The first thing I did today was to correct an error in the spreadsheet for the “Mixture Monty” problem. The SON H,D1 has likelihood 1/2, because if the contestant picks door #1 then Monty From Hell has two doors that he can open to show a goat, equal probabilities for each. We completed the spreadsheet and found that with Mixture Monty, contrary to our expectations, it is advantageous to switch, because the posterior probability is 2/3 that the door you didn’t pick has the prize, and only 1/3 that the door you picked has the prize.

I also did a simplified version of the N doors problem, where you pick a door and Monty opens all but 1 of the remaining doors, revealing all goats (he can do this because he knows where the prize is). Suppose you pick Door #1. Then, if the prize is behind Door #1, Monty can choose any one of the N-1 remaining doors to keep shut, since all of them have goats. So the probability that he opens a particular door k given that the prize is behind your chosen door is 1/(N-1). If the prize is behind Door #2 (which you didn’t choose), then he must open all remaining (N-2) doors, so he cannot leave door k>2 unopened, and the probability that he does is 0. Indeed, that must be the case for every door except for Door k. So, if you did not choose the right door at the outset, Monty is forced to keep the door that has the prize closed, and must open all the remaining doors. So, the probability that he keeps door k closed, given that the prize is behind door k, is 1, and all the other likelihoods (except for the door you chose) are 0. So the problem is actually quite easy from this point on as almost all the numbers we calculate are 0. We calculated the marginal probability as 1/(N-1), the posterior probability that our door is the right one as 1/N, and the posterior probability that door #k has the prize as (N-1)/N.

Then we looked at the problems. Most guesses were that the probability was 1/2 that the sibling is a boy. But the simulation showed that in twice as many of the tosses that had at least one head, the other toss was a tail (so, a girl), than the other toss being a head. One group did not keep the two tosses in each set together and so did not get reasonable results. The intent was for you to write down something like: First pair of tosses (H,T), Second pair (T,T), Third (H,T), Fourth (T,H), etc. There were several trees drawn. One of them had (K,Q) at the base of the probability tree. That’s not right, since you don’t know if there will even be a king or a queen until both coins are tossed. The random events are the coin tosses, so the tree should have at its base the outcome of the first coin toss (H1,T1) with probabilities 1/2 for P(H1) and 1/2 for P(T1). Then each of these branches has a pair of branches representing the second toss (H2,T2) again, and the conditional probabilities P(H2|H1)=P(T2|H1)=1/2, and similarly for the T1 branch. I remarked that since P(H2|H1)=P(H2) for example, the second toss is actually independent of the first. This is our first example of independence, which will be very important.

Finally I noted that the four outcomes are HH, HT, TH and TT (I displayed these in a table), and by counting, we see that there are twice as many cases that have at least one H and have a T, than cases where both are H.

On the dice problem, the easiest way is to draw a table with 36 boxes, in a 6×6 arrangement. Label the top and left with the number of spots showing (1-6) and then put the sum of the number at the top and the number on the left in the corresponding box. I noted that the number of boxes containing a ‘7’ is greater than for any other number, which is why ‘7’ is an important point in craps. We can just circle and count. We see five cases where there is an ‘8’ and circle them. Since there are 36 possibilities, the probability of getting an ‘8’ is 5/36. We then circled the cases where one of the dice is a ‘5’. This is the row labeled ‘5’ and the column labeled ‘5’, a total of 11 cases. Of these, two have a sum of eight, so the probability of rolling an eight, given that one of the dice is ‘5’, is 2/11. On the other hand, if we look at the five cases where the total is eight, two of them have a ‘5’, so the probability of seeing a ‘5’, given that you have rolled eight, is 2/5.

On the funny dice problem, all the probabilities are 2/3: P(B beats A)=2/3, P(C beats B)=2/3, P(D beats C)=2/3, P(A beats D)=2/3. There was a disagreement on this in one group, which wisely wrote down that fact and gave both methods of getting the result.

Finally, the answers to the last question are 1/8, 3/8, 1/7, 1/4.

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