The casino game of chuck-a-luck seems quite simple and that it would be a pretty good...
- The casino game of chuck-a-luck seems quite simple and that it would be a pretty good bet. In this game, three dice are rolled (usually inside a wire cage). You can bet a set amount, let's say $5, on a specific number 1-6. You win the amount of your bet for each appearance of your number on the dice. If the number does not appear, you lose your bet. For example, let's say you bet $5 on the number 3. If it appears on exactly one die, you win $5 (and keep your bet) with probability 75/216. If it appears on exactly two dice, you win $10 (and keep your bet) with probability 15/216. If it appears on all three dice, you win $15 (and keep your bet) with probability 1/216. If it appears on none of the dice, you lose your original $5 bet. Construct the probability distribution for chuck-a-luck and determine your expected winnings on a $5 bet.
Homework Answers
let x is winning amount ;below is probability distribution for chuck-a-luck:
P(X=5)=75/216 ( selected number appear on exactly one dice)
P(X=10)=15/216 ( selected number appear on two of dice)
P(X=15)=1/216 (selected number appear on three of dice)
P(X=0)=125/216 (selected number appear on none of dice)
therefore expected winnings on a $5 bet =xP(x)=5*75/216+10*15/216+15*1/216+0*125/216=$2.5
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