Suppose that you are to play three chess games against Princess and Hope, altern

Question

Suppose that you are to play three chess games against Princess and Hope, alternating between the two opponents after each game (i.e. Princess- Hope-Princess, or Hope Princess-Hope). Jillian, who is watching, bets you a $100 that you cannot win two consecutive games in a row. Assuming that Princess is a better chess player than Hope is, who should you play against first? (Hint: If it helps you, you can assume that you have 40% chance of beating Princess in a single chess game, and 60% chance of beating Hope.) Message

Explanation / Answer

Now the Win-Lose table in this case can be drawn as

Probability   Win   Lose
Princess        0.4   0.6
Hope   0.6   0.4

Now we can register two consecutive wins in the following ways:

P(Win-Win-Lose) OR P(Lose-Win-Win)

Scenario 1:If we start with pricess and follow the sequence Princess-Hope-Princess, the above two probabilities can be computed as:

P(WIn-Win-Lose) = 0.4*0.6*0.6 =0.144

P(Lose-Win-Win)= 0.6*0.6*0.4=0.144

Total Probability of Registering two wins = 0.144+0.144 = 0.288

P(Lose-Win-Win) = 0.6*0.6

Scenario 2:If we start with Hope and follow the sequence Hope-Princess-Hope, the above two probabilities can be computed as:

P(WIn-Win-Lose) = 0.6*0.4*0.4=0.096

P(Lose-Win-Win)= 0.4*0.4*0.6=0.096

Total Probability of Registering two wins = 0.096+0.096 = 0.192

Since Probability in case of scenario 1 is better than Scanario 2, you should start the game with playing with princess first.

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