Prob of consecutive loss

You can get a VERY close approximation by taking the probability of a single loss as 0.475.

But there's a mathematical quirk which makes the probability of two consecutive losses equal to something other than 0.475^2.

From any given starting point, you have an equal (0.5) chance of winning one OR MORE in a row and of losing one OR MORE in a row.

For this to be true, then the most likely single event is to win exactly one in a row (0.525 * 0.5); the second most likely event is a tie between winning or losing exactly two in a row (0.525 * 0.475) and (0.475 * 0.525), respectively, and the next most likely event is to lose exactly one in a row (0.475 * 0.5).

Thus, the probability of losing exactly three in a row is Pl * Pw^2, where "Pl" is the probability of losing a single hand and "Pw" is the probability of winning a single hand -- almost exactly 0.525 and 0.475, respectively.

If you put all this out on a spreadsheet, you'll see where the probability of losing "N" in a row accelerates MUCH more rapidly than the probability of winning "N" in a row, wherein lies the house edge.