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In many probability problems where there are a lot of parallel events, I have seen that the solution uses the approach/trick to consider events one at a time instead. E.g. as when throwing 6 dice, consider instead trowing one dice at a time.
Another example is the following problem:

How many people should we have in a room so that at least two of the people in the room have the same birthday with probability greater than $50\%$?

The solution given also considers the events serially

For the first individual in the room, he can have birthday on any day and still the statement: "all individuals in the room have different birthdays" would be true.
Hence the probability that "all individuals in the room have different birthdays" with just one individual is $1$.
Now we take the second individual: the probability that he has a birthday on a different day than the first individual is exactly $\frac{364}{365}$ as there are exactly $364$ non-conflict days available (we are assuming here all birthdays are equally likely). With three individuals the probability that "all individuals in the room have different birthdays" is:
$1\cdot\frac{364}{365}\cdot\frac{363}{365}$ and we can continue this process with the fourth, fifth etc

I think the solution is basically the same idea as in problems which state throwing of multiple dices, where we approach of what is going to happen instead of making the throw of N dice, but N sequential throws of $1$ dice.

So what are the per-requisites to be able to correctly use a thinking/solution approach from concurrent events to serial events? I.e. is there some twist of this problem that such an approach would fail?

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  • $\begingroup$ Is there anything I can do to improve the question to get some help/response? $\endgroup$
    – Jim
    Feb 22 '21 at 11:00
  • $\begingroup$ The reason you are not getting much of a response is that you question is more about opinions than mathematics. I would say that in mathematics in general and in probability in particular, there is usually more than one way to answer a question, getting to the same result. Sometimes one way is quicker than another for a particular problem, and sometimes they are much the same. Another example is in combinatorics: sometimes it is faster to work from first principles, sometimes to use a shortcut like stars and bars, sometimes to use a recurrence to produce a generating function or the reverse $\endgroup$
    – Henry
    Feb 25 '21 at 0:33
  • $\begingroup$ @Henry: Thank you for your feedback. It helps. I do understand that there are multiple ways to solve a problem. What I was trying to understand here is if the solution's reasoning process is more "standard"/intuitive in probability problems so I can have a kind of formulaic reasoning when encountering a problem. As part of that I was trying to understand the limitation if any of considering the events one after another as a solving principle. Does this make sense? Do you think I could update the post to reflect something non-opinion related? $\endgroup$
    – Jim
    Feb 25 '21 at 10:18
  • $\begingroup$ @Henry: I changed the post. Is it better/more specific now? $\endgroup$
    – Jim
    Mar 1 '21 at 10:33
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I think what you're getting at here is that the method you quote, is essentially assigning an arbitrary order to your random events, as a tool for thinking about them at one at a time.

The key scenario in which assigning an arbitrary order to events will fail, is when the order is important in affecting the outcome. This happens when event $x_n$ is not independent of event $x_{n-1}$ in some sequence.

For example, if you want to know the probability that a man had solved a Rubik's cube at least once before turn $m$ of the cube, you could no longer assign an arbitrary order to your random events. This is because once he has solved it on the $p^{th}$ turn, the probability is one for all $m\geq p$.

Indeed you can say more. Suppose he has a method of turning which ensures he never returns the cube to a previous state, the probability of solving the cube is a strictly monotonic increasing function of $n$.

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  • $\begingroup$ So basically as long as the events in a problem are independent I have the option to treat them as sequential without losing the generality right? $\endgroup$
    – Jim
    Mar 14 '21 at 17:14

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