Stat 35b: Introduction to Probability with Applications to Poker

Outline for the day:

1. Uniform random variables and R

2. Kaplan vs. Gazes,

3. More counting problems

4. Express, implied and reverse implied odds

5. Yang vs. Kravchenko

1. Uniform Random Variables and R

Continuous random variables are often characterized by their

probability density functions (pdf, or density):

a function f(x) such that P{X is in B} = ∫B f(x) dx .

Uniform: f(x) = c, for x in (a, b).

= 0, for all other x.

[Note: c must = 1/(b-a), so that ∫ab f(x) dx = P{X is in (a,b)} = 1.]

(For teams & examples for hw & computer projects, see

http://www.stat.ucla.edu/~frederic/35b/F09/diamond1.txt )

2) Poker After Dark: Kaplan vs. Gazes

3) More counting 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?

Suppose you’re 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) =

13*48 / 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 77 x, & there are 48 choices for x)

So P(flop 4-of-a-kind | pp) = 48/choose(50,3) = 0.245% = 1 in 408.

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.

4) Express, implied, and reverse implied odds.

From previous lecture: to call an all-in, need P(win) > B ÷ (B+pot).

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) > wager ÷ (wager + winnings),

where winnings = pot + amount you’ll 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 > pot or where wager > B, respectively.

5. Yang / Kravchenko.

Yang A 10. Pot is 19million. Bet is 8.55 million.

Needs P(win) > 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.

Yang / Kravchenko.

Yang A 10. 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(X=k)]

= [0 * 30%] + [27.55 million * 70%]

= 19.285 million.