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Alice plays repeatedly a game that has three results: win, lose, or tie. Each time she plays, she wins with probability $\frac{1}{3}$ and loses with probability $\frac{1}{3}$ (and therefore she ties with probability $\frac{1}{3}$) independently of whatever else happened before. Consider a situation in which Alice plays the game $10$ times.

(a) What is the probability that Alice wins exactly four times? (b) What is the probability that Alice wins two or more times?

For a) would the probability be $(\frac{1}{3})^4 * (\frac{2}{3})^6 = 4/81$ because she has to win exactly $4$ times and lose/tie $6$ times?

For b) would the probability be $(\frac{1}{3})^2 + (\frac{1}{3})^3 + ... + (\frac{1}{3})^{10}$ because we need to add up the probability for each case?

These are my ideas, so far but I could be wrong... Thanks!

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  • $\begingroup$ For your answer in a) you are not taking into account the different sequences that give rise to $4$ wins and $6$ losses/ties. Each such sequence will have that probability, but you then must multiply by the number of such sequences. Have you learned about the binomial distribution? $\endgroup$ – Rookatu Mar 27 '14 at 1:08
  • $\begingroup$ Do you know how to use LaTeX? Use $ around your math? $\endgroup$ – user122519 Mar 27 '14 at 1:09
  • $\begingroup$ No doesn't sound familiar... Oh that makes sense! So it would be $7 * (4/81)$? $\endgroup$ – user136038 Mar 27 '14 at 1:15
  • $\begingroup$ Put \frac{a}{b} for fractions, with dollar signs around it around for $\frac{a}{b}$. $\endgroup$ – Display Name Mar 27 '14 at 1:22
  • $\begingroup$ $\frac{4}{81} * 7$ Cool! $\endgroup$ – user136038 Mar 27 '14 at 1:25
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The probability that Alice wins exactly $4$ times is $\binom{10}{4}(1/3)^4(2/3)^6$.

Let W stand for a win, and N for stand for a non-win (loss or tie). Then for example the sequence WWWWNNNNNN has probability $(1/3)^4(2/3)^6$. But there are many other ways Alice can end up with $4$ wins. For example we could have NNWWNNNWNW. Each particular string with $4$ W and $6$ N has probability $(1/3)^4(2/3)^6$, but there are many of them.

The $\binom{10}{4}$ of the expression we gave counts the number of strings of length $10$ that have $4$ W and $6$ N.

For the probability that Alice wins $2$ or more times, it is much easier to compute first the probability of the complementary event that she wins $0$ times or $1$ time. This probability is $(2/3)^{10}+\binom{10}{1}(1/3)^1(2/3)^9$.

The required probability is then $1-(2/3)^{10}-\binom{10}{1}(1/3)^1(2/3)^9$.

We could find the probability of $2$ wins, the probability of $3$ wins, and so on, and add up. We will end up with the complicated sum $$\binom{10}{2}(1/3)^2(2/3)^8+\binom{10}{3}(1/3)^3(2/3)^7 +\cdots +\binom{10}{10}(1/3)^{10}(2/3)^0.$$ We end up with a computation that is much lengthier than the one with the "complementary event" approach.

Remark: suppose that Alice plays a game $n$ times. suppose that each time the probability of "success" is $p$, and therefore the probability of failure is $1-p$. suppose also that the outcomes of the games are independent. Then the probability that Alice has exactly $k$ successes is $$\binom{n}{k}p^k(1-p)^{n-k}.$$ Here $\binom{n}{k}$ is a binomial coefficient. It is sometimes denoted by $\text{C}_k^n$, or ${}^n\text{C}_k$, and by other related names.

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