The game Chuck-A-Luck is played as follows: A player bets $M on an integer chosen from 1-6 and then rolls three dice. If the number appears on exactly on die, then the player wins the amount bet. The player wins twice the amount bet if the umber appears on two dice and three times the amount bet if the number appears on all three dice. If the number does not appear at all, then the player loses the $M. What is the probability of winning at each level? At any level? For every n such bets of a fixed amount $M, what would be the House's average net gain? (In other words, find u such that u = average gain by the house).

Question

The game Chuck-A-Luck is played as follows: A player bets $M on an integer chosen from 1-6 and then rolls three dice. If the number appears on exactly on die, then the player wins the amount bet. The player wins twice the amount bet if the umber appears on two dice and three times the amount bet if the number appears on all three dice. If the number does not appear at all, then the player loses the $M. What is the probability of winning at each level? At any level? For every n such bets of a fixed amount $M, what would be the House's average net gain? (In other words, find u such that u = average gain by the house).

Answer 1

To find the probability of winning at each level and the House's average net gain, we need to determine the number of favorable outcomes and the total number of possible outcomes.

Let's start by calculating the probability of winning at each level:

1. Winning by appearing exactly once:
The number selected can appear on exactly one out of the three dice. There are three dice, and each can have six possible outcomes. So, the total number of possible outcomes is 3 x 6 x 6 = 108.

To calculate the number of favorable outcomes, we need to consider that only one die should show the selected number. There are three ways this can happen: the number appears on the first die while not appearing on the second and third dice, the number appears on the second die while not appearing on the first and third dice, or the number appears on the third die while not appearing on the first and second dice. Each of these scenarios has 1 x 5 x 5 = 25 possible outcomes since the remaining dice can show any of the five numbers other than the selected one. Therefore, there are 3 x 25 = 75 favorable outcomes.

The probability of winning at this level is 75/108.

2. Winning by appearing exactly twice:
The number selected can appear on exactly two out of the three dice. Using a similar approach as before, there are three ways for this to happen. The total number of possible outcomes remains 108.

To calculate the number of favorable outcomes, we consider that two dice should show the selected number while the remaining die should show a different number. This can happen in three different scenarios: the number appears on the first and second dice while not appearing on the third, the number appears on the first and third dice while not appearing on the second, or the number appears on the second and third dice while not appearing on the first. Each of these scenarios has 1 x 1 x 5 = 5 possible outcomes since the remaining die can show any of the five numbers other than the selected one. Therefore, there are 3 x 5 = 15 favorable outcomes.

The probability of winning at this level is 15/108.

3. Winning by appearing exactly three times:
The number selected needs to appear on all three dice. The total number of possible outcomes remains 108.

To calculate the number of favorable outcomes, we consider that all three dice should show the selected number. This can happen in only one scenario. Therefore, there is 1 favorable outcome.

The probability of winning at this level is 1/108.

Next, let's calculate the House's average net gain for every n such bets of a fixed amount $M:

To calculate the average gain for the House, we subtract the average payout to the players from the total amount of bets:

Average gain by the House = Total amount bet - Average payout to players

The total amount bet for n bets of amount $M each is n * M.

The average payout to players can be calculated by summing the payouts at each level and dividing by the total number of possible outcomes:

Average payout to players = (Payout at level 1 + Payout at level 2 + Payout at level 3) / Total number of possible outcomes

Payouts at each level can be calculated by multiplying the respective level's payout (1 times the amount bet, 2 times the amount bet, and 3 times the amount bet) with the probability of winning at that level.

Let's assume that the amount bet is $M.

Payout at level 1 = (M * probability of winning at level 1)
Payout at level 2 = (2M * probability of winning at level 2)
Payout at level 3 = (3M * probability of winning at level 3)

Now we can calculate the average payout to players and the House's average net gain.

Answer 2

Whatever number is chosen, there is a 1/6 chance of that showing up on each die.

If the events are independent, the probability of both/all events occurring is determined by multiplying the probabilities of the individual events.

P(win one) = 1/6 * 5/6 * 5/6 = ?

P(win none) = 5/6 * 5/6 * 5/6 = ?

I'll let you do the rest.