You work at a casino owned by Sheldon. Sheldon presents you with a roulette-like machine that spins around and randomly stops at a number from 1 to *n*, where *n* is fixed, each number having equal probability. Sheldon asks you to design a new game where the probability of winning is *p*. Can you do this? If so, explain how. If not, explain why not?

## Rockets Into Deep Space

The key to this puzzle is the famous “Rocket Equation” that you can find in Wikipedia and various other references, but is not very hard to derive by applying calculus to Newton’s Third Law. The momentum of an object with mass *M* and velocity *V* is *MV*. In a closed system, all changes in momentum must sum to 0 by Newton’s Third Law. So when a rocket of total mass *M* increases its velocity by *dV*, through burning an infinitesimal amount of rocket fuel *dM* with exhaust velocity *V _{e}*, the equation

*M*

*dV*= –

*V*

_{e}*dM*holds. If a rocket starts with total mass

*M*

_{0}and weighs

*M*after all its fuel is burned, then the total change in velocity it experiences can be expressed by the integral since

_{t}*M*is changing and

*V*is constant.

_{e}Rocket booster mass consists of structural mass *M _{s}* and fuel mass

*M*. Let the probe mass be

_{f}*M*. When

_{p}*n*boosters are fired in parallel

*M*

_{0}=

*M*+

_{p}*n*(

*M*+

_{s}*M*) and

_{f}*M*=

_{f}*M*+

_{p}*n*

*M*. To simplify calculations, we can choose units of mass so that

_{s}*M*= 1 and units of velocity so that

_{p}*V*= 1. Then from the statement of the puzzle we get two equations:

_{e}and .

These equations appear to not yield an algebraic solution, but numerical methods (like Solver in Excel) lead to the solution *M _{s}* = 0.852207 and

*M*= 11.4909. The ratio of Δ

_{f}*V*from firing

*n*boosters in parallel to Δ

*V*firing just one booster is . As

*n*→ ∞ this ratio increases and asymptotically approaches a limit of . So Wernher tells Walter that it is impossible to double the velocity of the probe, or even just to increase it by 36 percent, by adding boosters to be fired in parallel.

If a series of boosters is fired in sequence, discarding each after being fired, then Δ*V* from firing the *k*th from last booster is . See Table 1.

Consequently, Wernher tells Walter that 10 boosters fired in sequence will be needed to double the velocity of firing just one booster.

Solutions were also submitted by Bob Conger, Rob Kahn, Clive Keatinge, Jerry Miccolis and Brad Rosin.