Stat 35b: Introduction to Probability with Applications to Poker

Outline for the day:

1. Hand in hw4.

2. Review list

3. Tournament

4. Sample problems

* Final is Wednesday, 1-2:50pm, in class.

* Open note.

* Bring a calculator and a pen or pencil.

* about 20-25 multiple choice + 2-3 open-answers

+ an extra-credit with 2 open-answer parts.

1. Review List

1) Basic principles of counting

2) Permutations & combinations

3) Addition rule

4) Conditional probability

5) Multiplication rule

6) Independence

7) Odds ratios

8) Random variables (RVs)

9) Discrete RVs, and probability mass function (pmf)

10) Expected value

11) Pot odds calculations

12) Deal-making

13) Variance and SD

1. Review List, continued

14) Bernoulli RV [0-1. µ = p, = √(pq). ]

15) Binomial RV [# of successes, out of n tries. µ = np, = √(npq).]

16) Geometric RV [# of tries til 1st success. µ = 1/p, = (√q) / p. ]

17) Negative binomial RV [# of tries til rth success. µ = r/p, = (√rq) / p. ]

18) E(X+Y), V(X+Y)

19) Continuous RVs

20) Probability density function (pdf)

21) Uniform RV

22) Normal RV

23) Law of Large Numbers (LLN)

24) Central Limit Theorem (CLT)

25) Bayes’ Rule

Bayes’ Rule Example

If B1 , …, Bn are disjoint events with P(B1 or … or Bn) = 1, then

P(Bi | A) = P(A | Bi ) * P(Bi) ÷ [ ∑P(A | Bj)P(Bj)].

Q. Suppose your opponent looks at her cards and then looks at her chips

immediately. You know that she’d do that with 100% probability if she had AA

or KK, and with 50% probability if she had AK. Given only this information,

what is the probability that she has AA?

A. We want P(AA | looked at chips).

But we know P(looked at chips | AA) = 100%.

P(AA | looked at chips) = P(looked at chips | AA) P(AA) ÷

[P(looked at chips | AA)P(AA) +P(looked at chips | KK)P(KK) +P(looked at chips|AK)P(AK)]

= 100% x P(AA) ÷ [100% x P(AA) + 100% x P(KK) + 50% x P(AK)]

= 100% x 6/1326 ÷ [100% x 6/1326 + 100% x 6/1326 + 50% x 16/1326]

= 6/1326 ÷ [20/1326]

= 30.0%.

Rainbow flops.

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%.

Q: Out of 100 hands, expected number of rainbow flops? +/- what?

X = Binomial (n,p), with n = 100, p = 39.76%, q = 60.24%.

E(X) = np = 100 * 0.3976 = 39.76

SD(X) = √(npq) = sqrt(23.95) = 4.89.

So, expected around 39.76 +/- 4.89 rainbow flops, out of 100 hands.

More poker examples.

Suppose you’re all-in next hand, no matter what.

What is the probability that you will eventually make 4 of a kind? (In other

words, what is the probability that, after all 5 board cards are dealt, you will

have 4 of a kind? Note that this includes the possibility that there is 4 of a kind

on the board.)

Answer: Forget about which 2 cards are yours!

13 * choose(48,3) / choose(52,7) = .00168, or about 1 in 595.

4. Poker Examples, continued

Given that both of your hole cards are the same suit, what is the probability that

you will eventually make a flush? (Note that technically this includes the

possibility that the 5 board cards are all the same suit, even if this is not the

same suit as your two cards!)

P(exactly 3 of your suit or exactly 4 of your suit or 5 of your suit or 5 of other suit)

= [choose(11,3)*choose(39,2) + choose(11,4)*39 + choose(11,5) + 3*choose(13,5)]

choose(50,5)

~ 6.58%

(The worst possible beat.) You have pocket aces, and your opponent has 6

2. The first two cards of the flop are revealed, and they are both aces! At this

point, what is your opponent’s probability of winning the hand?

6 cards out. 46 cards left in the deck. The last 3 board cards must come:

3 4 5 , not nec. in that order. P(opponent wins) = 1 / choose(46,3) =

0.0000659 , or 1 in 15,180.

4. Poker Examples, continued

The definition of a jackpot hand is a hand that meets the following 2 conditions:

(a) You have a full house with 3 aces, or better. That is, you have a royal flush,

straight flush, 4 of a kind, or a full house with 3 aces.

(b) Your best 5-card hand must use both of your hole cards.

[Ignore the possibility of ambiguity as to whether your hole card plays,

such as where you have A5 and the board is AA553.]

Given that you make a hand satisfying Condition (a), what is the probability that

you satisfy Condition (b)?

Imagine ordering the 7 cards so that the first 5 are the ones actually used.

What’s the chance that your 2 cards are in those first 5?

= P(your first card is in those 5 AND your 2nd card is in those 5)

= 5/7 * 4/6

= 47.6%

Poker Examples, continued

Many 2-card hands that seem weak nevertheless have some chance of being two-

player jackpot hands. If you have 94, it is possible for you and your opponent to

both make jackpot hands, if for instance the board comes something like 44477,

and your opponent has 77.

Are there any 2-card hands you could have such that, no matter what your

opponent has and no matter what the board is, you and your opponent cannot

possibly both have jackpot hands? List them.

Only 32 offsuit. To satisfy (b), the board must be 22233 or 33322….

Hellmuth / Farha.

Given Hellmuth’s A Q and Farha’s A Q), P(≥ 3 s on the board)?

[ 48 cards left. 11 s and 37 non-s. ]

P(exactly 3 s ) + P(exactly 4 s ) + P(exactly 5 s )

= [ choose(11,3) x choose(37,2) + choose(11,4) x choose(37,1) + choose(11,5) ]

÷ choose(48,5)

= 7.16%.

Your opponent re-raises you all-in before the flop.

Suppose you think he would certainly do

that whenever she had AA, KK, or QQ, and would do that 50% of the time with AK

or AQ. Nothing else. Given only this (not Doyle’s cards), what’s P(he has AK)?

Given nothing, P(AK) = 16/C(52,2) = 16/1326. P(KK) = C(4,2)/C(52,2) = 6/1326.

Using Bayes’ rule,

P(AK | all-in) =. P(all-in | AK) * P(AK) ,

P(all-in|AK)P(AK) + P(all-in|AA)P(AA) + P(all-in|AA)P(AA) +

…

= . 50% * 16/1326 .

[50% * 16/1326] + [100% * 6/1326] + [1*6/1326] + [1*6/1326] + [50% * 16/1326]

(AK) (AA) (KK) (QQ) (AQ)

= 23.5%. Compare with 16/1326 ~ 1.21%.

P(SOMEONE has AA, given you have KK)? Out of your 8 opponents?

Note that given that you have KK,

P(player 2 has AA & player 3 has AA)

= P(player 2 has AA) x P(player 3 has AA | player 2 has AA)

= choose(4,2) / choose(50,2) x 1/choose(48,2)

= 0.0000043, or 1 in 230,000.

So, very little overlap! Given you have KK,

P(someone has AA) = P(player2 has AA or player3 has AA or … or pl.9 has AA)

~ P(player2 has AA) + P(player3 has AA) + … + P(player9 has AA)

= 8 x choose(4,2) / choose(50,2) = 3.9%, or 1 in 26.

What is P(SOMEONE has an Ace | you have KK)? (8 opponents)

(or more than one ace)

P(SOMEONE has an Ace) = 100% - P(nobody has an Ace).

P(nobody has an Ace) = P(pl2 doesn’t have one & pl.3 doesn’t & … & pl.9 doesn’t)

= P(pl.2 doesn’t) x P(pl.3 doesn’t | pl.2 doesn’t) x … x P(pl.9 doesn’t | 2-8 don’t)

= choose(46,2)/choose(50,2) x choose(44,2)/choose(48,2) x … x ch(32,2)/ch(36,2)

[ = choose(46,16)/choose(50,16) ]

= 20.1%.

So P(SOMEONE has an Ace) = 79.9%.