11 hours ago, cnemmick said:

It was pointed out to me by a friendly soul on reddit that having Han do two 3-die attacks per turn at deployment cost 10 may be too powerful. If I kept Han at 12, that dual attack would probably be just right.

If you do give him a way to reliabily interrupt to attack, it doesn't have to be his normal attack. They took a die away from Vader for his end of round attack, so that could work well here also. Interrupting with 2 greens and 2 accuracy would be good, but maybe not overpowering.

How about just giving him an extra white die? It amps up his luck factor, which I think is a huge part of his lore. The enemy can choose to ignore him (as they often do), but that would leave him free to run around, take some shots, take objectives etc.

Or, as has been suggested, just replace his Return Fire with "Always shoots first" - when an enemy declares an attack at Han, he can interrupt to declare an attack against them. He wouldn't even need the once per turn limit.

10 hours ago, burek277 said:

How about just giving him an extra white die?

Rolling Odds for Two White Dice Han Defense Results:

DODGE: 1/6 + (5/6 * 1/6) = 30.6%

4 BLOCK, 2 EVADE (using Cunning): 2/6 * 2/6 = 11.1%

3 BLOCK, 2 EVADE (using Cunning): (1/6 * 2/6) + (2/6 * 1/6) = 11.1%

3 BLOCK, 1 EVADE (using Cunning): (1/6 * 2/6) + (2/6 * 1/6) = 11.1%

When attacking Han defending with two white defense dice, you have a 63.9% chance of one of the four outcomes above. I think it wouldn't be fun playing against or with this Han.

You can throw out Cunning to make the BLOCK values more reasonable but that still doesn't fix the odds of DODGE increasing from 16.7% to 30.6% - and slightly higher if Han rerolls one of his white dice. If he doesn't have is multiattack ability, he's ignored by your opponent; if he can attack more than once per activation, he's exceptionally OP.

I wonder about 2 white die. First of all, the game doesn't, by default, include 2 white die (though die aren't limited components).

Also, what happens in the uncommon scenario of two dodges being rolled? I'd assume that it would simply count as a single dodge, but what about crazy scenarios (such as Deadly being applied)? There just isn't precedent for it right now.

Still, I do think that swinginess is good flavor to him. I just know the current meta kinda hates white dice alone, let alone adding a second.

I believe the chances of at least one dodge when rolling with two white dice is 1/6 + 1/6 = 1/3 or 33.3%

That would be annoying IMHO. I'd prefer a white and black die combo

59 minutes ago, subtrendy2 said:

I wonder about 2 white die. First of all, the game doesn't, by default, include 2 white die (though die aren't limited components).

Also, what happens in the uncommon scenario of two dodges being rolled? I'd assume that it would simply count as a single dodge, but what about crazy scenarios (such as Deadly being applied)? There just isn't precedent for it right now.

Still, I do think that swinginess is good flavor to him. I just know the current meta kinda hates white dice alone, let alone adding a second.

There's no problem with having multiple dodges or multiple white dice right now. There are ways in skirmish to add white dice to a white die defense. If you roll 2 dodges then you have 2 dodges. Only one dodge is required to make the attack miss, but if an effect removes 1 dodge the attack is still missing.

But 2 white dice would be an exceptionally annoying defense pool.

1 hour ago, VadersMarchKazoo said:

I believe the chances of at least one dodge when rolling with two white dice is 1/6 + 1/6 = 1/3 or 33.3%

That would be annoying IMHO. I'd prefer a white and black die combo

The actual probability is 1 - (5/6)^2 = 30.6%

One would certainly not expect that the chance of rolling at least one dodge when rolling with 6 white dice is 1/6 + 1/6 + 1/6 + 1/6 + 1/6 + 1/6=6/6=100% (Real prob would be 1 -(5/6)^6 = 66.5% )

@VadersMarchKazoo I make probability calculation mistakes a lot, too. The better I understand IA dice probabilities, the better I understand how to play my lists.

30 minutes ago, player1690582 said:

The actual probability is 1 - (5/6)^2 = 30.6%

One would certainly not expect that the chance of rolling at least one dodge when rolling with 6 white dice is 1/6 + 1/6 + 1/6 + 1/6 + 1/6 + 1/6=6/6=100% (Real prob would be 1 -(5/6)^6 = 66.5% )

I'm not understanding your logic. The probability of rolling a dodge is 1/6. If you do roll this 100 times you would expect dodges 16.6% of the time. The probability of two independent outcomes i.e. two dodges is an additive probability (two dependent outcomes are multiplicative). This is a very simple stats model. I'm not saying that if you roll a 6 sider six times that you will get the dodge. Your sample size is too low. But if you repeat this 1000 times you'll see the probability manifest.

Edit: Maybe I do understand. It looks like you are suggesting that you need to look at the chances of not rolling a dodge which is 5/6 in each case. So chance of not rolling a dodge is the probability in question.

Stats are weird, man. I don't even try.

8 minutes ago, VadersMarchKazoo said:

I'm not understanding your logic. The probability of rolling a dodge is 1/6. If you do roll this 100 times you would expect dodges 16.6% of the time. The probability of two independent outcomes i.e. two dodges is an additive probability (two dependent outcomes are multiplicative). This is a very simple stats model. I'm not saying that if you roll a 6 sider six times that you will get the dodge. Your sample size is too low. But if you repeat this 1000 times you'll see the probability manifest.

For a single die, the probability of dodge is 1/6 which is also equivalent to the complement of not rolling a dodge which is 1-(5/6)=1/6

Now, the question of rolling AT LEAST one dodge from 2 white dice is the complement of rolling no dodge at all, which is 1-(5/6)^2 = 30.6%

16 minutes ago, VadersMarchKazoo said:

I'm not understanding your logic. The probability of rolling a dodge is 1/6. If you do roll this 100 times you would expect dodges 16.6% of the time. The probability of two independent outcomes i.e. two dodges is an additive probability (two dependent outcomes are multiplicative). This is a very simple stats model. I'm not saying that if you roll a 6 sider six times that you will get the dodge. Your sample size is too low. But if you repeat this 1000 times you'll see the probability manifest.

In fact, if we list all the possible 36 combination of results of 2 white dice we get the following : (D=dodge, B=blank, BL-block, E=evade, BL/E=Block & Evade)

If you count all events where at least one D shows up you find there are 11 such cases out of 36 = 11/36 = 30.6% = 1-(5/6)^2

D - D
D - B
D - BL
D - E
D - BL/E
D - BL/E

B - D
B - B
B - BL
B - E
B - BL/E
B - BL/E

BL - D
BL - B
BL - BL
BL - E
BL - BL/E
BL - BL/E

E - D
E - B
E - BL
E - E
E - BL/E
E - BL/E

BL/E - D
BL/E - B
BL/E - BL
BL/E - E
BL/E - BL/E
BL/E - BL/E

BL/E - D
BL/E - B
BL/E - BL
BL/E - E
BL/E - BL/E
BL/E - BL/E

28 minutes ago, VadersMarchKazoo said:

I'm not understanding your logic. The probability of rolling a dodge is 1/6. If you do roll this 100 times you would expect dodges 16.6% of the time. The probability of two independent outcomes i.e. two dodges is an additive probability (two dependent outcomes are multiplicative). This is a very simple stats model. I'm not saying that if you roll a 6 sider six times that you will get the dodge. Your sample size is too low. But if you repeat this 1000 times you'll see the probability manifest.

I think the sentence in bold is where you're going wrong, in that you're saying that they add directly (i.e. 1/6 from one die + 1/6 from the other die = 1/3 when rolling both) when this isn't true.

To illustrate what Player was saying, try doing the same thing with a coin flip instead of a die roll. For each coin the odds of heads are 50%. If you flip two coins simultaneously, does that mean the odds of flipping at least one heads is 100% (50% + 50%)? Clearly it isn't.

Player had the correct formula above, but if you want an explanation for why it makes sense I'll put it in the spoiler tag here so that everyone that wants to can just skip over it

Makes sense. Though the phrase "not rolling no dodges" makes my skin crawl.

30 minutes ago, VadersMarchKazoo said:

Edit: Maybe I do understand. It looks like you are suggesting that you need to look at the chances of not rolling a dodge which is 5/6 in each case. So chance of not rolling a dodge is the probability in question.

The tricky part here is you need just one DODGE out of two dice. Here's how I think of it:

• Dodge on the first die: 1 in 6 chance.
• The result on the second die doesn't matter if you roll a DODGE on the first die. So the second die matters in 5 out of 6 times.
• In that second die roll, you have a 1 in 6 chance for that DODGE. However, we have to apply that chance to ALL POSSIBLE chances from the non-DODGE first die. We multiply the second DODGE probability (1/6) with the first non-DODGE probability (5/6) in order to represent that mathematically.
• So written out, the equation looks like this: "Probability of rolling a DODGE on the first die" + ("Probability of NOT rolling a DODGE on the first die" * "Probablity of rolling a DODGE on the second die")
• p_dodge = (1/6) + (5/6 * 1/6) = 30.6%

That's really interesting. The crux of the matter is that rolling at least one dodge is an independent roll so you would think additive probabilities. But the chance of not rolling any dodges is dependent and therefore multiplicative (5/6 * 5/6). 4th wall stuff:)

1 minute ago, cnemmick said:

The tricky part here is you need just one DODGE out of two dice. Here's how I think of it:

• Dodge on the first die: 1 in 6 chance.
• The result on the second die doesn't matter if you roll a DODGE on the first die. So the second die matters in 5 out of 6 times.
• In that second die roll, you have a 1 in 6 chance for that DODGE. However, we have to apply that chance to ALL POSSIBLE chances from the non-DODGE first die. That's why you multiply 1/6 with 5/6.
• So written out, the equation looks like this: "Probability of rolling a DODGE on the first die" + ("Probability of NOT rolling a DODGE on the first die" * "Probablity of rolling a DODGE on the second die")
• p_dodge = (1/6) + (5/6 * 1/6) = 30.6%

And that's the basis for the other stats formula that's used for this sort of thing! See, we don't need no fancy book learning, we've got it all figured out here!

14 minutes ago, player1690582 said:

In fact, if we list all the possible 36 combination of results of 2 white dice we get the following : (D=dodge, B=blank, BL-block, E=evade, BL/E=Block & Evade)

If you count all events where at least one D shows up you find there are 11 such cases out of 36 = 11/36 = 30.6% = 1-(5/6)^2

D - D
D - B
D - BL
D - E
D - BL/E
D - BL/E

B - D
B - B
B - BL
B - E
B - BL/E
B - BL/E

BL - D
BL - B
BL - BL
BL - E
BL - BL/E
BL - BL/E

E - D
E - B
E - BL
E - E
E - BL/E
E - BL/E

BL/E - D
BL/E - B
BL/E - BL
BL/E - E
BL/E - BL/E
BL/E - BL/E

BL/E - D
BL/E - B
BL/E - BL
BL/E - E
BL/E - BL/E
BL/E - BL/E

This is interesting written out because the double dodge essentially "steals" a potential dodge outcome. That's why it goes to 11. (instead of 12 like I'd expect)

I don't think that if Han is made to be worth his 12pts, that you are going to get away from the decision of "do i attack him" or "do i ignore him." Any viable 12 pt figure these days forces you make that decision....and its rarely easy.

2 White dice with cunning does seem to move his defense to the other extreme. However, if you didn't increase his health, the extra dodge potential helps offer multiple attacks, but maintains the luck aspect as well. I think I would be ok (in theory) with don't nothing but adding a white die to his defense and leaving it at that. He wouldn't be squishy and has potential for multiple attacks and forces the opponent to play around him as many other 12pt figures do.

13 minutes ago, wannabepudge said:

I don't think that if Han is made to be worth his 12pts, that you are going to get away from the decision of "do i attack him" or "do i ignore him." Any viable 12 pt figure these days forces you make that decision....and its rarely easy.

I'd argue both IG-88 and Jedi Knight Luke aren't defensively strong enough to withstand a round-long barrage from other hard-hitting uniques and the stronger multi-figure deployments (eWeequays, eImpJets, eAllianceRangers). IG w/ Focused on the Kill is very susceptible to Pierce and the attacker's surge abilities; Luke has the nice natural EVADE but still has to X-Man his way out taking the base rolled damage.

And that's the kind of balance you'd want for high-cost figures like Han, Chewie, Boba, AT-ST/General Weiss -- strong but not invincible.

My current Han fix ( back on page 1 ) reduces his point cost to 10, which I think fits Rebel builds more appropriately.

from http://adriantp.github.io/imperial-assault-dice/app/

probabilities of one or more dodges on 2 white dice is 30.56%. Carry on.

18 minutes ago, ManateeX said:

I think the sentence in bold is where you're going wrong, in that you're saying that they add directly (i.e. 1/6 from one die + 1/6 from the other die = 1/3 when rolling both) when this isn't true.

To illustrate what Player was saying, try doing the same thing with a coin flip instead of a die roll. For each coin the odds of heads are 50%. If you flip two coins simultaneously, does that mean the odds of flipping at least one heads is 100% (50% + 50%)? Clearly it isn't.

Player had the correct formula above, but if you want an explanation for why it makes sense I'll put it in the spoiler tag here so that everyone that wants to can just skip over it

Start with a more straightforward question - what is the chance of rolling two dodges? Most people would say, correctly, that it's the two probabilities multiplied together, or 1/6 * 1/6 = 1/36

Now what we're really looking here is a bit more complicated: the chance of rolling at least one dodge. But if we only know the formula above, how can we figure this out? We just need to rephrase the question a bit. Instead of asking "what are the chances of rolling at least one dodge" we can instead phrase it as "what are the chances of not rolling no dodges".

We know that the chance of rolling no dodges on 1 die is 5/6. Therefore the chance of rolling no dodges on both dice is 5/6 * 5/6 = 25/36. So if the chance of rolling no dodges is 25/36, that in turn means that the chance of not rolling no dodges is the "compliment" of that fraction, or 11/36 (30.6%).

Hopefully that helps a bit, as being able to do these sorts of calculations quickly can actually be of some help, even in a casual campaign setting. "What is the chance of passing this attribute test if I'm only rolling blue-green?" Think of it instead as the easier "what is the chance that I don't pass" - 1/2 (green) * 2/3 (blue) = 2/6 = 1/3 chance of failure. Meaning 2/3 chance of success - I think I'll go for it!

So applying the logic to your coin flip scenario the odds of at least one heads in a two simultaneous coin flip is: 1 minus the chances of not getting a heads (0.5 x 0.5) = 0.75 or 75% chance. or 1 - (0.5)^2 = 0.75

If you look at the outcomes it makes sense:

HH, HT, HT, TT (like a punet square)

Sorry if folks are tired of the stats but I'm just happy I got learned.

4 minutes ago, cnemmick said:

I'd argue both IG-88 and Jedi Knight Luke aren't defensively strong enough to withstand a round-long barrage from other hard-hitting uniques and the stronger multi-figure deployments (eWeequays, eImpJets, eAllianceRangers). IG w/ Focused on the Kill very susceptible to Pierce; Luke has the nice natural EVADE but still has to X-Man his way out taking the base rolled damage.

And that's the kind of balance you'd want for high-cost figures like Han, Chewie, Boba, AT-ST/General Weiss -- strong but not invincible.

My current Han fix ( back on page 1 ) reduces his point cost to 10, which I think fits Rebel builds more appropriately.

While dropping him to 10 is helpful for group building, I am a fan of these iconic characters costing the same and forcing you to be more creative with your group. Han certainly should bring something to the table that Luke doesn't and vice versa. So I say make him worth 12.

I do agree with your make them "strong but not invincible." but without play testing it, I'm not sure I am sold that 2 white dice is invincible. That said, perhaps a stipulation is added that, "when a dodge is rolled, if Return Fire is used, suffer 2 damage." That way you would be forced to decided if its worth taking the damage for the extra attack. I would guess that most times it would be and would help whittle away at his health.

8 minutes ago, VadersMarchKazoo said:

So applying the logic to your coin flip scenario the odds of at least one heads in a two simultaneous coin flip is: 1 minus the chances of not getting a heads (0.5 x 0.5) = 0.75 or 75% chance. or 1 - (0.5)^2 = 0.75

If you look at the outcomes it makes sense:

HH, HT, HT, TT (like a punet square)

Sorry if folks are tired of the stats but I'm just happy I got learned.

I'm pretty sure, Han Solo himself was the first person to leave this discussion ... I think I heard him say ".... never tell me the odds"

i still don't like return fire, but i'm ok if his ability to get a second attack isn't automatic (because his attack is pretty powerful). maybe he could exhaust 'never tell me the odds' to add a second white dice once per round. if not a dodge, you've also got a solid chance of getting a couple evades and 3-4 blocks out of a 2 white die attack. not too many figures will get damage through that defense. I also am a big fan of him getting 2 movement after his interrupt attack.

6 minutes ago, wannabepudge said:

I do agree with your make them "strong but not invincible." but without play testing it, I'm not sure I am sold that 2 white dice is invincible. That said, perhaps a stipulation is added that, "when a dodge is rolled, if Return Fire is used, suffer 2 damage." That way you would be forced to decided if its worth taking the damage for the extra attack. I would guess that most times it would be and would help whittle away at his health.

that doesn't make any sense. if you want a negative consequence, strain is the way to go.