  Each issue Jerry, O4R's puzzle guy, brings you a set of puzzles. Up until now you've had to wait until the following issue of Pro Facto to check your answers. Now you will be able to compare your answers with Jer's by checking the Puzzle Zone on our Web page (where you are now). Answers will be posted about two weeks after the newsletter goes out. However, if you have comments, questions or disagreements, you can still contact Jerry.

* Here's an easy one. There are two index cards, one is colored red on both sides, the other is red on one side and black on the other. They are in a hat held over your head so you can't see the cards. You reach up and pull out one card, viewing one side only. It is red. What's the probability the other side is also red?

* The new king of Lower Slobovia wants his likeness on the coin of the realm. He has some samples made and gives gives one to each of the N minters of the kingdom. They are to demonstrate the quality of their work by providing replicas for evaluation.

Each of the N minters brings a sack of coins. Let's number the sacks 1, 2, 3, ... N. The king's intelligence service notifies him that one of the sacks (they don't know which) contains coins either lighter or heavier than the standard, though within the sack all coins weigh the same. All the coins in all the other sacks have the same weight as the standard coin. What is the minimum number of weighings required to identify the errant sack and determine if the coins in that sack are lighter or heavier than the standard?

"How many coins in each sack?" you ask. As many as you need to solve the problem.

1) The probability the other side is red is 2/3. Think of the three red sides being numbered, R-1, R-2, for the card with two red sides and R-3 for the card with one black side. If a black face is viewed we just ignore it. The probability of this is 1/4. The probability of seeing a red card is 3/4. On the average, when a red card is seen (as in this problem) it will be R-1, R-2, R-3, each one third of the time a red card comes up. When either R-1 or R-2 is seen the other side will be red (R-2 or R-1 respectively). Only when R-3 is seen (1/3 of the time) will the other side be black. Hey, did you think the answer was 1/2 ("Well, it's either red or black on the other side")? Don't commit Hari Kari, you're in good company.

2) It takes two weighings.

On the first the king weighs 1 coin from each sack and the total will be N times the standard weight plus or minus the amount each coin in the "bad" sack is off. Let's call this amount e (for error). And we now know whether the bad coins are lighter or heavier than the standard.

On the next weighing one coin from sack #1, 2 from sack #2, 3 from sack #3,........n from sack #N are placed on the scale. If there were no bad sacks the total would be: 1 + 2 + 3 + 4 +...............n = (n/2)(n +1)* times the standard weight. But, because of the bad sack, call it the kth, the weighing will be off by k times the error or ke. Since we know e from the first weighing we can divide the total deviation from (n/2)(n + 1) by e and immediately find k, i.e., identify the kth sack as the bad sack.

*For example, if n = 5 we have 1 + 2 + 3 + 4 + 5 = (5/2)(5 + 1) = 15 