[...] So if you try to predict 6 results in a row using a device with 2 possible results (a coin), the formula is (1/2)^6 = 0.015625 = 1.5625 % . But after you get the first correct result, your goal is only for the 5 remaining predicted results, so the new formula is (1/2)^5 = 0.03125 = 3.125%.
If you are lucky enought to get the first 5 results as predicted, the last throw formula will be this one: (1/2)^1 = 0,5 = 50%.

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Btw:
This may be by far the most off topics argument I ever touched in this forum