Bounder game explanation needed

[Mrakvampire]

Hi All!

Maybe I'm not so good in English, so I need some help to understand rules of Bounder casino game presented in first adventure.

Example. We have 2 players and one dealer.
All make their stakes.

Step 1.
Player 1: rolls 1d20, gets "14"
Player 2: rolls 1d20, gets "2"
Some double stakes, some do not, it's not important.

Step 2.
Dealer rolls 3d6, gets "11"

Step 3.
Player 1: rolls 1d20, gets "2"
Player 2: rolls 1d20, gets "10"

Who wins?
As I understand winner should be Player 1, but I'm not 100% sure. :)

[Joana]

Quote:

To start, each player bets a stake (minimum 1 sp). Each player rolls his first d20, making his "point." After all players have rolled their points, each player may double his stake, if desired.

So here's your step 1.

Player 1: 14
Player 2: 2

Quote:

Then the dealer rolls 3d6. Anyone whose point the dealer matches loses his stake.

Step 2.

Dealer: 11

This matches no one's point, so no one is out of the game yet.

Quote:

Then each player rolls his second d20. If the player's two dice results are on either side of the dealer's result -- one greater than and one less than the dealer's number -- he "bounds" the dealer and wins an amount equal to the amount he bet. Otherwise, he loses his stake.

Step 3.

Player 1: 2
Player 2: 10

Player 1 has 14 and 2, which are greater than and less than the dealer's result of 11, so he wins.

Player 2 has 2 and 10, which are both less than the dealer's result, so he fails to "bound" the dealer and loses his stake. If his second roll had been a 12, he would also have won.

Second example

Player 1 1d20 ⇒ 9
Player 2 1d20 ⇒ 12
Player 3 1d20 ⇒ 8
Player 4 1d20 ⇒ 9

Dealer 3d6 ⇒ (3, 5, 3) = 11

Player 1 1d20 ⇒ 4
Player 2 1d20 ⇒ 3
Player 3 1d20 ⇒ 9
Player 4 1d20 ⇒ 7

Luck's with the house this round. Only player 2 "bounds" the dealer's result of 11, with a 12 that's greater than 11 and a 3 that's less than 11. Everyone else loses.

Does that make it clearer?