GeekDad Puzzle of the Week: How to Bullseye a Womp Rat

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DS-1. Image: Flickr/xoque

The Death Star has been all the rage lately, from the White House petition to the recent Kickstarter. So I thought it would be worth taking a look at the fate of the DS-1 Orbital Battle Station. Specifically, what was probability of success of Luke’s shot?

First, according to Wookieepedia, Death Star I was a sphere of diameter 160k. The rebel plan in the Battle of Yavin was to fly DS-1′s midhemisphere trench, which ran around the Star exactly half the distance between the equatorial trench and the command sector north, which sat like an evil Santa Claus at DS-1′s north pole.

Judging by the surface fail rate, the chance of surviving to even make a trench run was about 10 percent. Then once in the trench, the chance of surviving to pull the trigger decayed exponentially as a function of time in the presence of Vader. This was the mistake of the first wave of Gold Squadron Y-Wings: they failed to appreciate the precipitous decline in their chances of success due to the necessity of their slower but tougher Y-Wings simply taking too long in the trench. To a lesser degree, this was also the mistake of Red Leader flanked by Reds 10 and 12, who considered forward fire and not being caught from behind as the primary danger during their trench run.

This, of course, left Luke, Biggs and Wedge. Remember, Luke had only a 1/10 chance of making it to the trench and then his chance of firing further decayed over time. Imagine he had to fly 1/8 of the distance of the midhemisphere trench at the T-65 X-Wing’s stated maximum speed of 1050km/h. And imagine that time is measured in minutes and the decay constant due to the presence of Vader is 1.15. Now you have the probability of Luke surviving to take a shot. But what is the probability of that shot succeeding?

Recall that Luke used to bullseye womp rats in his T-16 back home and that a womp rate is not much bigger than 2 meters. For our purposes, we’ll call the area of a womp rat and thus that of the exhaust port opening that of a circle with diameter 2 meters. Now Luke is speeding along at 1050km/h and must pull the trigger at exactly the right time to land a photon torpedo anywhere along the length of the exhaust port opening (disregarding horizontal skew due to the power of the targeting computer). In what time window must he pull the trigger?

The average human reaction time is 0.2-0.25 seconds, which we’ll approximate at 0.22 seconds. Compare this 0.22s to the time window of Luke’s possible, successful trigger-pulling to discover the probability that he pulls the trigger within this window. Put it all together: the chance Luke survives to make a trench run, the probability he survives long enough in the trench to pull the trigger, and then the chance that he pulls the trigger in the short time in which a photon torpedo would enter the exhaust port and cause a chain reaction destroying the Death Star. What is the probability of Luke’s successful shot?

Of course, this disregards the influence of the Force. The Force’s effect on this probability depends very much on your interpretation of its powers. Can the Force preordain that Luke succeed? And then can the Force preordain that Luke choose to use the Force — in which case nothing from the moment the galaxy formed long, long ago could anything have any resemblance to probability? Or does the Force simply help to guide Luke’s shot?

There is a correct answer to the Force-less proposition of this problem. Please send it to Geekdad Central by Friday afternoon. And then I will happily enter your name a second time into this week’s drawing for a $50 ThinkGeek gift certificate if you can offer a well-reasoned interpretation of adjusted probability based on the influence of the Force. Note that well-reasoned and harebrained are not mutually exclusive.

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