Stat 35b: Introduction to Probability with Applications to Poker

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• Stat 35b Introduction to Probability with
  Applications to Poker
• Outline for the day
• Turn in Hw2. Hw3 assigned.
• Expected value and pot odds, continued
• Violette/Elezra example
• Yang / Kravchenko
• Deal-making and expected value
• Reminder project A is due by email by Wed, Feb
  6, 8pm.
• You may email me for your teammates email
  address.

? ? u ? ? ? u ?
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Teams team a Sohil Yara team b Chunlin
Matthew team c Kyle Vielka team d Bhavik
Shannon team e Andrew Megan team f Jasmine
Stefan team g Christopher Jon team h Richard
Michael team i Wing Shukki team j Alexia Siena
team k Casey Binh team l Uddhav Thanh team m
Kai Li team n Kaavya Yongxin team o Arash
Rocio team p Denise Wenhao
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• Turn in Hw2.
• Hw3 is due Thur Feb 28, 11am.
• 4.7, 4.8, 4.12, 4.16, 5.6, 6.2.
• Also, read chapter 5.
• Reminder project A is due by email by Wed, Feb
  6, 8pm.
• You may email me for your teammates email
  address.
• Just submit one email per team.

? ? u ? ? ? u ?
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2) Pot odds and expected value, continued. From
a 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. See p66.
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3) Example Poker Superstars Invitational
Tournament, FSN, October 2005. Ted Forrest 1
million chips Freddy Deeb 825,000 Blinds
15,000 / 30,000 Cindy Violette 650,000 Eli
Elezra 575,000 Elezra raises to
100,000  Forrest folds. Deeb, the small
blind, folds. Violette, the big blind with Ku
Ju, calls. The flop is 2u 7? Au
Violette bets 100,000. Elezra raises all-in
to 475,000. (pot 790,000) So, it's 375,000
more to Violette. She folds. Q Based on
expected value, should she have called? Her
chances must be at least 375,000 / (790,000
375,000) 32.3
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Violette has Ku Ju. The flop is 2u 7? Au. Q
Based on expected value, should she have
called? Her chances must be at least 375,000 /
(790,000 375,000) 32. vs. AQ 38.
AK 37 AA 26 77 26 A7 31
A2 34 72 34 TT 54
T9 87 73 50 Harrington's principle
always assume at least a 10 chance that
opponent is bluffing. Bayesian approach average
all possibilities, weighting them by their
likelihood. Maybe she's conservative.... but then
why play the hand at all? Reality Elezra had
7u 3?. Her chances were 51. Bad fold. What
was her prob. of winning (given just her cards
and Elezras, and the flop)? Of choose(45,2)
990 combinations for the turn river, how many
give her the win? First, how many outs did she
have? eight us 3 kings 3 jacks 14. She
wins with (out, out) or (out, nonout) or (non-u
Q, non-u T) choose(14,2) 14 x 31
3 3 534 but not (k or j, 7 or non-u 3)
and not (3u , 7 or non-u 3) - 6 4 - 1
4 506. So the answer is 506 / 990 51.1.
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4. 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.
See p49-53.
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4. 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.
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5. 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
(Theorems 7.6.6 and 7.6.7 on pp 151-152). For a
fair deal, the amount you win the expected
value of the amount you will win. See p61.
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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!
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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.
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Luck vs. skill. pp 71-79. Any thoughts? name1
c("gravity","tommy","ursula","timemachine","vera
","william","xena) decision1 list(gravity,
tommy, ursula, timemachine, vera, william, xena)
tourn1(name1, decision1, myfast1 2) run
quickly tourn1(name1, decision1, myfast1 0)
run slowly, showing key hands