Stat 35b: Introduction to Probability with Applications to Poker Outline for the day Review list.

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Stat 35b: Introduction to Probability with Applications to Poker Outline for the day Review list.

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Stat 35b: Introduction to Probability with Applications to Poker Outline for the day Review list.

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Stat 35b: Introduction to Probability with Applications to Poker

Outline for the day

- Review list.
- Gold, Farha, and Bayes’s rule.
- P(flop 2 pairs).
- P(suited king).
- Midterm is this Friday Feb 21.
- Homework 3 is due Feb 28.
- Project B is due Mar 8, 8pm, by email to frederic@stat.ucla.edu.
- Same teams as project A.

- 1. Review List
- Basic principles of counting.
- Axioms of probability, and addition rule.
- Permutations & combinations.
- Conditional probability.
- Independence.
- Multiplication rules.P(AB) = P(A) P(B|A) [= P(A)P(B) if ind.]
- Odds ratios.
- Random variables (RVs).
- Discrete RVs and probability mass function (pmf).
- Expected value.
- Pot odds calculations.
- Luck, skill, and deal-making.
- Variance and SD.
- Bernoulli RV. [0-1. µ = p, s= √(pq). ]
- Binomial RV.[# of successes, out of n tries. µ = np, s= √(npq).]
- Geometric RV.[# of tries til 1st success. µ = 1/p, s= (√q) / p. ]
- Negative binomial RV. [# of tries til rth success. µ = r/p, s= (√rq) / p. ]
- E(X+Y), V(X+Y) (ch. 7.1).
- Bayes’s rule (ch. 3.4).
- We have basically done all of ch. 1-5 except 4.6, 4.7, and 5.5.

2. Gold vs. Farha. Gold: 10u 7 Farha: Qu Q Flop: 9u 8 7

Who really is the favorite (ignoring all other players’ cards)?

Gold’s outs: J, 6, 10, 7. (4 + 4 + 3 + 2 = 13 outs, 32 non-outs)

P(Gold wins) = P(Out Out or Jx [x ≠ 10] or 6x or 10y [y≠Q,9,8] or 7z [z≠Q])

= [choose(13,2) + 4*28 + 4*32 + 3*24 + 2*30] ÷ choose(45,2)

= 450 ÷ 990 = 45.45%.

What would you guess Gold had?

Say he’d do that 50% of the time with a draw,

100% of the time with an overpair,

and 90% of the time with two pairs. (and that’s it)

Using Bayes’ rule, P(Gold has a DRAW | Gold raises ALL-IN)

= . [P(all-in | draw) * P(draw)] .

[P(all-in | draw) P(draw)] + [P(all-in | overpair) P(overpair)] + [P(all-in | 2pairs) P(2 pairs)]

= [50% * P(draw)] ÷ [50% * P(draw)] + [100% * P(overpair)] + [90% * P(2 pairs)]

3. P(flop two pairs).

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

This is a tricky 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.

4. What is the probability that you will be dealt a king and another card of the same suit as the king?

4 * 12 / C(52,2) = 3.62%.

The typical number of hands until this occurs is ...

1/.0362 ~ 27.6.

(√96.38%) / 3.62% ~ 27.1.

So the answer is 27.6 +/- 27.1.