Upgrade Disable Compromise

[Darkas]

In most case, you actually can bruteforce all possibilites, and as stochastic models only approximate results, I'd say a computer scientist try to bruteforce and if failed go for other solutions :P

--> Techgump

Offender's chance to win on first battle, 47%
Offender's chance to win on second battle having lost the first one, (1-0.47)*0.47=25%
47% + 25% = 72%
Defender chance of winning, that means winning both fight: (1-0.47)*(1-0.47)=28%
72% + 28% = 100%

Nothing lost in outerspace =)

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What does "alog" mean?

[techgump]

I''m sorry, but this does not work out this way IMO. Now you are sigularly looking at one roll by saying:

Offender's chance to win on first battle, 47%
Offender's chance to win on second battle having lost the first one, (1-0.47)*0.47=25%

Which just cannot be, as looking at each singular roll, as you agreed, produces the same % chance. Likewise, this is not (1-p)*(1-p).

Either way, I am done with this. We can go around in circles. I still believe that
a) 47% cannot be just trusted when brought in as fact with nothing to verify it.
b) The formula to output win % does leave 50% hanging in the wind, when you compute starting from each side's 47% vs 53% (attacker vs defender). There is no reason to change a static formula just because you are on the flip side of the coin (heads vs tails).

[Darkas]

a) Cannot convince you on that.
b) You're wrong on that =) No formula changed. You're confused because it's not symmetrical: defender have to win twice, offender just once.

No problem for me, I'm not a missionary, don't aim to convince you by all means.

[techgump]

Lastly, I can also state, based on game play experience, that if you would normally win the roll, then even with two, you will normally win. Likewise with losing.
And having equals, I would say my win rate as an attacker is just below 50%. Which means the odds are not changing any. It is just another roll with the same odds.

[Darkas]

What you state is true, except that if you get to have reroll your chance of winning the fight is increased (of course the second roll have the same probability as the first one, but you can either win the first, and you won, are lose the first and win the second, it seems to me obvious that you have bigger chance of winning the fight...)
It is just another roll with the same odds. <- totally, never said the opposite.

[techgump]

Offender's chance to win on first battle, 47%
Offender's chance to win on second battle having lost the first one, (1-0.47)*0.47=25%
47% + 25% = 72%
Defender chance of losing, that means losing both fight: (1-0.47)*(1-0.47)=28%

Right there, you are using two separate formulas for the same result. If this is not the case, how can one not look from the opposite coin by stating:
Attacker chance of losing, that means losing both fight: (1-0.53)*(1-0.53)=22%

You are in fact changing it! And in doing it this way, there is a 50% gap!
Gump, over and out.

[techgump]

if you get to have reroll your chance of winning the fight is increased

This is simply not true, as we both agreed, roll by roll the odds are the same. You are contradicting yourself. This is the exact fallacy that lotteries and casinos hope you believe. Your odds, however, simply do not change the next time around.

[Darkas]

Defender chance of losing, that means losing both fight: (1-0.47)*(1-0.47)=28%

--> Sorry about that, my mistake, I wanted to write "Defender chance of winning, that means winning both fight: (1-0.47)*(1-0.47)=28%"

Dude, if I let you have infinite reroll each time you lose, will you have the same chance to win the battle that if you don't or not ? No, you will definitely win in every fight (if your chance of winning are not 0).

The chance of one roll is the same, but having the opportunity to reroll increases your chance of winning the fight overall !

[techgump]

I am not disagreeing with that. But the formula to produce the outcome is very wrong. But this is about the number of rolls, not the % increase in win.

Here, let me give you a practical example. This bonus is no different than attacking the same piece twice in two different turns... it just happens in one turn.
But what you are saying is, if I attacked it once before in a previous turn, now my odds when I attack it again are 72%. And if I lose, then my next turn my odds are 85%, then 92%, and so on.

I am telling you, your odds are the same each time, every time. It cannot be 47% and 85% at the same time. Again, something is seriously wrong.

Ok, now am are really done. Good conversation.

[Darkas]

I never said that.

If you read any of my posts, you'll see that I'm talking about chance of winning the fight. And not winning a certain dice roll in the middle of it.

And the formula to compute the outcome is not wrong.

Have a good night.