Question

The second game You toss a fair coin until Head appears and you are paid the number of Tails received, for example, if we obtain TTTH you get 3 rubles. How much do you agree to pay the host for the game? The third and fourth games We have a deck of 52 cards. a) We randomly choose 5 cards (without repetition) and get 5 rubles for any ace or king chosen; b) We randomly choose 5 cards (with repetition, that is, we choose a card and return it to the deck before the next choosing) and get 5 rubles for any ace or king chosen. What are the fair stakes for the games?

Answer 1

Solution

Back-up Theory

Fair stakes for the game => no-profit-no-loss for the player => stake = expected return ................ (1)

If a discrete random variable, X, has probability function, p(x), x = x₁, x₂, …., x_n, then

Expected value = Mean (average) of X = E(X) = µ = Σ{x.p(x)} summed over all possible values of x…..…. (2)

A series of the form, a + (a + d)r + (a + 2d)r² + (a + 3d)r³ + …………to infinity is called an infinite AGP

(Arithmetico Geometric Progression), with common difference d and common ratio r.

Sum of an infinite AGP = S = {a + dr/(1 - r²)}/(1 - r).............................................................................. (3)

Now, to work out the solution,

Part (a) Second Game

Vide (1), we will equate the stake to the expected return

Let X = return per game = number rubles paid to the player = number of T’s preceding the first H. Then,

X = 0, => the outcome is: H => p(x) = ½

X = 1, => the outcome is: TH => p(x) = ½²

X = 2, => the outcome is: TTH => p(x) = ½³

And so on, mathematically, X can take any value 3, 4, 5,     to infinity.

So, vide (2),

E(X) = (0 x ½) + (1 x ½²) + (2 x ½³) + (3 x ½⁴) + ....... to infinity.

The above is an infinite AGP with common difference 1 and common ratio ½. So, vide (3), the sum

= [(1 x ½²) + {(½)/(1 - ½²)}]/(1 - ½)

= 13/4.

Thus, payment to the host for the game = 13/4 rubles. Answer 1

Part (b) Third Game

Here,

X = 0 => all 5 cards are non-ace-non-king => p(x) = (⁴⁴C₅)/(⁵²C₅) = 0.4178

X = 5 => 1 card is ace/king and 4 cards are non-ace-non-king => p(x) = (⁸C₁)(⁴⁴C₄)/(⁵²C₅) = 0.4178

X = 10 => 2 cards are ace/king and 3 cards are non-ace-non-king => p(x) = (⁸C₂)(⁴⁴C₃)/(⁵²C₅) = 0.1427

X = 15 => 3 cards are ace/king and 2 cards are non-ace-non-king => p(x) = (⁸C₃)(⁴⁴C₂)/(⁵²C₅) = 0.0204

X = 20 => 4 cards are ace/king and 1 card is non-ace-non-king => p(x) = (⁸C₄)(⁴⁴C₁)/(⁵²C₅) = 0.0012

X = 25 => all 5 cards are ace/king => p(x) = (⁸C₅)/(⁵²C₅) = 0.000022.

So, vide (2),

E(X) = 3.85

Thus, the fair stake is: 3.85 rubles Answer 2

DONE