Which Game is Better?

The probability of winning in the original version is $\frac{1}{8}$ .

In the revised version, there is a $\frac{1}{2}$ chance that you will have a $\frac{1}{6}$ chance of winning, and a $\frac{1}{2}$ chance that you will have a $\frac{1}{10}$ chance of winning. The probability of winning is therefore $\frac{1}{2} \times \frac{1}{6} + \frac{1}{2} \times \frac{1}{10}$ , which equals $\frac{2}{15}$ , and since $\frac{2}{15}>\frac{1}{8}$ , $\boxed{The Revised Version}$ is better.

The trick is we need the prob(win) = prob((win & heads) or (win & tails)) since there is no case of overlap between heads and tails outcomes, we add the probabilities of heads and 1/6 = 1/12 to tails and 1/10 = 1/20.

Total chance is 2/15 after simplification. This is better than 1/8.

New game is better!

In the original version the probability of winning is 1/8. In the other version, 1/2 * (1/6 + 1/10) = 2/15 and 2/15 > 1/8

wolfram
Basically saying the same thing as the guy below me, $\frac{1}{8}$ vs $\frac{1}{2}$ * $\frac{1}{6}$ + $\frac{1}{2}$ * $\frac{1}{10}$

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