Thanks for the response, but I’m still having trouble wrapping my brain around this. Consider the following lengthy clip from Cecil’s article:
Cecil is well aware the answer to the lottery question is “no”–if there are only two tickets left, they have equal odds of being the winner. The difference between this and the Monty Hall question is that we’re assuming Monty knows where the prize is, and uses that information to select a non-prize door to open; whereas in the lottery example the fact that the first 9,998 tickets are losers is a matter of chance. I put the question at the end of a line of dissimilar questions as a goof–not very sporting, but old habits die hard.
The answers given to Jordan Drachman’s questions–2 in 3 in both cases–were correct. The odds of the original card being an ace were 1 in 2 before it was placed in the hat. We are now trying to determine what card was actually chosen based on subsequent events. Here are the possibilities:
(1) The original card in the hat was an ace. You threw in an ace and then picked the original ace.
(2) The original card in the hat was an ace. You threw in an ace and then picked the second ace.
(3) The original card was a king; you threw in an ace. You then picked the ace.
In 2 of 3 cases, the original card was an ace. QED.
The second question is much the same. The possible gender combinations for two children are:
(1) Child A is female and Child B is male.
(2) Child A is female and Child B is female.
(3) Child A is male and Child B is female.
(4) Child A is male and Child B is male.
We know one child is female, eliminating choice #4. In 2 of the remaining 3 cases, the female child’s sibling is male. QED.
Now, in the Monty Hall scenario, Monty knows which doors give false answers, and can open them selectively. But in the above two scenarios, there is no omniscient “Monty” character to finagle the outcome, yet the odds are the same as with a Monty.
Assume a lottery with only three tickets (absurd, but then, what isn’t in higher mathematics?). The odds of my ticket winning are 1/3. The odds that the winner is one of the other tickets is 2/3. One ticket is scratched. Maybe it’s the winner, which makes further calculation moot. But, if it happens to be the loser, doesn’t that now make the other ticket a 2/3 favorite?
I’ll stop now – may brain is beginning to hurt…