On a TV game show, the contestant is asked to select a door and then is rewarded with the prize behind the door
selected. If the doors can be selected with equal probability, what is the expected value of the selection if the three doors have behind them a $40,000 foreign car, a $3 silly straw, and a $50 mathematics textbook?
expected value of car = (1/3)(40000) = 13333.33
expected value of silly draw = (1/3)(3) = 1.00
expected value of mathbook = (1/3)(50) = 16.67
so the expected value of event is the sum of these or $13531.00
If the grand prize is $1,000,000 and there is only one winner, what is a players expected value if he purchases only one ticket?
To determine the expected value of the selection, we need to calculate the weighted average of the possible outcomes, where each outcome is multiplied by its corresponding probability.
In this scenario, there are three doors with the following prizes:
1. Door 1: $40,000 foreign car
2. Door 2: $3 silly straw
3. Door 3: $50 mathematics textbook
Since the doors can be selected with equal probability, the probability of selecting any specific door is 1/3.
Now, we can calculate the expected value:
Expected value = (Prize 1 x Probability of selecting Door 1) + (Prize 2 x Probability of selecting Door 2) + (Prize 3 x Probability of selecting Door 3)
Expected value = ($40,000 x 1/3) + ($3 x 1/3) + ($50 x 1/3)
Expected value = $13,333.33 + $1 + $16.67
Expected value = $13,350
Therefore, the expected value of the selection is $13,350.