How to get 92 heads in a row, pt. 2

Two posts ago I showed how many coin flips it would take in order to have a 98% confidence of getting 92 heads in a row (à la Rosencratz and Guildenstern Are Dead). The answer turns out to be \(3.874\times 10^{28}\) coin flips, which if you tried to do by yourself, it would take 100 billion times longer than the current age of the universe. Since I mentioned that only Wowbagger, the Infinitely Prolonged could pull off such a stunt, I think it’s only fair that any other solution also be Adamsian, at least in practicality if nothing else.

My friend Spencer proposed a Dyson sphere to power a huge number of coin-flipping robots. I think he’s on the right track and I had similar thoughts, however my ideas are a bit larger in scale and less detailed. I won’t go into detail on what kind of Dyson Sphere is best or such, since even simplistic models are fraught with difficulties and instabilities (a nice page talking about Dyson Spheres and some simple analysis is here). Instead, let’s just say that we can build some kind of large Dyson network in order to capture a significant portion of the Sun’s energy. We’ll be conservative and say that after light capture, conversion to useful energy, and then maintenance, etc. we can use 10% of the Sun’s radiant energy to power an array of coin-flipping robots.

Spencer also mentioned the concern that once you have good enough robots, that coin flipping is no longer random: exactly precise robots flipping exactly precise coins in the exactly precise way will give the exact same result every time. That may be the case, but we’ll assume that the robots and coins are made imprecise enough that there will be enough random variance in between all the robots to make the system truly random and fair (this is in all reality probably impossible, but we’re in Adam’s universe so we’ll assume it can be done anyway).

Since the robots don’t have to do anything but flip coins and report the outcome, we’ll say each robot consumes about as much power as a toaster oven, or 1000 W. The sun’s luminosity is \(3.846\times 10^{23} W\), so assuming we can use 10% of the sun’s energy we have:
\(\displaystyle{(0.1)\left (3.846\times 10^{26} W \right)\left (\frac {1\;\text {robot}} {1000\;W} \right) = 3.846\times 10^{22}\;\text {robots}}\)
This many robots would give us the same number of flips every second, so that will give us the required number of flips in:
\(\displaystyle{\frac{\displaystyle{3.874\times 10^{28}\text{flips}}}{\displaystyle{3.846\times 10^{22}}\textstyle{\frac{\text{flips}}{s}}}=1.007\times 10^6s\approx 11\; \text{days and}\;14\; \text{hours}}\).
Now that is a considerable improvement.

This potential solution does have some problems though, the most obvious being whether there is enough useful material in the entire solar system to build \(3.846\times 10^{22}\) robots, plus the Dyson power grid to run the whole thing, plus a maintenance system to keep it all in good working order, etc. If we limit ourselves to just the easy to use material, like just the asteroid belt, that limits us to about \(\small{2\times 10^{21}}\) kg of mass. Assuming a total of 10 kg for each robot (including Dyson network power generation, infrastructure, maintenance, etc.), that limits us to just \(2\times 10^{20}\) robots. This number of coin-flipping robots would then take 6.14 years to get the required \(\approx 3.874\times 10^{28}\) coin flips, which still isn’t bad at all. It might take several millenia to build the coin-flipping robot Dyson network, but once it was up and running you’d have your 92 heads in a row in just a few short years!

So lets say we let our coin flipping Dyson array keep running, say, until the Sun becomes a red giant in about 5 billion years, destroying our Dyson array. We would have

\(\displaystyle{\begin{array}{c}\left( 2\times 10^{20}\text{robots}\right) \left( \frac{\displaystyle { 1\;\text{flip}}}{\displaystyle { \text{robot}\;s}}\right) \left( \frac{\displaystyle { 3.1557\times 10^{7}\; s}}{\displaystyle {1\; \text{yr}}} \right) \left( 5\times 10^{9}\; \text{yr} \right) \\ \approx 3.2 \times 10^{32} \text{coin flips} \end{array}}\)

From this can we calculate how many coin heads in a row we can expect to get during this time? Our initial equation is
\(\displaystyle{F = 1-\left( 1-2^{-n}\right)^{0.5\,f}}\)
where F the confidence probability we we desire (we’ve been using 0.98, or 98%), n is the number of heads in a row, and f is the number of coin flips. Rearranging this for n we have:
\(\displaystyle{n = \frac{1}{\text{ln}\,2}\: \text{ln} \left( \frac{-0.5\,f}{\text{ln}(1-F)} \right)}\)

For some various confidence probabilities we have these results:

\( \begin{array}{ l l } \textbf{\textit{F}} & \textbf{\textit{n}} \\ 0.9999 & 120 \\ 0.999 & 120 \\ 0.99 & 121 \\ 0.9 & 122 \\ 0.75 & 123 \\ 0.5 & 124 \end{array} \)

So what does this mean? You have a 99.99% chance of getting at least 120 heads in a row, pretty much guaranteed. However, you only have a 50% chance of getting up to 124 heads in a row. What gives? We go from flipping coins for 6 years to 5 billion years, and the only improvement we get is an additional 28 heads in a row? That’s because each additional head in a row has half the probability of occuring, so the probability decreases exponentially with a linear increase in number of heads required. Conversely, for an exponential increase in the number of coin flips, we see only a modest linear increase in number of expected heads in a row.