Question

2. (10 Pts) Shero pays $2 to play this game: Two cards are drawn at random from a standard deck of 52 cards. If both are black, Shero win $5. If both are red, she gets her $2 back. If one is black and one red, then she gets nothing back. Find the expected value, variance and standard deviation for this game.

Answer 1

Shero pays $2 to play the game.

If both of the two drawn cards are black, Shero wins $5. However he has to pay $2 to start the game, His net earning from the game = $(5-2)=$3

If both of the two drawn cards are red, Shero gets back $2. However he has to pay $2 to start the game, His net earning from the game = $(2-2)=$0

If of the two drawn one is black and 1 red , Shero wins $0. However he has to pay $2 to start the game, His net earning from the game = $(0-2)=-$2

The probability of drawing two black cards= $\frac{\binom{26}{2}}{\binom{52}{2}}$ $\approx$ 0.25

The probability of drawing two red cards= $\frac{\binom{26}{2}}{\binom{52}{2}}$ $\approx$ 0.25

A deck of 52 cards have 26 black and 26 red cards.

The probability of drawing 1 red and 1 black card = $\frac{\binom{26}{1}\binom{26}{1}}{\binom{52}{2}}$ $\approx$ 0.50

The required expected value= E(X) = 3*0.25+0*0.25+(-2)*0.5=0.75-1=-0.25

We know that

E(X²) = 3^2*0.25+0^2*0.25+(-2)^2*0.5 = 9*0.25+4*0.5 = 2.25+2=4.25

so, V(X) = 4.25-(-0.25)^2 = 4.25-0.0625=4.1875

The required standard deviation is given by,

$SD(X)=\sqrt {V(X)}=\sqrt { 4.1875}=2.046$