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Made in us
Pyromaniac Hellhound Pilot







So a 9 inch charge has about a 27% chance of happening.

With a reroll that improves to 52%? Not sure what a double dice( roll both dice instead of 1 since double dice seems to be hard to understand) reroll would be.

What if you get a +1 on the charge? Should I just count it as an 8 inch charge? Or am I not factoring something in?

Thanks!

If you want to tell me the answers I would not be upset <grin>

This message was edited 1 time. Last update was at 2019/05/09 17:32:19


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Made in us
Awesome Autarch






You cannot double re-roll (to start with). No dice can be re-rolled more than once.

Also, I don't mathhammer, but I believe it would be:

Chance to charge + (Chance to charge x Chance to fail initial roll), etc.

This message was edited 1 time. Last update was at 2019/05/09 17:14:52


 
   
Made in us
Regular Dakkanaut




Odds of a 9 inch charge: 27.78%
Rerolling 1 die: 52.31%
Rerolling both dice: 47.84%
Choice between rerolling 1 or both: 56.94%

For an 8 inch charge (i.e. with +1 to charge roll): 41.67%
Reroll 1 die: 68.06%
Reroll both: 65.97%
Choice between rerolls: 72.45%
   
Made in us
Da Head Honcho Boss Grot





Medicinal Carrots wrote:
Odds of a 9 inch charge: 27.78%
Rerolling 1 die: 52.31%
Rerolling both dice: 47.84%
Choice between rerolling 1 or both: 56.94%

For an 8 inch charge (i.e. with +1 to charge roll): 41.67%
Reroll 1 die: 68.06%
Reroll both: 65.97%
Choice between rerolls: 72.45%


Don't forget

Odds of a charge when you've stacked up every possible charge bonus, stratagem, reroll and modifier available within your codex: 0%.
   
Made in ca
Boom! Leman Russ Commander





London, Ontario

When checking for success, it is sometimes easier to determine the odds of failure, and subtract from one.

For example, going with a straight reroll of both dice, the odds of FAILING a 9" charge are 26/36 = 72.2%.

The odds of failing twice in a row are 0.7222 * 0.7222 = 0.5216%

Thus, the odds of success are 1.0000 - 0.5216 = 0.4784.

If you have the option to reroll a single die, that gets somewhat complicated. It becomes a situation where "What's the highest die I rolled? Ok, so having failed, what are the odds of success now that I know I have a 6,5,4 or 3 to now succeed? If my highest roll is 1 or 2, I can't succeed no matter what."

If you have the option to reroll one or two dice, that is extraordinarily complicated. Based on your highest die roll, you'd have to determine the odds of improvement of rerolling both (28%) vs comparing High Die. (ie: HD 6, rerolling 1 die gives a 67% success).

   
Made in fi
Decrepit Dakkanaut





 greatbigtree wrote:
If you have the option to reroll one or two dice, that is extraordinarily complicated. Based on your highest die roll, you'd have to determine the odds of improvement of rerolling both (28%) vs comparing High Die. (ie: HD 6, rerolling 1 die gives a 67% success).



When it comes to things like this I find it easier to simply do little bit of code that crunch it. Do the dice rolls 1000000 times, check how often you succeeded, odds there close enough.

Sure better ways but here's for example quick code to see odds of 8" charge reroll both or one dice:

Spoiler:

import random


successes=0
tries=1000000
for x in range(0, tries):
x1=random.randint(1,6)
x2=random.randint(1,6)
#print(str(x1)+" "+str(x2))
if (x1+x2)>=8:
successes=successes+1
#print("made it"
else:
if x1<4 and x2<4:
#print("rerolling both"
x1=random.randint(1,6)
x2=random.randint(1,6)
else:
#print("reroll 1"
if x1<x2:

x1=random.randint(1,6)
#print("rerolling x1 to:"+str(x1))
else:
x2=random.randint(1,6)
#print("rerolling x2 to:"+str(x2))

if (x1+x2)>=8:
#print("made it")
successes=successes+1

print(successes)
print(str((successes/tries)*100))

This message was edited 1 time. Last update was at 2019/05/10 06:39:36


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I already did the work for you. https://www.dakkadakka.com/dakkaforum/posts/list/0/766632.page#10225467

This message was edited 1 time. Last update was at 2019/05/10 06:45:14


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