[Sorcerer] A weaker opponent gives me fewer rollover dice?
rabindranath72:
Quote from: Ron Edwards on September 03, 2008, 01:25:41 AM
10, 9, 9, 9, 8, 7, 5, 2, 1, 1 vs. 10, 9, 9, 9, 7, 5, 4, 3, 3
A wins with one victory
Hi Ron,
shouldn't the above example be 3 victories for A? The 10, 9, 9, 9 cancel.
Then we match 8 vs. 7, which is one victory for A. But then, we need to look next to see if B "stops the fall" of victories, and we find 7 vs. 5, for another victory for A. Then, 5 vs. 4, another victory for A. Then 2 vs. 3, so the process stops. ?-
rabindranath72:
No ok, my bad. I misread the sequence.
Ron Edwards:
It seems like you're still doing Risk, or reminiscent thereof. I should state it right out:
Once you've hit the relevant high value for the losing side, stop.
Does that help?
Best, Ron
rabindranath72:
I guess I have it right.
The algorithm would be:
1) Sort the values for both sides from highest to lowest
2) Discard ties
3) One of the two sides has an highest die; that side is the winner.
4) Count the victories: count the number of dice that the winning side has which are larger than the highest die of the losing side.
So:
A = 10, 9, 9, 7, 5, 1
B= 10, 9, 8, 1
Remove ties:
A = 9, 7, 5, 1
B= 8, 1
A wins, with one victory (only 9 is larger than all the dice remaining to B)
Is this correct?
Cheers,
Antonio
rabindranath72:
Quote from: rabindranath72 on September 03, 2008, 04:27:25 AM
A wins, with one victory (only 9 is larger than all the dice remaining to B)
Well, actually the sentence above is redundant, we only need check the highest die rolled by B.
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