1. Stat 35b: Introduction to Probability with Applications to Poker
Outline for the day:
Midterms back
E(X+Y+Z) example, corrected.
Rainbow flops.
New WSOP champion.
The fundamental theorem of poker
Luck vs. skill
Random walks.
Reflection principle.
Ballot theorem.
  u    u 

2. 2. E(X+Y+Z) example again.
Deal til the 2nd king. X = # of cards til the 2nd king. E(X)?
Consider any ordering of the deck. e.g. 7,Q,Ju,2, 9,K,3,9,7,A, K,…
For any such ordering, there is another with the cards before the first king exchanged with the cards between the 1st king and the 2nd king.
Similarly, for the cards between the 2nd king and 3rd king, and for the cards between the 3rd king and 4th king, and between the 4th king and the end of the deck.
So, P(# of cards before 1st king = 5) = P(# of cards bet. 1st and 2nd king = 5), etc.
Therefore if X1 = the # of cards before the 1st king,
and X1 = # of cards between 1st and 2nd king, then
E(X1) = ∑k k P(X1 =k)
= ∑k k P(X2 =k)
= E(X2).
And, X1 + X2 + X3 + X4 + X5 = 48, so EX1 + EX2 + EX3 + EX4 + EX5 = 48,
so E(X1) = E(X2) = E(X3) = E(X4) = E(X5) = 48/5 = 9.6
We want E(X1 + X2 + 2) = E(X1) + E(X2) + 2 = 21.2.

3. 3. Rainbow flops.
P(Rainbow flop) = choose(4,3) * * 13 * ÷ choose(52,3)
choices for the 3 suits numbers on the 3 cards possible flops
~ 39.76%.
Alternative way: conceptually, order the flop cards. No matter what flop card #1 is, P(suit of #2 ≠ suit of #1 & suit of #3 ≠ suits of #1 and #2)
= P(suit #2 ≠ suit #1) * P(suit #3 ≠ suits #1 and #2 | suit #2 ≠ suit #1)
= 39/51 * 26/50 ~ 39.76%.
Q: Out of 100 hands, expected number of rainbow flops? +/- what?
X = Binomial (n,p), with n = 100, p = 39.76%, q = 60.24%.
E(X) = np = 100 * = 39.76
SD(X) = √(npq) = sqrt(23.95) = 4.89.
So, expected around / rainbow flops, out of 100 hands.

4. 4. New World Series of Poker Champion Joe Cada.
5. The Fundamental Theorem of Poker.
David Sklansky, The Theory of Poker, 1987.
“Every time you play a hand differently from the way you would have played it if you could see all your opponents’ cards, they gain; and every time you play your hand the same way you would have played it if you could see all their cards, they lose. Conversely, every time opponents play their hands differently from the way they would have if they could see all your cards, you gain; and every time they play their hands the same way they would have played if they could see all your cards, you lose.”
Meaning?
LLN: If X1, X2 , etc. are iid with expected value µ and sd s, then X ---> µ.
Equity = pot * your probability of winning. Equity – your bets = your expected profit.
Your expected profit can go up or down because of the cards dealt (luck) or because of what happens during the betting (skill?).

5. 6. Luck versus skill.
Example. Minieri vs. Lederer on Poker After Dark.
Blinds 800/1600. Minieri Ac Jc, Lederer Ah 9h. Minieri %, Lederer %.
Minieri raises to 4300, Lederer calls 2700.
Flop 6c Ts Tc. Minieri 85.91%, Lederer 14.09%.
Lederer checks, Minieri 6500, Lederer folds.
a) Preflop dealing (luck). Minieri
Minieri is dealt a % probability of winning the pot, so his
increase in expected profit is % * =
Lederer is dealt %, so his increase in expected profit is % * =
b) Preflop betting (skill). Minieri
The pot gets increased to =5400.
Minieri has paid an additional 2700 but has % * 5400 = additional equity, so Minieri's expected profit due to betting is =
Correspondingly, Lederer's profit due to betting is , since % * =

6. Blinds 800/1600. Minieri Ac Jc, Lederer Ah 9h. Minieri 70
Blinds 800/1600. Minieri Ac Jc, Lederer Ah 9h. Minieri %, Lederer %.
Minieri raises to 4300, Lederer calls 2700.
Flop 6c Ts Tc. Minieri 85.91%, Lederer 14.09%.
Lederer checks, Minieri 6500, Lederer folds.
c) Flop dealing (luck). Minieri
On the flop, Minieri's equity went from % of the 8600 pot to 85.91% of it. So by luck Minieri increased his equity by (85.91% %)*8600 =
d) Flop betting (skill). Minieri
Because of the betting on the flop, Minieri's equity went from 85.91% of the 8600 pot to 100% of it, so Minieri increased his expected profit by (100% %)*8600 =
So, by luck, Minieri got = By skill, Minieri got =
The total = = 4300, which is the number of chips Minieri won from Lederer in the hand.
GRAND TOTAL for Minieri over 27 hands they showed, Luck Skill
On the first 19 hands, Minieri gained 20, due to skill, On hands 20 and 21, Minieri tried two huge unsuccessful bluffs, and on those two hands combined, Minieri lost 40, in equity due to skill.

7. 7. Random walks.
Suppose that X1, X2, …, are iid and that Yk = X1 + … + Xk for k = 1, 2, …. Y0 = 0.
Then the totals {Y1, Y2, …} is a random walk.
The classical example is when each Xi is 1 or -1 with probability ½ each.
For the classical random walk, what is P(Yk = j)?
0 if k is even and j is odd. Or vice versa. Or if j is not an integer, or if |j| > k.
But otherwise, P(Yk = j) = P{go up (k+j)/2 times and down (k-j)/2 times}
= Choose{k, (k+j)/2} (1/2)k.

8. 8. Reflection principle.
If j and k are > 0, then the number of paths from (0,j) to (n,k) that touch the x-axis = the number of paths from (0,-j) to (n,y).
Reason: Any path from (0, -j) to (n,y) must go through the x-axis, and for any path from (0,-j) to (m,0) to (n,y), there is a corresponding path from (0,j) to (m,0) to (n,y), by reflecting each step of the path from (0,-j) to (m,0).

9. 9. Ballot theorem.
Suppose that in an election, candidate X gets x votes, and candidate Y gets y votes, where x > y.
The probability that X always leads Y throughout the counting is
(x-y) / (x+y).
Proof. We know that, after counting n = x+y votes, the total difference in votes is x-y = a.
We want to count the number of paths from (1,1) to (n,a) that do not touch the x-axis.
By the reflection principle, the number of paths from (1,1) to (n,a) that do touch the x-axis equals the total number of paths from (1,-1) to (n,a).
So the number of paths from (1,1) to (n,a) that do not touch the x-axis equals the number of paths from (1,1) to (n,a) minus the number of paths from (1,-1) to (n,a)
= Choose(n-1,x-1) – Choose(n-1,x)
= (n-1)! / [(x-1)! (n-x)!] – (n-1)! / [x! (n-x-1)!]
= (n-1)! x / [x! (n-x)!] – (n-1)! (n-x) / [x! (n-x)!]
= {n! / [x! (n-x)!]} (x/n) – {n! / [x! (n-x)!]} (n-x)/n
= (x – y) / (x+y) {n! / [x! (n-x)!]}
= (x – y) / (x+y) {Number of paths from (0,0) to (n,a)}. And each path is equally likely.

10. For a classical random walk,
P(Y1 ≠ 0, Y2 ≠ 0, …, Y2n ≠ 0) = P(Y2n = 0).
Proof. The number of paths from (0,0) to (2n, 2r) that don’t touch the x-axis at positive times
= the number of paths from (1,1) to (2n,2r) that don’t touch the x-axis at positive times
= N2n-1,2r-1 – N2n-1,2r+1.
Let pn,x = P(Yn = x).
P(Y1 > 0, Y2 > 0, …, Y2n-1 > 0, Y2n = 2r) = ½[p2n-1,2r-1 – p2n-1,2r+1].
Summing from r = 1 to ∞,
P(Y1 > 0, Y2 > 0, …, Y2n-1 > 0, Y2n > 0) = (1/2) p2n-1,1 = (1/2) P(Y2n = 0).
By symmetry, the same is true for P(Y1 < 0, Y2 < 0, …, Y2n-1 < 0, Y2n < 0).
So, P(Y1 ≠ 0, Y2 ≠ 0, …, Y2n ≠ 0) = P(Y2n = 0).