How much should one pay, answer using expectation of random variable?

Question

In a gambling game, a woman is paid dollar 3 if she draws a jack or a queen and dollar 5 if she draws a king or an ace from an ordinary deck of 52 playing cards. If she draws any other card, she loses. How much should she pay to play if the game is fair?

i am new to probability and i am stuck on this question , couldn't find an answer with a good explanation ,please help ........thanks in advance

Answer 1

To put you on track:

Let's say that she pays $l$ dollars to play.

Three things can happen and all with a certain probability: she wins $3$, she wins $5$ or she looses. Denoting the corresponding probabilities with $p_3$, $p_5$ and $p_l$ there is an expectation of: $$3p_3+5p_5-lp_l$$ A fair game means that this expectation equals $0$. Now start finding $l$.

Answer 2

Suppose C is the dollar cost of a fair game then expected gain/loss must be zero for a fair game.

P(win $\$3) = \dfrac8{52}\;\;$, P(win $\$5) = \dfrac8{52}$, thus

$E[X] = C - \left(3\cdot\dfrac{8}{52} + 5\cdot\dfrac{8}{52}\right) = 0$

Solving, we get $C = \dfrac{24 +40}{52} = \$1.23$

Answer 3

Finally i solved this problem , and thought it would be helpful for someone who needs a well explained answer , so here it goes.....

Understand the language of question:

"How much should she pay to play if the game is fair?" This simply means that on the average there should be no profit no loss in playing the game , in other words if the lady plays the game n times (n is very large no ) her money spent on the game should be equal to 0.

Solving the problem:

let z = " money paid by the lady "

and let X is a random variable , which is the total money spent on one game.

so now support of X (all the values that x can take) = (z-3), (z-5), z

(z-3) when she draws jack or queen

(z-5) when she draws ace or king

z when she draws some other card

now find probability of all these values of X

P(z-3) = 2/13 , since there are 8 cards for jack and queen which gives probability = 8/52 = 2/13

p(z-5)= 2/13 since there are 8 cards for ace and king which gives probability = 8/52 = 2/13

p(z) = 9/13 since there is 36 other cards than ace, king, queen and jack

now we find expected value of X and set it equal to 0 to find value of z

E(X) = (z-3)*2/13 + (z-5)*2/13 + z*9/13 = 0

    2z-6 +2z -10 + 9z = 0

    13z= 16

      z=1.23

Answer 4

This is kinda late, but hopefully this helps other people:

1. Let the discrete random variable $X$ denote the outcome of a single game, where we define "outcome" to be either the receipt of money or the loss of money.

2. This discrete random variable has an associated probability distribution function $f(x)$. We can find the values of $f(x)$ using standard counting methods/probability.

Let us define $p$ to be the amount of money you spent to play the game. Then there are three possible outcomes:

$-p:$ you won nothing, and you lost the amount of money you paid

$3-p:$ you gained this much money because you drew either a jack or a queen

$5-p:$ you gained this much money because you drew either a king or an ace

We also define "fair" to be a game whose average result is $\geq 0$. In other words, on average, a player either doesn't lose money.

Okay, on to the probabilities:

$f(-p) = \frac{{36 \choose{1}}}{ {52 \choose 1}} = \frac{36}{52}$

Repeat this process for the others, and you get $\frac{8}{52}$ for each.

All that remains is applying what you know about expected (average) values, setting the expected value expression to be greater than or equal to $0$, and solving for an appropriate $p$.

The game will only be fair if the amount $p$ that they're asking you to pay is less than or equal to some value.