Selected Answers to Problem Set 3
3. The math approach: if I choose a door at random, it is clear that I have a 33% chance at guessing the right door. Now, if I do not change my guess, then the door I chose still has a 33% chance at being right. If I switch, then that leaves one other door. The door I had guessed at the beginning had a 33% chance of being right and the opened door now has a 0% chance of being right. Therefore, the other unopened door as a 67% chance of being the right door. It is in my best interest to always switch.
Here's a link to a script that I wrote to demonstrate this.4a) Aahh...one of my favorite problems of the year. The probability of interviewing all the candidates and taking the best one is, of course, 1. The probability of simply taking the first candidate, if there are six, is 1/6.
4b) If we interview the first half of the candidates and then take the next best from the second half, then two things must be true. First, the best candidate must be in the second half. Second, the best candidate must come before any other candidates that are better than the best of the first half. That is, if the order were (1 4 2 5 6 3) then candidate #5 would be chosen because it is the first candidate better than the best of the first half, #4. Taking these two clauses into consideration, if candidate #5 is in the first half and #6 is in the second half we are guaranteed to get #6, for no other candidate is better than #5 except for #6. The odds of #5 being in the first half is 50% (3/6), and if this is true then the odds of #6 being in the second half is 60% (3/5), since #6 can be in three of five possible positions. Taking the product of these two yield a 30% chance, which is greater than 25%. This same process can be repeated to see the odds of #4 being the best candidate of the first half and #6 occuring before #5 in the second half, and likewise to see the odds of #3 being the best candidate of the first half and #6 occuring before both #4 and #5.
5. A family is more likely to have three of one sex and one of the other rather than two of each sex. An easy view of this problem is to look at the binary sequence from 0000 to 1111 and to see that eight out of the 16 times three 0's or 1's exist, whereas only six times do two 0's and two 1's exist. It can also be said that two of these 16 times yield four children of the same sex.