Comparing strings from the back?

[Johannes Bauer]

On 04.09.2012 04:17, Steven D'Aprano wrote:

> On average, string equality needs to check half the characters in the 
> string.

How do you arrive at that conclusion? When comparing two random strings,
I just derived

n = (256 / 255) * (1 - 256 ^ (-c))

where n is the average number of character comparisons and c. The
rationale as follows: The first character has to be compared in any
case. The second with a probability of 1/256, the third with 1/(256^2)
and so on.

Best regards,
Johannes

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