Biased coin

The result only depends on the first trial.
If the first is a head, Alice wins. If it is a tail, Bob wins.
(Why?)
Answer $=60$

Alice can only win if the flips before the head-tail pattern are heads, otherwise Bob would have won before Alice (the consecutive head pattern would be broken by a tails, e.g. $HH \underline{HT} HH$ ). Alice can therefore only win in the trials HT, HHT, HHHT, and so on. The probability of Alice winning is therefore $\displaystyle \left(\frac{3}{5}\right)\left(\frac{2}{5}\right)+\left(\frac{3}{5}\right)^{2}\left(\frac{2}{5}\right)+\left(\frac{3}{5}\right)^{3}\left(\frac{2}{5}\right)+... = \frac{\left(\frac{2}{5}\right)\left(\frac{3}{5}\right)}{1-\frac{3}{5}} = \frac{3}{5}$ .

Bob can only win if the flips before the tail-head pattern are tails, otherwise Alice would have won before Bob (the consecutive tails pattern would be broken by a heads, e.g. $TT \underline{TH} TT$ ). Bob can therefore only win in the trials TH, TTH, TTTH, and so on. The probability of Bob winning is therefore $\displaystyle \left(\frac{2}{5}\right)\left(\frac{3}{5}\right)+\left(\frac{2}{5}\right)^{2}\left(\frac{3}{5}\right)+\left(\frac{2}{5}\right)^{3}\left(\frac{3}{5}\right)+... = \frac{\left(\frac{2}{5}\right)\left(\frac{3}{5}\right)}{1-\frac{2}{5}} = \frac{2}{5}.$

Alice has the greatest chance of winning with a probability of $\boxed{60\%}$ .