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Suppose, I'm performing a certain experiment with many outcomes corresponding to $4$ events - $A,B,C,D$.

We have been given the following data : $$p(A)>p(B),p(C),p(D)$$

$$p(A)<p(A^c)$$

This means, in a single trial $A$ is more likely to occur than other events $B,C,D$, since it has a higher probability.

However, if we carry out the same experiment many-many times, we'd get a lot of different results, and from these results we'd expect $A$ to be the mode, but there will be more results where $A$ does not happen, than those where $A$ does happen. This is because, even though the individual probabilities of $B,C,D$ are less than that of $A$, the combined probability is larger. Thus, in the final tally, $A$ would happen more than $B,C,D$ individually, but there will be more outcomes where $A$ does not happen.

From this perspective, $A$ is less likely to occur.

So, in a single trial $A$ is more likely to occur than others, but from the perspective of multiple trials, $A$ is less likely to occur.

An example is an event of $100$ coin tosses. The probability of getting $50$ heads is roughly $0.08$. This is more than any other probability. Does this then imply, that in a single trial we are most likely to obtain an equal number of heads and tails, but if we repeat the experiment many times, there will simply be many more cases of not getting equal amounts of heads and tails? So, from the perspective of many trials, getting an equal amount of heads and tails is less likely, even though in a single trial it is most likely?

In a single trial, we compare the probabilities of individual events to determine what is most likely. However, in many trials, we compare the probability of something happening against that same thing not happening, to determine if it is likely to happen or not.

Is this a correct interpretation of what is meant by more or less likely in probability? Can someone verify this for me?

The answers to this [question]: (Feynman random walk) seem to talk about this thing - but I'm not being able to interpret this correctly)

Can anyone tell me, if the above interpretation is correct?

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    $\begingroup$ Events, at least in the sense that the word is generally used on stats.SE and in probability theory for that matter, are collections of outcomes, and an event is said to have occurred if the outcome is any member of the event. What you seem to implying is that $A, B, C, D$ are disjoint or mutually exclusive events: at most one of them can occur on a trial (and it is possible that none of them might occur on a trial). Does this match your understanding of whatever you are analyzing? $\endgroup$ Sep 16 at 15:25
  • $\begingroup$ @DilipSarwate I should then add $p(A)+p(B)+p(C)+p(D)=1$. If you check the example of the coins, and the link that I've provided, maybe that would explain what my question really is about. For example, in the link, the answers to the question talks about two alternate views of 'more' likely. I'm just asking if my explanation makes sense. $\endgroup$ Sep 16 at 15:30
  • $\begingroup$ That still doesn't answer @Dilip's question: are your events mutually exclusive or not? Since you aren't sure the link you provide is even discussing the same things, it's of little use to resolve this ambiguity. $\endgroup$
    – whuber
    Sep 16 at 16:01
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    $\begingroup$ It seems like you're conflating two different things (A vs B vs C vs D, or A vs not-A). This can be cleared up by more precise terminology. When you say "more likely" or "less likely", you need to include the "than"—what you're comparing to. $\endgroup$ Sep 16 at 16:49
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    $\begingroup$ @whuber thanks. I suppose the entire faulty interpretation stems because of changing the comparison in the middle of the question. In a single trial I was comparing the possibility of an even happening against other similar events. In the multiple trial I was just checking the possibility of an event questions. These appear to be two separate questions, that are meaningless unless I specify what I'm comparing with. $\endgroup$ Sep 16 at 17:04
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Is this a correct interpretation of what is meant by more or less likely in probability? Can someone verify this for me?

In short, no, this is not a correct interpretation. Properly used, the word "likely" should refer to the likelihood function: $$\mathcal L(\theta|x) = p(X=x|\theta)$$ where $X$ is a random variable with distribution $p$ which is parameterized by some parameter $\theta$. So the likelihood is a function of the distribution parameter $\theta$, not the random variable $X$ or the specific observation $x$.

For the specific example there is not a distribution with a parameter, so I think that the example is not using "likely" in its technical sense. I think that here they are using it simply as a synonym for "probable". So in that sense $A$ is more likely (meaning "probable") than any of the other values.

The issue that you are running into is not in the meaning of the word "likely" or "probable" but in the use of the words "more" or "less". The words "more" or "less" imply a comparison between two things. So saying that "$A$ is more probable" is incomplete unless you specify what it is more probable than.

$A$ is more probable than $B$. $A$ is more probable than $C$. $A$ is more probable than $D$. $A$ is less probable than ($B$ or $C$ or $D$). $A$ is less probable than (not $A$). All of those are standard probability statements and apply regardless of the number of trials or observations being made.

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  • $\begingroup$ In the link I've provided, why are the answers talking about two types of 'more likely' ? $\endgroup$ Sep 16 at 15:50
  • $\begingroup$ You would have to ask them why they answered that way. I don't think that is correct usage. $\endgroup$
    – Dale
    Sep 16 at 15:55
  • $\begingroup$ I think OP is referring to colloquial rather than formal meaning of “likely”. $\endgroup$
    – Tim
    Sep 16 at 17:51
  • $\begingroup$ @Tim Yes, which is why I said "I think that here they are using it simply as a synonym for 'probable'" $\endgroup$
    – Dale
    Sep 16 at 17:54

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