In a game of dice, the probability of rolling a 12 is 1/36. The probability of rolling a 9, 10, or 11 is 9/36. The probability of rolling any other number is 26/36. If the player rolls a 12, the player wins $5. If the player rolls a 9, 10, or 11, the player wins $1. Otherwise, the player loses $1. What is the expected value of this game? If the game is played 100 times what are the expected winnings (or losses) of the player?
Please show more than one approach.
do you divide 1/36 and times it by 5. It would make it five over thirty six. For the amounts of 9, 10, 11, you do the same thing, which makes it 18/36. You add them together 23/36? That is where I get lose.
To find the expected value of the game, we need to multiply the value of each outcome by its probability, and then sum those products.
Approach 1:
Let's break down the possible outcomes and their associated payouts:
1. Rolling a 12: Probability = 1/36, Payout = $5
2. Rolling a 9, 10, or 11: Probability = 9/36, Payout = $1
3. Rolling any other number: Probability = 26/36, Payout = -$1 (loss of $1)
Now, let's calculate the expected value using the formula:
Expected value = (Probability of outcome 1 * Payout of outcome 1)
+ (Probability of outcome 2 * Payout of outcome 2)
+ (Probability of outcome 3 * Payout of outcome 3)
Expected value = (1/36 * $5) + (9/36 * $1) + (26/36 * -$1)
Expected value = $5/36 + $9/36 - $26/36
Expected value = $-12/36
Expected value = -$1/3 or -$0.33 (rounded to the nearest cent)
So, the expected value of this game is -$0.33. This means that, on average, the player is expected to lose $0.33 per game.
Approach 2:
Another way to calculate the expected value is by using a probability distribution table.
Outcome Probability Payout
---------------------------------------------------------
Rolling a 12 1/36 $5
Rolling 9, 10, 11 9/36 $1
Any other number 26/36 -$1
In the table, we multiply the probability by the payout to get the contribution to the expected value for each outcome.
Expected value = (1/36 * $5) + (9/36 * $1) + (26/36 * -$1)
Expected value = $5/36 + $9/36 - $26/36
Expected value = $-12/36
Expected value = -$1/3 or -$0.33 (rounded to the nearest cent)
So, regardless of the approach used, the expected value of this game is -$0.33 per play.
To find the expected winnings or losses if the game is played 100 times, we can multiply the expected value by the number of times the game is played:
Expected winnings/losses = Expected value * Number of times played
Expected winnings/losses = -$0.33 * 100
Expected winnings/losses = -$33.00
Therefore, if the game is played 100 times, the player is expected to lose $33.00 on average.