@Unknown:
Odds are, if you missed in R1, you’ll get two hits in R2 and the defender will miss.
.67*.67=45% chance he scores two hits in a row.
I hope this is a joke.
If not, you might want to have a look at http://en.wikipedia.org/wiki/Gambler’s_fallacy
Actually, your link does not apply.
My statement is statistically correct. The odds of getting two hits with two dice when your target number is 4 or less is .67 * .67 which is approximately 45%
We could extrapolate further if you wish. The odds of a battleship scoring 10 hits in a row is .67*.67*.67*.67*.67*.67*.67*.67*.67*.67 = 1.62% (approximately.)
What your link refers too is a gambler saying that BECAUSE he missed in round 1, he WILL hit in round 2. What I am saying is that the odds of getting two hits in a row is 45% and the odds of getting 10 hits in a row is 1.62%. Doesn’t matter if you roll two dice at the same time or one die and then the same die a second time. Statistics are statistics.
My statement is statistically correct.
No, it absolutely is not. Even though you’re probably going to give me -karma again for pointing out your error, I will anyway.
The odds of getting two hits with two dice when your target number is 4 or less is .67 * .67 which is approximately 45%
No argument, this is obv true.
What your link refers too is a gambler saying that BECAUSE he missed in round 1, he WILL hit in round 2.
Right. And this is what you said:
Odds are, if you missed in R1, you’ll get two hits in R2
So how does the link not apply? You’re suggesting that if you get “bad” luck in R1 (no hits), you will have “good” luck in R2 (2 hits) to balance things out. This is the gambler’s fallacy: odds don’t “even out” on future rounds because you got a bad result. Probabilities don’t work that way, because the dice have no memory of what happened. The odds of hitting with both fighters in R2 is exactly the same as they were in R1.
What I am saying is that the odds of getting two hits in a row is 45%
OK, but that doesn’t matter. The first hit already happened. Its in the past, so it’s not a probability event anymore, it either happened or it didn’t. Given that the BB hit in R1, the odds of it hitting again at the start of R2 are 67% (just like they were at the start of R1), not 45%.
Statistics are statistics.
Yes, yes they are. But we’re discussing probability here, not statistics.
You didn’t reference that part of my statement in conjunction with your link to wiki. You linked the part of the statement where I said that odds of the battleship getting two hits was 45%
Yes, technically it’s true that you do not have better odds of getting two hits with your two fighters in the second round if they missed in the first.
As for your personal bad karma, don’t look at me bud. I have one person I give bad karma to and that’s it. You gotta be the world’s biggest jack-ere, donkey to get me riled up to the point you get BK from me.
You didn’t reference that part of my statement in conjunction with your link to wiki.
Well, if you read the original quote I took from your post you’ll see that I did. I should have taken out the part about the probability of BB hitting twice, though, as it was irrelevant to the point I was making. Sorry about the confusion.
As for your personal bad karma, don’t look at me bud. I have one person I give bad karma to and that’s it. You gotta be the world’s biggest jack-ere, donkey to get me riled up to the point you get BK from me.
My apologies then.
I hit SZ56 with 2 FGT and %3 with the DD and 2 FGT. In non combat I put the CVs and the surviving fighters in SZ51. I also withhold the CA from SZ50 using the BB only. I position the CA with the fleet in SZ51. I have had the US get cocky and attack this fleet. Yeah it cost me a CV I could replace and it cost the US a wad of units that were going to take several rounds to replace.
