clearly increases the chance I have to reroll my 20-sided die; but if I *do* have to reroll my die, the former has a much better expected value of the lower die.
Or am I giving it away even by phrasing it this way?
(oh, let me point out the previous post was only published today, even though it says "October 30" -- the draft was dumped onto the site way back then, but I didn't get around to reviewing it until just today!)
Viewing Game #621844
This game was last modified on Mon Dec 17, 2007 16:13:27
* Next Turn * Player: Anders *ONE MILLION POINTS* *Fanatic* * Next Turn *
Button Man: Yuranosuke Score: 10 (-22 sides) Rounds Won/Lost/Tied: 1 / 1 / 0 (out of 3 round(s))
Captured Dice: Poison 16-sided die, X Swing 6-sided die
| Dice | 12-sided die | Focus 12-sided die | |
| Value | 6 | 10 | |
| Dice | 8-sided die | 8-sided die | 20-sided die |
| Value | 2 | 7 | 20 |
Button Man: Shepherd Score: 43 (22 sides) Rounds Won/Lost/Tied: 1 / 1 / 0
Captured Dice: X Swing 13-sided die, 8-sided die, 4-sided die
Results from last turn:
ElihuRoot is performing a power attack with a 8-sided die showing 2,
targeting Anders's 4-sided die showing 2.
The 8-sided die is rerolled and the new value is 7.
3 comments:
How much better is the expected value of your lower die if you reroll the [6/12] vs the [10/12]? Ignoring the [7/8], it looks like 6.25 vs 4.75, so I guess that is pretty significant, although it only translates to a 7.5% increase in your chance to catch the re-rolled [20]. (The difference is obviously more significant if the die you're trying to catch is smaller.)
Of course, you can't ignore the [7/8]... Is the right adjustment to multiply the expected value times the chance of keeping the re-rolled die? Because if so, the numbers are 2.60 (6.25 times 5/12) vs 3.96 (4.75 times 10/12), which gives a substantial edge to rerolling the higher die.
Is this the right math, though? Thinking about it further...
maybe expected value is a bit misleading here, especially since I voiced it based
on different conditional circumstances.
look at it this way:
if he rerolls the 12-sided die with value 6:
7/12 of the time he rolls 1-7 and I take it with my 8-sided die, game over.
1/12 of the time he rolls 8, and has 8/20 chance of winning
1/12 of the time he rolls 9 and has 9/20 chance
3/12 of the time, he rolls 10,11,12, his smallest die is a 10, he has a 50-50 shot.
overall: his chance of winning is: 1/12*(8/20 + 9/20) + 3/12* 10/20
that's (8 + 9 + 30)/240 = 47/240 (19.58%)
on the other hand, if he rerolls the 12-sided die with value 10, capturing my
8-sided die with value 7:
2/12 of the time he rolls 1-2, game over
1/12 of the time he rolls 3, he has 3/20 chance of winning
1/12 of the time he rolls 4, he has 4/20 chance of winning
1/12 of the time he rolls 5, he has 5/20 chance of winning
7/12 of the time he rolls 6-12, his smallest die is 6, he has 6/20 chance
that's 1/12 *(3/20 + 4/20 + 5/20) + 7/12*6/20
(3+4+5+42)/240 = 54/240 (22.5%) -- this is the better choice, but it's pretty darn close!
if my 8-sided die with value 2 instead had value 4 at the beginning-- or if HIS 12-sided die with value 6 had value 5 -- his two choices would have been exactly equal good for him!
oops! coming back to the problem last night, I completely spaced and wrote nonsense --
the computations above are all correct computations of the probability of being able to capture the 20 sided die on the very next turn -- but -- of course -- that's not the same thing as winning at all!
(A) even that DOES happen, he might reroll too low and get captured by the last remaining 8 sided die
(B) even if that DOESN'T happen, he could capture the remaining 8 sided die and reroll high enough to capture the 20 sided die, if it at least rolled less than 11.
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