Credit Card Roulette Reverse Draw Questions

#1
My friends and I have been debating the statistics of a credit card roulette game (reverse draw) we play when the dinner bill arrives to see who pays for dinner. This is how the game works: Each friend puts in a different number of credit cards into a hat based on the results from previous draws. For example, there are 3 players in this game. Player 1 puts in 1 card, Player 2 puts in 2 cards and Player 3 puts in 7 cards. Cards are drawn 1 at a time out of the hat until the last card in the hat pays for dinner. What are the odds for each player to be the last card remaining in the hat? What are the odds at each stage of the draw? Is there a statistical formula for this problem? Thanks for your help!
 
#2
Exactly the same as if the first card drawn pays. There are 10 cards, the players have 1, 2 and 7 chances. So 1/10 2/10 7/10
Exactly the same reasoning applies part way through only use the number of cards of each sort remaining in the hat.
 
#5
No, just a debate between friends because every time we play the game it seems like the person who has more cards in the hat doesn't have to pay the bill.