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When rerolling one or both dice, how do you know? 2d6 Charge


d36williams

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It looks like when rolling 2d6 for charges, Black Templars will be able to reroll one or both dice.

 

I'm trying to work out the best strategy for knowing what to reroll. This is what I came up with:

  • if you get a 3 & 2, rerolling the 2 gets you another 17% chance, and rerolling both gets you another 28% chance
  • if you get, 4 & 3, rerolling 3 gets you another 33%, rerolling both gets you another 28% chance
  • if you get 5 & 3, rerolling 3 gets you another 50%, rerolling both gets you another 28% chance
  • if you get a 4 & 4, rerolling 1 gets you another 33% chance, rerolling both gets you another 28% chance

It looks like 4 is the magic number, where it's better to reroll any dice less than 4, but not equal to or greater than 4. Except when both are 4, then reroll 1 of them.

 

If you're curious how the odds improved, rerolling both dice per the old Chapter Tactic was a 48% chance affair. Rerolling one or both improves that to 56%

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I wrote a little anydice simulation for that: https://anydice.com/program/1709e 
 

You want to look at the "combo" function. It's the result that gets show at the bottom.

First it checks whether you rolled high enough to make the charge, in this case a 9 (to change that simply replace every 9 in the output with whatever you require to roll to make the charge). If you didn't roll high enough, it checks whether both are a 3 or less, in which case it re-rolls both dice. If only one dice is a 3 or less it re-rolls the lowest.

This will give you the highest chance of success, which is 56.94% without any other modifier.

Edited by sfPanzer
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A 7" charge is actually easy to calculate, because it's the number you get 50% of the time when rolling 2d6.

 

  • if you have 3 & 3, rerolling one or both give the same 50% second chance
  • don't reroll a 4
  • don't reroll a 3 if you don't have too, otherwise do

 

In general, you can take the charge range, half it, round down, that is the magic number.

  • 10" charge, 5 magic number
  • if you have a 4 & 4, rerolling 1 gives you 16.5% chance, rerolling both 16.67% (so close its uncanny) (so same difference)
  • reroll anything less than a 5

 

3" if you failed a 3" charge you got double 1s and need to reroll both

4" reroll anything less than a 2

5" rereoll anything less than a 2 and a single 2 if you have to

6" reroll anything less than a 3

7" charge reroll anything less than 3 and a 3 if you have to

8" charge reroll anything less than 4

9" charge reroll anything less than 4, and 1 4 if you have to

10" charge, reroll anything less than 5.

11" charge, reroll anything less than a 5, and 1 5 if you have to

12" charge, reroll anything less than a 6.

 

That's a neat AnyDice script sfPanzer! Where did you learn it's syntax?

Edited by d36williams
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That's a neat AnyDice script sfPanzer! Where did you learn it's syntax?

 

Well I'm a programmer so the syntax wasn't hard to figure out from some other examples I've found online. ^^

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Really just depends on what you roll the first time and going off the results and the situation you are trying to focus on. The one BT player at my store is seriously considering planning on running a few BT Drop Pod units, so at the very least he is going to need to roll a 10 in order to make a Turn 1 Charge.
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Why 10? It's 9" for a deep strike charge.

 

Edit' I found the FAQ which refers:

 

"Q: A number of abilities allow a unit to arrive during the battle and be set up more than 9" from any enemy models. If I use such an ability to set up as close as allowed towards an enemy unit and then select it as the target of the charge, what is the minimum charge distance I need to roll to make a successful charge (assuming no modifiers)? A: 9."

Edited by Brother Adelard
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This seems gamey to me, but can you re-roll one dice and then afterwards decide to re-roll the other?

 

Eg. You need to make a 9 on a charge roll.

First roll is 2 and 5. Re-roll the 2, get a 3.

Now you have 3 and 5. Can you re-roll the 5 at that point and hope you get a 6?

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This seems gamey to me, but can you re-roll one dice and then afterwards decide to re-roll the other?

 

Eg. You need to make a 9 on a charge roll.

First roll is 2 and 5. Re-roll the 2, get a 3.

Now you have 3 and 5. Can you re-roll the 5 at that point and hope you get a 6?

I think that counts as two separate rerolls, so I'm going to go with no.

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d36williams, your maths are slightly off, actually...

 

A 7" charge is actually easy to calculate, because it's the number you get 50% of the time when rolling 2d6.

  • if you have 3 & 3, rerolling one or both give the same 50% second chance
  • don't reroll a 4
  • don't reroll a 3 if you don't have too, otherwise do
Your chances of rolling 7+ on 2d6 are slightly above 50%, actually (21/36 = 58.33%), which means you should reroll both dice if your highest is anything less than 4.

 

3" if you failed a 3" charge you got double 1s and need to reroll both

4" reroll anything less than a 2

5" rereoll anything less than a 2 and a single 2 if you have to 3

6" reroll anything less than a 3 4

7" charge reroll anything less than 3 and a 3 if you have to 4

8" charge reroll anything less than 4

9" charge reroll anything less than 4, and 1 4 if you have to

10" charge, reroll anything less than 5.

11" charge, reroll anything less than a 5, and 1 5 if you have to

12" charge, reroll anything less than a 6.

 

That's a neat AnyDice script sfPanzer! Where did you learn it's syntax?

The easiest way to look at it is by looking at your odds:

3+ : Odds on 2d6 are 35/36. There's no way to beat those odds on 1 dice, so always re-roll both.

4+ : Odds on 2d6 are 33/36. There's no way to beat those odds on 1 dice, so always re-roll both.

5+ : Odds on 2d6 are 30/36 (5/6). Keeping a 3 and re-rolling the other dice gives the exact same odds.

6+ : Odds on 2d6 are 26/36. Keeping a 3 brings your odds down to 24/36, so don't keep anything below 4.

7+ : Odds on 2d6 are 21/36. Keeping a 4 raises your odds to 24/36.

8+ : Odds on 2d6 are 15/36. Keeping a 4 raises your odds to 18/36.

9+ : Odds on 2d6 are 10/36. Keeping a 4 raises your odds to 12/36.

10+: Odds on 2d6 are 6/36. Keeping a 4 gives you exact same odds.

11+: Odds on 2d6 are 3/36. Keeping a 5 raises your odds to 6/36.

12+: Odds on 2d6 are 1/36. Keeping a 6 raises your odds to 6/36.

 

So overall: If your charge target is less than 5, re-roll both. If it's 5 or above, keep a 4 unless it's not enough to make it.

 

Note on 5 and 10:

When your highest is 3 or 4 respectively, re-rolling one or both dice gives you the same odds. I would suggest keeping the high dice and re-rolling only 1, that way if you still fail you could use another CP to use the re-roll one dice strat to re-roll the one dice you originally didn't...

Edited by Spinsanity
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You're not allowed to reroll the second die after you've rerolled the first because you've already rerolled the charge result and would thus be rerolling a reroll, which is not allowed.

Isn’t this limitation on a ‘per-dice’ basis?

 

+Edit: the Rulebook FAQ simply says “You can never re-roll a dice more than once“

Edited by Spinsanity
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