Last week I presented some problems:

1. You have a normal chessboard and a set of dominoes. The dominoes are just the right size to cover two squares on the chessboard. Assume you have enough dominoes to completely cover the chessboard. Now assume the center four squares are missing. Can you still cover the board with dominos and not have any empty spaces or overhanging half dominos? Now start with a whole board again the remove the two squares diagonally across from each other on the corners. Can you cover this board with not empty space and no overhanging dominos?

2. What is the lowest number of people in a room such that the probability of two of them having the same birthday is about 50%?
3. A constant-skill archer in competition shoots an arrow. He shoots another arrow. It is further out than the first. What is the probability his third shot will also be worse than the first one?
4. Two gamblers flip coins. One does 1000 tosses. The other does only 100. Which gambler has the better chance of having equal number of heads and tails? Hint: consider the limits of the number of tosses each could make and have equal numbers of heads and tails in principle (i.e. more than one toss…).

Here are some answers:

1. The two squares that are removed are the same color. Dominoes cover two squares each; one of either color. Therefore the board cannot be covered because the board has more square of one color than the other.

2. It’s lower than you probably think. The first person can have any date. The second person has 1 chance out of 365 to match it, but from there on is gets a bit messier. Because this is similar to the puzzle I posted yesterday, I will hold off on the answer.
3. This is one of the problems that can be easily solved by making a payoff matrix. Try writing down the six possible ways the three arrows could be scored. The conditions of the puzzle eliminates three of them. Of the three possible configurations, the third arrow wins twice. So the probability of the third arrow being the best is 2/3. Most people guess 1/3.
4. The one who only throws 100 has the better chance. If they only threw two coins, the odds are fifty/fifty. As they throw more (always and even number of total tosses), the odds decrease, going to zero at an infinite number of tosses. The paradoxical thing about this little puzzle is that as the total number of tosses increases, the fractional difference between the number of heads and tails also decreases and approaches zero is the number of tosses approaches infinity. Strange things happen as we go to infinity.

For those who wish to delve further into decision theory without wading through a lot of equations, I have posted a tutorial on elementary decision theory. It shows examples of faulty physicians’ diagnoses (important for those considering surgery) and how to evaluate anti-terrorist activities (important for everyone). That tutorial can be found here.