24. Winning Poker A very good poker player is expected to earn $1 per hand in $1

Question

24. Winning Poker A very good poker player is expected to earn $1 per hand in $100/$200 Texas Hold'em. The standard (a) What isthe probablity a very good poker player carm a profit (more than $0) after playing 50 hands in $100/$200 Texas Hold'em? (b) What is the probability a very good poker player loses (earns less than $0) after playing 100 hands in $100/$200 Texas Hold'em? (c) What proportion of the time can a very good poker player expect to earn at least $500 after playing 100 hands in $100/$200 Texas Hold'em? Hint: $500 after 100 hands is a mean of $5 per hand. (d) Would it be unusual for a very good poker player to lose at least $1000 after playing 100 hands in $100/$200 Texas Hold'em? (e) Suppose twenty hands are played per hour. What is the probability that a very good poker player earns a profit during a twenty-four hour marathon session?

Explanation / Answer

Let X = earning per hand of a very good poker player in $100/200 Texas Hold’em.

We assume X ~ N(µ, 2), where we are given µ = 1 and = 32

Back-up Theory

If a random variable X ~ N(µ, 2), i.e., X has Normal Distribution with mean µ and variance 2, then, pdf of X, f(x) = {1/(2)}e^-[(1/2){(x - µ)/}2] …………………………….(A)

Z = (X - µ)/ ~ N(0, 1), i.e., Standard Normal Distribution ………………………..(1)

P(X or t) = P[{(X - µ)/ } or {(t - µ)/ }] = P[Z or {(t - µ)/ }] .………(2)

X bar ~ N(µ, 2/n),…………………………………………………………….…….(3),

where X bar is average of a sample of size n from population of X.

So, P(X bar or t) = P[Z or {(n)(t - µ)/ }] …………………………………(4)

Probability values for the Standard Normal Variable, Z, can be directly read off from

Standard Normal Tables or can be found using Excel Function……………………..(5)

Property of symmetry: P(Z > t) = P(Z < - t)………………………………………….(6)

Part (a)

Probability of earning profit (more than $0) after 50 hands

= Probability of earning an average profit (more than $0) per hand

= P(Xbar > 0)

Now, X ~ N(1, 322) and hence by (3) above, based on a sample of 50, Xbar ~ N[1, (322/50)]

So, P(Xbar > 0) = P[Z > {50(0 - 1)/32}] [vide (4) above]

= P[Z > {50(0 - 1)/32}]

= P(Z > - 0.221)

= 0.413 [using Excel Function of Normal Distribution] ANSWER

Part (b)

Probability losing (less than $0) after 100 hands

= Probability losing (less than $0) per hand

= P(Xbar < 0)

Now, X ~ N(1, 322) and hence by (3) above, based on a sample of 100, Xbar ~ N[1, (322/100)]

So, P(Xbar < 0) = P[Z < {100(0 - 1)/32}] [vide (4) above]

= P[Z < {100(0 - 1)/32}]

= P(Z < - 0.313)

= 0.377 [using Excel Function of Normal Distribution] ANSWER

Part (c)

Proportion of time poker player is expected to earn at least $500 after playing 100 hands

= Probability earning a profit of at least $500 after 100 hands

= Probability earning an average profit of at least $5 per hand

= P(Xbar 5)

Now, X ~ N(1, 322) and hence by (3) above, based on a sample of 100,

Xbar ~ N[1, (322/100)]

So, P(Xbar 5) = P[Z {100(5 - 1)/32}] [vide (4) above]

= P[Z {100(4/32)}]

= P(Z 1.25)

= 0.106 [using Excel Function of Normal Distribution] ANSWER

Part (d)

Probability losing at least $1000 after 100 hands

= Probability losing an average of at least $10 per hand

= P(Xbar - 10)

Now, X ~ N(1, 322) and hence by (3) above, based on a sample of 100,

Xbar ~ N[1, (322/100)]

So, P(Xbar - 10) = P[Z {100(- 10 - 1)/32}] [vide (4) above]

= P[Z {100(- 11/32)}]

= P(Z - 3.44)

= 0.0003 [using Excel Function of Normal Distribution]

This probability being very low, we conclude that it is unusual for a very good poker player to lose at least $1000 after playing 100 hands. ANSWER

Part (e)

20 hands per hour for 24 hours => 480 hands. So, we want to find probability of earning profit (more than $0) after 480 hands

= Probability earning an average profit (more than $0) per hand

= P(Xbar > 0)

Now, X ~ N(1, 322) and hence by (3) above, based on a sample of 480,

Xbar ~ N[1, (322/480)]

So, P(Xbar > 0) = P[Z > {480(0 - 1)/32}] [vide (4) above]

= P[Z > {480(0 - 1)/32}]

= P(Z > - 0.6847)

= 0.247 [using Excel Function of Normal Distribution] ANSWER

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