### Twisty Temperature

Problem:

When Waldo recently did a conversion of a positive integral Celsius temperature $$c = 275$$ to its Fahrenheit equivalent $$f$$ (which turned out to be $$527$$ ), he noticed to his amazement that he could have simply moved the last digit of $$c$$ to the front to obtain $$f$$. Doing some intense calculations he failed to discover the next largest such example. Does one exist, and if so, what is it?

Solution:

Let $$c = x_{n}\cdot 10^{n-1} + ... + x_{1}\cdot 10^1 + x_{0}$$ with $$x_{n} > 0$$, then $f = x_{0}\cdot 10^{n-1} + (c-x_{0})/10.$ We also have that $f = (9/5)\cdot c + 32.$ Notice that in order for $$f$$ to be integral $$c$$ must be divisible by 5; this implies that $$x_0=5$$ since it cannot equal 0 (since as a number $$f>c$$). Our equation then becomes $(9/5)\cdot c + 32 = 5\cdot 10^{n-1} + (c-5)/10$ implies $c = 5\cdot (10^n - 65)/17.$ Now it turns out that 10 is a primitive root modulo 17, and so it follows that $$c$$ is integral if and only if $$n$$ is of the form $$16\cdot m + 3$$. When $$m=0$$ we get $$c=275$$; when $$m=1$$ we get the next highest such temperature, which is $5\cdot (10^{19}-65)/17 = 2941176470588235275.$

### 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 sho…

### What's the Value of a Win?

In a previous entry I demonstrated one simple way to estimate an exponent for the Pythagorean win expectation. Another nice consequence of a Pythagorean win expectation formula is that it also makes it simple to estimate the run value of a win in baseball, the point value of a win in basketball, the goal value of a win in hockey etc.

Let our Pythagorean win expectation formula be $w=\frac{P^e}{P^e+1},$ where $$w$$ is the win fraction expectation, $$P$$ is runs/allowed (or similar) and $$e$$ is the Pythagorean exponent. How do we get an estimate for the run value of a win? The expected number of games won in a season with $$g$$ games is $W = g\cdot w = g\cdot \frac{P^e}{P^e+1},$ so for one estimate we only need to compute the value of the partial derivative $$\frac{\partial W}{\partial P}$$ at $$P=1$$. Note that $W = g\left( 1-\frac{1}{P^e+1}\right),$ and so $\frac{\partial W}{\partial P} = g\frac{eP^{e-1}}{(P^e+1)^2}$ and it follows \[ \frac{\partial W}{\partial P}(P=1) = …