Easy Math help (for some of you)
 

Can someone explain how to calculate the probability of the following events?

Let's say there are 10 people and they are going to draw a card number 1 to 10. 1 wins, 2 comes in second, 3 third on so on.

Being that there is no skill involved I think it is correct to say that all players have a 1 in 10 chance or 10% chance of winning. (Or finishing in any specific spot) right?

If the same players where to repeat the same game would the odds of the same player winning be 1 in 100 or 1%? (10 * 10)

If the second time there were 20 players and 20 cards would the odds of the first player winning again 1 in 200 or .5%?

If there is a third or forth game do you just keep multiplying?

Thanks

Re: Easy Math help (for some of you)
 

You have to be very careful how you phrase questions like this.

If the same game were played a second time the person who won the first game would have a 10% chance of winning because his first win is already a given.
Even if you asked "What are the odds that one player wins twice in a row" the answer would STILL be 10% because someone must win the first game.

To get the 1% you're looking for you must ask "starting now, what are the odds player 1 wins the next two games?"
If there were 20 cards and 20 people playing, your odds on each game is 1/20 and yes, you keep multiplying that number by itself to determine streaks. So game two you'd win 1/400 times.