First big post-launch patch hits COH2

Cool thing is that if he uses sniper in the M3 then there is a 50% chance for each member to die when M3 is destroyed. Thus there is 25% chance that both die when M3 explodes. There is also a 75% chance that one of the two snipers will die, thus making it easy to chase down the remaining member with the german 222(or whatever unit) you killed it with.

Interesting math...if you say the chance of both dieing is 25% and either one of them dieing is 75%, then what happened to the chance none of them dies..?

Indeed, the chance either one of them dies is 50%, the chance they both die is 25% and the the chance none of them die is 25%.

Interesting math...if you say the chance of both dieing is 25% and either one of them dieing is 75%, then what happened to the chance none of them dies..?

Indeed, the chance either one of them dies is 50%, the chance they both die is 25% and the the chance none of them die is 25%.

Yes, the chance that none of them die is indeed 25%. In the other 75% one or both of them die.(bold is what i should have added) Thanks for correcting me.

I look at it with binary logic:
say 1 = live and 0 = die.

The chances for each sniper is 50% thus 50% to end up 1(alive) and 50% to end up 0(dead).

The combinations of possible outcomes are:

0 0 (both die) Outcome 1
0 1 (sniper #1 dies) Outcome 2
1 0 (sniper #2 dies) Outcome 3
1 1 (both live) Outcome 4

In outcomes 1 to 3 at least one sniper dies. (These three rows have one or more zeros in them).

Thus 3/4 of the time there dies at least one sniper which is 75% of the time. The likelihood of the fourth row is the remaining 25% (both live).

Depending on your country of residence I give you an A/100%/10.

Hehe, it pays to do the math in rts games

Good thinking - although how does the weighting on each outcome affect it? If at all?

0 0 (Both dead) 25% chance on destruction
0 1 (1 dead) 50% chance on destruction
1 0 (1 dead) 50% chance on destruction
1 1 (Both live) 25% chance [not sure of that one though] on destruction

So isn't there a greater chance overall that 1 of them will die than both dying and both living? If we give each outcome a number out of 4:

0 0 allocate 1
0 1 allocate 2
1 0 allocate 2
1 1 allocate 1

So for at only 1 of them to die, that would be 4/6 - 66.6% chance. For both to die would be over a 16.6% chance and for both to survive would also be 16.6%.

My maths comprehension can be way off though, so am prepared to be told it's nonsense

in your example, the probabilties add up to 150%.
the way it was explained above is correct.
as a more general way of calculating it: P=0.5^x , with x being the number of squadmembers to die (or survive, with no one dieing), and P being the probabilty for that to happen. so the probability for a whole conscript squad to perish on explosion would be P=0.5^6=0.015625 or a 1.5625% chance.
the inverse (not 6 members die, but between 0 and 5) is obviously 1-P=0.984375

in your example, the probabilties add up to 150%.
the way it was explained above is correct.
as a more general way of calculating it: P=0.5^x , with x being the number of squadmembers to die (or survive, with no one dieing), and P being the probabilty for that to happen. so the probability for a whole conscript squad to perish on explosion would be P=0.5^6=0.015625 or a 1.5625% chance.
the inverse (not 6 members die, but between 0 and 5) is obviously 1-P=0.984375

Agreed, chance for a whole script/guard squad to die will be slim, for engi's it will be a bit better

Yeh, I understand the probability of a whole squad dying as that's pretty straightforward (said as much myself on a very similar thread http://www.coh2.org/topic/5122/comments-on-the-july-10-patch-notes/page/2#post_id51187 ), what threw me is trying to understand and apply weighting to the thing above from Mauser that was going a level deeper. Perhaps it doesn't need weighting. Which is kind of what I was asking.

Yeh, I understand the probability of a whole squad dying as that's pretty straightforward (said as much myself on a very similar thread http://www.coh2.org/topic/5122/comments-on-the-july-10-patch-notes/page/2#post_id51187 ), what threw me is trying to understand and apply weighting to the thing above from Mauser that was going a level deeper. Perhaps it doesn't need weighting. Which is kind of what I was asking.

I don't think it needs any "weighting", my example was just explaining it through binary, cr4wler's way uses normal mathematics which is a more general approach. If I understand correctly you are asking whether you need to "weigh" each row's averages? You don't need to because the chance is 50% for every squad member to die. So in my example the chance is 50% that it ends up 0 and 50% that it ends up 1. See also that in each coloumn sniper #1 is value 1(alive 50%) of the time and sniper 2 is value 0(dead) 50% of the time.

Cr4wler's way of calculating the percentages is easier if you want to calculate larger squads which he explains correctly above. If the squad size is 6 you would need 6 coloumns of binary thus

000000
000001
000010
000011
000100
etc
etc..

Which will take much longer than just using P = 0.5^6

Also you asked earlier if "there isn't a greater chance for (one of the two to die) than (both of them dying and none of them dying). The answer to that question is no. This is because:

01 (#1 dies)
10 (#2 dies)
Above combinations happen 2 out of 4 times.

00 (both die)
11 (both live)
Above combinations happen 2 out of 4 times.

The only reason I used binary for the 2 snipers is because it visually shows you the probabilities of 1's and 0's which is nice.

Note: if the chance was 30% for each squad member to die (or any number other than 50%) you hve to use P = 0.3^x because the binary will only work for 50%. Each bit is a '0' 50% of the time and a '1' 50% of the time. This cannot change.

okay, lets go over it

0 0 (Both dead) 25% chance on destruction
0 1 (1 dead) 50% chance on destruction this is wrong, only 25% chance for entity 2 to die
1 0 (1 dead) 50% chance on destruction 50% is the chance for EITHER of them dieing
1 1 (Both live) 25% chance [not sure of that one though] on destruction

So isn't there a greater chance overall that 1 of them will die than both dying and both living? If we give each outcome a number out of 4:

i have no idea what this is supposed to mean
0 0 allocate 1
0 1 allocate 2
1 0 allocate 2
1 1 allocate 1

So for at only 1 of them to die, that would be 4/6 - 66.6% chance. For both to die would be over a 16.6% chance and for both to survive would also be 16.6%.

there is 4 possible outcomes, as you mentioned yourself. each has the same probability. if i DO understand that allocate thing correctly, it should look like this:

0 0 allocate 1 (no one dies)
0 1 / 1 0 allocate 2 (either of them dies)
1 1 allocate 1 (both die)

which would give you the 25%/50%/25% chance again

Probably best to forget that nonsense and just go with the standard way of calculating prob, which I understand. Forget it.

Your red text is not entirely correct though.

it is though :-P

mathematically:

P(entity1 (not) dead and entity2 (not) dead) = P(entity1 (not) dead) * P(entity2 (not) dead) = 0.5 * 0.5 = 0.25 ==> 25%

P(at least 1 entity dead) = 1 - (P(no entities dead) + P(all entities dead)) = 1 - (0.25+0.25) = 0.5 ==> 50%

also remember that the probability of an event and its inverse always have to be 100% or something in your calculation is wrong (the inverse of "at least 1 entity dead" would be "not exactly 1 entity dead", which would be 50% for 2 entities).

if you want to delve deeper into this, feel free to PM me.

Bottom line is if you got snipers in the m3 you will only have both standing after the explosion 25% of the time.

I Like the way they made it risk vs reward.

Does anyone know how to counter an m3 followed by a t70? maybe two 222's?

see also http://www.reddit.com/r/CompanyOfHeroes/comments/1i56cd/t70s_as_follow_up_to_m3_is_hard_to_beat/

Would really like some advice cos this is build i lose to most often.

Am i in a math class

Am i in a math class

You will be tested on this material next week.

I can't be bothered. I'm ok with a 25% chance for both to die and 50% chance for one or the other to die. That's all I really need to know.

Strafe.

The german halftrack base damage of the MG42 has been reduced? (or have I miss read) Was that necessary, the soviet satchel charge needs to be looked into, it's faster then MG42 pack up time and so even if they pack up they die before getting out of its range. Also thanks for the info