A biased coin has probability 0.6 of turning up heads. You win $x$ dollars if a head comes up and you lose $y$ dollars if a tail comes up. If your expected winnings is $0$ dollars, what is the relationship between x and y?

So far I have:

$P(x) = 0.6$

$P(y) = 0.4$

$E[x] = $expected winnings$ = 0$

$E[x] = x(0.6) + y(0.4) = 0$

$x = -y(0.4)/(0.6)$

I'm kind of stumped on how to continue from here, how can I use the information given to express the relationship between x and y?

  • 1
    $\begingroup$ You correctly got $0.6x + 0.4y = 0$ which leads to $x=-\frac{2}{3} y$ as you got. That is the final answer. There is nothing more to do, this equation describes the relationship between $x$ and $y$ in full detail. Any alternate ways of expressing this information is just that, an alternate way of expressing the same information. $\endgroup$ – JMoravitz Feb 25 at 17:31
  • $\begingroup$ @JMoravitz I think it is not correct. $\endgroup$ – callculus Feb 27 at 16:55
  • $\begingroup$ @callculus how so? Just the difference in the negative sign? That merely varies based on whether you consider a positive value of y to refer to a loss or whether a positive value of y to refer to a gain. Both are valid interpretations and ways of setting up the problem that have little to no impact on the final results beyond a sign change. $\endgroup$ – JMoravitz Feb 27 at 20:03


You´re on the right track, but you haven´t taken into account that you $\texttt{lose}$ $y$ dollars if a tail comes up. Therefore the equation is $0.6\cdot x-0.4\cdot y=0$

Now solve the equation for $\frac{x}y$.


You want to look at the ratio $\frac {x}{y}$.

If $x = -y (0.4) / (0.6)$, then $x / y = - (0.4) / (0.6)$

$\frac {x}{y} = - \frac {2}{3}$


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