I think it’s the same as flipping a coin and getting 550 consecutive heads.

50%. The first ballot would have to be for A, and that is a 50/50 chance.

@zenvelo

??

Note “the ballots are counted” and “throughout the ballot count”. This is not about counting only the first ballot.

Also I think my first answer was wrong, but want to hear more possibilities before sharing my new reasoning.

@Call_Me_Jay But the first ballot must be A. I guess the second has to be A also. But after that it doesn’t matter as long as B’s don’t catch up and pass the A’s.

but wait….you don’t mention if the ballots are being counted on a moving train going 87mph and approx. 32 minutes until its arrival at the station with the ballot counting completion due in 26.5 minutes and whether or not the ballot counter has slight ADD and a limp.

“the first ballot must be A”

??

Why would they be in any particular order?

@Call_Me_Jay Because …A will be in the lead throughout the ballot count.

Throughout means it must be from the start to the end.

@zenvelo I see, thank you.

Also I rescind my first answer, it’s totally wrong. For some reason I leaped to the idea of counting 550 consecutive “A” votes, not simply being in the lead.

I now think @zenvelo is correct, 50%.

I would guess the chances are very slim one in many thousand. B would have a great many opportunities to get ahead even if he loses in the end.

The answer turns out to be 10%. My intuition was that the probability would be very small.

In the general case where A gets p votes and B gets q votes, the chances of keeping the lead is the same as the winning percentage, (p – q)/(p+q). Here is the Wikipedia article on it.

There is no intuitive way that I can think of as to why the formula is true, although if you know basic combinatorics, the mathematics is not too bad using the reflection method. This article gives several proofs, only the first of which I bothered with. Unlike the Wikipedia article, it includes a useful picture showing how the reflection principle is used. You might find the reasoning of interest even if you can’t follow all the math.

Here is a brief summary.
As @zenvelo points out, A has to get the first vote. We first compute the total number of outcomes where A gets that first vote.

Then an interesting graph is produced, showing A’s lead after each vote, which will be 0 if the vote is tied and negative if B takes the lead.

We want to subtract all the cases where A gets the first vote, but the vote graph touches the 0 line. Using the graph, the article shows that this is exactly the same as the total number of cases where the final vote is the same but where B gets the first vote. We can compute this number and subtract from the first and, after a little algebraic manipulation (not shown), end up with the final answer.

One thing that is intuitive is that, for a fixed number of total votes, the chances of maintaining the lead should increase when the winning percentage increases. Someone who wins by 90% should have a better chance of maintaining the lead than someone who wins by 1%. It is just unintuitive that the numbers should always be the same.

9.9 percent chance

Well, this was interesting, thank you @LostInParadise .

It is just unintuitive that the numbers should always be the same.

Are you saying they are? Because using the formula below with A:B = 900:100 I get 80%.

(p – q) / (p + q)

(900–100) / (900+100)

(800) / (1000) = 80%

Yes that is correct.