In the Monte Hall door scenario, if Monte picked the door randomly like you did in the card game then the player would lose the game 2/3 and win 1/3 of the time when switching. Why? The player can lose because of the the first pick or if not, then by swicthing. The player would instantly lose the 1/3 of the time Monte randomly picked the prize door first. When Monte didn't pick it first then the player now has a one of two chances when switching. So overall, 2/3 of the time the player would still have a 1/2 chance of loss=1/3 for a total of 1/3 + 1/3=2/3 total chance of loss.
When Monte specifically picks a non-prize door first (as in the non-random orginal version classic problem) it eliminates the "first pick" loss chance and leaves the player with only the 1/3 chance of loss when switching.
In your card game the player would instantly lose if you happened to pick an Ace. You left the player the chance of the "first pick" loss along with the potential loss from switching so it is equivalent to the Monte Hall random pick scenario.This is NOT the original version of the problem. Under your card game scenario, the chances of loss at the end must be 50%. You could re-configure the card game so there was no immediate player loss from you picking an Ace. So if you found an Ace, you would replace it without showing. Then when it got down to one card, the player would have 90% chance of winning by switching.
