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The Hell of Stalingrad» Forums » Strategy

Subject: Empirical Probabilities in Hell of Stalingrad rss

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David Weiss
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I wanted to know what the probability of victory was on any given Break Test for a specified number of German and Soviet dice. Tried to figure out a closed-form algebraic expression but got bogged down. So instead I did a Monte Carlo Simulation (yeah, there's a joke in there somewhere) and came up with the following (tried various cut-and-pastings but couldn't preserve table format, sorry):

Soviet German Dice
Dice 1 2 3 4 5 6 7 8
1 41% 61% 70% 76% 80% 83% 86% 88%
2 25% 41% 52% 59% 65% 71% 74% 78%
3 17% 30% 40% 48% 54% 60% 64% 68%
4 13% 23% 32% 39% 46% 51% 56% 61%
5 10% 17% 25% 32% 39% 44% 49% 54%
6 8% 14% 21% 27% 33% 39% 44% 48%

BTW, this does not include the possibility of a Bloody Slaughter--too many other variables.

For example, if the Germans and Soviets both roll 2 dice then the German wins approximately 41% of the time, but if the Germans can get to roll even one more die (so that it's now 3 to 2) the probability jumps to about 52%.

What I glean from this table is the following:

(a) the Germans are going to have a ruddy tough time of it in general. Parity is far from enough; they need to be rolling at least 25% more dice than the Soviets just to have a 50% of winning (on the first roll, anyway--see below. Once the Soviets are rolling 5 dice you're basically just spinning the chambers and waiting to blow your own brains out.
(b) the Germans might want to prioritize buildings with the lowest innate defence, because each additional Soviet die makes it that much harder to storm;
(c) inferno inferno inferno (or bucket bucket bucket if you're the Soviets);
(d) Russian heros that steal Break Test dice are EVIL;
(e) maintaining a plurality of Shielded Units in buildings is a key way to boost your attack dice in the aftermath of a Bloody Slaughter.

Anyhoo, I welcome anyone checking my data and/or disputing my advice; it's a free Geekdom.
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