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Citing from a book:

First, count the number of possible results that satisfy your desired condition. Then, count the total number of possible results (including those that satisfy the condition and those that do not). Divide the first number by the second and that is the probability that the condition happens on any roll.

If yes, you agree with this definition, here is my second question. Consider two problems:

Problem 1 If the coin is tossed 3 times what is the probability of getting only one heads?

Problem 2 The same as Problem 1, but coins are biased: Probability of heads is 85% Probability of tails is 15%

If the definition above is right, then probably you should agree that these two problems have the same answers?

Narek
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    The implicit assumption is that when you count the # of possibilities, each possibility is equally likely. Since this is no longer the case in Problem 2, you cannot simply divide #successes/#total – Zubin Mukerjee Mar 17 '22 at 14:34
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    The basic counting definition is a great way to introduce the concept of probability to beginners. There are, of course, situations in which two different possible events have different probabilities. – lulu Mar 17 '22 at 14:35
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    That quotation is based on the classical interpretation of probability, where all the outcomes of an experiment are equally likely. I made references to some common interpretations of probability in this answer. – ryang Mar 17 '22 at 14:42
  • This is only possible in the Laplacian case (every elementary event has the same probability to occur). This should have been clearly pointed out. Perhaps it is done somewhere else in the book . If not, it is not good book. – Peter Mar 17 '22 at 14:54
  • The little bit of this book visible on Google Books doesn’t seem to do a good job of underlining the important assumption that each possible result is equally likely. https://books.google.com/books?id=2Fs1EAAAQBAJ&pg=PT393&lpg=PT393#v=onepage&q&f=false – Steve Kass Mar 17 '22 at 14:58
  • @SteveKass how did you find the book? :) I am searching Google the lines I have quoted, I find nothing. – Narek Mar 17 '22 at 15:08
  • @ZubinMukerjee "equally likely" is a dangerous word. It is the same as equally probable. And you can't define concept probability recursively using probability in that definition. What you think? – Narek Mar 17 '22 at 15:25
  • @Narek That's an excellent point! Hmm, then how about "where no outcome of an experiment is more favourable than another"? "Favourable" seems more qualitative—while still objective—than "likely"/"probable". – ryang Mar 17 '22 at 15:45
  • @Narek I googled the exact first sentence of the quote, and this was the only result. (The whole book is not available, but a few pages near the excerpt were.) https://www.google.com/search?q=%22First%2C+count+the+number+of+possible+results+that+satisfy+your+desired+condition.%22 – Steve Kass Mar 17 '22 at 17:00

2 Answers2

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As mentioned in the comments, the definition mentioned makes sense when there are a finite number of outcomes, each of which is equally likely. This is so in the first case, but not so in the second case. You can recast the second problem as a new experiment where outcomes are equally likely and proceed.

(1) The favorable cases are $TTH,HTT,THT$ out of a totality of eight cases: $HHH,HHT,HTH,THH,TTH,HTT, THT, TTT$. So the probability is $3/8$.

(2) To apply the above definition, let us suppose a single toss is equivalent to drawing a ticket out of a box containing 100 slips, out of which 85 are marked heads (and are further labeled as 1st head, 2nd head etc) and 15 are marked tails (and are further labeled 1st tail, 2nd tail etc). Our experiment consists of drawing a slip, replacing it and then drawing again. This is equivalent to having 3 such boxes and drawing a slip from each. Now everything is equally likely. A counting argument shows that there are a totality of $100^3$ cases out of which $3\cdot 15^2\cdot 85=57375$ are favorable. So the probability, in this case, ought to be $\frac{57375}{100^3}$.

Concerning the comment '"equally likely" is a dangerous word.': Actually, the underlying philosophy for defining probability followed by your book is the Principle of Insufficient Reason. According to this principle, if we have no evidence of the propensity of any particular outcome of the experiment over others, we should accord an equal measure of certainty to all the outcomes. In other words, if there are $n$ outcomes possible for our experiment and we have no other information, then the probability of each outcome should be $\frac{1}{n}$. As a corollary, if the the set of our favorable cases $A$ has $m$ outcomes in it and there are a total of $n$ outcomes, we should define $P(A)=\frac{m}{n}$.

This principle makes sense in the first case, but not so in the second case where we have some evidence regarding the coin's biasedness that we ought to incorporate. To apply the principle we can recast the second problem as a new experiment where we have no information again.

Note that this principle is not a mathematical law and we are at liberty to accept it or not. It's just that it suits a large number of situations.

Narek
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  • Thanks for the answer. Would you please clarify some points in your answer? 1) What you mean by saying "Now everything is equally likely." What that "everything" means? 2) " there are a totality of 1003 cases" - 100 cases of what? – Narek Mar 17 '22 at 15:20
  • By everything is equally likely, I mean that all outcomes are equally likely. It is as likely as drawing a combination (3rd heads, 1st tails, 14th tails) as is drawing (4th tail, 83rd head, 10th tail). There are 100^3 such outcomes. –  Mar 17 '22 at 15:23
  • Got you, thanks! – Narek Mar 17 '22 at 15:26
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Already mentioned in the comments but formally, a probability model for an experiment with finitely many outcomes is a tuple: $(S, p)$ where $S$ is the set of all possible outcomes (assumed finite and often called sample space) and $p$ is a function $S \to [0, 1]$ satisfying that $\sum\limits_{o \in S} p(o) = 1.$ When such a tuple is given, then by definition, the probability of a result $R$ (simple or compounded) is $\mathbf{P}(R) = \sum\limits_{o \in R} p(o),$ that is, you sum the individual probabilities assigned to each outcome that is possible in the result. It cannot be stressed enough but the assignation $p$ is what determines the probability and not $S$ (i.e. not the results of the experiment). William Feller in his book An Introduction to Probability Theory and Its Application, vol. 1, discusses situations that arose in physics where much confusion happened. Physicists thought that the results of an experiment where the ones determining $p$ while this is not so and there are physical experiments that have the same underlying space of outcomes ($S$) but different probability distributions ($p$). Anyway, Since $S$ is finite, it has some number $k$ of elements. You can check that $p(o) = \frac{1}{k}$ is a valid assignation (out of many) and this results in the classical definition of probability $$ \mathbf{P}(R) = \sum\limits_{o \in R} \frac{1}{k} = \frac{\text{number of positive outcomes}}{\text{total number of outcomes}}, \quad \text{when} \quad p = \frac{1}{k} \text{ on } S. $$

Other common assignations widely used in practice are Binomial distribution (more generally, Multinomial distribution), Hypergeometric, truncated versions of Poisson, Geometric and Negative Binomial, etc.

William M.
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