Pick your favorite number from 1 to 6. How many dice must be thrown simultaneously to have at least a 50% probability of at least one die coming up with your number?

After much philosophizing and exploration in the physical world, today we return to the relative safety of gambling. From the questions I get, some readers are having trouble with probabilities that depend on several things happening. The puzzle given above is a simple problem that illustrates again the principle that you can be psychologically drawn into making the wrong decision by the way the problem is presented, particularly if it is presented in such a way that the wrong answer appeals to common sense.

In the dice puzzle, if you quickly said the answer is three dice, you were probably thinking along the lines that the probability of any side coming up is 1/6, so three dice thrown would give a 50% probability. That is a common sense way of looking at the problem, but as you have also probably guessed by now, not the correct way.

As is the case with many problems of this type, the best way to approach it is to start with the observations that the probability of any side (whichever one you picked is not relevant) coming up is indeed 1/6, but more important, the probability of your chosen number not coming up is 5/6. Then, since the multiple dice come up as independent events, the probability of your number not coming up with two dice is the product of the probabilities of the two events, or 25/36 = 0.6944… >0.5

Continue the process with three dice. The probability of not getting your number is (5/6) (5/6) (5/6) = 124/216 = 0.5787… >0.5. So with three dice, the probability of not getting your number is more than 50%, which means the probability of having your number come up is less than 50%, and we need to keep going.

Consider 4 dice: (5/6)(5/6)(5/6)(5/6) = 0.4823…> 0.5.

So to meet the conditions of the puzzle, one must throw 4 dice, not 3.

There is a non-brute force way to solve this problem as stated. I will outline the method and leave it to you to prove if it works. (Interestingly enough, we have already discussed what is the most probable number when multiple dice are thrown, and this puzzle has an interesting relation to that puzzle.) Consider that all numbers have an equal probability of coming up on a single die. So if you threw a die 6 times, on which of the throws would the chosen number be most likely to appear? Well, right in the middle is the most likely spot, but the middle of 1, 2, 3, 4, 5, 6 is between 3 and 4! Therefore to meet the condition of throwing whole dice, we must round up to 4 to get more than a 50% probability. Does this look hokey - or have I given a valid analysis and dressed it up to look hokey? After all, psychological misdirection can work both ways.

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.