is not possible?
• How does this formula or the probably in general change when there are objects with know unique outcomes are at play? Examples:
• a 3+ sided die with the same outcome on more than one side side?
• flipping two coins but 1 coin is a trick coin ( has 2 heads)

Sorry this is my first question, and I'm not sure how the platform words to use the tools. If someone can edit the question to format is better, I would appreciate it.

• Welcome to MSE. Please type your questions rather than posting images. Images can't be browsed, and are not accessible to those using screen readers. If you need help formatting math on this site, here's a tutorial To begin with, surround all math expressions (including numbers,) with $ signs. Use ^ for exponents and _ for subscripts. $x_1^{2/3}$ shows up as$x_1^{2/3}\$. Commented Jun 8, 2021 at 3:18
• The coin flip formula is for two equal probability outcomes. Other cases will have similar looking results but not as simple as your guess. Commented Jun 8, 2021 at 3:27
• @NikhilKumarSingh I am not sure what you mean by your comment? Commented Jun 8, 2021 at 3:31
• What would a 3-sided die look like? :-D
– user169852
Commented Jun 8, 2021 at 3:48
• A triangular prism with rounded ends. Commented Jun 8, 2021 at 4:13

When you have $$n$$ trials each with an independent and identical rate of success $$p$$, the count of success among them will have a probability mass function of$$\mathsf P(X\,{=}\,x;n,p)=\dfrac{n!~ p^x~(1-p)^{n-x}}{x!~(n-x)!}$$

This is the pmf a binomial distribution. In particular the distribution of counts of heads and tails, $$X_H, X_T$$, on $$n$$ fair die tosses would be$$\mathsf P(X_H=x, X_T=n-x)=\dfrac{n!}{x!~(n-x)!~2^x~2^{n-x}}$$

(Where of course, $$2^x2^{n-x}$$ simplifies to $$2^n$$)

You seek to extend this to multinomial case where, rather than success or failure, you have a set of disjoint states each trial may take.

Say, a die with faces one 1, two 2, and three 3, with counts $$X_1,X_2,X_3$$ in $$n$$ tosses. Then we expand the above to

$$\mathsf P(X_1=x, X_2=y, X_3=n-x-y) = \dfrac{n!}{x!~y!~(n-x-y)!}\dfrac{1^x2^y3^{n-x-y}}{6^n}$$

In general, when we have a set of $$m$$ trial results with rates $$(p_i)_{i=1}^{m}$$ (where $$\sum_{i=1}^m p_i = 1$$) then : \begin{align}\mathsf P(\vec X = \vec x; n, \vec p) ~&=~n!\prod_{i=1}^m\dfrac{p_i^{x_i}}{x_i!}\\[3ex]&=~\dbinom n{x_1,\ldots,x_m}\cdot p_1^{x_1}\cdots p_m^{x_m} \end{align}

This is the probability mass function for a Multinomial Distribution.

• More properly we should indicate the support. $$(\sum_{i=1}^m x_i=1)\wedge(\forall i\in[[1..m]]~.(x_i\in[[1..n]])$$

Flipping two coins, but one flip uses a trick coin with two heads.

In cases like this, you do not have identical success rates on each trial. So the formula is not quite applicable. Here you have the count for heads being the sum of two (independent) binomial distributions with differing success rates; $$X_H\sim\mathcal{Bin}(1,1/2)+\mathcal{Bin}(1,1)$$. $$\mathsf P(X_H=x, X_T=2-x) = \dbinom 1 {x-1}\dfrac{1}{2}\quad\big[x\in\{1,2\}\big]$$

Which is trivial in this case, but a rather more complicated for larger trial amounts and different success rates.