The drunkard between home and the cliff

Question

A drunkard stands on a straight path. Home is $3$ steps to his right; a cliff edge is $7$ steps to his left. Each step he takes is equally likely to be one step left or one step right, independently. He stops when he reaches home (safe) or the cliff (not).

What is the probability he reaches home before falling off the cliff?

Accepted Answer

Put the drunkard at position 0, home at +3, cliff at -7. Each step is \pm 1 with equal probability, so his position is a fair random walk, a martingale. Optional stopping conserves the expected position: it starts at 0 and must equal its expected value at the moment of absorption. Let p be the probability he reaches home (+3) first; then with probability 1 - p he reaches the cliff (-7). Setting the expected stopping position to the starting position: (+3)\,p + (-7)(1 - p) = 0 \implies 3p - 7 + 7p = 0 \implies 10p = 7 \implies p = \frac{7}{10}. He reaches home with probability 	frac{7}{10} an…

The drunkard between home and the cliff

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