### 8.4 Expecting the Unexpected

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• Math Help

Be sure you see in the solution to Example 1 that after the contestant chooses a door, the host reveals the goat behind one of the remaining doors. At this point, the contestant has only two doors to consider. Assuming the contestant chooses door 1, here are the situations.

1. In this situation, the host can reveal the goat behind either door 2 or door 3. The contestant can only win by staying with door 1.

2. In this situation, the host only has one choice and must reveal the goat behind door 3 (the other goat is behind door 1, the door the contestant picked). The contestant can only win by switching.

3. In this situation, the host only has one choice and must reveal the goat behind door 2 (the other goat is behind door 1, the door the contestant picked). The contestant can only win by switching.

Use the Monty Hall Game to answer the Checkpoint question.

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• Checkpoint Solution

When I played the Monty Hall Game 20 times by staying, I won 5 times and lost 15 times. This is a winning percentage of . The expected winning percentage is or about 33.3%, which is 8.3% greater than my winning percentage. Although my winning percentage was a little less than expected, it still supports the claim that staying is not the best strategy.

When I played the Monty Hall Game 20 times by switching, I won 14 times and lost 6 times. This is a winning percentage of . The expected winning percentage is or about 66.7%, which is 3.3% less than my winning percentage. Although my winning percentage was a little greater than expected, it still supports the claim that switching is the best strategy.

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