Question

1. Daniel Reisman of Niverville, NY submitted the following question to Marilyn vos Savant's December 27, 1998, Parade Magazine column, "Ask Marilyn." At a monthly 'casino night,' there is a game called Chuck-a-luck. Three dice are rolled in a wire cage. You place a bet on any number from 1 to 6. If any one of the three dice comes up with your number, you win the amount of your bet. (You also get your original stake back). If more than one die comes up with your number, you win the amount of your bet for each match. For example, if you had a $1 bet on number 5, and each of the dice came up with 5, you would win $3. It appears that the odds of winning are 1 in 6 for each of the three dice for a total of 3 out of 6 or 50%. Adding the possibility of having more than one die come up with your number, the odds would seem to be in the gambler's favor. What are the odds of winning this game? I can't believe that a casino game would favor the gambler. Daniel computed the probabilities incorrectly. There are four possible outcomes. (The selected number can match 0, 1, 2, or 3 of the dice.) The random variable X represents the profit from a $1 bet in Chuck-a-Luck. The table below summarizes the probabilities of earning a profit of x dollars from a $1 bet. Use this table to answer the following questions. Profit $-1 Number of dice matching the chosen number O dice 1 die 2 dice 3 dice Probability P(X=x) 125/216 75/216 15/216 1/216 S1 a. Verify that this is a discrete probability distribution (i.e., what two conditions are met?).

Answer 1

216 let X be the profit from a $2 bet in chuck a lucki The probability distribution of x is. Number of dice profit Rrobability apca) 2²p(at) matching the (a) P(X=2) chosen number o dice $-1 125 = 0.5787 -0.5787 1 dice $1 7/16=0.3472 0.3472 2 dice $2 15/216=0.0694 011388 3 dice $3 /216=0.0046 0.0138 0.9999 -1.00 -0.0789 This is a valid discrete probability distribution since, is os P(x=125i 11) EP(XIX) = 1 * for all a. Total I b) The total area of the bars is 2.00. i since area under first bar=0.5787 second bar=0.3472, third bar= 0.1388 and fourth bar it is 0.0138 The part is in questin 1 a, shows the sum of all probability is l. - - - -

is I c) The mean of a random variable x us E2.P(x=x) N=-1,1,2,3 - u=-0.0789 The mean of random variable y represents that on an average marilyn loss $0.0789 in casino bet. - - a) Based on the expected value (mean value) we recommend not to play this game as their is an average loss of $0.0789 - - - - = e) The variance of x is given by VCX) = Ex?PC*=*) - 42 =-1*-0.57877 180.34724280.1388 +3x0.0138 -(-0.0789) = 0.578770-3472 +0.2776 +0.0414 -(0.006225) = 1.2449-0.006225 fv(x= 1.2387)

f) The standard deviation of x is 6 = fvcx) 2.6= $1.2387 [6= 1.11307 6-1.1130
g. The probability that a player who plays once will lose this game is given by:

P(Y= 0 dice)=0.5787

h. The probability that a player will match all three of the dice is given by:

P(X= 3 dice)=0.0138

i. The probability that a player win match at least one of the dice is given by:

PY > Idice) = 1 - P(Y = Odice)

PY > Idice) = 1 -0.5787

PY > Idice) = 0.4213