Author Topic: Rerolling: Some Numbers  (Read 4502 times)

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nerd65536

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Re: Rerolling: Some Numbers
« Reply #20 on: June 25, 2011, 03:29:40 AM »
That's way more complicated than it needs to be.
Choosing the highest die out of ndm, the probability of getting a k is:

Which makes the Expected Value (average):


While it looks complicated, we can just plug in to Wolfram Alpha, like so:
2d20=13.825
3d20=15.4875
4d20=16.4833375

4d6=5.2446...

Bauglir

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Re: Rerolling: Some Numbers
« Reply #21 on: June 25, 2011, 01:12:43 PM »
Huh, so it is. Well, that's handy, anyway. Odd, I could've sworn I tried that before and got probabilities that didn't add up to 1, but I must've had a typo when I tried to check it. Or something, since it clearly works.
So you end up stuck in an endless loop, unable to act, forever.

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nijineko

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Re: Rerolling: Some Numbers
« Reply #22 on: June 25, 2011, 01:25:30 PM »
[spoiler][/spoiler]

irrelevant perhaps, but i was tolerably amused to see that "dnd" at the beginning of the formula.
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Shiki

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Re: Rerolling: Some Numbers
« Reply #23 on: June 25, 2011, 01:37:11 PM »
^Intentional for sure would be my guess. :p
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Bauglir

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Re: Rerolling: Some Numbers
« Reply #24 on: June 25, 2011, 02:39:55 PM »
[spoiler][/spoiler]

irrelevant perhaps, but i was tolerably amused to see that "dnd" at the beginning of the formula.

I also was, but I believe that was actually a coincidence arising from the notation chosen. She didn't seem to have noticed when I pointed it out the other day, anyway.
So you end up stuck in an endless loop, unable to act, forever.

In retrospect, much like Keanu Reeves.

Saxony

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Re: Rerolling: Some Numbers
« Reply #25 on: June 25, 2011, 05:16:48 PM »
Interesting quirk:

+4 works best when you're working hard or hardly working to succeed and reroll works best when you're close to 50% chance of winning or when you only fail on a natural 1 and natural 1 = automatic failure. But the methods are roughly equal in terms of how often they'll work best, just in different ways. If you expect either Team Player or Team Monster to face heavy opposition, go with +4. If you expect roughly even chances, go with reroll.

A reroll works better when your target number is in the middle of the RNG (in this case, 5.52 to 14.47, or 5-6 to 14-15), and a +4 works better on the ends of the RNG (in this case, 1 to 5-6 and 14-15 to 20). However, the two ranges are roughly equal, taking up half of the RNG each. The +4 allows you to win in circumstances where you normally couldn't (you'd have to roll a 22, for instance) and the reroll can get you out of natural 1s. So, where natural 1 = failure, the reroll method is better than the +4 method in a slightly larger range.

Example: Let's say you win on a 19 or 20. Give yourself a +4. Now you win on a 15,16,17,18,19, or 20. Your 10% chance of winning improved to 30%. Give yourself a reroll. Now your chance of success is (1-[18/20]2) = 19%. 30 > 19.

Another example: Let's say you would have to roll a 21 to succeed. With a reroll, your chance of success stays at 0%. With a +4, now you have to roll a 17,18,19, or 20, your chance of success improved from 0% to 20%. 20 > 0.

Another example: Let's say you win on a 2 or better (95% chance to win). Give yourself a +4. Now you win 100% of the time. Give yourself a reroll. Now you win 99.75% of the time. 100 > 99.

Another example: Target number = 10. Normal winning chance = 50%. Give yourself a reroll. Now you have a 75%. Give yourself a +4. Now you have a 70%.

Justification / Proof thing:

Let's find out the probabilities of success for these two methods and solve for where they provide the same benefit.

Where x = number of die faces you can roll and win,

Normal probability of success = x/20

Probability of success for reroll = 1- ([20-x]/20)2 = 1 - (20-x)2/400

Probability of success for +4 = (x+4)/20

Set them equal and solve for x
1 - (20-x)2/400 = (x+4)/20
400 - (20-x)2 = 20x + 80
400 - (20-x)(20-x) = 20x + 80
400 - 400 +40x - x2 = 20x + 80
40x - x2 = 20x + 80
20x - x2 = 80
20x - x2 - 80 = 0
-x2 + 20x - 80 = 0
x2 - 20x + 80 = 0
(x-5.52)(x-14.47) = 0

The above equation is true when x = 5.52 or 14.47.

Thus, the two methods provide equal chance of success when the target number is 5-6 or 14-15. Reroll works best when the target number is in the middle of the RNG and +4 works best on the ends of the RNG.
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Bauglir

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Re: Rerolling: Some Numbers
« Reply #26 on: June 26, 2011, 05:08:38 PM »
Yeah, an interesting quirk that charging for a +4 overlooks is that every additional die you can roll skews the numbers ever more toward high values, which has its own merits that are lost in averaging to a flat number. It's a bit of a naive assessment, but it seemed fair enough for the resulting pricing, at least subjectively. But as your analysis suggests, the +4 is better in some cases (it makes anything lower than the equivalent of a 5 flat-out impossible and increases the range of possible successful numbers at the top end of the scale).
So you end up stuck in an endless loop, unable to act, forever.

In retrospect, much like Keanu Reeves.