> He will not know the day of the hanging until the executioner knocks on his cell door at noon that day. I have placed a ball under one of five cups. You can

> He will not know the day of the hanging until the executioner knocks on his cell door at noon that day. I have placed a ball under one of five cups. You can check only sequentially if the ball is under each cup. Do you know which cup the ball is under? If you start turning over the cups one by one, will you be surprised when you find it? You won’t be surprised after you’ve turned over all but one cup, if you still haven’t found it, since you know the ball is under at least one of them. You would be surprised if the ball was found under any of the other cups though. And so there are only 4 days on which you would be surprised. The executioner knocks on Thursday and provides more data than on the prior 3 days. The judge makes a promise he cannot keep, that no matter what day he schedules the execution that it will be a surprise. But how can you schedule something for the end of a finite series and still have it be a surprise? You can’t. So then does it follow if the usable tail of my sequence is marked out of range, what then becomes of the 2nd to last item in the sequence? This assumes a sequential cascade leads to the state where all possible days have been marked out-of-range. The simple reality is that the judge can only promise you a surprise Monday thru Thursday. By Thursday at noon the surprise will be spoiled. And there’s a chance that you will sit for a day knowing exactly when you will die.