# 2048 and algebra (Part 10)

In this series of posts, I used algebra to show that 114,795 moves were needed to produce the following final board. This board represents the event horizon of 2048 that cannot be surpassed.

I reached after about four weeks of intermittent doodling. It should be noted that the above game board was accomplished in practice mode, and I needed perhaps a couple thousand undos to offset the bad luck of a tile randomly appearing in an unneeded place.

For what it’s worth, my personal best in game mode was reaching the 8192-tile. I’m convinced that, even with the random placements of the new 2-tiles and 4-tiles, the skilled player can reach the 2048-tile nearly every time and should reach the 4096-tile most of the time.  However, reaching the 8192-tile requires more luck than skill, and reaching the 16384-tile requires an extraordinary amount of luck.

So what are the odds of a skilled player reaching the event horizon in game mode, without the benefit of undoing the previous move? I will employ Fermi estimation to approach this question. Of the approximately 100,000 moves, I estimate that about 5% of the moves require a certain 2-tile or 4-tile to appear at a certain location on the board. For example, in the initial stages of the game, the board is wide open and really doesn’t matter a whole lot where the new tiles appear. However, when the board gets quite crowded, it’s essential that new tiles appear in certain places, or else even a highly skilled player will get stuck.

What is the probability of getting the right tile on each of these occasions? Usually it’s quite high (over 90%). But sometimes it’s necessary to get a 4-tile in exactly the right place when there are four blank spaces (estimated probability of 3%). So let’s estimate 10% to be the probability for getting the right tile for all of these occasions. Let’s also assume that the random number generator is indeed random, so that the tiles appear independently of all other tiles.

With these estimates, I can estimate the probability of reaching the event horizon in game as $\displaystyle \left( \frac{1}{10} \right)^{5000} = \displaystyle \frac{1}{10^{5000}}$. While this analysis isn’t foolproof, it sure beats playing the game about $10^{10,000}$ times and then dividing by the number of times the event horizon is reached by the total number of attempts!

How small is $\displaystyle \frac{1}{10^{5000}}$? Since $2^3 \approx 10$, this is approximately equal to $\displaystyle \frac{1}{2^{15,000}}$, and that’s a probability so small that it was reached (and surpassed) when the Heart of Gold spaceship activated the Infinite Improbability Drive in The Hitchhiker’s Guide to the Galaxy. By way of comparison:

• $\displaystyle \frac{1}{2^{276,709}}$ is the probability that someone stranded in the vacuum of space will be picked up by a starship within 30 seconds.
• $\displaystyle \frac{1}{2^{100,000}}$ is the probability of skidding down a beam of light… or having a million-gallon vat of custard appearing in the sky and dumping its contents on you without warning.
• $\displaystyle \frac{1}{2^{75,000}}$ is the probability of a person turning into a penguin.
• $\displaystyle \frac{1}{2^{50,000}}$ is the probability of having one of your arms suddenly elongate.
• $\displaystyle \frac{1}{2^{20,000}}$ is the probability of an infinite number of monkeys randomly typing out Hamlet.

These are the events to which the probability of reaching the event horizon in 2048 without any undos should be compared.

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