Data Session 8: Solutions

A B C D
Homework

Solutions for Session 8: Homework

See solutions for Problems: H1 | H2 | H3 | H4 | H5

Problem H1

Here is the table of the possible outcomes of this game:

Red Die

Blue Die

+

1

2

3

4

5

6

1

2

3

4

5

6

7

2

3

4

5

6

7

8

3

4

5

6

7

8

9

4

5

6

7

8

9

10

5

6

7

8

9

10

11

6

7

8

9

10

11

12

•
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.

<< back to Problem H1

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.
b.	Here is one possible way to change the game to make it fair:

Red Die

Blue Die

+

1

2

3

4

5

6

1

0

1

2

3

4

5

2

1

0

1

2

3

4

3

2

1

0

1

2

3

4

3

2

1

0

1

2

5

4

3

2

1

0

1

6

5

4

3

2

1

0

•	Player A wins when the difference is 1 or 2.
•	Player B wins when the difference is 0, 3, 4, or 5.

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

<< back to Problem H2

Problem H3

Here is the table of sums for this game:

Player B

Player A

+

1

2

3

1

2

3

4

2

3

4

5

3

4

5

6

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.

<< back to Problem H3

Problem H4

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

Player B

Player A

+

2

1

3

2

4

3

5

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).

<< back to Problem H4

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:

Player B Picks

Player A Picks

Odd

Even

Odd

4/9

2/9

Even

2/9

1/9

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):

Player B Picks

Player A Picks

Odd

Even

Odd

2/6

2/6

Even

1/6

1/6

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.

<< back to Problem H5

Session 8 | Notes | Solutions | Video