Uniform Distribution

“X is uniformly distributed over the integers between a and b, inclusive.”

Expected value = (a + b)/2

Binomial Distribution

“X has distribution Binomial (n, p)” =“Pick X = x. In n tosses of a coin with probability p of coming up heads, what is the probability of x heads?”

Expected value = np

Geometric Distribution

“X has distribution Geometric (p)” =“Pick X = i. You are tossing a coin with probability p of turning up Heads until it turns up Heads. What is the probability that the ith toss turns up Heads (= what is the probability that all the previous tosses were Tails?)”

Expected value = 1/p

Poisson Distribution

“X has distribution Poisson (λ)” =“X has distribution Binomial (n, λ/n)” =“Pick X = i. In n tosses of a coin with probability p (= λ/n) of coming up heads, what is the probability of i heads? Assume n is very large and p is very small.”

Expected value = λ

Door-“nom”-ial Distribution

An infinite line of cats are walking through an open catflap, which has a probability p of shutting behind any cat. How many cats can make it through the catflap, on average?

Assume ideal conditions: the cats are moving at equal speed through the catflap (what.) and the catflap does not close on a cat: just before or after. However, at least one cat makes it through.

Door-“nom”-ial DistributionLet X be a random variable such that if X = i, then i cats have passed through.So, if X = 1, one cat has passed through; if X = 2, two cats have passed through; and so on.

Pr[X = 1] = p,Pr[X = 2] = (1 – p)p,Pr[X = 3] = (1 – p)2p, and so on.

X has a geometric distribution! E[X] = 1/p.

(1/p) cats will pass through on average.

Door-“nom”-ial Distribution II

Infinite cats? That’s so unrealistic! Let’s restrict it to a line of p cats. The catflap has an equal probability of closing after any cat, after which no other cats can pass through.

What is the probability distribution of the number of cats that pass through the catflap?

Door-“nom”-ial Distribution IIIf there are p cats, then the catflap has a probability of q = 1/p of shutting between cats.

Again, let X be a random variable that count how many cats have passed through.

Pr[X = 1] = qPr[X = 2] = (1 – q)qPr[X = 3] = (1 – q)2q …Pr[X = (p – 1)] = (1 – q)(p – 2)qPr[X = p] = (1 – q)p-1 (All of the cats make it through).Pr[X > p] = 0.

What do you “mean”?

Now, how many cats pass through on average? (Feel free to leave your answer in terms of a summation.)

Schrödinger’s Expectations

You now have n = 40 cats, and you stuff them each into their own box. Each box has a device containing poisonous hydrogen sulfide; each box has 40% chance of breaking after five minutes and poisoning the cat, but 60% otherwise.

After five minutes, you open the boxes: how many cats will still be alive, on average?

Schrödinger’s Expectations

Let X be the number of cats that survive. Notice that this experiment is similar to tossing a coin with probability p = 0.6 of coming up Heads, and counting the number of Heads in n = 40 trials.

X has a binomial distribution!E[X] = np = 24 cats will be alive, on average.

(No cats were harmed in this experiment.The previous statement is a lie, and so is this one.)

Monty Hall v2.0

In this variant, Contestant A picks a door from the three. Contestant B then picks a door and opens it. Finally, Contestant A is given the option to stick or switch.

Given that Contestant B’s door did not contain the prize, what is Contestant A’s best strategy?

Monty Hall v2.0

1 2 3

Assume Contestant A picks Door 1Sample Space = (Door with Cake, Contestant B’s Door)

Ω = {(1, 2), (1, 3), (2, 2), (2, 3), (3, 2), (3, 3)}

Monty Hall v2.0

X = Contestant A has the winning doorX = {(1, 2), (1, 3)}

Y = Contestant B chose a losing doorY = {(1, 2), (1, 3), (2, 3), (3, 2)}