Mr-Klipp
Joined: Aug 02, 2003
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 Posted:
Sep 13, 2003 - 10:16 |
 
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| Rimmer wrote: | | How you came to the conclution of one treemen being on the pitch (first half) in 75% of the cases I do not know. But that is utterly wrong since the dice roll is 50%. |
No such conclusion was made, read it again. What I said was that a treeman will play for 75% of all of the games his is in, not make the first half 75% of the time. This accounts for him missing 50% of 50% of his games. |
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sceadeau
Joined: Aug 02, 2003
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| Posted:
Sep 13, 2003 - 17:59 |
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| Rimmer wrote: | About the treemen dÃscussion.
First of all I am an mathematician so trust me that this is correct:
For a single treeman:
You have a 50% dice roll to field him the first half.
For 2 treemen:
Each treeman has a 50% dice roll of playing the first half.
This means that you have 4 different outcomes of the dice rolls:
(Succes-Succes), (Succes-failure), (Failure-Succes), (Failure-Failure) each of these four combination has the same probobility (i.e. 25%).
So to conclude:
1 Treeman is on pitch (first half) in 50% of the cases.
2 Treemen: 25% of the time you have them both, 25% of the time you have 0 and 50% of the time you have 1 (during the first half).
How you came to the conclution of one treemen being on the pitch (first half) in 75% of the cases I do not know. But that is utterly wrong since the dice roll is 50%. |
As has been stated multiple times...this math is so blindingly easy, an 8th grader could figure it out. Each treeman must be taken seperately.
A niggled played plays 5/6th of the time, mathematically. A double niggled player plays 5/6th of 5/6th of the time, so therefore, a niggled player's strength is discounted (5/6)^n, where n = number of niggles.
A treeman has only two possible outcomes when it rolls at the beginning. Either it will play for both halfs (100% of the game) or for the second half (50%) of the game. There is an equal chance per roll for it to miss the first half, and will always be there the second. Therefore, a treeman is discounted 3/4 of strength, as it will be on the pitch for 3/4 of your turns (on average, which is the only way to work out a global str formula).
The problem I have with most people and bloodbowl is that everything thinks it all depends on the die rolls. The die rolls don't matter, mathematically. It doesn't matter if you roll a 1 for take root, or a 6. What matters is what you should statistically roll over the life of the team.
It's a little like poker...you can win or lose each hand, but you should be playing for the statistical long term. |
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Falc
Joined: Aug 02, 2003
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Sep 15, 2003 - 04:37 |
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You're all correct, it's just that you're not talking about the same thing.
Rimmer correctly points out that your treeman will only play 50% of all FIRST halves. The rest otoh is talking about how much time a Treeman could spend on the field on average over many games. 'Could spend' because of course injuries and tactics might dictate otherwise, but that number is indeed 75%. |
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Rimmer
Joined: Aug 19, 2003
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| Posted:
Sep 18, 2003 - 15:50 |
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| Quote: | | No such conclusion was made, read it again. What I said was that a treeman will play for 75% of all of the games his is in, not make the first half 75% of the time. This accounts for him missing 50% of 50% of his games. |
A treeman will play 100% of all games. But he will only play an average of 75% of the gametime of all games. It is a big difference and that is why I misinterpreted you. |
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Pardus
Joined: Aug 02, 2003
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| Posted:
Sep 18, 2003 - 16:26 |
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the throw-ra and blitz-ra are calculated for when they still had ag3
simple |
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HunterX
Joined: Aug 02, 2003
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| Posted:
Sep 18, 2003 - 16:56 |
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