Stat 35b: Introduction to Probability with Applications to Poker

Outline for the day:

0. Collect hw2, return hw1, give out hw3.

1. Project A competition.

2. Bernoulli, binomial.

3. Geometric random variable.

4. Negative binomial

5. E(X+Y)

6. Harman / Negreanu , P(flop 2 pairs)

7. Running it twice

8. “Unbeatable” strategy

9. Play (and lose) like the Pros

10. P(rainbow flop)

11. Nguyen vs. Szenkuti

2. 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 = √(pq).

Binomial Random Variables.

If 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.

Key idea: X = Y1 + Y2 + … + Yn , where the Yi are independent and Bernoulli (p).

pmf: P(X = k) = choose(n, k) * pk qn - k.

If X is Binomial (n,p), then µ = np, and = √(npq).

3. 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 = (√q) ÷ p.

e.g. Suppose X = the number of hands til your next pocket pair. P(X = 12)? E(X)? ?

X = Geometric (5.88%).

P(X = 12) = p1 q11 = 0.0588 * 0.9412 ^ 11 = 3.02%.

E(X) = 1/p = 17.0. = 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.

4. Negative Binomial Random Variables.

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 = (√rq) ÷ p.

e.g. Suppose X = the number of hands til your 12th pocket pair. P(X = 100)? E(X)? ?

X = Negative Binomial (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. = 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.

5) E(X+Y) = E(X) + E(Y). Whether X & Y are independent or not!

Similarly, E(X + Y + Z + …) = E(X) + E(Y) + E(Z) + …

And, if X & Y are independent, then V(X+Y) = V(X) + V(Y).

so SD(X+Y) = √[SD(X)^2 + SD(Y)^2].

Example 1: Play 10 hands. X = your total number of pocket aces. What is E(X)?

X is binomial (n,p) where n=10 and p = 0.00452, so E(X) = np = 0.0452.

Alternatively, X = # of pocket aces on hand 1 + # of pocket aces on hand 2 + …

So, E(X) = Expected # of AA on hand1 + Expected # of AA on hand2 + …

Each term on the right = 1 * 0.00452 + 0 * 0.99548 = 0.00452.

So E(X) = 0.00452 + 0.00452 + … + 0.00452 = 0.0452.

E(X+Y) = E(X) + E(Y). Whether X & Y are independent or not!

Similarly, E(X + Y + Z + …) = E(X) + E(Y) + E(Z) + …

And, if X & Y are independent, then V(X+Y) = V(X) + V(Y).

so SD(X+Y) = √[SD(X)^2 + SD(Y)^2].

Example 2: Play 1 hand, against 9 opponents. X = total # of people with pocket aces.

Now what is E(X)?

Note: not independent! If you have AA, then it’s unlikely anyone else does too.

Nevertheless, Let X1 = 1 if player #1 has AA, and 0 otherwise.

X2 = 1 if player #2 has AA, and 0 otherwise.

etc.

Then X = X1 + X2 + … + X10.

So E(X) = E(X1) + E(X2) + … + E(X10)

= 0.00452 + 0.00452 + … + 0.00452 = 0.0452.

Harman vs. Negreanu….

6. Harman/Negreanu

If you’re sure to be all-in next hand, what is P(you will flop two pairs)?

Tough one. Don’t double-count (4 4 9 9 Q) and (9 9 4 4 Q)!

There are choose(13,2) possibilities for the NUMBERS of the two pairs.

For each such choice (such as 4 & 9),

there are choose(4,2) choices for the suits of the lower pair,

and same for the suits of the higher pair.

So, choose(13,2) * choose(4,2) * choose(4,2) different possibilities for the two pairs.

For each such choice, there are 44 [52 - 8 = 44] different possibilities for your fifth card, so

that it’s not a full house but simply two pairs. So,

P(flop two pairs) = choose(13,2) * choose(4,2) * choose(4,2) * 44 / choose(52,5)

~ 4.75%, or 1 in 21.

7) Running it twice. Harman has 10 7 . Negreanu has K Q . The flop is 10 7 K .

Harman’s all-in. $156,100 pot. P(Negreanu wins) = 29%. P(Harman wins) = 71%.

Let X = amount Harman has after the hand.

If they run it once, E(X) = $0 x 29% + $156,100 x 71% = $110,831.

If they run it twice, what is E(X)?

There’s some probability p1 that Harman wins both times ==> X = $156,100.

There’s some probability p2 that they each win one ==> X = $78,050.

There’s some probability p3 that Negreanu wins both ==> X = $0.

E(X) = $156,100 x p1 + $78,050 x p2 + $0 x p3.

If the different runs were independent, then p1 = P(Harman wins 1st run & 2nd run)

would = P(Harman wins 1st run) x P(Harman wins 2nd run) = 71% x 71% = 50.4%.

But, they’re not quite independent! Very hard to compute p1 and p2.

However, you don’t need p1 and p2 !

X = the amount Harman gets from the 1st run + amount she gets from 2nd run, so

E(X) = E(amount Harman gets from 1st run) + E(amount she gets from 2nd run)

= $78,050 x P(Harman wins 1st run) + $0 x P(Harman loses first run)

+ $78,050 x P(Harman wins 2nd run) + $0 x P(Harman loses 2nd run)

= $78,050 x 71% + $0 x 29% + $78,050 x 71% + $0 x 29% = $110,831.

HAND RECAP: Harman 10 7 . Negreanu K Q . Flop 10 7 K .

Harman’s all-in. $156,100 pot. P(Negreanu wins) = 29%. P(Harman wins) = 71%.

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The standard deviation changes a lot! Say they run it once.

V(X) = E(X2) - µ2.

µ = $110,831, so µ2 ~ $12.3 billion.

E(X2) = ($156,1002)(71%) + (02)(29%) = $17.3 billion.

So V(X) = $17.3 billion - $12.3 billion = $5 billion.

The standard deviation = sqrt($5 billion) ~ $71,000.

So if they run it once, Harman expects to get back about $110,831 +/- $71,000.

If they run it twice? Hard to compute, but approximately, if each run were independent, then

V(X1+X2) = V(X1) + V(X2),

so if X1 = amount she gets back on 1st run, and X2 = amount she gets from 2nd run,

then V(X1+X2) ~ V(X1) + V(X2) ~ $1.3 billion + $1.3 billion = $2.6 billion,

The standard deviation = sqrt($2.6 billion) ~ $51,000.

So if they run it twice, Harman expects to get back about $110,831 +/- $51,000.

8) “Unbeatable Texas Holdem Strategy”

http://www.freepokerstrategy.com : 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 65-35 underdog, so you

win 11 x 30% x 35% x $100 = $115.50

lose 11 x 30% x 65% x $100 = $214.50.

Total: exp. to win $77 + $115.50 - $55 - $214.50 = -$77/ 100 hands.

9) Phil Helmuth, Play Poker Like the Pros, Collins, 2003.

Strategy for beginners: AA, KK, QQ, or AK.

P(getting one of these hands)?

3 x choose(4,2)/choose(52,2) + 4x4/choose(52,2)

= 3 x 6/1326 + 16/1326 = 3 x 0.45% + 1.21% = 2.56% = 1 in 39.

Say you play $100 NL, table of 9, blinds 2/3, for 39x9 = 351 hands.

Pay 5 x 39 = 195 dollars in blinds.

Expect to play 9 hands.

Say P(win preflop) ~ 50%, and in those hands you win ~ $8.

Other 50%, always vs. 1 opponent, 60% to win $100.

So, expected winnings after 351 hands

= -$195

+ 9 x 50% x $8

+ 9 x 50% x 60% x $100

+ 9 x 50% x 40% x -$100

= -$69.

That is, you lose $69 every 351 hands on average

= $20 per 100 hands.

10) Rainbow flop = all 3 cards are of different suits.

P(Rainbow flop) = choose(4,3) * 13 * 13 * 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%.

11) 11/4/05, Travel Channel, World Poker Tour, $1 million Bay 101 Shooting Star.

4 players left, blinds $20,000 / $40,000, with $5,000 antes. Avg stack = $1.1 mil.

1st to act: Danny Nguyen, A 7. All in for $545,000.

Next to act: Shandor Szentkuti, A K. Call.

Others (Gus Hansen & Jay Martens) fold. (66% - 29%).

Flop: 5 K 5 . (tv 99.5%; cardplayer.com: 99.4% - 0.6%).

P(tie) = P(55 or A5)

= (1 + 2*2) ÷ choose(45,2) = 0.505%. 1 in 198.

P(Nguyen wins) = P(77) = choose(3,2) ÷ choose(45,2) = 0.30%. 1 in 330.

[Note: tv said “odds of running 7’s on the turn and river are 274:1.”

Given Hansen/Martens’ cards, choose(3,2) ÷ choose(41,2) = 1 in 273.3.]

* Szentkuti was eliminated next hand, in 4th place. Nguyen went on to win it all.

Turn: 7. River: 7!