1. A person shows you a bag that has one red ball. He then adds another ball, which has a 50/50 chance of being either red or green. He then randomly takes out a ball, and it turns out to be red. What is the probability that the other ball is red as well?
  2. A hockey team has 10 games left in the season. The probability that they win each game is 0.4, and are independent of each other. A turnover is defined as occurring when the team goes from a win to a loss, or a loss to a win. What is the probability that the team ends up with one turnover in the last 10 games.
  • $\begingroup$ Do we know the color of the ball in the bag of part 1? $\endgroup$ – A.Sh Dec 14 '15 at 16:51
  • $\begingroup$ Oops, the first line should read: "A person shows you a bag that has one red ball." $\endgroup$ – Richard Yang Dec 14 '15 at 16:52
  • $\begingroup$ In 2. you want "exactly 1 turnover" or "at least 1 turnover"? $\endgroup$ – Jimmy R. Dec 14 '15 at 17:00
  • $\begingroup$ Exactly one turnover. As a bonus, can you find E(turnover)? $\endgroup$ – Richard Yang Dec 14 '15 at 17:02

For 1. Denote with $R_2$ the event that the added ball was red (then $R^c_2$ is the complementary event, that the second ball was green) and with $D$ the event that the ball you draw is red. Then we want to calculate $P(R_2 \mid D)$ We have by Bayes' rule that $$P(R_2 \mid D)=\frac{P(D\mid R_2)P(R_2)}{P(D)}=\frac{1\cdot \frac12}{P(D)}$$ Now, we can calculate $P(D)$ with the Law of total probability $$P(D)=P(D\mid R_2)P(R_2)+P(D\mid R_2^c)P(R_2^c)=1\cdot\frac12+\frac12\cdot\frac12=\frac34$$ Hence $P(R_2 \mid D)=\frac{\frac12}{\frac34}=\frac23$

For 2. We want exactly 1 turnover. Let's take first the case that the turnover occurs as WL (win then loss). This can happen in the following ways:

  1. Matches $1$ and $2$ and then all loses. String: (WL)LLLLLLLL.
  2. In the middle. It must be precented by a win and followed by a loss (otherwise more turnovers will occur). String: ...W(WL)L... This can occur in places $(2,3),(3,4),\ldots,(8,9)$, i.e. starting in matchday $2, 3, 4, \ldots, 8$. ($7$ possibilities).
  3. Matches $9$ and $10$ and before all wins. String: WWWWWWWW(WL).

The first string has probability $0.4\cdot0.6^9$ the other $7$ strings different probabilities which are $0.4^k\cdot0.6^{10-k}$ for $k=2,3,\ldots 8$ and the last string $0.4^9\cdot0.6^1$. So, the total probability of this scenario is $$\sum_{k=1}^{9}0.4^k\cdot0.6^{10-k}=0.6^{10}\sum_{k=1}^{9}\left(\frac{2}{3}\right)^k=0.6^{10}\cdot\left(\frac23\frac{1-(2/3)^{9}}{1-(2/3)}\right)=2\cdot0.6^{10}\left(1-(2/3)^9\right)$$ The summation was simplified by the formula of the geometric sum $$\sum_{k=1}^{n}p^k=p\frac{1-p^{n}}{1-p}$$ for $|p|\neq 1$. Here $p=2/3$.

Repeat for the case that the turnover occurs as LW, which due to symmetry gives the same probability. So, the result (the probability of exactly $1$ turnover) is given by $$2\sum_{k=1}^{9}0.4^k\cdot0.6^{10-k}=4\cdot0.6^{10}\left(1-(2/3)^9\right)\approx 0.02356$$

  • $\begingroup$ This is where I get confused. I understand the use of Baye's rule, but aren't the individual probabilities different depending on which ball (red or green) is added? Just to clarify as well, P(D) is the probability that the first ball drawn is red, and P(R2) is the probability that the second ball drawn is red. $\endgroup$ – Richard Yang Dec 14 '15 at 17:16
  • $\begingroup$ We use conditional probabilities $P(A \mid B)$ and they take into account these differences. Now, $P(D)$ is the probability that the single ball that is drawn is red (there is no first and second drawn) and $P(R_2)$ is the probability that the added ball is red. $\endgroup$ – Jimmy R. Dec 14 '15 at 17:19
  • $\begingroup$ Ah I finally understand. Could we solve it another way that's a big easier by just listing the possible outcomes and then finding the ones that fit? Ex: Original red ball drawn, other red ball remains Added red ball drawn, original red ball remains Original red ball drawn, added green ball remains Therefore 2/3 of the events have 2 red balls drawn. Or is this just a coincidence $\endgroup$ – Richard Yang Dec 14 '15 at 17:30
  • $\begingroup$ No, it is not a coincidence. You certainly can do it the way you propose. It works well because the "list" of possible outcomes is short. It would work anyway, but sometimes listing all possible outcomes is not so easy. $\endgroup$ – Jimmy R. Dec 14 '15 at 17:40
  • $\begingroup$ For question 2, how exactly did you simplify the summation? I got lost. Also, wouldnt the two probabilities be the same? Since WWWWWWWWWL is the same probability as LWWWWWWWWW but backwards in sequence. How would you find the expected value of E(T) without having to solve for all values of T? $\endgroup$ – Richard Yang Dec 14 '15 at 18:34

For the second question, observe that once a turnover happens, the last outcome must be repeated until the last game (e.g. If a team won before the turnover, then the turnover is a loss, and the team loses each subsequent game, or else there would be another turnover). Also note that a turnover cannot happen on the first game.

Denote the sequence of outcomes by a string of W's and L's, representing wins and losses. The sequences we are looking for take the form WLLLLLLLLL (1W, 9L), WWLLLLLLLL (2W, 8L), ..., WWWWWWWWWL (9W, 1L), and LWWWWWWWWW (1L, 9W), LLWWWWWWWW (2L, 8W), ..., LLLLLLLLLW (9L, 1W).

The probability of a given sequence is $P(W)^{\text{Number of wins}}*P(L)^{\text{Number of losses}}$. So the probability that any of these sequences occurs is $\sum_{k=1}^9 2(0.4)^k(0.6)^{10-k}$.


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