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A coin is biased in such a way that on each toss the probability of heads is $\frac{2}{3}$ and the probability of tails is $\frac{1}{3}$. The outcomes of the tosses are independent. A player has the choice of playing Game A or Game B. In Game A she tosses the coin three times and wins if all three outcomes are the same. In Game B she tosses the coin four times and wins if both the outcomes of the first and second tosses are the same and the outcomes of the third and fourth tosses are the same. How do the chances of winning Game A compare to the chances of winning Game B?

A The probability of winning Game A is $\frac{4}{81}$ less than the probability of winning Game B.

B The probability of winning Game A is $\frac{2}{81}$ less than the probability of winning Game B.

C The probabilities are the same.

D The probability of winning Game A is $\frac{2}{81}$ greater than the probability of winning Game B.

E The probability of winning Game A is $\frac{4}{81}$ greater than the probability of winning Game B.

Let H denote getting heads and T denote getting tales. For game A, the probability of winning is either getting HHH or TTT. The first outcome has a probability of $(\frac{2}{3})^3$ and the second has a probability of $(\frac{1}{3})^3$. Adding them together gives P(Winning A) = $\frac{27}{81}$.

For game B, the player can win by getting either HHTT or TTHH. So we can do $(\frac{2}{3})^2 \cdot (\frac{1}{3})^2 \cdot 2$ (since we can just count for one of HHTT and double it to get the other), which gives $\frac{8}{81}$. None of the answers A-E seems to fit what I have, though. Where did I go wrong?

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    $\begingroup$ In the second game (B), you can also win by getting HHHH or TTTT, in addition to the two possibilities you wrote. $\endgroup$ – Zubin Mukerjee Feb 4 at 2:35
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For Game A, winning corresponds to getting $3$ heads in a row or $3$ tails in a row. These have probabilities $(2/3)^3$ and $(1/3)^3$, respectively. This means that the probability of winning Game A is, as you correctly computed,

$$\left(\frac{2}{3}\right)^3 + \left(\frac{1}{3}\right)^3 = \frac{1}{3}$$


For Game B, winning corresponds to one of these four combinations: HHHH, TTTT, HHTT, TTHH. These have probabilities $(2/3)^4$, $(1/3)^4$, $(2/3)^2(1/3)^2$ and $(2/3)^2(1/3)^2$, respectively. This means that the probability of winning Game A is $$\left(\frac{2}{3}\right)^4 + \left(\frac{1}{3}\right)^4 + \left(\frac{2}{3}\right)^2\left(\frac{1}{3}\right)^2+ \left(\frac{2}{3}\right)^2\left(\frac{1}{3}\right)^2 = \frac{16+1+4+4}{81} = \frac{25}{81}$$


The difference between the probability of winning Game A and that of winning Game B is therefore $$\left|\frac{25}{81}-\frac{1}{3}\right| = \boxed{\frac{2}{81}\,}$$

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