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**Problem H1
**Use the table of sums for the sum of two dice to determine the more likely winner of this game of chance:

**• **Player A wins when the sum of the two dice is an even number.

**• **Player B wins when the sum of the two dice is an odd number.

**Problem H2
**Consider another game in which two players roll a pair of dice and look at the magnitude of the difference of the outcomes:

**• **Player A wins when the difference is 0, 1, or 2.

**• **Player B wins when the difference is 3, 4, or 5

**a. **Determine whether this game is fair by considering the possible outcomes for two dice. You may want to use the outcomes generated in Part B.**
b.** If the game is fair, come up with a similar game that seems fair but is not. If the game is unfair, change the game in some way to make it fair.

**Problem H3
**Here is another game that two players can play, which is similar to the game Rock, Paper, Scissors.

Two players each hold between one and three fingers behind their backs, then hold out their hands at the same time:

**• **Player A wins if the sum of the number of fingers is even.

**• **Player B wins if the sum of the number of fingers is odd.

Suppose that each player selects randomly among the three choices. Determine whether this game is fair by constructing the possible outcomes (there are a total of nine).

**Problem H4
**Player B realizes that the game is unfair and changes strategies: Now Player B will always choose two fingers.

**Problem H5
**Is there a strategy that Player B can use to make the game fair, regardless of what Player A tries to do?

Think about the way players win the game: with an “odd” sum or an “even” sum. With each player picking all three choices with equal probability, why aren’t “odd” and “even” sums equally likely to occur?

**Suggested Readings:**

**Kader, Gary and Perry, Mike (February, 1998). Push Penny — What Is Your Expected Score? Mathematics Teaching in the Middle School, 3 (5), 370-377.**

Reproduced with permission from

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Push Penny — What Is Your Expected Score?

Continued

Continued

Continued

**Perry, Mike (Spring, 1999). Push Penny: Are You a Random Player? Teaching Statistics, 21 (1), 17-19.**

This article first appeared in

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Push Penny: Are You a Random Player?

**Problem H1
**Here is the table of the possible outcomes of this game:

• Player A wins when the sum is even. There are 18 (out of 36) possible outcomes in which the

sum is even.

**• **Player B wins when the sum is odd. There are 18 (out of 36) possible outcomes in which the

sum is odd.

Since each player wins 50% of the time, this game is fair; neither player is more likely to win.

**Problem H2
a. **This game is not fair. Player A wins in 24 of the 36 possible outcomes, while Player B wins in only 12 outcomes.

**• **Player A wins when the difference is 1 or 2.

**• **Player B wins when the difference is 0, 3, 4, or 5.

**Problem H3**

In this case, both Player A and Player B win in 18 possible outcomes.

Here is the table of sums for this game:

**Problem H4**

Player A wins in five of the nine possible outcomes. Therefore, if each player selects randomly from the three choices, the game favors Player A.

**a.** If Player B always chooses two fingers, here is what the sample space becomes:

Since two of the possible three outcomes now result in a win for Player B, Player B is more likely to win if Player A does not change strategies.

**b.** We’d like to assume that Player A is sure to notice this strategy and will start choosing two fingers each time as well (resulting in an even total, and a win for Player A).

**Problem H5**

Yes. The reason the game is unfair is that there are more odd than even choices. Each player picks an odd number two-thirds of the time and an even number one-third of the time, which results in the following probabilities:

Therefore, the probability that the total will be even is 4/9 + 1/9 = 5/9, making the game unfair in favor of Player A.

To equalize things, Player B should change strategies, and pick odd and even numbers 50% of the time. The easiest way to do this is for Player B to alternate between one and two (or two and three) fingers, but there are other ways to accomplish this, as long as Player B chooses two fingers half the time. This results in the following probabilities (if Player A does not change strategies):

Therefore, the probability that the total is even is now 2/6 + 1/6 = 3/6, or one-half. Regardless of Player A’s strategy, the probability that the total is even will always be one-half, and the game is made fair by Player B’s new strategy.