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Showing posts from October, 2013

Lotto Urn with a Twist.

Q: An urn contains lotto tickets numbered 1 to 100. You play the game by paying a fee. The rules of the game are as follows, you need to draw two tickets, you are paid the value of the lower number of the tickets. What is a fair price to play this game?

Probability Theory: The Logic of Science

A: The first ticket drawn could be any of the tickets from 1 to 100. The second draw sets up the payoff which will be the lower of the two. Assume the first drawn ticket has the number \(i\) on it. The second ticket drawn could be either greater or lesser with probabilitie as shown in the figure below.
 If the chosen number is lesser, the expected value is \(\frac{i-1}{2}\). Likewise, if the chosen value is greater the expected value is \(\frac{N-i}{2}\). Thus, the total expected value can be computed as
E = \frac{i-1}{N}\times \frac{i-1}{2} + \frac{N-i}{N}\times\frac{N-i}{2}
Note, the above expression is just for the case when \(i^{th}\) ticket is selected. This happe…

Pirates and Pigeons

Q:Six pirates meet on an island. Prove that at least three of them are all mutual friends or at least three of them are complete strangers?
Discrete Mathematics with Applications

A:This is yet another case for applying the Pegionhole principle. Though the way it can be applied is quite subtle. Lets start by enumerating the many possible connections that are possible. For example, if we name the pirates A,B,C,D,E and F, then the possible connections are AB, AC, AD,...,EF.
A would have 5 connections, B four and so on until E has one. This enumerates out to 15 possible connections. The connections for just one pirate is shown in the picture below.
If there are \(n\) pirates, the total number of connections possible is given by \(\frac{n(n-1)}{2}\), the sum of integers to \(n-1\). Each of these connections falls into two states (i.e. our pegionholes). They are either "friends" or "strangers". As an example, assume 10 connections fall into the "…

Handshakes and Pigeons

Q: A party has a hundred guests. The guests mingle freely and randomly shake hands with one another (no repeat handshakes). What is the probability that any two persons will have the same number of handshakes?
Discrete Mathematics with Applications

A: In order to solve this, a powerful yet simple tool available is the "Pigeonhole Principle". At its heart its a rather simple statement. If you place \(n\) pigeons in \(m\) holes where \(n \ge m\) at least one hole must have more than one pigeon. This is simple to grasp if you look at the picture below which shows the case with 3 pigeons in 2 boxes.
A guest at the party can shake hands 0 to \(n - 1\) times. Each of these counts can be thought of as states a guest can get into and the resulting distribution is quite arbitrary. For example, there could be guests who shake hands with none, with 3 other guests and so on. Notice how this fits nicely with the pigeon hole principle. In such a setting, a guest can fi…

Acute Angles and Wall Clocks

Q: If you were look at a wall clock between 12noon and 1pm at random, what is the probability that the minute hand and the hour hand make an acute angle?
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A: If you rush this one, you will be tempted to answer 50%, but it is not so. Assume we are at 12 noon. The angle swept by the hour hand of a clock in an hour is \(\frac{30^{o}}{60}=(\frac{1}{2})^{o}\) per minute. If \(t\) minutes have passed in an hour, the hour hand would sweep out \(\frac{1}{2}t\) degrees. Likewise the minute hand of the clock sweeps out \(\frac{360^{o}}{60} = 6^{o}\) per minute. The condition required for an acute angle to form between the hour and minute hand can be phrased as
\theta_{m} - \theta_{h} \le 90^{o} \\
6t - \frac{t}{2} \le 90 \\
The above simplifies to
t \le \frac{180}{11} \approx 16
However, there is one more case that creates an acute angle between the hour and minute hand of the clock. Thi…