to get the exact chance of winning with each decision, but is there a general principle you can extract that would allow you to more easily figure out the best move in similar situations?
Button Men Game #599998
Tournament Legal challenge, copying communication from game 599661
ElihuRoot is performing a power attack with a 20-sided die showing 9,
targeting Desdinova's 12-sided die showing 7.
The 20-sided die is rerolled and the new value is 14.
ElihuRoot takes Desdinova's die.
| Your Dice: | Opponent's Dice: |
| 20-sided die showing 14 20-sided die showing 1 | 8-sided die showing 6 Option 10/20 10-sided die showing 7 |
Rounds Won/Lost/Tied: 1 / 0 / 0 (Out of 3 wins)
Your Score: 72 (15.3 sides)
Opponent's Score: 49
5 comments:
Since you need to take both his remaining dice to win, he wants the highest number possible on his lower die.
If he takes your die with the 8-sider:
25% chance he rolls a 7 or 8, giving him a 35% chance of winning;
12.5% chance he rolls a 6, giving him a 30% chance of winning;
12.5% chance he rolls a 5, giving him a 25% chance of winning;
and 12.5% chance each of 1-4, with the obvious 5% to 20% chance of winning for each one.
This gives him slightly less than a 22% chance of winning (if I did the calculations correctly).
If he takes with the 10-sider:
50% chance he rolls a 6 or higher, giving him a 30% chance of winning;
10% chance he rolls a 5, giving him a 25% chance of winning;
10% chance he rolls a 4, giving him a 20% chance of winning;
and so on fo 3 through 1.
This gives him a 22.5% chance of winning.
While not a much better chance (0.625%), taking your die with the 10-sider is the better move.
As a general set of rules for this sort of situation, I believe that if both dice have at least a 50% chance of rolling the lower number or higher, the die showing the lower number should be rerolled; otherwise, the one with the higher chance of rolling the lower number or higher should be rerolled.
This is a critical game move here; it happens over and over in the end game. Your goal is to have the highest possible lower value.
What's a good general principal? Consider each die as coming up with the average value. A D10 will reroll as 5.5, and a D8 will reroll as 4.5.
So your choices are:
reroll D8: 7 & 4.5
reroll D10: 6 & 5.5
Rerolling D10 gives a higher lowest value.
Both comments have the essential thing right -- D ought to to maximize the expected value of his second-largest value. (that expected value is directly proportional to his chance of winning here -- if my 20-sided die had a value of 8 or 9, i.e. smaller than the maximum possible value of the reroll, the situation would be a little more complex, though it wouldn't change the decision in this case)
this basic idea is well-known, see for example, the very first strategy article
http://www.cheapass.com/bpu/strategy1.html
gryphon does the full computation, quite correctly (no need for me to repeat it, is there? except my own obsessive demons:
if you reroll the 8:
the second lowest number will be 1-6 each with probability 1/8, and 7 with 2/8 probability. expected value: 35/8, i.e. 4.375
(and your chance of winning is 4.375/20, 21.875%)
if you reroll the 10: the second lowest number willbe 1-5 each with probability 1/10 and 6 with probability 1/2, for an expected value of 4.5 (and your chance of winning is 4.5/20, 22.5%)
devious is right about the goal, and, as an approximation, his formula of using the expected value of the rerolled die as a stand-in for the die isn't too bad -- and it's certainly a lot quicker than the full computation! -- but it's not 100% accurate.
For example, even if the D8 were showing a 5 in the original example, instead of a 6... that rule of thumb would still say to reroll the 10-sided die. But, in fact, in this case the 8 sider is the one to reroll! (it still gives the same 21.875% chance of victory, but rerolling the 10 sided die only gives 20%)
these numbers are all pretty darn close! We can probably cook up a more extreme example, but in practice, I think, the hard decisions come when they ARE close...
Oh... I should point out that the last paragraph of Gryphon's comment (it beings... "as a general rule") is also just an approximation -- the first part (everything up to the word "otherwise") is fine, but the second part is not totally accurate (and my counterexample above applies in this case, too).
actually, maybe even the first part isn't entirely accurate! example:
opponent has a die showing 1 and a 30-sider showing 21. you have a 4-sider showing 3 and a 20-sider showing 4. You need to capture both his dice to win...
The condition:
"both dice have at least a 50% chance of rolling the lower number or higher" is met.
yet, if you reroll the 4-sided die, your lower number will be 1-4 with probability 1/4 each, expected value 2.5; if you reroll the 20-sider your lower number will be 1-2 with probability 1/20 each and 3 with probability 18/20... expected value 2.85.
so the conclusiong "the die showing the lower number should be rerolled" is not true in this case
If you try to strengthen this by something like: if both dice have a t least a 50% chance of rolling HIGHER than the lower number, then reroll the lower die, it's closer, but still not quite true, though to make the point clear it's easiest to use outlandishly large dice: say you have a 20-sided die showing 10, and a 1000-sided die showing 11. both dice have at least a 50% chance of rolling 11 or more; yet if you reroll the 20-sided die, your expected value for the smaller die is only 33/4, i.e. 8.25. if you reroll the 1000-sided die, your expected value for the smaller die is 9.955.
actually, you could replace that 1000-sided die by a 26 sided die (or bigger) showing 11 and the point would still be true.
smaller examples that mirror this phenomenon:
a 10 sided die showing 5 and a 20-sided die (or bigger) showing 6
a 12 sided die showing 6 and a 20-sided die (or bigger) showing 7
a 24-sided die showing 12 and a 30-sided die (or bigger) showing 13...
[the formula for this very special case is a 2n-sided die with value n and any die with at least 2n*(n-1)/(n-3) sides, and value n+1.]
But, I do think, as an approximation, it's still accurate, these exceptions are fairly pathological -- a few of them MIGHT come up rarely -- but even in those cases, the "general rule" is only a little bit suboptimal.
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