Probability in a game of chess

Question

In a game of chess, you can win, draw, or lose.

 Gary plays two games of chess against Mijan.

The probability that Gary will win any game against Mijan is 0.55

The probability that Gary will draw any game against Mijan is 0.3

 

a) Work out the probability that Gary will win exactly one of the two games against Mijan.

 

 

In a game of chess, you score 1 point for a win, 1/2 point for a draw, 0 points for a loss.

 

b) Work out the probability after two games, Gary's total score will be the same as Mijan's total score.

 

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 Please explain how to answer this! :)

Answer 1

Hi Tahlia,

 

The sum of all possibilities must always add up to 1. Since we know that:

 

    P(win) = 0.55

    P(draw) = 0.3

 

Then:

 

    P(loss) = 1 - 0.55 - 0.3

    P(loss) = 0.15

 

a)

 

We want to know the probability that Gary will win EXACTLY ONE game out of two. The general format for calculating a specific number of successes out of a certain number of trials is:

 

    _(#trials)C_(#successes) * (Probability of success)^(#successes)*(Probability of failure)^(#failures)

 

I realize this looks complex, but we simply need to plug in the correct numbers into the right places:

 

    #trials = 2 (he plays two games)

    #successes = 1 (he must only win one game)

    P(success) = P(win) = 0.55

    P(failure) = P(draw) + P(loss) = 0.3 + 0.15 = 0.45

    #failures = 1 (if he only wins one game, he must lose or draw the other game)

 

So, the overall probability of winning exactly one game would be:

 

    ₂C₁ * (0.55)¹ * (0.45)¹

 

    = 0.495

 

b)

 

In this case, we want to figure out the probability that Gary's score would be the same as Mijan's score after two games. The only ways that this could happen are:

 

    1)  Gary wins one game, and Mijan wins the other

 

    2)  Both games are a draw

 

Any other combination would NOT result in the scores being equal. Whenever there are MULTIPLE ways of achieving the goal, we will calculate the individual probabilities of each of the options, and then add them for the overall probability.

 

    For Gary to win one game, and Mijan win the other, the probability would be:

 

    _(#games)C_(#wins) * (Probability of a win)^(#wins) * (Probability of a loss)^(#losses)

 

        #games = 2

        #wins = 1

        P(win) = 0.55

        P(loss) = 0.15

        #losses = 1

 

    So:

 

    ₂C₁ *(0.55)⁽¹⁾ * (0.15)⁽¹⁾

 

    = 0.165

 

Now let's calculate the probability that both games are a draw:

 

    _(#games)C_(#draws) * (Probability of a draw)^(#draws) * (Probability of non-draw)^(#non-draws)

 

    #games = 2

    #draws = 2

    P(draw) = 0.3

    P(non-draw) = 1 - P(draw) = 1 - 0.3 = 0.7

    #non-draws = 0

 

    ₂C₂ * (0.3)² * (0.7)⁰

 

    = 0.09

 

So, the total probability of either of these two things happening (since either of them would result in a tie score) is:

 

    0.165 + 0.09

 

    = 0.255

 

I hope this helps!

 

Andy