Recall the standard Monty Hall scenario where a presenter hides one prize with high value $h$ behind one closed door (e.g., in the classical version: the one high prize is a car), and, equal prizes with equal low value $l$ behind all other closed doors (e.g., in the classical version: all other low prizes are goats).
The total amount of initial closed doors is $c \ge 3$ (e.g., in the classical version: $3$). The presenter explains (the standard Monty Hall rules): The player will be asked to choose one closed door as hiding the prize to receive (but it stays closed), and, next, the presenter (who knows what's behind each door) will open $1 \le o \le c-2$ doors different from the player's choice (e.g., in the classical version: $1$) which he can choose, and also effectively chooses, to all reveal an equal low value prize.
Then, the presenter will offer the player an alternative: "You get a second chance: either you prefer, and stick to, your first choice, or, you prefer to change your mind and chose another, remaining, closed, door as hiding the prize you receive.".
Everything happens according to the rules, and, the player, who, like many people, heard about the Monty Hall puzzle, informs the presenter to switch and prefer to chose another remaining closed door. "Good choice ... " says the presenter, "compared to alternative, you increased your average prize value with multiplicative factor $1.24$ and with additive term $2400000$ dollars.". The player plays as preferred (and, doing so, indeed can expect exactly higher average prize value as presenter mentioned).
The (one) high prize value ($h$) was the three times the (all other equal) low prize value ($l$).
That is to say: $h=3l$
The amount ($o$) of doors opened by the presenter was even, and, the initial amount ($c$) of closed doors was equal to three times amount of opened doors divided by two.
That is to say: $c=3o/2$
How many doors were there and what was the value of the one high valued prize?