Thursday, June 28, 2012

A Bayes' Solution to Monty Hall

For any problem involving conditional probabilities one of your greatest allies is Bayes' Theorem. Bayes' Theorem says that for two events A and B, the probability of A given B is related to the probability of B given A in a specific way.

Standard notation:

probability of A given B is written \( \Pr(A \mid B) \)
probability of B is written \( \Pr(B) \)

Bayes' Theorem:

Using the notation above, Bayes' Theorem can be written: \[ \Pr(A \mid B) = \frac{\Pr(B \mid A)\times \Pr(A)}{\Pr(B)} \]Let's apply Bayes' Theorem to the Monty Hall problem. If you recall, we're told that behind three doors there are two goats and one car, all randomly placed. We initially choose a door, and then Monty, who knows what's behind the doors, always shows us a goat behind one of the remaining doors. He can always do this as there are two goats; if we chose the car initially, Monty picks one of the two doors with a goat behind it at random.

Assume we pick Door 1 and then Monty shows us a goat behind Door 2. Now let A be the event that the car is behind Door 1 and B be the event that Monty shows us a goat behind Door 2. Then
\begin{aligned}
\Pr (A \mid B) &= \frac{\Pr(B \mid A)\times \Pr(A)}{\Pr(B)} \\
&= \frac{1/2\times 1/3}{1/3\times 1/2+1/3\times 0+1/3\times 1} \\
&= 1/3.
\end{aligned}The tricky calculation is \( \Pr(B) \). Remember, we are assuming we initially chose Door 1. It follows that if the car is behind Door 1, Monty will show us a goat behind Door 2 half the time. If the car is behind Door 2, Monty never shows us a goat behind Door 2. Finally, if the car is behind Door 3, Monty shows us a goat behind Door 2 every time. Thus, \[ \Pr(B) = 1/3\times 1/2+1/3\times 0+1/3\times 1 = 1/2. \]The car is either behind Door 1 or Door 3, and since the probability that it's behind Door 1 is 1/3 and the sum of the two probabilities must equal 1, the probability the car is behind Door 3 is \( 1-1/3 = 2/3 \). You could also apply Bayes' Theorem directly, but this is simpler.

So Bayes says we should switch, as our probability of winning the car jumps from 1/3 to 2/3.

Wednesday, June 27, 2012

The Infamous Monty Hall

A (Brief) History:

The Monty Hall problem is arguably the most infamous probability puzzle in recent history. It was originally proposed in its current form in 1975, but only really surged into the public spotlight in 1990 when in appeared in a Parade column written by Marilyn vos Savant. For more details on the history of this problem see Wikipedia: Monty Hall problem.

The Original:

As given in Marilyn's column the problem read:
Suppose you're on a game show, and you're given the choice of three doors: Behind one door is a car; behind the others, goats. You pick a door, say No. 1 [but the door is not opened], and the host, who knows what's behind the doors, opens another door, say No. 3, which has a goat. He then says to you, "Do you want to pick door No. 2?" Is it to your advantage to switch your choice?
The phrasing here is ambiguous. Does Monty always show you a goat? Does he only show you a goat when switching would lead to a win? Does he only show you a goat when switching would lead to a loss? In fact, Monty could assign you any probability \( p \) of winning when switching that he wanted. All Monty needs to do is with probability \( p \) only show you a goat if switching wins; otherwise, he'd only show you a goat if switching loses. Some of the unstated assumptions undoubtedly led to additional confusion in what's already a conceptually difficult problem for many people.

Unambiguous Phrasing (Krauss and Wang 2003):
Suppose you're on a game show and you're given the choice of three doors [and will win what is behind the chosen door]. Behind one door is a car; behind the others, goats [unwanted booby prizes]. The car and the goats were placed randomly behind the doors before the show. The rules of the game show are as follows: After you have chosen a door, the door remains closed for the time being. The game show host, Monty Hall, who knows what is behind the doors, now has to open one of the two remaining doors, and the door he opens must have a goat behind it. If both remaining doors have goats behind them, he chooses one [uniformly] at random. After Monty Hall opens a door with a goat, he will ask you to decide whether you want to stay with your first choice or to switch to the last remaining door. Imagine that you chose Door 1 and the host opens Door 3, which has a goat. He then asks you "Do you want to switch to Door Number 2?" Is it to your advantage to change your choice?
Solution:

Standing, of course, gives us a \( 1/3 \) probability of winning. But what if we switch? Well, the car is behind one of the doors, and if it's behind Door 1 with probability \( 1/3 \) it must be behind one of the remaining two doors with probability \( 2/3 \). But it's not behind Door 3 because Monty just showed us a goat. Aha, so the probability that the prize is behind Door 2 must be \( 2/3 \) now, so we should switch! Another way of seeing that this is correct is to pretend that you close your eyes and cover your ears right before Monty Hall reveals the goat. Now by switching you'll win the prize if it's behind either Door 2 or Door 3, so your probability of winning increases from \( 1/3 \) to \( 2/3 \) by switching.

My next entry will be a solution using Bayes' Theorem.

Monday, June 25, 2012

Another Nice Application of Difference Equations

Here's a nice problem I encountered in a course on applied stochastic processes.

Problem:

Show that the set of all pairs of positive integers can be placed into one-one correspondence with the positive integers by giving an explicit one-one mapping between the two sets.

Solution:

This can be done expediently using the theory of partial difference equations. A standard diagonalization can be characterized by the following relations:
  1. \( f(1,1)=1 \)
  2. \( f(x,y) = f(x-1,y)+x+y \)
  3. \( f(x,y) = f(x,y-1)+x+y-1 \)
These can be written in difference notation as:
  1. \( f(1,1)=1 \)
  2. \( \frac{\Delta f}{\Delta x} = x+y \)
  3. \( \frac{\Delta f}{\Delta y} = x+y-1 \)
Equation 2 gives \( \frac{{\Delta}^2 f}{{\Delta x}{\Delta y}} = 1 \) and equation 3 gives \( \frac{{\Delta}^2 f}{{\Delta y}{\Delta x}} = 1, \) so this is an exact partial difference equation. Summing using equation 2, \[ f(x,y) = x(x+1)/2 + xy + g(y). \] Differencing and setting this equal to equation 3, \[ x+\frac{\Delta g}{\Delta y} = x+y-1 \] and so \( g(y) = y(y+1)/2 - y + C \) . Finally \( f(1,1)=1 \) implies that \( C=-1 \) , and so \[ f(x,y) = x(x+1)/2 + xy + y(y+1)/2 - y - 1. \]

Difference Equations

Difference equations are frequently covered in courses on algorithms due to the relationship to recursion. Believe it or not, I first learned the basics of difference equations from George Boole's book (of Boolean algebra fame). This is freely available either as a download or to read online.

The methods that are the most useful for solving linear difference equations are use of the characteristic equation or applying the Z-transform. These are the discrete analogues of the characteristic equation and Laplace transform in differential equations.

References:

Testing Android Blogger

Hopefully this works!

\[ y' = \frac{dy}{dx} = \frac{1}{x^2} \]

Sunday, June 24, 2012

A Dicey Problem: Solution

Problem: Waldo rolls one standard (fair) 6-sided die repeatedly, keeping a running total of what he rolls. To three decimal places, what's the probability that he eventually reaches a total of exactly 1,000,000?

Solution: On average Waldo will be adding \( \mu = (1+2+3+4+5+6)/6 = 3.5 \) per roll, so we would "intuitively" or "naively" expect that the probability of hitting a particular total that's large enough would be close to \( 1/3.5 \). This turns out to be correct.

More generally, if we have any finite set of positive integers with no common factor, then the probability of hitting a particular total for large enough numbers is \( 1/\mu \), where \( \mu \) is the expected value for a single roll. Note that the probabilities associated with the positive integers in our set don't need to be the same, either, just greater than zero and summing to \( 1 \).

Proof: I'll show how you can prove this rigorously for the case where the outcomes are \( 1 \) and \( 2 \), each with probability \( 1/2 \). Let \( p(n) \) be the probability of totaling the value \( n \). The only possible ways to total \( n \) are to reach \( n-2 \) and immediately roll a \( 2 \), or reach \( n-1 \) and immediately roll a \( 1 \). This implies that \( p(n) = 1/2\cdot p(n-1) + 1/2\cdot p(n-2) \), which is a linear, second order, homogeneous difference equation (recurrence relation). I'm going to apply the theory here, but for more details see Wikipedia: Recurrence relation.

In this case the associated characteristic polynomial is \( r^2 = r/2 + 1/2 \) with roots \( r_1 = 1, r_2 = -1/2 \). So we have \( p(n) = c_1 + c_2\cdot (-1/2)^n \) for constants \( c_1, c_2 \). The boundary conditions are \( p(0) = 1 \) and \( p(1) = 1/2 \); solving we get \( c_1 = 2/3 \) and \( c_2 = 1/3 \). This implies that \( p(n) \) converges to \( 2/3 \) very rapidly; exponentially, in fact.

Why Is the Statistician Angry?

Because all the good blog names were taken. And all the bad blog names were taken. In fact, all of the blog names were taken.

Attempts that were made:

  • threequarks (taken)
  • the initial letters of all six quarks in order (taken)
  • amanaplanacanalpanama (taken)
  • hyperreal (taken)
  • octonion (taken)
  • overthruster (taken)
  • oscillationoverthruster (taken)
  • poincare (taken)
  • galois (taken)
  • logistic (taken)
  • seaquarks (taken; really?)
  • monkeyrage (taken; really?)

Dicier and Dicier

The Easy Way: Waldo and Basil are playing a game involving a single normal die. They take turns rolling it, adding the number appearing to their score, and the winner is the first person to total 100 points or more (totals greater than 100 are counted as 100). During the course of a game, which number is the least likely to appear among either Waldo's or Basil's subtotals?

The Hard Way: Waldo and Basil get tired of that game, and instead decide to play a similar game involving throwing two regular dice with the marathon goal of scoring 10,000 points or more. Estimate the probability Waldo, Basil, or both score exactly 9,000 points during the course of the game.

Testing MathJax

Right click on any equation for additional options.

\[ e = mc^2 \]
\[ \frac{r^2}{r^2+s^2} \]
\[ \sum_{n=1}^{\infty} \frac{1}{n^2} = \frac{{\pi}^2}{6} \]
\[ \int_{0}^{\infty} e^{-x}\, dx = 1 \]
\[ \int_{x^2 + y^2 \leq R^2} f(x,y)\,dx\,dy \]
\[ P(E) = {n \choose k} p^k (1-p)^{n-k} \]
\[ J_\alpha(x) = \sum\limits_{m=0}^\infty \frac{(-1)^m}{m! \, \Gamma(m + \alpha + 1)}{\left({\frac{x}{2}}\right)}^{2 m + \alpha} \]
\[\mathbf{V}_1 \times \mathbf{V}_2 = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ \frac{\partial X}{\partial u} & \frac{\partial Y}{\partial u} & 0 \\ \frac{\partial X}{\partial v} & \frac{\partial Y}{\partial v} & 0 \end{vmatrix} \]

A Dicey Problem

Problem: Waldo rolls one standard (fair) 6-sided die repeatedly, keeping a running total of what he rolls. To three decimal places, what's the probability that he eventually reaches a total of exactly 1,000,000?

Hint: If you use your intuition the answer is a single division.

Bonus: Rigorously prove that what you did is correct.