Q: you are dealt the 52 cards in a random sequence.

Each red card earns you $360

Each black card costs you $360

Game ends at end of the 52 draws or any time you quit. How do you optimize your earning and how much is admission ticket worth?

Let’s denote the amount of money you take home as H, a random variable. Your net profit/loss would be H minus admission price. If 555 reasonable/intelligent people play this game, then there would be five hundred and fifty-five values of H. What’s the average? That would be the answer.

p or “+” denotes the # of positive cards earned so far

n or “-” denotes the # of negative cards suffered so far

Exp(H|p=25,n=26) = Exp(H|p=26,n=26) = $0.

There’s an intrinsic value in our hands, Defined as i=(p-n). The e-value or Extrinsic value Evaluated from Expectation, may be higher or lower. Whenever i-value > e-value, we should exit. This is our strategy/policy.

Whenever intrinsic value <$0, E(H) is out of the money but always non-negative because we can wait till maturity and get all 52 cards.

E(p24,n26) = p(next is p)E(p25,n26) = 100%E(p25,n26) = $0

E(p26,n25) = $360 because we will exit

E(p25n25) = Prob(next is p)E(p26,n25) + Prob(next card is n)E(p25,n26) = 50%$360+0 = $180

E(p24n25) = p(p)E(p25,n25) + p(n)E(p24,n26)= p(p)E(p25,n25)+$0 = 66%$180= $120

E(p25n24)= p(p)E(26,n24)+p(n)E(p25,n25)= 33%$360×2 + 66%$180= $360

E(p24,n24)= half of E(25,24)+E(24,25)= half of $360+$120= $240

E(p23,n25)= p(p)E(24,25)

+ p(n)E(p23,n26) # $0

=3/4x$120= $90

E(p23,n24)= p(p)E(24,24)+p(n)E(23,25)= 3/5 .$240+2/5 .$90

### Like this:

Like Loading...

*Related*