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Showing posts from December, 2013

### Santa's Dice Game

Q: Santa offers you to play a game of dice. You get to roll a dice six times. You can stop rolling whenever you wish and you get the dollar amount shown on that roll. What is an optimal strategy to maximize your payoff?

Probability and Statistics (4th Edition)

A: Let us take a moment and think through this. At each point in the sequence of rolls you make, you have a decision to make. Do you keep rolling or do you stop and walk away with what is being "offered" to you? You also need to bear in mind that if you keep pushing your luck you will reach a point (the 6th roll) where you would have to be content with whatever comes out for the last roll. So lets start with the simple case of what the expected payoff is for the last roll. Lets call this $$E_{6}$$. To compute it, simply take the payoff multiplied by the respective probability.
$$E_{6} = \frac{1 + 2 + 3 + 4 + 5 + 6}{6} = \frac{7}{2} = 3.5$$
The general strategy to be followed is to check what the…

### A Box of Apples and Oranges

Q: A box contains 6 apples and 5 oranges. You know the number of apples are 6. You pick them out one at a time. What is the probability that the box will be empty by the time you have all 6 apples out?

A: It is tempting to say that the answer is $$50\%$$, but its not, as explained further below. Let us start with the number of ways to pull all 11 out. Since you know that there are six apples, you would keep drawing until you see all six apples. The six apples can come out in any order. So there are $$\binom{11}{6}$$ ways to do this.The number of favorable cases can be computed by making the following observation. If the box has to be empty when the last apple is drawn, then the finishing of the draws must end with 1, 2, 3, 4, 5 or 6 apples. These are the only scenarios. It can never end in an orange being drawn. Lets first consider the case when the last fruit drawn is an apple and the one prior is an orange. That's two d…

### The Numbers on a Dice

Q: A die is rolled and the numbers on all the visible faces is multiplied together $$=S$$. How would you choose a highest number such that it is guaranteed to divide the product.

A: The puzzle rests on a simple fact of prime factorization. Every number can be expressed as a product of prime numbers. For the given example, assume the face that is not visible is 1. The product of the remainder of numbers is $$6 \times 5\times\ldots \times 2$$ or equivalently $$6!$$. If the face that is invisible is 2, the number $$S$$ is $$\frac{6!}{2}$$, if it is 3 the number $$S$$ is $$\frac{6!}{3}$$ and so on. The number $$6!$$ can be expressed as prime numbers as $$2^{4}\times3^{2}\times 5$$. Let us enumerate the cases for each of the faces that is invisible, the product $$S$$ expressed in terms of prime numbers are as follows
$$2^{4}\times3^{2}\times5$$$$2^{3}\times3^{2}\times5$$$$2^{4}\times3^{1}\times5$$ $$2^{2}\times3^{2}\times5$$$$2^{4}\times3^{2}$$ \(2^{3}\times3^{1}\tim…

### A Cow, A Monkey and a Tree

Q: A Cow is tethered to a tree by a rope that has half the length of the circumference of the tree's stump. A monkey with a stone in hand jumps onto the tree and starts hopping around the tree's branches which covers a circular area of radius five times that of the stump. This causes the cow to run about at random and monkey drops the stone at random. What is the probability that the cow would get hit?