104

You are mistaken in thinking that what you perceive as "the massive importance that is afforded to Bayes' theorem in undergraduate courses in probability and popular science" is really "the massive importance that is afforded to Bayes' theorem in undergraduate courses in probability and popular science." But it's probably not your fault: ...


80

This is a classical problem. Without changing the problem, we can let the digits on the dice be $0, \ldots, 5$ instead of $1, \ldots, 6$ to make our notation easier. Now we make two polynomials: $$ P(x) = \sum_{i=0}^5 p_ix^i,\qquad Q(x) = \sum_{i=0}^5q_ix^i. $$ Now we can succinctly phrase your condition on $p_i, q_i$: it is satisfied if and only if $$ P(x)Q(...


69

While I agree with Michael Hardy's answer, there is a sense in which Bayes' theorem is more important than any random identity in basic probability. Write Bayes' Theorem as $$\text{P(Hypothesis|Data)}=\frac{\text{P(Data|Hypothesis)P(Hypothesis)}}{\text{P(Data)}}$$ The left hand side is what we usually want to know: given what we've observed, what should our ...


53

Let's assume that this is possible. We can derive a contradiction from this assumption. The probability of rolling a total of $2$ must be $1/11$, and the probability of rolling a total of $12$ must also be $1/11$, so $$ \begin{align} p_1 q_1 &= 1/11,\ \text{and}\\ p_6 q_6 &= 1/11. \end{align} $$ The probability of rolling a total of $7$ must also be $...


51

I imagine the argument may go like this... Let's assume you have two identical wards A and B in the hospital, both having nurseries, in each nursery there are $3$ boys and $k$ girls. Then a woman in ward A gives birth to a boy and another woman in ward B gives birth to a girl. Now there are $4$ boys in ward A's nursery, but still $3$ boys in ward B's. ...


49

I would probably start by breaking into cases based on $A$ and $C$. Conditioned on $A$ and $C$ having different signs, there are always real roots (because $4AC\leq 0$, so that $B^2-4AC\geq0$). The probability that $A$ and $C$ have different signs is $\frac{1}{2}$. Conditioned on $A\geq0$ and $C\geq 0$, you return to the problem solved in the link above. ...


47

You ask a very insightful question that I wish were emphasized more often. EDIT: It appears you are seeking reputable sources to justify the above. Sources and relevant quotes have been provided. Here's how I would explain this: In probability, the emphasis is on population models. You have assumptions that are built-in for random variables, and can do ...


40

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37

Buffon's needle: rule a surface with parallel lines a distance $d$ apart. What is the probability that a randomly dropped needle of length $\ell\leq d$ crosses a line? Consider dropping any (continuous) curve of length $\ell$ onto the surface. Imagine dividing up the curve into $N$ straight line segments, each of length $\ell/N$. Let $X_i$ be the indicator ...


35

It's because if the series is convergent but not absolutely convergent, you can rearrange the sum to get any value. Any good notion of "mean" or "expectation" should not depend on the ordering of the $x_i$'s. For a more abstract reason, note that we define the expectation $E[X]$ of a random variable $X$ defined on a probability space $(\...


22

As lulu mentioned in a comment, the fact that a uniformly random permutation $\pi\colon\{1,2,\dots,n\}\to\{1,2,\dots,n\}$ has in expectation one fixed point is a quite surprising statement, with a one-line proof. Let $X$ be the number of fixed points of such a uniformly random $\pi$. Then $X=\sum_{k=1}^n \mathbf{1}_{\pi(k)=k}$, and thus $$ \mathbb{E}[X] = \...


22

You might know only $\Pr[A\mid B]$ and not $\Pr[B\mid A]$, not because someone "adversarially told you the wrong one", but because one of those is a natural quantity to compute, and the other is a natural quantity to want to know. I am about to teach Bayes' theorem in an undergraduate course in probability. The general setting I want to consider is ...


20

Let $X_i$ be an indicator random variable that is $1$ if card $i$ being shuffled back to its original position, otherwise $0$. We see $E[X_i] = 1/52$ since card $i$ has an equal chance of each of 52 positions where it could be permuted to. Another way to see this: there are 51! permutations of the cards with card $i$ shuffled to its original position and the ...


20

There are two main issues here. One is that on a Bayesian interpretation of probability (this term doesn't reference the theorem, but they're both named for Bayes), probability quantifies how well we know individual events, not detailed available frequency statistics. The best-of-both-worlds hope, if you combine Bayesian and frequentist perspectives, is that ...


19

We know from the quadratic formula that the polynomial $Ax^2 + Bx + C$ has real roots if $B^2 - 4AC \geq 0$. We can think of this problem in terms of volumes. To do so, it's easier if we rename the coefficients as $x \equiv A$, $y \equiv C$, and $z \equiv B$. Hence, in order to have real roots we require that $z^2 \geq 4xy$ for $x,y,z \in (-1,1)$. The ...


19

We can find the exact answer by recursion. In all final states of the board, the probability is either $0$ or $1$. In every other state $S$, the probability is the average of the probabilities you get in states obtained by a single move from $S$. There are fewer than $3^9 = 19683$ board states, so a computer has no trouble with this. Here is some Mathematica ...


18

What's going on in the example you quote is different from the constant-function issue. We don't actually mean that $s$ is either the constant function $0$ or the constant function $1$. We mean $s$ to be a proper random variable, we just haven't defined it fully. Random variables without a sample space Formally, a random variable needs a sample space to be ...


16

You have a geometric series, $$\frac12+\frac1{2^3}+\frac1{2^5}+\frac1{2^7}+\ldots=\sum_{n\ge 0}\frac12\cdot\left(\frac14\right)^n=\frac{\frac12}{1-\frac14}=\frac23\;.$$ Alternatively, if $p$ is the desired probability, then $p=\frac12+\frac14p$: with probability $\frac12$ you get a head on the first toss, and with probability $\frac14$ you start with two ...


16

You are right. It is not that surprising. Each specific pattern has probability $\frac{1}{2^{100}}$. The probability to obtain $100$ heads is the same as obtaining any other particular outcome. But usually we do not ask for a specific pattern but ask for the probability to obtain e.g. $k$ heads and this makes the difference. There is just $\color{blue}{\...


15

The first definitions you gave are correct and standard, and statisticians and data scientists will agree with this. (These definitions are given in statistics textbooks.) The second set of quantities you described are called the "sample mean" and the "sample variance", not mean and variance. Given a random sample from a random variable $...


15

The probability generating function of the dice are $P(x) = \sum_{i=1}^6 p_i x^i$ and $Q(x) = \sum_{i=1}^6 q_i x^i$. The probability generating function for their sum is $R(x) = P(x)Q(x)$. You want all possible sums $2, \ldots, 12$ to have the same probability, or equivalently you want $R(x) = \frac{1}{11} x^2 (1+x+\cdots+x^{10})$. Hence an equivalent way to ...


14

Here's a way to think about it: imagine initializing a "total damage counter" as $D_0=0$, and on the $n$th turn, when $d_n$ damage is done to the boss, increment it to $D_n=D_{n-1}+d_n$ and color the interval $(D_{n-1},D_n]$ according to which player attacked. If we allow this to continue forever, it will color the whole number line with a random ...


14

If the sequences that you consider "irregular" constitute the vast majority of all possible sequences, then the probability that the sequence you get is irregular is very high, and it's surprising if you get one of the relatively few "regular" sequences. One could make the concept of "irregular" precise using something like ...


14

I think any explanation is going to make reference to the fact that without absolute convergence, the value of an infinite sum or an improper Riemann integral depends on the order in which the "pieces" are summed up. That alone may satisfy you, but it didn't fully satisfy me. The more specific reason that satisfied me is "without absolute ...


13

You are right that there are $\binom k2$ (or if you prefer $C_{k,2}$) possible pairs, and each pair has a $\frac{364}{365}$ probability of NOT having the same birthday. But these events are not independent, and so we can't just multiply their probabilities! For an example of how they're not independent: suppose you are given that A and B have the same ...


13

Here's a fun fact: you might know that the probability that a random permutation has no fixed points is approximately $e^{-1}$ (and approaches this value quite rapidly). This generalizes: the probability that a random permutation has no cycles of length $\le \ell$ is approximately $$e^{-H_{\ell}} = e^{- \left( 1 + \frac{1}{2} + \dots + \frac{1}{\ell} \right)}...


13

The probability of an integer $x$ (where $|x| < n$ for a fixed $n$) being even, when uniformly randomly chosen, in the limit as $n \to \infty$, is equal to $1/2$. The reason for the phrasing of this is that you need to be very careful about how you define this. "Uniformly randomly" essentially ensures each integer is equally likely. Hopefully ...


13

There's a relatively nice way to do this by hand: pretend they play until the board is full, then grade it. There are $\binom 9 5$ such ending boards, most of which have a unique winner or no winner, and each multiple winner (ambiguous) case will have the same analysis. So, first off are the multiple winner (I'll call these ambiguous) cases: These must be $2$...


11

Examine the $6$ permutations of $123$: $$\color{red}1\color{red}2\color{red}3$$ $$\color{red}1\color{black}3\color{black}2$$ $$\color{black}2\color{black}1\color{red}3$$ $$\color{black}2\color{black}3\color{black}1$$ $$\color{black}3\color{black}1\color{black}2$$ $$\color{black}3\color{red}2\color{black}1$$ with red indicating a correct position. Then each ...


11

Before the new baby is born, there are 2 boys (Andy and Bob). The new baby (NB) is born, then we pick a baby at random and get a boy. In order for this to happen, one of the following scenarios occurred: NB is a boy and we picked Andy NB is a boy and we picked Bob NB is a boy and we picked NB NB is a girl and we picked Andy NB is a girl and we picked ...


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