# Sample space for 4 consecutive coin flips vs 4 coins flipped at once

I am trying to firm up my understanding of sample space. I am playing with the idea of two random processes, where one realization of each process is

(A): flip a coin 4 times, record the results

(B): flip 4 coins simultaneously, record the results.

The wording suggests that the number of elements in the sample space for experiment (A) is $$2^4$$ because order of the sequence matters and sample space would look like $$\{(TTTT), (HTTT), (THTT) ....\}$$

For (B), there is no order, because the coins are flipped simultaneously, so you have no way of imposing an order. So the number of elements in the sample space is 5? $$\{ \{T,T,T,T\}, \{H,T,T,T\}, \{H,H,T,T\}, \{H,H,H,T\}, \{H,H,H,H\} \}$$

Are these correct interpretations of sample space?

Thanks

However, it is certainly possible for you to partition the sample space into the five categories you suggested in (B), then assign the appropriate probability to each of the five partitions based on the relative frequencies of the finding each state, which is obtained by analyzing the part (A) results in detail. In fact, we have a name for that: the binomial distribution, where $$n$$ is the number of coins but $$m$$ is the number of heads $$(m=\{0,\dots,n+1\}$$ has $$n+1$$ possible outcomes) and each outcome is weighted by its binomial frequency $$\binom{n}{m}$$ multiplied by the simple (case B) probability of $$m$$ heads and $$n-m$$ tails.