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I noticed that a lot of people will roll two dice as a shortcut when the rules state that you should roll one dice and then reroll if you get a bad result.

example: I fire a twin-linked lascannon from a land raider. Instead of rolling 1D6 and rolling it again if I miss, I roll 2D6 and if either die is a 3+ the lascannon hits.

Innocent enough, right?

I am no statistical analysis expert, but I suspect that it is actually cheating. It seems akin to something they call the Monty Hall problem. It seems to become more apparent with larger numbers.

If I can fire 10 twin-linked shots that hit on a 3+, and I just roll 20 dice to speed things up, there are 2/3 odds and I will average 13 hits rounding up. I roll my dice. There are 13 hits. Since there are only 10 guns shooting, I only count 10 (what sane opponent would let you count all 13?).

If I can fire 10 twin-linked shots that hit on a 3+, and I roll 10 dice, there are 2/3 odds and I will average 7 hits rounding up. If I reroll the 3 misses, 1 will still miss on average.

I'm pretty sure the problem is the same with 1 or 2 dice.

Am I right? Did I miss something?

wow...I've never seen anyone do that before but I see the concern you propose. It definately looks to me that the odds would skew; however, they should have the odds of skewing both for and against a player doing this. Personally, I'll just stick to rolling and then re-rolling if I miss.

It only works if it's a 1-shot weapon, like a lascannon. If it's multi-shot, like a heavy bolter, you can't roll 6 dice at once, you have to roll 3 dice and then re-roll.

Well if someone is firing a 1 shot weapon, like a T/L Lascannon, then most people I know do roll the 2 dice at the same time and count a hit if either rolled a hit.

However once you get into the realm of anything multiple shot weapon that is also T/L, then they roll them one at a time.

For instance, a unit of 8 Space Marine Bikes. They would roll 8 dice (or 16 if shooting twice with Rapid Fire). They then would take the misses and re-roll. They would not roll 32 dice and then only allow 16 to hit.

EDIT: Beaten to it!

Like dietrich said, it only works on 1 shot weapons. I don't have a problem with either.

A twin linked lascannon with bs 4 has an 89% chance to hit anyway. Although it does throw off the reroll at a large scale, if you roll them on a gun by gun basis it shouldn't be a problem.

edit: dang...

Mike Leon wrote:If I can fire 10 twin-linked shots that hit on a 3+, and I just roll 20 dice to speed things up, there are 2/3 odds and I will average 13 hits rounding up. I roll my dice. There are 13 hits. Since there are only 10 guns shooting, I only count 10 (what sane opponent would let you count all 13?).

As was already mentioned, this only works when rolling for a one shot weapon. However, my guess is that that's what your opponents have been doing and you're still confused, so here's a little bit more in depth explanation.

Let's say you're firing that twin linked las cannon. You roll two dice, both are above 3+, do you get two hits? Of course not, you only fired one shot. You know you only fired one shot and you intend to count it as a hit if either die or both dice roll a 3+. So, obviously, this is why you can't do it with a multi shot weapon, you don't know when both dice would hit for the same shot. It would work if you had multi colored dice, but that's more effort than it's worth.

So, say you fired that twin linked heavy bolter. You have two blue dice, two green dice, and two red dice. Each color corresponds to a single shot the bolter fires. You roll them all together. Both of the blue dice miss, so that shot misses. One red die hits and one misses so the red shot hits. Both green dice hit, so the green shot hits. This gives you a total of two hits, even though you rolled a 3+ on 3 dice, only 2 count as hits since only two of the color groups hit. If you had simply rolled six together of one color without designating which dice go to which shot then you would be right, you end up with more hits than you deserve. But since you kept the groups of two separate by color coordination, you didn't change the odds any. You got a total of two hits and it would not have been possible to get more than three since double hits on any one color is only counted as one hit.

I think this illustrates that the problem is still apparent even when you fire only 1 weapon.

Rolling 2 dice at the same time for 1 shot:

D1 | D2
miss hit
hit miss
miss miss <------ This is the only possibility of failure out of 4 options. 25% chance to miss.
hit hit

Rolling a dice and then rerolling:

D1 | reroll
hit
miss hit
miss miss <---- This is the only possibility of failure out of 3 options. 33% chance to miss.

Mike Leon wrote:I think this illustrates that the problem is still apparent even when you fire only 1 weapon.

Rolling 2 dice at the same time for 1 shot:

D1 | D2
miss hit
hit miss
miss miss <------ This is the only possibility of failure out of 4 options. 25% chance to miss.
hit hit

Rolling a dice and then rerolling:

D1 | reroll
hit
miss hit
miss miss <---- This is the only possibility of failure out of 3 options. 33% chance to miss.

You need to go back and study probability some more, because you're working the numbers all wrong. To put this in the simplest terms possible....

Let's say all you have to do is flip a coin to see if you hit. heads hits, tails misses.

If you flip one coin, and then a second coin if the first one fails, you have the following probabilities:

1st coin - heads, 50% of the time. No need to flip a second coin.
2nd coin - heads, 50% of 50% of the time, so the total chance of hitting is 75% (50%+25%)

If you flip two coins simultaneously, there are four potential outcomes:
25% - heads/heads
25% - heads/tails
25% - tails/heads
25% - tails/tails.

Total chance of hitting: 75%

Comprende?

Double post.

Centurian99 wrote:
If you flip two coins simultaneously, there are four potential outcomes:
25% - heads/heads
25% - heads/tails
25% - tails/heads
25% - tails/tails.

Total chance of hitting: 75%

Comprende?

You are correct in that there are four potential outcomes of this method, but there are still only 3 for the other method even in your example.

or if you must roll lots of them at the same time why not get some of these ...

http://www.rpgshop.com/default/25mm-double-die.html

Mike Leon wrote:I think this illustrates that the problem is still apparent even when you fire only 1 weapon.

Rolling 2 dice at the same time for 1 shot:

D1 | D2
miss hit
hit miss
miss miss <------ This is the only possibility of failure out of 4 options. 25% chance to miss.
hit hit

Rolling a dice and then rerolling:

D1 | reroll
hit
miss hit
miss miss <---- This is the only possibility of failure out of 3 options. 33% chance to miss.

You are seriously missing something here. Your second chart is flawed because you assume that if you roll a hit, there is no need to reroll. And, although this is true in game terms, it is statistically relevant. If, after rolling a hit, if you rolled a second die, just to see what the result of a reroll would have been, would it change your odds of hitting any? So, you hit with that twin linked las on the first roll but just for fun you roll again to see if the reroll would have been a hit as well, can we both agree this doesn't increase your odds of hitting?

Ok, hopefully we are agreed. So now your chart should look like this:

Two dice:

D1 | D2
miss hit
hit miss
miss miss
hit hit

Reroll:


D1 | reroll
hit hit
hit miss
miss hit
miss miss

Mike Leon wrote:

Centurian99 wrote:
If you flip two coins simultaneously, there are four potential outcomes:
25% - heads/heads
25% - heads/tails
25% - tails/heads
25% - tails/tails.

Total chance of hitting: 75%

Comprende?

You are correct in that there are four potential outcomes of this method, but there are still only 3 for the other method even in your example.

You really need to study math more. It's not the number of outcomes, its the probability of those outcomes occuring. Flipping the coins sequentially, the probabilities are:

First Coin Heads - 50% (no need to flip a second coin)
First Coin Tails, Second Coin Heads - 25%
First Coin Tails, Second Coin Tails - 25%

Do you understand how the math works here? You're trying to treat each outcome as equally likely.

whocares wrote:

You are seriously missing something here. Your second chart is flawed because you assume that if you roll a hit, there is no need to reroll. And, although this is true in game terms, it is statistically relevant. If, after rolling a hit, if you rolled a second die, just to see what the result of a reroll would have been, would it change your odds of hitting any? So, you hit with that twin linked las on the first roll but just for fun you roll again to see if the reroll would have been a hit as well, can we both agree this doesn't increase your odds of hitting?

Ok, hopefully we are agreed. So now your chart should look like this:

Two dice:

D1 | D2
miss hit
hit miss
miss miss
hit hit

Reroll:


D1 | reroll
hit hit
hit miss
miss hit
miss miss

I underlined that part because I think it gets to the heart of the issue. The way I see it, when you roll 2 dice at the same time you ARE rolling to hit again after you already rolled a hit, AND the 2nd roll is becoming statistically relevant. It is a roll that never should have been made and it could be altering the outcome slightly.

Also, if you do the chart like this I think you end up with this problem:

D1 | reroll
hit hit
hit miss <----this is now a missing shot (remember you rerolled so the original hit is no longer on the table),
miss hit
miss miss <----this is now a missing shot too. Your odds of hitting are now 50%.

I'll take the 75% chance of rolling 2 dice together over the 50% any day.

Tri wrote:or if you must roll lots of them at the same time why not get some of these ...

http://www.rpgshop.com/default/25mm-double-die.html

That's neat.

Another method is just using sets of 2 with different coloured/sized dice.

So if you are firing a twin linked heavy bolter you roll two red, two white and two black dice. Only a die of the same colour can replace another one.

Here's the difference as far as I can tell:

2 dice method

Hit l Hit
Hit l Miss (hit)
Miss l Hit (hit)
Miss l Miss (miss)

1/4 sets will yield a miss

Rerolls due to Twin-Linked

Hit
Miss l Hit
Miss l Miss

1/3 sets will yield a miss

I may not have set this up properly as I am no statistician.

Mike Leon wrote:hit miss <----this is now a missing shot (remember you rerolled so the original hit is no longer on the table)

See, this is your problem right here. If either the original roll hit, or the reroll hit, the shot hits.

If it would make you happy, leave the original shot on the table and roll a second die.

Mike Leon wrote:
I underlined that part because I think it gets to the heart of the issue. The way I see it, when you roll 2 dice at the same time you ARE rolling to hit again after you already rolled a hit, AND the 2nd roll is becoming statistically relevant. It is a roll that never should have been made and it could be altering the outcome slightly.

Also, if you do the chart like this I think you end up with this problem:

D1 | reroll
hit hit
hit miss <----this is now a missing shot (remember you rerolled so the original hit is no longer on the table),
miss hit
miss miss <----this is now a missing shot too. Your odds of hitting are now 50%.

I'll take the 75% chance of rolling 2 dice together over the 50% any day.

no, its a sign that either the original poster you quoted doesn't understand probability, or you don't understand probability, and to be honest, I'm not all that interested in figuring out which it is. I'm not saying that to be mean or harsh, but that's just a fact based on your question.

In the situation we're talking about (twin-linked weapons) there's never a reason to re-roll a hit, so it will never happen. This is different from say, Tau Decoy Launchers or Venerable Dread damage results, which have to be re-rolled sequentially because the second roll supercedes the first. In the case of twin-linked weaponry, if the first roll hits, the second roll would never be made. Therefore, you can safely roll 2d6, and if either die hits, take the result. As long as the probabilities are identical (and they are, whether you roll sequentially or roll 2dy and take the highest) its fair.

sourclams wrote:Here's the difference as far as I can tell:

2 dice method

Hit l Hit
Hit l Miss (hit)
Miss l Hit (hit)
Miss l Miss (miss)

1/4 sets will yield a miss

Rerolls due to Twin-Linked

Hit
Miss l Hit
Miss l Miss

1/3 sets will yield a miss

I may not have set this up properly as I am no statistician.

No, you did not. See my response to Mike Leon.

It makes no difference if you roll two dice one at a time or two dice together, which is really all that this issue boils down to.

sourclams wrote:Here's the difference as far as I can tell:

<snip>
I may not have set this up properly as I am no statistician.

You didn't set it up properly, and this is probability, not statistics.

sourclams wrote:Here's the difference as far as I can tell:

2 dice method

Hit l Hit
Hit l Miss (hit)
Miss l Hit (hit)
Miss l Miss (miss)

1/4 sets will yield a miss

Rerolls due to Twin-Linked

Hit
Miss l Hit
Miss l Miss

1/3 sets will yield a miss

I may not have set this up properly as I am no statistician, but as far as I can tell there is significantly greater chance to miss if we follow the rules as written versus the two dice speed play method.

Bottom line, don't let your opponent throw extra dice.

Um.

Are you reading what the other people said at all?

2 dice method assuming 4+:
Hit l Hit (25%)
Hit l Miss (hit) (25%)
Miss l Hit (hit) (25%)
Miss l Miss (miss) (25%)

So 75% chance of hitting.

Rerolls due to Twin-Linked
Hit (50%)
Miss l Hit (25%, as there is a 50% chance you don't roll the second die at all)
Miss l Miss (25%)

So 75% chance of hitting.

That's all fine and dandy, but I don't see how any of it is relivant to a Bs4 TL LasCannon.

3-6 = hit 66%
1-2 = miss 33%
reroll miss, 66% of 33% = rerolled hit 22%
hit + rerolled hit = 88%

the coin theory is great and all, but clearly doesn't work here as we are not talking about Bs3.

Centurian99 wrote:

Mike Leon wrote:
I underlined that part because I think it gets to the heart of the issue. The way I see it, when you roll 2 dice at the same time you ARE rolling to hit again after you already rolled a hit, AND the 2nd roll is becoming statistically relevant. It is a roll that never should have been made and it could be altering the outcome slightly.

Also, if you do the chart like this I think you end up with this problem:

D1 | reroll
hit hit
hit miss <----this is now a missing shot (remember you rerolled so the original hit is no longer on the table),
miss hit
miss miss <----this is now a missing shot too. Your odds of hitting are now 50%.

I'll take the 75% chance of rolling 2 dice together over the 50% any day.

no, its a sign that either the original poster you quoted doesn't understand probability, or you don't understand probability, and to be honest, I'm not all that interested in figuring out which it is. I'm not saying that to be mean or harsh, but that's just a fact based on your question.

In the situation we're talking about (twin-linked weapons) there's never a reason to re-roll a hit, so it will never happen. This is different from say, Tau Decoy Launchers or Venerable Dread damage results, which have to be re-rolled sequentially because the second roll supercedes the first. In the case of twin-linked weaponry, if the first roll hits, the second roll would never be made. Therefore, you can safely roll 2d6, and if either die hits, take the result. As long as the probabilities are identical (and they are, whether you roll sequentially or roll 2dy and take the highest) its fair.

If either of you had read my post, you would have seen that I said "roll again after you hit just to see what the result of a reroll would have been and see that it in no way changes your odds" I did not say to apply that result. I said that because that's essentially what you're doing when you roll two dice at once, since you don't know which one was rolled first or second is irrelevant, if either hit, then the shot hits.

Taking this back to dice, lets say you have a twin linked las cannon at BS4, so you need 3+ to hit. Rolling two dice at the same time in the hopes to get at least one 3+ can be shown as:

1-((2/6)*(2/6))= .89 (89%)

This formula works regardless of if you roll one at a time, or both at once. It takes into account that up to two dice can be rolled and their probable outcomes. Rolling two at the same time (assuming its a once shot weapon) is the same as rolling one then the other.

Thanks, Axyl, nice to see I'm not alone in using stats in context.

delete

Now I might get cruxified for this comment but when it is a multi shot wep. Like a twin linked heavy bolter I just roll all the dice at once however i roll 2 diffrent colors of dice and declare which ones are ment be the originals and which ones are ment to be for the reroll.

If I roll 10 pairs of dice (20 total), it's the same as rolling 10 dice and re-rolling.

But not the same as rolling 20 identical dice. Once the dice are paired-up as re-roll pairs, it's fine.

Emrab wrote:Now I might get cruxified for this comment but when it is a multi shot wep. Like a twin linked heavy bolter I just roll all the dice at once however i roll 2 diffrent colors of dice and declare which ones are ment be the originals and which ones are ment to be for the reroll.

Edit: I misread the question. See my very first reply.