Nine upticks, four downticks, always in the green

Question

Over a session a stock records 9 up-ticks (each $+1$ ) and 4 down-ticks (each $-1$ ). The 13 ticks arrive in a completely random order.

What is the probability the running cumulative total is strictly positive after every tick?

Accepted Answer

The cumulative total is a random walk of 9 up-steps and 4 down-steps, and we want it to stay strictly above zero from the first tick onward. This is the ballot problem, with the up-count and down-count playing the roles of the two vote totals: P(	ext{always strictly positive}) = \frac{u - d}{u + d} = \frac{9 - 4}{9 + 4} = \frac{5}{13} \approx 0.385. The first tick must be an up-tick (chance 9/13) just to start positive, and holding above zero afterward reduces it to 5/13. Takeaway. Any +1 and -1 sequence obeys the ballot theorem: staying strictly positive throughout has probability (u - d)/(…

Nine upticks, four downticks, always in the green

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