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?



1 Answer 1


The question is whether the coins are truly identical, or not. If there is any way we can tell the coins apart (scratches, year of minting, weight, asymmetries from being in circulation, spectroscopy of absorption, spectrometry of isotope composition, etc.) they are not identical. For real world coins, it is always possible to tell them apart, so the counting rule in experiment (A) is correct. (Order doesn't matter, here, but identity does. You used order as a surrogate for identity in part A.)

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.

But what if they are truly identical? I hear you cry. Truly, truly, truly! As in, nothing we use, nothing we can manipulate, nothing we can identify, not even quantum mechanics, could not tell them apart? Well, then (B) is correct. This is called Bose-Einstein statistics, and yes, it does describe the statistics of photons, among other things.

But real world coins are not that way, and if you insist on counting them using experiment (B) methods, you will be shocked to see that each of the five cases is not represented 1/5 of the time. Experiment B cannot explain that, and the only way you could do it is if you went back to our experiment (A) counting methods and used the binomial distribution.

  • 1
    $\begingroup$ Ah interesting, I had not thought about the possibility of probability being equal in case (B), but your answer about transforming the sample space from (A) to (B) helps sort things out for me. $\endgroup$ Mar 9, 2019 at 17:48

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.