Stat 35b: Introduction to Probability with Applications to Poker

1

• Stat 35b Introduction to Probability with
  Applications to Poker
• Outline for the day
• More counting combinations problems
• Yang / Kravchenko expected value
• Deal making expected value
• Odds ratios, revisited, Gold/Hellmuth
• Variance and SD
• Bernoulli random variables
• Binomial random variables
• Geometric random variables
• Projects due by email by Saturday.
• You may email me for your teammates' email
  addresses.

? ? u ? ? ? u ?
2
1) Combinations problems -- How likely is it
to make 4-of-a-kind? 1 in ___ ? -- What
about the probability of flopping 4-of-a-kind?
-- What about the prob. of flopping
4-of-a-kind, given that you have a pocket
pair? -- Is it less likely than
flopping an ace-high flush? flopping a
straight-flush?
3
Suppose youre all in next hand, no matter what
cards you get. P(eventually make 4-of-a-kind)?
including case where all 4 are on board
Trick just forget card order, and consider all
collections of 7 cards. Out of choose(52,7)
different combinations, each equally likely, how
many of them involve 4-of-a-kind? 13 choices
for the 4-of-a-kind. For each such choice, there
are choose(48,3) possibilities for the other 3
cards. So, P(4-of-a-kind) 13 choose(48,3) /
choose(52,7) 0.168, or 1 in 595. P(flop
4-of-a-kind) 1348 / choose(52,5) 0.024
1 in 4165. P(flop 4-of-a-kind pocket pair)?
No matter which pocket pair you have, there are
choose(50,3) possible flops, each equally likely,
and how many of them give you 4-of-a-kind? 48.
(e.g. if you have 7? 7?, then need to flop 7u 7?
x, there are 48 choices for x) So P(flop
4-of-a-kind pp) 48/choose(50,3) 0.245 1
in 408.
4
P(flop an ace high flush)? where
the ace might be on the board -- 4 suits
-- one of the cards must be an ace. choose(12,4)
possibilities for the others. So P(flop ace high
flush) 4 choose(12,4) / choose(52,5)
0.0762, or 1 in 1313. P(flop a straight
flush)? -- 4 suits -- 10 different
straight-flushes in each suit. (5 high, 6 high,
, Ace high) So P(flop straight flush) 4 10 /
choose(52,5) 0.00154, or 1 in 64974.
5
1) Pot odds and expected value, continued. From
previous lecture to call an all-in, need P(win)
gt B (Bpot). Expressed as an odds ratio, this
is sometimes referred to as pot odds or express
odds. If the bet is not all-in another
betting round is still to come, need P(win) gt
wager (wager winnings), where winnings
pot amount youll win on later betting
rounds, wager total amount you will wager
including the current round later rounds,
assuming no folding. The terms Implied-odds
/ Reverse-implied-odds describe the cases
where winnings gt pot or where
wager gt B, respectively.
6
2. Yang / Kravchenko. Yang A? 10u. Pot is
19million. Bet is 8.55 million. Needs P(win) gt
8.55 (8.55 19) 31. vs. AA 8.5. AJ-AK
25-27. KK-TT 29. 99-22 44-48. KQs
56. Bayesian method average these
probabilities, weighting each by its likelihood.

7
2. Yang / Kravchenko. Yang A? 10u. Pot is
19.0 million. Bet is 8.55 million. Suppose
that, averaging the different probabilities,
P(Yang wins) 30. And say Yang calls. Let X
the number of chips Kravchenko has after the
hand. What is E(X)? Note, if Yang folds, then X
19.0 million for sure. E(X) ? k
P(Xk) 0 30 27.55 million 70
19.285 million.
8
3. 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.
9
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!
10
Another example. Before the Wasicka/Binger/Gold
hand, Gold had 60M, Wasicka 18M, Binger
11M. Payouts 1st place 12M, 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/(60M18M11M) 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.
11
4. 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(BA), but you cant multiply odds
ratios. Example Gold vs. Hellmuth on High
Stakes Poker
12
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.
13
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(AB) P(A)P(BA) 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. Moral you cant multiply odds ratios!
14
5. Variance and SD. Expected Value E(X) µ ?k
P(Xk). Variance V(X) s2 E(X- µ)2. Turns
out this E(X2) - µ2. Standard deviation s
v 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.
15
6. 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 pq 100. If X is Bernoulli (p),
then µ E(X) p, and s v(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
v(.0588 0.9412) 0.235.
16
7. 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 n10, k3 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 1s, 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
v(pq). If X is Binomial (n,p), then µ np, and
s v(npq). e.g. Suppose X the number of
pocket pairs you get in the next 100
hands. Whats P(X 4)? Whats 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 v(100 0.0588 0.9412) 2.35. So, out of
100 hands, youd typically get about 5.88 pocket
pairs, /- around 2.35.
17
8. 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 8. pmf P(X k) p1 qk - 1.
e.g. say k5 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 (vq)
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,
youd typically expect it to take 17 hands til
your next pair, /- around 16.5 hands.