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2019/05/09 17:08:42

Subject: Help me with my Mathammer: Chances of a charge

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

.Only a fool believes there is such a thing as price gouging. Things have value determined by the creator or merchant. If you don't agree with that value, you are free not to purchase.

2019/05/09 17:13:59

Subject: Help me with my Mathammer: Chances of a charge

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%.

2019/05/10 03:59:06

Subject: Help me with my Mathammer: Chances of a charge

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).

2019/05/10 06:38:01

Subject: Help me with my Mathammer: Chances of a charge

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: