Simple question for this one:

In Riddler City, the city streets follow a grid layout, running north-south and east-west. You’re driving north when you decide to play a little game. Every time you reach an intersection, you randomly turn left or right, each with a 50 percent chance.

After driving through 10 intersections, what is the probability that you are still driving north?

I have an answer, and I can’t see what’s wrong with it, but its obviousness makes me a bit skeptical. The calculation is in a Google Sheet.

One part of this that I am confident of is that, after driving through an even number of intersections, you must be driving either directly North or directly South.

Aaaactualllyyyy… thinking about it, this means that I am confident in my result after all. Think about intersection one: whichever way you turn, when you get to intersection two, there’s a 50% chance you’ll turn back to face North, and a 50% chance you’ll turn to face South. The same is true after any even number of intersections. So most of the stuff in my spread sheet is actually superfluous!

In which case I think I can see how to do the extra credit. If you do a pair of turns, both of them either left or right with equal probability, then you have a 50% chance of coming back to face North at the end. If you do an *odd* number of left or right turns, there’s *no* chance you’ll end up facing North again at the end.

Use the binomial theorem to calculate the odds of there being ten “straight-ons” over the interesections, a lower even number, or an odd number. If there are ten straight-ons, you’re obviously still going North. If there are a lower number of even straight-ons, you have a 50% chance of heading North, by the argument of the two previous paragraphs. And if you have an odd number of straight-ons, you have zero chance of heading North.

That comes out to just a little over 25%, which seems intuitively correct. The spread sheet gives an answer of 0.250 012 701 316, to 12 d.p.

(My intuition that the chance should be > 25% is driven by two things: firstly, the small but positive possibility of not making any turns at all, and secondly, in the case of at least one turn being made, the slightly greater than 50% chance of making an even number of turns.)