It still seems like this is a little flawed... Im assuming the example is given in round 2. The example you gave doesn't really make sense because theres no reason that player A would bet less then 4 since even if both players bet 2 and he lost he'd be eliminated. So he might as well bet 4.
Yes, Player A should have placed 4 instead of 3, but the example is there to show the math of the Raise system, and not to give a likely situation.
Then player B would also know he was betting 4 and could calculate the best possible outcome for him.
This is correct, but there are many cases in the event when players need to judge the strength of their Draw in order to make an optimal bet. The Draw influences the Raise in all of Round 1 and some of Round 2 and Round 3.
I also don't get why you lose more then bet. Unless im reading it wrong.
Players can win or lose more than their Raise, but the current Raise system is the most balanced method.
Basically, there are 8 other possible options, but none of them work as well as the current method.
If players won their Raise and lost their Raise, then their opponent's Raise would have no impact on them.
If players won their Raise and lost their opponent's Raise, then the best option would always be to place the maximum Raise.
If players won their Raise and lost the sum of the two Raises, then players would run out of Elements too quickly.
If players won their opponent's Raise and lost their Raise, then the best option would always be to place the minimum Raise.
If players won their opponent's Raise and lost their opponent's Raise, then their own Raise would have no impact on them.
If players won their opponent's Raise and lost the sum of the two Raises, then players would run out of Elements too quickly.
If players won the sum of the two Raises and lost their Raise, then players would gain Elements too quickly.
If players won the sum of the two Raises and lost their opponent's Raise, then players would gain Elements too quickly.
What if....
Since it still seems like its usually better to bid the minimum...
Well, not really. In Round 1 and sometimes in Round 2/3, the optimal Raise depends on the Draw, and can sometimes be the maximum Raise for that round.
what if for each raise you are above your opponent you get 1 upped card, but they still have the advantage of getting the sum of the bids.
example: player A bids 1, player B bids 2 player B would get the advantage of 1 upped card
player A has the advantage of only bidding 1 while he could win 3
Just something I was thinking about anyways, feel free to correct me if I read that wrong.
In that situation, Player A would not have any advantage. Even though Player A can win 3, Player A can also lose 3, despite only Raising 1.
