1. Stat 35b: Introduction to Probability with Applications to Poker
Outline for the day:
Luck and skill in poker.
Odds ratios, revisited, and Gold vs. Hellmuth on HSP.
Variance and SD.
Bernoulli random variables
Binomial
Geometric
“Unbeatable” strategy.
More sample code.
  u    u 

2. Reminder: project A is due by email by Wed, Feb 8, 8pm.
You may me for your teammate’s address.
Hw3 is due Thur Mar 1, 12:30pm.
4.7, 4.8, 4.12, 4.16, 5.6, 6.2.
Also, read chapter 5 this week.
Luck and skill in poker, p71-79.
  u    u 

3. 2. Odds ratios, revisited:
Odds ratio of A = P(A)/P(Ac)
Odds against A = Odds ratio of Ac = P(Ac)/P(A).
An advantage of probability over odds ratios is the multiplication rule:
P(A & B) = P(A) x P(B|A), but you can’t multiply odds ratios.
Example: Gold vs. Hellmuth on High Stakes Poker….

4. Gold: A K. Hellmuth: A K. Farha: 8 7. Flop: 4 7 K.
Given these 3 hands and the flop, what is P(Hellmuth makes a flush)?
43 cards left: 9 s, 34 non-s. Of choose(43, 2) = 903 eq. likely turn/river combos,
choose(9,2) = 36 have both s, and 9 *34 = 306 have exactly 1 . 342/903 = 37.9%.
So, P(Hellmuth fails to make a flush) = 100% % = 62.1%.

5. Moral: you can’t multiply odds ratios!
Gold: A K. Hellmuth: A K. Farha: 8 7. Flop: 4 7 K.
P(Hellmuth fails to make a flush) = 100% % = 62.1%.
Alt.: Given these 3 hands and the flop, P(neither turn nor river is a )
= P(turn is non- AND river is non-)
= P(turn is non-) * P(river is non- | turn is non-) [P(A&B) = P(A)P(B|A)]
= 34/43 * 33/42 = 62.1%.
Note that we can multiply these probabilities: 34/43 * 33/42 = 62.1%.
What are the odds against Hellmuth failing to make a flush?
37.9% ÷ 62.1% = 0.61 : 1.
Odds against non- on turn = 0.26 :1.
Odds against non- on river | non- on turn : 0.27 : 1.
0.26 * 0.27 = Nowhere near the right answer.
Moral: you can’t multiply odds ratios!

6. 3. Variance and SD.
Expected Value: E(X) = µ = ∑k P(X=k).
Variance: V(X) = s2 = E[(X- µ)2]. Turns out this = E(X2) - µ2.
Standard deviation = s = √ V(X). Indicates how far an observation would typically deviate from µ.
Examples:
Game 1. Say X = $4 if red card, X = $-5 if black.
E(X) = ($4)(0.5) + ($-5)(0.5) = -$0.50.
E(X2) = ($42)(0.5) + ($-52)(0.5) = ($16)(0.5) + ($25)(0.5) = $20.5.
So s2 = E(X2) - µ2 = $ $ = $ s = $4.50.
Game 2. Say X = $1 if red card, X = $-2 if black.
E(X) = ($1)(0.5) + ($-2)(0.5) = -$0.50.
E(X2) = ($12)(0.5) + ($-22)(0.5) = ($1)(0.5) + ($4)(0.5) = $2.50.
So s2 = E(X2) - µ2 = $ $ = $ s = $1.50.

7. 4. Bernoulli Random Variables.
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 * ) =

8. 5. Binomial Random Variables.
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).

9. 5. 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.

10. 6. Geometric Random Variables.
Suppose now X = # of trials until the first occurrence.
(Again, each trial is independent, and each time the probability of an occurrence is p.)
Then X = Geometric (p).
e.g. the number of hands til you get your next pocket pair.
[Including the hand where you get the pocket pair. If you get it right away, then X = 1.]
Now X could be 1, 2, 3, …, up to ∞.
pmf: P(X = k) = p1 qk - 1.
e.g. say k=5: P(X = 5) = p1 q 4. Why? Must be Prob. = q * q * q * q * p.
If X is Geometric (p), then µ = 1/p, and s = (√q) ÷ p.
e.g. Suppose X = the number of hands til your next pocket pair. P(X = 12)? E(X)? s?
X = Geometric (5.88%).
P(X = 12) = p1 q11 = * ^ 11 = 3.02%.
E(X) = 1/p = s = sqrt(0.9412) / = 16.5.
So, you’d typically expect it to take 17 hands til your next pair, +/- around 16.5 hands.

11. 7) “Unbeatable Texas Holdem Strategy”
: all in with AK-AT or any pair.
P(getting such a hand) = 4 x [16/1326] + 13 x [6/1326]
= 4 x 1.2% + 13 x 0.45% = 10.7%.
Say you’re dealt 100 hands. Pay ~11 blinds = $55.
Expect 10.7 (~ 11) such good hands.
Say you’re called by 88-AA, and AK, for $100 on avg.
P(player 1 has one of these) = 7 x 0.45% + 1.2% = 4.4%.
P(of 8 opponents, someone has one of these) ~ 1 - (95.6%)8 = 30%.
So, you win pre-flop 70% of the time. (Say $10 on avg.)
= 11 x 70% x $10 = $77 profit.
Other 30%, you’re on avg about a underdog, so you
win 11 x 30% x 35% x $100 = $115.50
lose 11 x 30% x 65% x $100 = $
Total: exp. to win $77 + $ $55 - $ = -$77/ 100 hands.

12. if((crds1[1,1] == crds1[2,1]) || ((crds1[1,1] > 13.5) &&
8. More example code for project A
unbeatable1 = function(numattable1, crds1, board1, round1, currentbet, mychips1, pot1, roundbets, blinds1, chips1, ind1, dealer1, tablesleft){
## any pair, or AT-AK,
a1 = 0
if((crds1[1,1] == crds1[2,1]) || ((crds1[1,1] > 13.5) &&
(crds1[2,1]>9.5))) a1 = mychips1 a1
} ## end of unbeatable

13. timemachine = function(numattable1, crds1, board1, round1, currentbet, mychips1,
pot1, roundbets, blinds1, chips1, ind1, dealer1, tablesleft){
## any pair 7 or higher
## AK, AQ
## AJ if nobody's all in yet.
## if less than 3 times bb & one card is Ten or higher, then 75%
a1 = 0
x = runif(1)
if((crds1[1,1] == crds1[2,1]) && (crds1[1,1] > 6.5)) a1 = mychips1
if((crds1[1,1] == 14) && (crds1[2,1] > 11.5)) a1 = mychips1
if((crds1[1,1] == 14) && (crds1[2,1] == 11) &&
(currentbet <= blinds1)) a1 = mychips1
if((mychips1 < 3*blinds1) && (crds1[1,1] >= 10) && (x<.75)) a1 = mychips1 a1
} ## end of timemachine

14. zebra = function(numattable1, crds1, board1, round1, currentbet, mychips1, pot1,roundbets, blinds1, chips1, ind1, dealer1, tablesleft){
## if pair of 10s or higher, all in for sure, no matter what. If AK or AQ, all in with probability 75%.
## if pair of 7s or higher and there are 6 or fewer players at your table (including you), then all in.
## if your chip count is less than twice the big blind, go all in with any cards.
## if nobody's raised yet ... and there are 3 or fewer players left behind you, go all in with any pair or ace.
## and there's only 1 or 2 players behind you, then go all in with any cards.
a1 = 0
x = runif(1) ## x is a random number between 0 and 1.
y = max(roundbets[,1]) ## y is the maximum bet so far.
big1 = dealer1 + 2
if(big1 > numattable1) big1 = big1 - numattable1
z = big1 - ind1
if(z<0) z = z + numattable1
## the previous 4 lines make it so z is the number of players left to act behind you.
if((crds1[1,1] == crds1[2,1]) && (crds1[2,1] > 9.5)) a1 = mychips1
if((crds1[1,1] == 14) && (crds1[1,2]>11.5) && (x<.75)) a1 = mychips1
if((crds1[1,1] == crds1[2,1]) && (crds1[2,1] > 6.5) && (numattable1 < 6.5)) a1 = mychips1
if(mychips < 2*blinds1) a1 = mychips1
if(y <= blinds1){ if((z < 3.5) && ((crds1[1,1] == crds1[2,1]) || (crds1[1,1] == 14))) a1 = mychips1
if(z < 2.5) a1 = mychips1 }
a1} ## end of zebra