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I was just thinking of one situation of the coin toss problem in my mind. Suppose I am given two identical coins. Just by looking at them, I cannot distinguish which one is which. Both of them look similar. Now if I perform a random experiment by taking both the coins in my right hand and throwing it on a table in front of me. Then for this experiment, the outcomes on each coin shall matter, but I guess the order shall not matter, unlike head appearing on the first coin and tail appearing on the second coin. Simply the outcome should be $\{head,tail\}$. [Using set notation to express unordered pair].

That being said, I guess the sample space can be reported as:

$$S=\{\{head,head\},\{head,tail\},\{tail,tail\}\}$$

So our sample space has $3$ possible outcomes, contrary to the experiment when toss the coins one after the other. (In the later case, the ordering matters and as such the sample space is given as :

$$S_{new}=\{(head,head),(head,tail),(tail,head),(tail,tail)\}$$

So in this later experiment we have $4$ possible outcomes in our sample space).

In general if a question asks about tossing two coins simultaneously, then is my way of thinking flawed? Because in most textbooks I find the problem solved as the one having $4$ outcomes even if the question is stated as "two identical coins tossed simultaneous". Or is it so that logically, for tossing, we can toss only one coin at a time using one hand, as as such, when two coins are being tossed, we are using out left and right hand simultaneously, and this is what is causing the ordering? Outcome on coin tossed by left hand and that on coin tossed by right hand...

One a similar note, if I roll two die simultaneously, just as in the case of coins, I feel that sample space shall contain unordered pairs as: $$S_{die}=\{\{x,y\} \mid 1\leq x \leq 6\quad and \quad 1\leq y \leq 6\}$$ where $\{x,y\}$ is an unordered pair or set. And $|S_{die}|=21$.

Please can anyone confirm whether I am flawed or not?

Abhishek Ghosh
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    The version with four outcomes works better because all outcomes are equally probable... – David C. Ullrich Aug 12 '22 at 09:05
  • @DavidC.Ullrich yes, I understand that, in the $4$ outcome version each outcome has probability $\frac{1}{4}$, and in my case $Pr{{H,H}}=Pr{{T,T}}=\frac{1}{4}$ and $Pr{{H,T}}=\frac{2}{4}$ Is it just because of this? Just a matter of convention? – Abhishek Ghosh Aug 12 '22 at 09:11
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    Your inability to observe the distinction between the two coins may affect the event space, and so the overall probability space, but does not need to affect the sample space (the coins are in fact distinct). There is a risk that, by changing the sample space, you may make errors such as thinking two heads is a likely as a heads and a tails. – Henry Aug 12 '22 at 09:49
  • @Henry got the possibility of error which you are talking about. And yes, the coins are in fact distinct... When I calculate $Pr{{H,T}}$ I am indeed sort of used the fact that the coins are distinct... $(H,T)$ and $(T,H)$.... So it sort of justifies. Thanks – Abhishek Ghosh Aug 12 '22 at 10:29

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