Suppose someone gives you 12 to 3 odds that you cannot roll two even numbers wit
ID: 3197767 • Letter: S
Question
Suppose someone gives you 12 to 3 odds that you cannot roll two even numbers with the roll of two fair dce. This means you win $12 if you succeed and you lose $3 if you fail. What is the expected value of this game to you? Should you expect to win or lose the expected value in the first game? What can you expect if you play 100 times? Explain. (The table will be helpful in finding the required probabilities.) Click the icon to view the table. What is the expected value of this game to you? Data Table Table of outcomes and sums for the roll of two dice 1+1=2 1+2.3 1+3.4 1+4=5 1+5=6 1+6-7 2 2+1-3 2+2-4 2+3-5 2+4-6 25-7 2 6-8 3 3+1-4 3+2-5 3+3-6 3+4-7 3+5-8 3 6-9 4 4+1-5 4+2-6 4+3-7 4+4-8 4+5-9 4+6-10 5 5+1-6 5+2-75+3-8 5+4-9 5+5-10 5+6-11 6 6+1-7 6+2-8 6+3-9 6+4-10 6+5-11 6+6-12 Print DoneExplanation / Answer
Here we are rolling two fair dice and we are looking for two even numbers.
Here are all possibilities as (2,2)(2,4)(2,6)(4,2)(4,4)(4,6)(6,2)(6,4)(6,6). So there are 9 possible ways for this to happen.
When rolling two fair dice, we can get any of 6 outcomes on dice 1 and any of 6 outcomes on dice 2. So sample space ( all possibilities)has 36 values.
This means our chance of rolling two even number is 9/36. So the probability of Winning is 9/36=0.25.
Probability of losing is 1-.25 = 0.75.
The expected value is (probability of winning)×(amount we get one we win) + (Probability of losing)(amount of value we loss, which is negative as it is a loss)
So the expected value is (0.25)(12) + (0.75)(-3) = $ 0.75
So the expected value of this game is $ 0.75 .........(answer)
Since the expected value is positive, we can expect to win in the first game. The more we play ,the more we win.
If we play 100 times, the expected value is 100(0.75) = $75.
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