1. Stat 35b: Introduction to Probability with Applications to Poker
Outline for the day:
Bayes’s Rule
Bernoulli random variables.
Binomial random variables.
Farha vs. Antonius, expected value and variance.
Midterm is Feb 21, in class. 50 min.
Open book plus one page of notes, double sided. Bring a calculator!
  u    u 

2. 1. Bayes’s rule, p49-52.
Suppose that B1, B2 , Bn are disjoint events and that exactly one of them must occur.
Suppose you want P(B1 | A), but you only know P(A | B1 ), P(A | B2 ), etc.,
and you also know P(B1), P(B2), …, P(Bn).
Bayes’ Rule: If B1 , …, Bn are disjoint events with P(B1 or … or Bn) = 1, then
P(Bi | A) = P(A | Bi ) * P(Bi) ÷ [ ∑P(A | Bj)P(Bj)].
Why? Recall: P(X | Y) = P(X & Y) ÷ P(Y) So P(X & Y) = P(X | Y) * P(Y).
P(B1 | A) = P(A & B1 ) ÷ P(A)
= P(A & B1 ) ÷ [ P(A & B1) P(A & B2) … P(A & Bn) ]
= P(A | B1 ) * P(B1) ÷ [ P(A | B1)P(B1) + P(A | B2)P(B2) + … + P(A | Bn)P(Bn) ].

3. Bayes’s rule, continued.
Bayes’s rule: If B1 , …, Bn are disjoint events with P(B1 or … or Bn) = 1, then
P(Bi | A) = P(A | Bi ) * P(Bi) ÷ [ ∑P(A | Bj)P(Bj)].
See example 3.4.1, p50. If a test is 95% accurate and 1% of the pop. has a condition, then given a random person from the population,
P(she has the condition | she tests positive)
= P(cond | +)
= P(+ | cond) P(cond) ÷ [P(+ | cond) P(cond) + P(+ | no cond) P(no cond)]
= 95% x 1% ÷ [95% x 1% + 5% x 99%]
~ 16.1%.
Tests for rare conditions must be extremely accurate.

4. Bayes’ rule example.
Suppose P(your opponent has the nuts) = 1%, and P(opponent has a weak hand) = 10%.
Your opponent makes a huge bet. Suppose she’d only do that with the nuts or a weak hand, and that P(huge bet | nuts) = 100%, and P(huge bet | weak hand) = 30%.
What is P(nuts | huge bet)?
P(nuts | huge bet) =
P(huge bet | nuts) * P(nuts)
P(huge bet | nuts) P(nuts) + P(huge bet | horrible hand) P(horrible hand)
= % * 1%
100% * 1% % * 10%
= 25%.

5. 2. Bernoulli Random Variables, ch. 5.1.
If X = 1 with probability p, and X = 0 otherwise, then X = Bernoulli (p).
Probability mass function (pmf):
P(X = 1) = p
P(X = 0) = q, where p+q = 100%.
If X is Bernoulli (p), then µ = E(X) = p, and s = √(pq).
For example, suppose X = 1 if you have a pocket pair next hand; X = 0 if not.
p = 5.88%. So, q = 94.12%.
[Two ways to figure out p:
(a) Out of choose(52,2) combinations for your two cards, 13 * choose(4,2) are pairs.
13 * choose(4,2) / choose(52,2) = 5.88%.
(b) Imagine ordering your 2 cards. No matter what your 1st card is, there are 51 equally
likely choices for your 2nd card, and 3 of them give you a pocket pair. 3/51 = 5.88%.]
µ = E(X) = SD = s = √(.0588 * ) =

6. 3. Binomial Random Variables, ch. 5.2.
Suppose now X = # of times something with prob. p occurs, out of n independent trials
Then X = Binomial (n.p).
e.g. the number of pocket pairs, out of 10 hands.
Now X could = 0, 1, 2, 3, …, or n.
pmf: P(X = k) = choose(n, k) * pk qn - k.
e.g. say n=10, k=3: P(X = 3) = choose(10,3) * p3 q7 .
Why? Could have , or , etc.
choose(10, 3) choices of places to put the 1’s, and for each the prob. is p3 q7 .
Key idea: X = Y1 + Y2 + … + Yn , where the Yi are independent and Bernoulli (p).
If X is Bernoulli (p), then µ = p, and s = √(pq).
If X is Binomial (n,p), then µ = np, and s = √(npq).

7. Binomial Random Variables, continued.
Suppose X = the number of pocket pairs you get in the next 100 hands.
What’s P(X = 4)? What’s E(X)? s? X = Binomial (100, 5.88%).
P(X = k) = choose(n, k) * pk qn - k.
So, P(X = 4) = choose(100, 4) * * = 13.9%, or 1 in 7.2.
E(X) = np = 100 * = s = √(100 * * ) = 2.35.
So, out of 100 hands, you’d typically get about 5.88 pocket pairs, +/- around 2.35.

8. 4) Farha vs. Antonius, expected value and variance.
E(X+Y) = E(X) + E(Y). Whether X & Y are independent or not!
Similarly, E(X + Y + Z + …) = E(X) + E(Y) + E(Z) + …
And, if X & Y are independent, then V(X+Y) = V(X) + V(Y).
so SD(X+Y) = √[SD(X)^2 + SD(Y)^2].
Also, if Y = 9X, then E(Y) = 9E(Y), and SD(Y) = 9SD(X). V(Y) = 81V(X).
Farha vs. Antonius.
Running it 4 times. Let X = chips you have after the hand. Let p be the prob. you win.
X = X1 + X2 + X3 + X4, where X1 = chips won from the first “run”, etc.
E(X) = E(X1) + E(X2) + E(X3) + E(X4)
= 1/4 pot (p) + 1/4 pot (p) + 1/4 pot (p) + 1/4 pot (p)
= pot (p)
= same as E(Y), where Y = chips you have after the hand if you ran it once!
But the SD is smaller: clearly X1 = Y/4, so SD(X1) = SD(Y)/4. So, V(X1) = V(Y)/16.
V(X) ~ V(X1) + V(X2) + V(X3) + V(X4),
= 4 V(X1)
= 4 V(Y) / 16
= V(Y) / So SD(X) = SD(Y) / 2.