Question

Use the binomial distribution to determine which of the following three games has the highest probability of
winning. What is the minimum rational price you should charge to play this game if you must pay a winner
$5.00?
Game 1: throw a single die 6 times. You win if you roll a 1 at least once.
Game 2: throw a single die 12 times. You win if you roll a 1 at least twice.
Game 3: throw a single die 18 times. You win if you roll a 1 at least three times.

Answer 1

Game 1

A die has six sides (namely 1,2,3,4,5 and 6) of which all the numbers are equally likely. Thus, the probability that a single roll of a die will result in a 1 is equal to 1/6.

[Using Probability = No. of favorable outcomes/Total no. of outcomes]

Now, let X denote the number of times we roll a 1 on 6 rolls of a die.

Since, each roll of a die is independent

=> X ~ Binomial(n = 6, p = 1/6)

Now, Probability of winning under this game:

Minimum Rational Price

The minimum rational price that we should charge for this game is equal to the amount, that we expect to payout per game. This is due to the fact that if we play the game long enough then the average winnings per player will be equal to the expected payout and we should charge atleast this amount to be in no loss, no profit condition. If we charge lower then we will make a loss in the long run and if we charge higher then we will make a profit in the long run.

Thus,

Minimum rational price = Expected winnings per game

= (Winning amount)*P(winning) + (Losing amount)*P(losing)

= 5*0.665102 + 0*(1-0.665102) [The losing amount is zero because we don't have to pay anything after we lose]

= 3.32551

= $3.33

Game 2

Now, let X denote the number of times we roll a 1 on 12 rolls of a die.

=> X ~ Binomial(n = 12, p = 1/6)

Now, Probability of winning under this game:

Minimum Rational Price

Following the logic as in the case of Game 1, we get:

Minimum rational price = Expected winnings per game

= (Winning amount)*P(winning) + (Losing amount)*P(losing)

= 5*0.618667 + 0*(1-0.618667) [The losing amount is zero because we don't have to pay anything after we lose]

= 3.09335

= $3.09

Game 3

Now, let X denote the number of times we roll a 1 on 18 rolls of a die.

=> X ~ Binomial(n = 18, p = 1/6)

Now, Probability of winning under this game:

Minimum Rational Price

Following the logic as in the case of Game 1, we get:

Minimum rational price = Expected winnings per game

= (Winning amount)*P(winning) + (Losing amount)*P(losing)

= 5*0.597346 + 0*(1-0.597346) [The losing amount is zero because we don't have to pay anything after we lose]

= 2.98673

= $2.99

Now, the probability of winning under the three games is given by:

Game	P(winning)
Game 1	0.665102
Game 2	0.618667
Game 3	0.597346

Thus, clearly Game 1 has the highest probability of winning and the minimum rational price we should charge for this game is $3.33 [Answer]

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