But wait, that's only part of the story. So far all we know is that the player will lose at least one crew member a little over half the time. However, there's another layer to the mechanic. The rule states that the player loses one crew for each monster showing. So although we know the player will lose some crew 55.6% of the time, as thorough game designers we should really calculate the average number of crew lost. Put simply, we must account for the fact that sometimes the player will lose two crew at the same time (by rolling two monsters).

To calculate the average crew lost, we need to do an "Expected Value", or EV calculation. This will be familiar to anyone who studies Game Theory, or perhaps anyone who reads a lot of poker books! (Sadly, I fall into both categories.)

EV is essentially the chance of something happening multiplied by the value of that result. When you're talking about poker, you are multiplying the chance of a hand showing up by the money you'll win if it does, but you are also including the chance of the hand not showing up multiplied by the amount of money you lose. If the EV is positive, it's a worthwhile value bet; if it's negative, then math says you'll lose money in the long run. Sure, you might get lucky here and there, but if you make the bet 1,000,000,000,000 times, you will come out behind.

For Longship, I just want to know the EV in relation to how many crew will be lost. Maybe after I do the calc it will make more sense, because all this writing is confusing me, too:

• EV = ChanceOfEvent1 * value1 + ChanceOfEvent2 * value2 + ChanceOfEvent3 * value 3 + etc.

  • Also seen as: SIGMAAAAAAAAAAAAAAAA

• EVaveragecrewloss = (pExactly1Monster * -1 crew) + (p2Monsters * -2 crew)

• pExactly2Monsters = 2/6 * 2/6 = .111 = 11.1%

• pExactly1Monster = .555 - .111 = .444 = 44.4% (we know we will hit at least one Monster 55.55% of the time and we know p2Monsters = 11.1%, so p1Monster is what's left.)

• EVaveragecrewloss = (.444 * -1.0) + (.111 * -2.0) = -0.667

Therefore, on average the player will lose 2/3 (66.7%) of a crew member whenever they raid. This is significantly different than the 55.6% chance of losing at least 1 crew. It should be accounted for during game balancing.

The Tragedy of the 2d6 Commons

This example illustrates one of the easiest probability mistakes made when assessing 2d6 combinations and results. On one hand, you can calculate the chances of, say, at least one "6" showing up on two dice (a). But if each individual "6" actually counts for something, then you really need to make sure that you account for it (b). If all you are looking for is "at least one six", then it doesn't matter (similar to the sea monster check earlier in this article).

To drive the point home and use another game example, let's say we have a pirate game (because we all should) and in the game, cannons hit long distance targets on a "6″. You have two cannons on your ship, so you roll two dice when firing.

• pChanceOfHittingAtLeastOnceWithTwoCannons = 1 - (5/6 * 5/6) = 36/36 - 25/36 = 11/36 = 30.6%.

In other words, there are 11 out of 36 combos for which at least one hit ("6″) occurs. You could have arrived at this number by brute force by counting of the combos instead of the converse prob calc. But converse probability was a bit faster.

Now let's calculate the "EV" of two cannons at long range, since this is what we'll use for game balancing. To do this, we need to know how those 11 special combos break down. As it turns out, 10 of the combos are hits with 1 cannon, and only one combo is a hit with both cannon. See, there are 10 ways to roll 2d6 and get a "6" on only one die, but there is one way to roll 2d6 and get a "6" on both dice.

• Average Number of Hits Scored = EVcannonhit = (comboswith1hit / 36 * 1 hit) + (comboswith2hits/36 * 2 hits) = [(5+5)/36*1] + [1/36 * 2] = 10/36 + 2/36 = 12/36 = .333 hits

You'll hit 30.6% (.306) of the time on average, but the average number of hits you will cause is .333. Sometimes ya hit 'em with both barrels!

_{Boom-chikka-Boom: 1/36^th of the time!}

If this is still confusing, then I guess I succeeded at least in illustrating the deceptively easy mistake you can make when looking at 2d6 combos. Most of the time, it's not a huge deal. But make a couple percent error when building a satellite, and it will pancake on Mars instead of nicely landing. Bye bye billions.

Back to the Longship!

The game, that is. Another rule:

Razing Territories:
Roll 1 die. The territory is razed if any axes show.

After taking the last treasure from a territory, there is a chance that it gets razed. This means that the place is temporarily exhausted of all valuable relics, coins, peasants, and so on. Razed territories take longer for the territory to refill with treasures. This makes some territories suddenly less valuable for a while, which creates some dynamic game situations.

I decided I wanted a simple 50/50 chance here. Design-wise, I didn't want it to raze all the time (playtested that, didn't like it) and I also didn't want things not to raze because I like the dynamic aspect.

Getting 50/50 was fortunately the easiest calculation yet. For this rule I went with a simple one die resolution to speed things up and also to differentiate the roll from the raid check immediately preceding it. If any axes show at all, the territory is razed:

• pRaze = any axes at all on 1 die = 2/6 (single axes) + 1/6 (double axe) = 3/6 = 50%

You've just been Plwnd3rd

As is tradition in my articles, if you've made it this far then you should either be commended or committed. Hopefully, this feature has helped illustrate some of the design process that goes into something as simple as a dice roll.

If you enjoyed this article (or didn't!), please be sure to let myself or the awesome Gamasutra editors know. It may affect how often they let me out of the box in the future.

Also, if Longship sounds interesting and you want to know about a potential commercial release, please drop me an email.

--TS

LEGALESE: Longship: Viking Raiders is ©2007 and TM Sigman Creative Services LLC. All rights reserved.