# Alice plays a game with $\frac{1}{3}$ odds. Probability Question

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!

• 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? – Rookatu Mar 27 '14 at 1:08