Question

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 ​

Answer 1

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

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.