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8 November, 09:11

Many everyday decisions, like who will drive to lunch or who will pay for the coffee, are made by the toss of a (presumably fair) coin and using the criterion "heads, you will; tails, I will." This criterion is not quite fair, however, if the coin is biased (perhaps due to slightly irregular construction or wear). John von Neumann suggested a way to make perfectly fair decisions, even with a possibly biased coin. If a coin, biased so that P (x) equals 0.4700 and P (t) equals 0.5300 , is tossed twice, find the probability

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  1. 8 November, 09:28
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    P (hh) = 0.2209

    P (ht) = 0.2491

    P (th) = 0.2491

    P (tt) = 0.2809

    John von Neumann suggested that if both tosses results in same outcome then discard the result and start again. If each result is different then accept the first one

    Step-by-step explanation:

    We are given that a coin is unfair and the probabilities of getting a head and tail are

    P (h) = 0.47

    P (t) = 0.53

    John von Neumann suggested a way to make perfectly fair decisions, even with a possibly biased coin.

    He suggested to toss the coin twice, so the possible outcomes are

    Sample space = {hh, ht, th, tt}

    The probabilities of these outcomes are

    P (hh) = P (h) * P (h)

    P (hh) = 0.47*0.47

    P (hh) = 0.2209

    P (ht) = P (h) * P (t)

    P (ht) = 0.47*0.53

    P (ht) = 0.2491

    P (th) = P (t) * P (h)

    P (th) = 0.53*0.47

    P (th) = 0.2491

    P (tt) = P (t) * P (t)

    P (tt) = 0.53*0.53

    P (tt) = 0.2809

    He suggested that if both tosses results in same outcome then discard the result and start again.

    If each result is different then accept the first one, for example,

    if you get heads on the first toss and tails on the second toss then result is heads.

    if you get tails on the first toss and heads on the second toss then result is tails.

    If you notice the probability of P (ht) and P (th) are same therefore, this strategy allows to make fair decision even when the coin is biased.
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