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A drunkard is trying to reach his home, at the top of a steep hill. He is currently three steps from the door.

Each time he attempts to take a step uphill, he has a $\frac 16$ probability of sliding $N$ steps back down instead. Initially, $N=1$.

But each time the drunkard slides backward, $N$ increases by $1$.

What is the probability that the drunkard gets home?

MJD
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  • The only way I can think of to approach this is as a random walk on a quarter-infinite plane, where the $y$ coordinate represents the value of $N$, and we attempt to compute the probability that $D$ reaches $x=0$. But it seems complicated. – MJD Jun 27 '22 at 03:02
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    Lovely question! What is the context? How did you arrive at this question? – Yuval Peres Jun 27 '22 at 03:02
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    Thanks. There's no context, I just thought it up and couldn't think of a good way to proceed. It seems like an interesting variation on a typical one-dimensional random walk question like this one. The question is almost the same but the increasing $N$ seems to change it substantially. – MJD Jun 27 '22 at 03:06
  • Running a python script gives an answer that he reaches home approximately $97%$ of the time, buy I have no clue how to actually solve it – Moko19 Jun 27 '22 at 13:10
  • That seems pretty unlikely to me. The base probability of getting home in three steps is only 0.579, and once he starts sliding, his chances should start to drop rapidly. Are you sure? – MJD Jun 27 '22 at 14:31
  • I got 97% success also. The drunkard's chances of recovering from missteps are much better than I guessed. But I suppose I should have realized this, because he can expect to get five good steps before every bad step, so $N$ must increase to $5$ before he can expect to stop making progress. – MJD Jun 27 '22 at 14:54
  • Here's a thought. Let $P(i)$ be the probability that the drunkard gets home when $N$ is capped so that it cannot increase beyond $i$, and let $P$ be the probability when $N$ increases as in the question. Clearly $P < … < P(i+1) < P(i) < … < P(1) $ for all $i$. $P(1)$ is straightforward to calculate. It might not be difficult to express $P(i+1)$ in terms of $P(i)$, and if so we have $P$ bounded above by a monotonically decreasing sequence, which must converge, presumably to $P$. – MJD Jun 27 '22 at 14:59
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    I thought I had a formula but the counting is trickier than I thought. The 97% seems likely to me.

    Here's my counting.

    no bad throws - occurs 1 way - 57.9% 1 bad throws - occurs 3 ways (bad throw being 1, 2, or 3) - 24.1% 2 bad throws - occurs 9 ways (bad being 12, 13, 14, 15, 23, 24, 25, 34, 35) - 8.4% 3 bad throws - occurs 37 ways (bad being 123, 124, 125, 126, 127, 128, 134, 135, 136, 137, 138, 145, 146, 147, 148, 156, 157, 158, 234, 235, 236, 237, 238, 245, 246, 247, 248, 256, 257, 258, 345 346, 347, 348, 356, 357, 358) - 3.3%

    Completing after up to 3 bad throws I have 93.7%.

     – Mike O'Connor Jun 27 '22 at 19:03

1 Answers1

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This is more of a comment than an answer - but it's long and doesn't fit as a comment. It does respond to your comment "The only way I can think of to approach this is ...".

  1. Only certain numbers of attempts can get you home. If 0 backwards slides occur, then you arrive home in 3 steps. If 1 backwards slide occurs you get home in 5 attempts (the original 3 + 1 backward slide of 1 step + 1 step you need to recover). If 2 backwards slides occur you get home in 8 attempts (the original 3 + 2 backward slides of 3 steps total + 3 steps you need to recover).

Generally if k backward slides occur the number of attempts to get home is 3 + k + (k*(k+1)/2).

  1. The backwards slides can't occur in just any spot in your safe journey. For instance if there's 1 backwards slide total, it can't occur in attempt 4 or 5. Two ways of thinking of that - a) if it occured in attempt 4, you must have been at least one step away anyway, so the backward slide in 4 would place you 2 steps away and attempt 5 couldn't get you home; b) if it occurred in 4 or 5 that means attempts 1, 2, and 3 were all forward so you would be home already after attempt 3.

  2. The number of possible different attempt paths that get you home after k backslide attempts would be overestimated by using (3 + k + (k*(k+1)/2)) choose k. The top represents the total number of attempts while the bottom represents the backslides during those attempts. What you need to do is remove the overcounting as some of those attempt paths would have resulted in arriving at home already.

  3. The way to remove them is to subtract the following for each lower number k of backslides:

the number of valid attempt routes for i backslides * the number of possible choices in the remaining attempts for k-i choices.

  1. Examples - call f(k) the number of valid attempt routes for k backslides

f(0) = 1 (attempts 1,2,3 are all positive)

f(1) = (5 choose 1) - f(0) * ((5-3) choose 0) = 3

f(2) = (8 choose 2) - f(0) * ((8-3) choose 2) - f(1) * ((8-5) choose 1) = 9

Continuing in this manner I arrived at

f(3) = 37, f(4)=173, f(5)=1402, f(6)=14096.

  1. Multiplying f(k) * probability of k backslides * probability of 3 + (k*(k+1)/2) forward attempts gives us the probability of achieving a safe path with k backslides.

I got the following probabilities for arriving home after exactly k backslides

0 - 0.5787037037

1 - 0.2411265432

2 - 0.08372449417

3 - 0.03319836982

4 - 0.01247627397

5 - 0.006772193099

6 - 0.003800480282

  1. Summing those together I got the following cumulative probabilties of a safe trip after up to k backslides

0 - 0.5787037037

1 - 0.8198302469

2 - 0.9035547411

3 - 0.9367531109

4 - 0.9492293849

5 - 0.956001578

6 - 0.9598020583

  1. I didn't actually program this up but used a calculator and a spreadsheet so I might have made mistakes along the way. If you're interested try coding this technique and let k run much higher. I'm sure you'll see close to the 97% mentioned in the comments.
Mike O'Connor
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