I have not found the type of person who asks trick questions to be the type of person who finds it interesting to have the trick questions they've posed to be prodded.
Completely tangential, but something I enjoyed reading that feels in the same realm: https://blog.plover.com/math/logic/annoying-boxes-solution.h...
I find it depends entirely on whether the person is asking a trick question to try prove themselves smart (and are sensitive about it), or as in this case, are confident in their own intelligence, and want to assess yours.
I had to reread that a few times to figure out what he was saying. All that comes down to is the fact that in his presentation technically there's nothing linking the propositional value of the box labels to the box contents. In most puzzles this linkage specified "outside the puzzle world" but in this case it's specified "inside the puzzle world" and so nothing can be deduced from it. But any sane person would assume the two align (especially in the setting of a puzzle), and so there's the gotcha.
Seems very different from the kind of "trick" questions in interview which are closer to one-way questions where the problem is trivial with some key insight but quite hard otherwise.
The random selection omission is intentional.
This trick actually makes the problem easier, you always need 7 tries, and the payoff, and not just the expected value is $940. Of course, if these are not whole numbers, the payoff is $0.
Now you can continue the interview whether you would cheat if you are guaranteed not to get caught, as it is the assumption you made for Ballmer here ;)
Do the payout between 0 and 1 as the percentage of the amount.
With a range of 1 to 1 to pay off is obviously one
With a range of 1 to 2 the payout is .5
At three values it becomes more interesting. There are two strategies for the candidate either a binary search for the endpoints.
At four values you still have one level of binary search possible but after that it devolves down to the two value problem.
At five values. If the interviewer thinks the candidate would choose binary search and it becomes too too value problems on each side after removing the middle element.
There's definite problems with this but I wonder if he's already possible pay off matrix
Assuming you've found such a strategy for N, when you go to N+1 you still need to find the (N+1) element vector representing the probability that you select each number as your first guess, and you likewise need to know your opponent's probability vector for adversarially choosing a number. Once you guess at those vectors you can use your recursively built up DP sub-solutions to get the value of the game, but you are still stick with solving the optimization problem of finding those mixed strategy vectors for N+1, and will probably need something like CFRM or a similar technique to find them.
