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JackassRampant
Last seen 5 hours ago
Human
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280/131/119
Win Percentage
65%
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109/45/60
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61%
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[R]
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730/275/314
Win Percentage
66%
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2014-02-27 02:15:42
6 votes, rating 4.3
Bloodweiser Babes and Monty Hall
I didn't do a web search for this: if someone else has observed the parallel first, please send me there.

But I was doing the math on some inducements and I had a revelation. People like to think of Babes as if they were really additive, but they're not. See, on the one hand it is true that one Babe adds +1 and two adds +2, and that one Babe will, in the aggregate, save 1/6 of a KO'ed player per drive, and two will double that to 1/3 of a player. Fine. But that's not what Babes are used for.

See, the salient math is this: 0 Babes means each KO is a 50/50 shot that the player doesn't return. At the time this 50/50 shot is generated, if it fails, it generates a d3 result.
* If you took 0 Babes, this d3 result goes unnoticed, just like the implicit d3+3 result generated from a success (4-6) on the recovery roll.
* If you took 1 Babe, she effectively acts as a 5+ save (3+ on 1d3, but the math is the same).
* If you took 2 Babes, they act as a 3+ save (2+ on 1d3, as above).

But wait! One 3+ roll is better than 5+ with a re-roll: in fact, a 3+ is equal to a 5+ with a re-roll at 4+. So if you take a second Babe, the second one is 50% better relative to the first one when it comes to the math you actually care about.

This is the Monty Hall Problem, writ for Blood Bowl. If you are considering two Babes relative to a 100k inducement, then it's one thing: the Babes represent a 3+ save or 1/3 recovery, fine. But if you are considering two different packages and one has one Babe while the other has two, consider that the second Babe is a noticeably better bargain than the first one. Not enough to throw you off what you know you want, but it's better to know that if one Babe really is worth 50k, the other one is worth 75k.
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Comments
Posted by pythrr on 2014-02-27 07:52:18
bollocks!
Posted by SzieberthAdam on 2014-02-27 07:52:41
recovery: 0: 3/6; 1: 4/6; 2: 5/6
rel_improvement(1; 0): (4/6-3/6)/(3/6) = 1/3
rel_improvement(2; 1): (5/6-4/6)/(4/6) = 1/4
First one is better. :)
Posted by fidius on 2014-02-27 08:34:09
baa
Posted by bigGuy on 2014-02-27 09:20:44
2 babes are better than others 100k inducements (in most situations)
And WA is better than Igor
Posted by jamesfarrell129 on 2014-02-27 09:51:16
I wish I could pass the "2 Babes" test :-(
Posted by Verminardo on 2014-02-27 11:17:13
I don't get the math but I get how one babe is good but the second babe is much more awesome.
Posted by gjopie on 2014-02-27 12:08:55
Sorry, don't understand "But wait! 5+/5+ is not 3+." Where does 5+/5+ come from? First roll is a 4+, second roll is either a 5+ or a 3+. So surely you should compare 4+/5+ to 4+/3+? Might have totally missed the point though.
Posted by the_Sage on 2014-02-27 15:14:04
why should the relevant improvement be the relative one Szieberth? I don't agree with you there at all.
Posted by JackassRampant on 2014-02-27 16:33:19
Edited for clarity.
Posted by JackassRampant on 2014-02-27 16:38:42
@ gjopie: A KO fail (before Babes) generates a random number between 1 and 3. One Babe gives you a 3+ save on this number. 3+ on 1d3 is identical in difficulty to 5+ on 1d6. Two Babes gives you a 2+ on that d3, which is identical to a 3+. 5+ is exactly half the success odds of 3+, and twice the fail odds. So the second Babe functions like a 4+ save should the first one (a 5+ save) fail.

@ Szieberth: Because you don't make a proper sample of KO rolls, the additive math doesn't matter. If you were rolling dozens of dice and only caring about your success rate, the impact on the aggregate result is the salient reading, and in that case, +1 on a 4+ is more bang than +1 on a 3+. And if you only cared about your total number of successes, +1 is +1 and it doesn't matter what the base is, as long as it's no greater than the size of the die. But here, you're not making enough rolls to just aggregate it all out: you're making just one or a few rolls most likely, and you're really interested in the failure odds, because recovery fails on an individual basis are the major killer.
Posted by Jeffro on 2014-02-27 17:23:20
So is the Monty Hall effect anything like the Benny Hill effect, where when you have one babe playing with herself that's all well and good, but two babes making out is more than twice as good...?

That math makes perfect sense to me :)
Posted by JackassRampant on 2014-02-27 18:58:09
What this really means I guess is that you can't think of 2 Babes "one Babe, taken twice." One Babe is a rounding error, an alternative to a card for low-AV teams or teams with no reserves. Two Babes are an alternative to a WApo, and their value depends on the odds of taking multiple KOs in the first half. That is, if you think you'll have three KOs, which means 5 guys out including Casualties, in the first half, two Babes are clearly better than an Apoth, and that threshold falls to 2 KOs if you think you'll get two recovery rolls. So, I'd say: Babes over Igor always except in league games at high level, Babes over WApo on AV7 teams or when inducing lots of stars. For AV9, always WApo over Babes, don't be afraid to use on KO (if just one guy goes out, a WApo is better than 3 Babes).
Posted by pythrr on 2014-02-28 01:09:46
jeffro

YES