Transcript (p).

Stat 35b: Introduction to Probability with Applications to Poker
Outline for the day:
1. Deal-making and expected value
2. Odds ratios, revisited
3. Variance and SD.
4. Bernoulli random variables
5. Binomial
6. Geometric
7. Negative binomial
Projects due by email by Monday, Oct 26, 8pm.
u 
u
1. Deal-making. (Expected value, game theory)
Game-theory: For a symmetric-game tournament, the probability of winning is approx.
optimized by the myopic rule (in each hand, maximize your expected number of chips),
and
P(you win) = your proportion of chips.
For a fair deal, the amount you win = the expected value of the amount you will win.
For instance, suppose a tournament is winner-take-all, for $8600.
With 6 players left, you have 1/4 of the chips left.
An EVEN SPLIT would give you $8600 ÷ 6 = $1433.
A PROPORTIONAL SPLIT would give you $8600 x (your fraction of chips)
= $8600 x (1/4) = $2150.
A FAIR DEAL would give you the expected value of the amount you will win
= $8600 x P(you get 1st place) = $2150.
But suppose the tournament is not winner-take-all, but pays
$3800 for 1st, $2000 for 2nd, $1200 for 3rd, $700 for 4th, $500 for 5th, $400 for 6th.
Then a FAIR DEAL would give you
$3800 x P(1st place) + $2000 x P(2nd) +$1200 x P(3rd)+$700xP(4th) +$500xP(5th) +$400xP(6th).
Hard to determine these probabilities. But, P(1st) = 25%, and you might roughly estimate the
others as P(2nd) ~ 20%, P(3rd) ~ 20%, P(4th) ~ 15%, P(5th) ~10%, P(6th) ~ 10%, and get
$3800 x 25% + $2000 x 25% +$1200 x 20% + $700x 15% + $500x 10% +$400x 5% = $1865.
If you have 40% of the chips in play, then:
EVEN SPLIT = $1433. PROPORTIONAL SPLIT = $3440.
FAIR DEAL ~ $2500!
Another example. Before the Wasicka/Binger/Gold hand,
Gold had 60M,
Wasicka 18M,
Payouts: 1st place $12M,
Binger 11M.
2nd place $6.1M,
3rd place $4.1M.
Proportional split: of the total prize pool left, you get your proportion of chips in play.
e.g. $22.2M left, so Gold gets 60M/(60M+18M+11M) x $22.2M ~ $15.0M.
A fair deal would give you
P(you get 1st place) x $12M + P(you get 2nd place) x $6.1M + P(3rd pl.) x $4.1M .
*Even split:
Gold $7.4M,
Wasicka $7.4M,
Binger $7.4M.
*Proportional split: Gold $15.0M,
Wasicka $4.5M,
Binger $2.7M.
*Fair split:
Gold $10M,
Wasicka $6.5M,
Binger $5.7M.
*End result:
Gold $12M,
Wasicka $6.1M,
Binger $4.1M.
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….
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% - 37.9% = 62.1%.
Gold: A K. Hellmuth: A K. Farha: 8 7. Flop: 4 7 K.
P(Hellmuth fails to make a flush) = 100% - 37.9% = 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 = 0.07. Nowhere near the right answer.
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 = $20.5 - $-0.502 = $20.25. 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 = $2.50 - $-0.502 = $2.25. s = $1.50.
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) = .0588.
SD = s = √(.0588 * 0.9412) = 0.235.
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 1 1 1 0 0 0 0 0 0 0, or 1 0 1 1 0 0 0 0 0 0, 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).
5. Binomial Random Variables.
pmf: P(X = k) = choose(n, k) * pk qn - k.
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).
e.g. 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) * 0.0588 4 * 0.9412 96 = 13.9%, or 1 in 7.2.
E(X) = np = 100 * 0.0588 = 5.88. s = √(100 * 0.0588 * 0.9412) = 2.35.
So, out of 100 hands, you’d typically get about 5.88 pocket pairs, +/- around 2.35.
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 0 0 0 0 1. 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 = 0.0588 * 0.9412 ^ 11 = 3.02%.
E(X) = 1/p = 17.0. s = sqrt(0.9412) / 0.0588 = 16.5.
So, you’d typically expect it to take 17 hands til your next pair, +/- around 16.5 hands.
7. Negative Binomial Random Variables.
Recall: if each trial is independent, and each time the probability of an occurrence is p,
and X = # of trials until the first occurrence, then:
X is Geometric (p),
P(X = k) = p1 qk
- 1,
µ = 1/p,
s = (√q) ÷ p.
Suppose now X = # of trials until the rth occurrence.
Then X = negative binomial (r,p).
e.g. the number of hands you have to play til you’ve gotten r=3 pocket pairs.
Now X could be 3, 4, 5, …, up to ∞.
pmf: P(X = k) = choose(k-1, r-1) pr qk - r.
e.g. say r=3 & k=7: P(X = 7) = choose(6,2) p3 q4.
Why? Out of the first 6 hands, there must be exactly r-1 = 2 pairs. Then pair on 7th.
P(exactly 2 pairs on first 6 hands) = choose(6,2) p2 q4. P(pair on 7th) = p.
If X is negative binomial (r,p), then µ = r/p, and s = (√rq) ÷ p.
e.g. Suppose X = the number of hands til your 12th pocket pair. P(X = 100)? E(X)? s?
X = Geometric (12, 5.88%).
P(X = 100) = choose(99,11) p12 q88
= choose(99,11) * 0.0588 ^ 12 * 0.9412 ^ 88 = 0.104%.
E(X) = r/p = 12/0.0588 = 204.1. s = sqrt(12*0.9412) / 0.0588 = 57.2.
So, you’d typically expect it to take 204.1 hands til your 12th pair, +/- around 57.2 hands.