### Solving a Bar Bet Using at Most Three Different Operations

A bar bet as presented in the YouTube video The HARDEST Puzzle Yet! involves starting with three of the same number from 0 through 9, then adding mathematical operations that result in an evaluation of 6. For example, if we start with three of the number 6, one solution could be $6+6-6=6 .$ I'll demonstrate a method for solving this bar bet puzzle starting with three of any number, say $$N$$, which involves using at most three different mathematical operations (although some of these may be used many, many times).

If $$0 \leq N \leq 2$$ we have the solutions
\begin{eqnarray}
(0! + 0! + 0!)! &= 6 \\
(1! + 1! + 1!)! &= 6 \\
2+2+2 &= 6.
\end{eqnarray} If $$N\geq 3$$, concatenate the three numbers. Repeatedly applying the square-root operation we'll eventually end up with a result $$x$$ with $$3 \leq x < 9$$.  If we now take the greatest integer $$\lfloor x \rfloor$$ we have an integer $$n$$ with $$3 \leq n \leq 8$$. If we can exhibit solutions for each of these cases that use only square-roots, greatest integers and one other operation, we'll be done. Using factorial for the third operation, some possibilities are
\begin{eqnarray}
\href{http://www.wolframalpha.com/input/?i=3%21}{3!} &= 6\\
\href{http://www.wolframalpha.com/input/?i=%5Cleft%5Clfloor+%5Csqrt%7B%5Csqrt%7B%5Cleft%5Clfloor+%5Csqrt%7B%5Csqrt%7B%5Csqrt%7B%5Csqrt%7B%5Csqrt%7B%284%21%29%21%7D%7D%7D%7D%7D+%5Cright%5Crfloor%21%7D%7D+%5Cright%5Crfloor+%21}{\left\lfloor \sqrt{\sqrt{\left\lfloor \sqrt{\sqrt{\sqrt{\sqrt{\sqrt{(4!)!}}}}} \right\rfloor!}} \right\rfloor !} &= 6\\
\href{http://www.wolframalpha.com/input/?i=%5Cleft%5Clfloor+%5Csqrt%7B%5Csqrt%7B5%21%7D%7D+%5Cright%5Crfloor+%21}{\left\lfloor \sqrt{\sqrt{5!}} \right\rfloor !} &= 6\\
\href{http://www.wolframalpha.com/input/?i=6}{6} &= 6 \\
\href{http://www.wolframalpha.com/input/?i=%5Cleft%5Clfloor+%5Csqrt%7B%5Csqrt%7B%5Csqrt%7B%5Cleft%5Clfloor+%5Csqrt%7B%5Csqrt%7B7%21%7D%7D%5Cright%5Crfloor+%21%7D%7D%7D+%5Cright%5Crfloor+%21}{\left\lfloor \sqrt{\sqrt{\sqrt{\left\lfloor \sqrt{\sqrt{7!}}\right\rfloor !}}} \right\rfloor !} &= 6\\
\href{http://www.wolframalpha.com/input/?i=%5Cleft%5Clfloor+%5Csqrt%7B%5Csqrt%7B%5Csqrt%7B8%21%7D%7D%7D+%5Cright%5Crfloor+%21}{\left\lfloor \sqrt{\sqrt{\sqrt{8!}}} \right\rfloor !} &= 6.
\end{eqnarray}
I've added Wolfram Alpha links so you can verify that these do indeed evaluate to 6.

As an illustrative example, when $$N=1337$$ we have the solution $\href{http://www.wolframalpha.com/input/?i=%5Cleft%5Clfloor+%5Csqrt%7B%5Csqrt%7B%5Cleft%5Clfloor+%5Csqrt%7B%5Csqrt%7B%5Csqrt%7B%5Csqrt%7B%5Csqrt%7B%28++%5Cleft%5Clfloor+%5Csqrt%7B%5Csqrt%7B%5Csqrt%7B%5Csqrt%7B133713371337%7D%7D%7D%7D+%5Cright%5Crfloor+%21%29%21%7D%7D%7D%7D%7D+%5Cright%5Crfloor%21%7D%7D+%5Cright%5Crfloor+%21}{\left\lfloor \sqrt{\sqrt{\left\lfloor \sqrt{\sqrt{\sqrt{\sqrt{\sqrt{\left( \left\lfloor \sqrt{\sqrt{\sqrt{\sqrt{133713371337}}}} \right\rfloor !\right)!}}}}} \right\rfloor!}} \right\rfloor !} = 6.$

### 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) = … ### Solving a Math Puzzle using Physics The following math problem, which appeared on a Scottish maths paper, has been making the internet rounds. The first two parts require students to interpret the meaning of the components of the formula $$T(x) = 5 \sqrt{36+x^2} + 4(20-x)$$, and the final "challenge" component involves finding the minimum of $$T(x)$$ over $$0 \leq x \leq 20$$. Usually this would require a differentiation, but if you know Snell's law you can write down the solution almost immediately. People normally think of Snell's law in the context of light and optics, but it's really a statement about least time across media permitting different velocities. One way to phrase Snell's law is that least travel time is achieved when \[ \frac{\sin{\theta_1}}{\sin{\theta_2}} = \frac{v_1}{v_2},$ where $$\theta_1, \theta_2$$ are the angles to the normal and $$v_1, v_2$$ are the travel velocities in the two media.

In our puzzle the crocodile has an implied travel velocity of 1/5 in the water …