# Another Probability and Statistics Question

• Oct 3rd 2007, 07:47 PM
ballinisahobby
Another Probability and Statistics Question
I saw a specific problem on here earlier yet we have to answer one with more specifications:

In a gambling game a woman is paid \$3 if she draws a jack or a queen and \$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?

Here we want c to be the cost to play each game
Let R.V. x = net gain per game = payoff -(cost to play the game).

So, X = { 3- c if she jack or queen
5-c if king or ace
-c other wise
E[x] expected net gain per single. Play of each game. If game is fair, then e[x] = 0. So, in this problem you want to find the value of c so that e[x]=0.

Note: No gambling company would run a fair game. for the gambling complany, the expected gain to player per single game is always negative i.e. gain to company is positive
• Oct 3rd 2007, 08:21 PM
feiyingx
This problem requires the use of Expectation. So first let us figure out all the possible scenarios of the game.

First, you pay \$c to play the game. The results of the game can be of the following:

Case 1: You draw a jack or a queen. You get \$(3-c). In order for case 1 to happen, you must draw a jack or a queen. This implies that the probability of case 1 happening is equal to the probability of you drawing a jack or a queen out of the deck of 52 cards. This probability is equal to (4 jacks + 4 queens)/(52 total cards) = 8/52.

Case 2: You draw a king or an ace. You get \$(5-c). In order for this to happen, you must draw either a king or an ace out of the 52 cards. The probability of that happening is also 8/52. (Why?)

Case 3: You didnt draw a queen, jack, king, nor ace. You get \$(0-c) in return. The probability of this happening is 36/52. (Why?)

Now we can calculate the expected value of the reward of game:
E[reward of the game] = (3-c)*(8/52) + (5-c)*(8/52) + (0-c)*(16/52)

We've established that in order for this game to be a fair game, the Expected reward must = 0. So therefore set the above eqn to zero and solve for c.
E[reward of the game] = (3-c)*(8/52) + (5-c)*(8/52) + (0-c)*(16/52) = 0
• Oct 3rd 2007, 08:35 PM
ballinisahobby
thanks!!!!!!!!!!
we've been sitting on here arguing over this problem for hours so we appreciate it!!!
• Oct 3rd 2007, 09:19 PM
CaptainBlack
Quote:

Originally Posted by ballinisahobby
I saw a specific problem on here earlier yet we have to answer one with more specifications:

In a gambling game a woman is paid \$3 if she draws a jack or a queen and \$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?

Here we want c to be the cost to play each game
Let R.V. x = net gain per game = payoff -(cost to play the game).

So, X = { 3- c if she jack or queen
5-c if king or ace
-c other wise
E[x] expected net gain per single. Play of each game. If game is fair, then e[x] = 0. So, in this problem you want to find the value of c so that e[x]=0.

Note: No gambling company would run a fair game. for the gambling complany, the expected gain to player per single game is always negative i.e. gain to company is positive

See here

RonL