p(th) = p(t) p(h)
p(ht) = p(h) p(t)
Hence p(th) = p(ht) regardless of coin imbalance as long as both events actually will happen. QED.
p(th) = p(t) p(h)
p(ht) = p(h) p(t)
Hence p(th) = p(ht) regardless of coin imbalance as long as both events actually will happen. QED.