Quick puzzle: how long to get to heaven?

This is math team math all the way: if you were asking any question I might have to think, but since it involves repetitive calculations of expected value (probability times result), infinite length, and a trick allowing hand calculability, I know it has to decompose to sum of an infinite geometric series somehow.

Some variants left as an exercise to the reader:

a) Three door game as above with doors where A = heaven, B = 2 days + retry, and C = 4 days plus retry. If you get the lesson, this should be trivial.

b) Write a Satan's version of the three door game, where the rules look "simple enough" but the expected time to reach heaven is not finite. (This would be a great extra credit problem for a high school advanced math course -- and full points and a smiley face for "Doors A, B, and C all say 'come back tomorrow'.")

  import random
  def one():
    n = 0
    while True:
      k = random.randint(0,2)
      if k==0: return n
      n += k
  def many(k):
    return sum(one() for i in range(n)) / float(n)
  [many(1000) for i in range(10)]
  many(100000)

The previous-but-one posting on the same blog has a distinctly more difficult question: repeatedly pick independent random numbers uniform on [0,1]; stop when your latest number is smaller than some previous one; what's the average number of numbers you have to pick? Exercise for the reader: provide the appropriate definition of one() to make the calculation above work. It's only a few lines.

(Mathematically expert readers will be able to solve both these problems rigorously and quickly without simulating. But if such a problem comes up in a context where you actually care about the answer you'd do well to do the simulation anyway, since it's an easy way to check your reasoning.)

I calculated that from a scale of 0 and 1, we have a 0.33 chance of going in in 0 days.

The probability of going in within 1 day is equal to the probability of choosing the second door (0.33) multiplied to the probability of choosing the first door the second time (0.33). 0.33 * 0.33 is equal to 0.1089, so we have a 10% chance of getting in with 1 days.

To get in in 2 days, we would have to choose the second door, and then the first door. Again, this is another 10%

At this point I thought that we have a 53% chance of getting in within two days (33% + 10% + 10%), and 47% of getting in with more than 2 days.

The probability of getting in within 3 days is 0.33 * 0.33 * 0.33, which is around 3%. Considering that the probability of getting in within 4 days would be 1% (0.33 * 0.33 * 0.33 * 0.33), I somehow imagined that of the remaining 47% of chances of getting in within more than two days, the 75% was represented by the probability of getting in within 3 days. The 75% of 47% is 35.2%, which is slightly higher than the chance of getting in within 0 days. The answer is 3.

Given that I lack math background, I wonder if my way of solving this problem is correct (hopefully someone can follow my reasoning).

I've run into a few problems like this. The simplest solution exploits the property that there is an ordering on the states, so each "later" state can depend only on "earlier" states or itself. A more general EV problem would require Gaussian elimination instead of DP. Fun stuff, all in all.