Easy | Probability |

We have a weighted coin which shows a Head with probability p, (0.5<p<1). How do we get a fair toss from this? That is, how do we toss this coin in such a way that we can have probability of winning = loosing = 50%?

Hint

Clearly we cannot have a fair toss in a single flip of this coin. But by tossing this coin 2 times, we can assign the outputs to "win" or "loose", such that we have a 50% chance for both.

Answer

Toss 2 times, mapping HT to "win", TH to "loose" and repeat the process otherwise.

Solution

Toss the coin twice. If consecutive Heads-Tails appears (HT), we "win". In case of (TH), we "loose". If (TT) or (HH) appears, repeat the process. Probability of infinite repetition (p*p*.....) + (1-p)*(1-p)*.... = 0 + 0 = 0;

Probability of H-T and T-H is equal hence it's fair now.

I proposed a faster method, "lets keep tossing the coin to form a sequence of H's & T's . I win if HT appears before TH" . Was I bluffing?

Actually yes, the probability of HT before TH is just p in second game.

Probability of H-T and T-H is equal hence it's fair now.

I proposed a faster method, "lets keep tossing the coin to form a sequence of H's & T's . I win if HT appears before TH" . Was I bluffing?

Actually yes, the probability of HT before TH is just p in second game.

Source: Common

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