Question

The Play 4 is won by selecting 4 digits. These can repeat themselves but the order does matter. You could win $5000 before taxes.

answer the following questions

How many ways can winning tickets be selected for this scenario? How did you determine this?

• What is the probability of winning this scenario? What formulas did you use to calculate this number?
• Do you think it is worth the risk if you pay $5 to participate? What about $10? $100?

Answer 1

The first digit can be selected as any one of the digit from the 10 possible digits.

The second digit can be selected as any one of the digit from the 10 possible digits.

The third digit can be selected as any one of the digit from the 10 possible digits.

The fourth digit can be selected as any one of the digit from the 10 possible digits.

All the four digits are independent of each other.

Thus number of ways winning tickets can be selected for this scenario = 10101010 = 10000

.

There is only one combination out of the 10000 possible combinations of the four digits which wins the prize.

Thus the probability of winning in this scenario = 1/10000 = 0.0001

.

The winning prize money is $5000.

If $5 is paid to participate, average profit amount

= $5000(Probability of winning) - $5(Probability of losing) = $50000.0001 - $5(1-0.0001) = -$4.4995

So, on average, the participant will suffer a loss. So it is not worth the risk to play.

If $5 is paid to participate, average profit amount

= $5000(Probability of winning) - $5(Probability of losing) = $50000.0001 - $10(1-0.0001) = -$9.499

So, on average, the participant will suffer a loss. So it is not worth the risk to play.

If $5 is paid to participate, average profit amount

= $5000(Probability of winning) - $5(Probability of losing) = $50000.0001 - $100(1-0.0001) = -$99.49

So, on average, the participant will suffer a loss. So it is not worth the risk to play.