Probability problem, from 50 to 100

[fredrikj]

Let's hope someone here can solve this :)

Start at 50. Repeat: Subtract or add 1 depending on the outcome of throwing a 2-side die.

How many times will you likely have to throw it to reach 0 or 100?

Testing with a computer program gave an average of 2500 throws, but what's the exact number where the probability of getting 0 or 100 reaches or exceeds 1/2 (is that correctly expressed, btw)?

[wblipp]

Those 2-sided dice are hard to find - I usually substitute a coin flip.

2500 is correct for the average number of trials to first exceed the bounds.

There is some ambiguity in your question. Is it sufficient to have exceeded the bounds at least once, or do you want the probability of actually exceeding the bounds at the final point in time.

To clarify. consider a scenario that was at 110 on turn 600, but is at 85 on turn 1200. When we count up the probability of your event at 1200, do you want to include this because the bounds were exceeded, or do you want to exclude it because the bounds are not presently exceeded?

For the included case - the "sticky boundary" problem where ever reaching the boundary means that you have ever-afterward reached that boundary, the probabilty in 1893 turns is 0.4998 and the probability in 1894 turns is 0.5003

However, the probability is pretty high that at exactly 1894 turns things will have returned to inside the bounds - 97.7%. If you want the other problem - the point at which you can first say the probability of being outside the boundaries ON THIS TURN is greater than 50%, that answer is 21,546.

All values figured using MathCad to represent the transition matrix and playing with number of transitions.

[fredrikj]

Those 2-sided dice are hard to find - I usually substitute a coin flip.

I prefer to be abstract ;)

Now to clarify: what I'm looking for is simply the mathematical explanation why 2500 is the average number of trials it takes to first exceed the bounds.

[smh]

Those 2-sided dice are hard to find - I usually substitute a coin flip.

But a coin could fall on it's edge :D

[cheesehead]

Originally Posted by smh

But a coin could fall on it's edge :D

Use an edgeless coin. :)

Originally Posted by fredrikj

what I'm looking for is simply the mathematical explanation why 2500 is the average number of trials it takes to first exceed the bounds.

The coin's "random walk" takes it sqrt(number of steps) away from the origin, on average.

If the bounds were -60 and +60 and you started at 0, the average number of trials would be 3600.

[fredrikj]

Originally Posted by cheesehead

The coin's "random walk" takes it sqrt(number of steps) away from the origin, on average.

If the bounds were -60 and +60 and you started at 0, the average number of trials would be 3600.

Ah, that makes sense. Thanks!