Celyn
Joined: Aug 02, 2003
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 Posted:
Jun 01, 2010 - 16:11 |
 
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Ropetus: Well, you are allowed to make blocks before the Blitz, no leap needed. The other plan is solid, though, and I use it often. |
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Purplegoo
Joined: Mar 23, 2006
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Jun 01, 2010 - 17:50 |
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I'm going to spend some time either tonight or tomorrow playing with some of these suggestions and see what I think is the best in terms of rolls (on the base 'no skills Woodies with 11 v 11 problem). It's an interesting topic! I like how there isn't an answer that someone can point to for 100% yet.
Thanks to all for the input so far. I think we should have more of these chats!
On my original setup I used to start the ball rolling, the more I look at it, the more the empty row needs to go and the fingers crossed no QS needs to be taken into account! I have trouble getting it in diagrams, need to see them in the client to fully visualise... |
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maysrill
Joined: Dec 29, 2008
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Jun 01, 2010 - 18:07 |
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It might be interesting to translate this into some sort of contest form:
One person proposes their setup (11 generic ST3 players, no skills) versus an 11-player rookie woody team (no rerolls).
Then everyone posts their "solution" and its theoretical chance of success.
We can repeat the process with other defensive setups for the same players, and at the end come up with a "best" defense based on the lowest "best" response by the woodies.
Math for taking QS into account might make it tough on people getting a numeric answer, but if some people want to go the extra mile, it would be more accurate. |
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Author of Firehurler (Twinborn Trilogy Book #1), Aethersmith (Book #2), Sourcethief (Book #3) |
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Purplegoo
Joined: Mar 23, 2006
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Jun 01, 2010 - 18:08 |
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That's a fine idea! There have been some good ones suggested, where should we begin? |
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maysrill
Joined: Dec 29, 2008
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Jun 01, 2010 - 19:18 |
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To work out the bugs, probably start with the base assumption: 3 LOS (packed together) and 8 spaced along the back, 1 square before the end zone. |
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Author of Firehurler (Twinborn Trilogy Book #1), Aethersmith (Book #2), Sourcethief (Book #3) |
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Purplegoo
Joined: Mar 23, 2006
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Jun 01, 2010 - 19:39 |
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So, that's no PD, no Blitz, two pushes, 4+, 3+, 2+, 2+, 2+, with Dodge RR and team RR, correct? We'll ignore PD or TAR, since we don't know FF, etc.
I sense the need to find my calculator, normally I just try and do it, working out how likely it is is a bit of a pain! 
Who loves this stuff and is less lazy than me?!  |
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maysrill
Joined: Dec 29, 2008
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Jun 01, 2010 - 19:44 |
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Well, that's a basic solution to the basic problem. Others could look to see if 3 or 4 pushes to eliminate the GFI makes it better or worse. |
_________________
Author of Firehurler (Twinborn Trilogy Book #1), Aethersmith (Book #2), Sourcethief (Book #3) |
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James_Probert
Joined: Nov 25, 2007
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Jun 01, 2010 - 19:59 |
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I'm going to start by calculating for no RRs, as it make the rest a bit easier to see.
Assuming you can get 2 dice for both blocks, it's 20/36 to get a push on one dice.
The second block it doesn't matter if it's a push or a drop, which makes it 32/36, or 8/9.
Then there's 4+, 3+, 2+ sequence of dodges, and 2 2+ GFIs.
This comes to 625/6561 probability of not needing a RR at all to succeed.
I'll be back to you with the RR adjusted probabilities. |
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Purplegoo
Joined: Mar 23, 2006
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Jun 01, 2010 - 20:17 |
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I think that's OK, I mean, if we assume everything is the same, I guess it doesn't matter so long as we get an order. The order doesn't change with RRs, it just gets easier, right?
Good job on the maths, I can never be arsed! |
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maysrill
Joined: Dec 29, 2008
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Jun 01, 2010 - 20:39 |
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If the second push doesn't include a knockdown, you'll have to dodge away from the LOS to start with.
Also there's the small matter of getting the ball into the hands of the scorer. Do we want to assume a touchback to make things simpler? It would highlight just how much easier it is for certain teams to make the OTT work. A skaven team retrieving the ball with one gutter and pushing another for the score are actually a lot less likely to succeed than a wood elf thrower with Sure Hands and a catcher being pushed. |
_________________
Author of Firehurler (Twinborn Trilogy Book #1), Aethersmith (Book #2), Sourcethief (Book #3) |
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Purplegoo
Joined: Mar 23, 2006
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Jun 01, 2010 - 20:43 |
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| Reisender wrote: | oh yeah it actually would. placement of the other guys still to be discussed -> will need to work put push possibilities)
_ _ _ _ | _ X _ X _ X _ | _ _ _ _
_ X _ _ | X _ _ _ _ _ X| _ _ X _
_ _ _ X | _ _ X _ X _ _| X _ _ _ |
I made that easier than the standard, assuming I could move the front row, centre guy. As I think you mention later.
| Reisender wrote: |
_ _ _ _ | X X_ X _ _ _ | _ _ _ _
_ X _ _ | X _ _ X _ _ T| _ _ _ _
_ _ _ _ | _ _ _ _ _ _ _| _ _ _ _
_ X _ _ | X _ _ X _ _ T| _ _ _ _ |
That, however, is tougher. Makes it a 3, 3, 3, 2 Dodge run instead of 4, 3, 3, and you have to move the guy on the LOS not touching, and you need a 4 to fill a hole on the pushes, by my count? Is it easier than that via a different route?
| maysrill wrote: | | Also there's the small matter of getting the ball into the hands of the scorer. |
Good point, lets assume a touchback, but you have to leave player #11 back to pass the ball incase it's not, so he can't contribute to the pushes. |
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Chingis
Joined: Jul 09, 2007
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Jun 01, 2010 - 21:17 |
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| Purplegoo wrote: | | I think that's OK, I mean, if we assume everything is the same, I guess it doesn't matter so long as we get an order. The order doesn't change with RRs, it just gets easier, right? |
'Fraid not.
Consider a chain of 1/6, 3/6 and 4/6 rolls. (Case 1)
Then a chain of 2/6, 2/6 and 3/6 rolls. (Case 2)
Both have the same probability: 12/216, or 1/18.
Then if we have re-rolls:
Case 1:
Re-roll needed first roll: (5/6 x 1/6) x 3/6 x 4/6 =60/1296
Re-roll needed second roll: 1/6 x (3/6 x 3/6) x 4/6 =36/1296
Re-roll needed third roll: 1/6 x 3/6 x (2/6 x 4/6) =24/1296
No re-roll needed: 12/216 =72/1296
Total =192/1296
Case 2:
Re-roll needed first roll: (4/6 x 2/6) x 2/6 x 3/6 =48/1296
Re-roll needed second roll: 2/6 x (4/6 x 2/6) x 3/6 =48/1296
Re-roll needed third roll: 2/6 x 2/6 x (3/6 x 3/6) =36/1296
No re-roll needed: 12/216 =72/1296
Total =204/1296
Good luck!  |
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MadTias
Joined: Jun 19, 2004
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Jun 01, 2010 - 21:25 |
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| CircularLogic wrote: | | I feel the MadTias2 is especially easy to crack. |
Yea, I worked through my setups at work today.  They are surprisingly easy to crack! Just shows that I haven't given this pretty important topic enough attention in my 2000 games.
Actually, having spent some time on this today, I couldn't find a set-up that was better than "3 on the LOS, rest in the back". Nothing I came up with forced harder dodges than into 2TZ-1TZ-0TZ, like that one does. My current analysis is that a rookie team might as well set up "3 on the LOS, rest in the back". Give me a SF and a couple of Guard and Tackle though... |
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James_Probert
Joined: Nov 25, 2007
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Jun 01, 2010 - 21:40 |
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Long Post Warning
feel free to skip to the bottom for the results if you can't be bothered to read it (or don't understand )
sequence: Block 1, Block 2, 4+ dodge, 3+ dodge, 2+ dodge, 2+ GFI, 2+ GFI
P(block 1 succeeds)=20/36
P(Block 1 RRd Succeeds)= 20/81
P(Block 2 Succeeds)=8/9
P(Block 2 RRd Succeeds)= 8/81
P(4+ succeeds) = 1/2
P(4+ dodgeRR used) = 1/4
P(3+ succeeds) = 2/3
P(3+ RRd to success) = 2/9
P(2+ dodge succeeds) = 5/6
P(2+ RRd success) = 5/36
P(2+ succeeds) = 5/6
P(2+ RRd success) = 5/36
P(2+ succeeds) = 5/6
P(2+ RRd success) = 5/36
these are the probabilties of the individual steps succeeding, the it becomes a matter of considering the different permutations of success with only allowing ourselves 1 RR, plus 1 of the dodges RRd
I'll start with the dodges, as that's a stand alone permutation tree due to it's dodge RR:
P all dodges, no RR = 5/18
P(dodges, RR 4+) = 5/36
P(dodges, RR 3+) = 5/54
P(dodges, RR 2+) = 5/108
P(total, no TRR used) therefore= 5/9
(I did this on paper to make it easier to see, so just the result here)
P(TRR used)=55/648
So then: our different nodes now look like this:
P(block 1 succeeds)=20/36
P(Block 1 RRd Succeeds)= 20/81
P(Block 2 Succeeds)=8/9
P(Block 2 RRd Succeeds)= 8/81
P(dodges no TRR)= 5/9
P(dodges TRR)= 55/648
P(2+ succeeds) = 5/6
P(2+ RRd success) = 5/36
P(2+ succeeds) = 5/6
P(2+ RRd success) = 5/36
then, for the chance of success, we simply choose 1 or 0 from the cases where RR is used, and sum all those permutations.
block 1 RRd: 5000/59049
Block 2 RRd: 1250/59049
dodges RRd: 0.029107
GFI 1 RRd: 625/19683
GFI 2 RRd: 625/19683
no TRR used: 1250/6561 (more than before because of dodge RR factoring in)
the sum of this method of getting a catcher to the end-zone is therefore: 0.3890
that is 38.9% success if you have a TRR available
else it's 0.191, 19.1% if you don't have one available (which is entirely possible at the end of a half) |
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Purplegoo
Joined: Mar 23, 2006
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| Posted:
Jun 01, 2010 - 21:43 |
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Yes, I see your point, however, so long as we assume the case is the same every time (for instance the player has Dodge and the team a RR left - or none of the above, or whatever), surely we can just take a worst case for every setup, ergo the order will remain the same? Or have I misunderstood your point?
As for James - I definitely think I need to be awake with a cup of tea to follow that through, not on the back of a day's work. However, at first glance, yep, agreed (no smirking you at the back, really! ). |
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