Well, assuming it’s true, and assuming the best case – that the cleanse occurs before Death Mark would otherwise kill the troop – what that means is there’s a 10% chance for the troop to have 100% chance to survive, and 90% chance for the troop to have 90% chance to survive. That means that on turn 1, any given troop afflicted with Death Mark would have a chance of (0.1 * 1) + (0.9 * 0.9), or 91% chance to survive, instead of the 90% I was assuming before.
That means that the chance of all 4 troops surviving one round of Death Mark is (0.91)^4, or about 68.5%. Still works out to about 31.5% chance of losing one troop before you can do anything about it, which is much worse than the odds you get with just about everything else.
If you don’t lose a troop on turn 1 and you cannot cleanse, then the odds for all troops to survive round 2 become: ((0.1 * 1) + (0.9 * ((0.2 * 1) + (0.8 * 0.9))))^4, or about 0.74, which is 74%. That means the total odds of surviving two rounds without losing a troop are (about) 50%. There’s a 50% chance you’ll lose at least one troop in two rounds.
EDIT: Forgot to factor in the “cleansed in turn 1” factor to the turn 2 survival rate.
EDIT 2: Was corrected on Death Mark’s 10% odds not increasing over time. Lowers the number a lot, but still 50%.