An Oddity in the Gamma World Artifact Use Charts

Yesterday, I discussed how to estimate the odds for the Gamma World technology charts using the RandBetween function on an Excel spreadsheet. 

But today, I want to look at an oddity in the charts themselves--or specifically at an oddity in Chart A. I have no idea if anyone else has remarked on this oddity. It may be quasi-common knowledge among some in the community or not. The only thing I can say is that I haven't seen it discussed, and I have been jumping around the blog posts on this issue a fair amount recently. 

Here's the oddity:

Without explicitly computing the probabilities or doing random simulations, it would seem obvious that the longer you spend in sustained concentration trying to figure out an item, the greater your chance of success. The rules also state that while you can fiddle for as long as you want, any interruption means you must begin again at the start. It would seem, therefore, that you don't want to stop or be interrupted. However, while it may seem that way, it's actually not the case. For Chart A, at least, it's actually better to stop or be interrupted after just a few rolls (and thus begin again at the start) than it is to try to continue rolling.

Now, it might be objected that this is a trivial observation, at least in certain cases. If you keep on getting high rolls, then you're probably moving closer to that skull and crossbones (again, henceforth A) and so in that kind of a case it's better to start over than to keep going. While this might be true, it's not precisely what I'm getting at. The falsity of the claim does not depend on the player knowing what his rolls are. The claim would be false even if the referee were rolling behind a screen.

Here's the basic point: while each roll gets you potentially closer to F, it also gets you potentially closer to A. However, it takes fewer rolls to get to F than to A. That implies that after a certain number of rolls (whatever they are) it might be better to start again at the beginning than to continue.

Remember the results of our first sample of 1000 attempts at rolling ten times:

F (success): 503
A (accident): 32
No Result: 465 

Now let's look at what happens when we roll only five times:

F (success): 324
A (accident): 5
No Result: 671

The 5 accidents make sense, since we saw yesterday that we know for certain that after only five rolls there is only a 0.63% chance of getting to A.

So, while taking ten rolls as opposed to five increases the odds of success by 50% or so (503 as opposed to 324) , it also increases the chance of catastrophic failure by perhaps 600% (32 to 5). The chances are still small, but when it comes to, say, dying, I'd rather have a really small chance--0.63%--than a small chance--3.0%.

But by stopping at five rolls, don't you also sacrifice your chance of success? So, isn't there a trade-off between risk and reward?

No, actually, there isn't.

How about this strategy: Make five rolls. If you don't get to F (or A), stop, go back and try again from scratch.
The results of this strategy will roughly approximate those below:

F (success): 543
A (accident): 8
No Result: 449

(We get 543 by using this formula: 324 + (.324 * 671). We get 8 by using this formula: 5 + (.005 * 671).)

So, it's better to start again after five rolls, than to continue on to ten. It's better to be interrupted.

In choosing two sets of five rolls (if we need them), rather than ten, we have roughly the same chance of success (or even more of a chance according to one simulated set of 1000 iterations) but a much lower chance of catastrophic failure.

But, of course if we had been thinking clearly, we should have already had a hint of this phenomenon. To see this, consider the results of only four rolls:

F (success): 227
A (accident): 0
No Result: 773

We had 0 occurrences of A. Should we roll 1000 more times to see if this was a fluke? No. As we saw yesterday, it takes a minimum of 5 rolls to get to A. It is impossible to get to A in only four rolls. We should have known that. We could have known that by simply looking at the chart.

So, how about a strategy of choosing three sets of four rolls (if we need them)? Here are the results:

F (success): 539
A (accident): 0
No Result: 461

So, again, the chances of success are about the same, but now we have completely eliminated the chance for catastrophic failure.

Hurrah, we've just come up with a foolproof scheme for sussing out simple artifacts without risk of breakage, injury or death!

We've also shown how Chart A is slightly broken.

Does this sort of thing also apply to Charts B and C? I'll leave that for another time. For now, though, how can Chart A be fixed?

Three possibilities come to mind. One is to make the minimum path to A shorter than the minimum path to F. Another (though this is potentially far more lethal) is to give some chance of getting to A from any position or at least from more of them. Finally, we could simply decree that being interrupted or voluntarily stopping does not mean that you go back to S. Rather, you always start where you left off from.

We'll discuss these in another post.

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Follow Save Versus All Wands on Twitter at the Twitter home of its author: @OakesSpalding

Gamma World: How to Determine Probabilities for Chart A

I've been doing work on a post-apocalyptic or far-future supplement to Seven Voyages of Zylarthen. The idea is sort of to do for Zylarthen what Mutant Future did for Labyrinth Lord--not a clone of Mutant Future (since that would be pointless) nor a precise clone of Gamma World (since that would be pointless and probably a violation of copyright) but a variation on Zylarthen, incorporating much of the Gamma World vibe.

Among other things, I've been looking at the 1st edition Gamma World technology charts--the mechanism by which the game simulates the process by which player-characters figure out the function and use of ancient technological artifacts.

There are three sorts of increasingly complex flow charts representing different sorts of artifacts--from laser pistols (Chart A, the simplest chart) to, say, permanent cybernetic installations (Chart C, the most complicated chart). The idea is that you roll a ten-sided die to negotiate through, with possible bonuses or penalties based on intelligence and how many others are available to help. The longer you take (the more die rolls you make) the greater the chance you'll figure out the artifact out but also the greater the chance you'll end up breaking it or even harming yourself or your party. Chart A looks like this:

So, after discovering an artifact--"a sort of jumble of sticks or tubes containing many grooves and colored knobs"--you start at S and want to get to F. In the process, you want to avoid getting to the skull and crossbones (henceforth A for accident). Depending on the artifact and the kindness or lack thereof of the referee, an "A" result could mean anything from "you break the trigger rendering it permanently inoperable" to "you shoot yourself in the face with a laser at point-blank range--roll 10d6 damage."

Now, I think the consensus is that the "chart system" is original and in theory very cool but it is usually a disappointment in practice. At worst it takes something that should be quite exciting--learning the purpose of some wondrous and powerful device--and turns it into something boring--rolling a die over and over again to track an abstract mechanic with no opportunity for player choice or referee creativity.

What can be done about that is a question for another time. And of course since Gamma World was published, all sorts of tweaks have been offered, in places ranging from the early issues of Dragon Magazine to the latest blog posts.

What I want to do first, however, is to look at the charts (or rather Chart A) from the point of view of probabilities. Given x number of rolls, what is the probability that you will get to F, or get to A or simply get nowhere at all?

For low numbers of rolls, the probabilities are easily calculated. Then it gets tougher:

From S to F (the number of rolls is the number on the left):

1: 0%

2: 0%

3: 10.5%

4: Complicated

Why 10.5%? Because there is only one path that will get you there in three rolls. Using a d10, you need to roll a 1-7, then a 1-5, then a 1-3. The chance of doing that is 70% * 50% * 30% or 10.5%.

From S to A:

1: 0%

2: 0%

3: 0%

4: 0%

5: 0.63%

6: Complicated

Why 0.63%? Because there is only one path that will get you there in five rolls. Using a d10, you need to roll a 1-7, then a 1-5, then a 8-10, then a 9-10 and then finally an 8-10. The chance of doing that is 70% * 50% * 30% * 20% * 30%, or 0.63%.

Though it might be complicated to precisely compute the odds at higher number of dice rolls for getting to F or A, we can roughly calculate them by performing simulations in Microsoft Excel. I imagine some of you mathematics wonks have done this or something like it in the past, or even have a better way of doing it using Excel or some other program. But I thought a few of you might be interested to see how I did it.

First you assign a number to each of the nine squares, circles and diamonds. I assigned them numbers from 101 to 109 so as not to get them confused with the pips 1 to 10 on a d10. I started at the top, going row by row down, and then left to right in each row where there was more than one shape. Thus, the top circle is 101, S is 102, F is 105, the diamond is 108 and A is 109.

We can then render the flow chart into this nested IF/THEN formula, where A1 is where you are and B1 is a random number from 1 to 10:  

 =IF(A1=102,IF(B1<=7,103,102),IF(A1=103,IF(B1<=5,104,IF(B1<=7,103,101)),IF(A1=101,IF(B1<=2,104,101),IF(A1=104,IF(B1<=3,105,IF(B1<=7,104,107)),IF(A1=106,IF(B1<=2,103,IF(B1<=4,102,106)),IF(A1=108,IF(B1<=3,106,IF(B1<=7,102,109)),IF(A1=107,IF(B1<=1,105,IF(B1<=5,106,IF(B1<=8,107,108))),IF(A1=105,105,IF(A1=109,109,0)))))))))

Once you get to 105 or 109, you stop, or rather, you "move" nowhere.

So, to simulate one set of rolls, you would create these cells from left to right:

• A1: "102" (that's because you always start at S)
• B1: "=Randbetween(1,10)" (that gives you a random integer from 1 to 10)
• C1:"=IF(A1=102,IF(B1<=7,103,102),IF(A1=103,IF(B1<=5,104,IF(B1<=7,103,101)),IF(A1=101,IF(B1<=2,104,101),IF(A1=104,IF(B1<=3,105,IF(B1<=7,104,107)),IF(A1=106,IF(B1<=2,103,IF(B1<=4,102,106)),IF(A1=108,IF(B1<=3,106,IF(B1<=7,102,109)),IF(A1=107,IF(B1<=1,105,IF(B1<=5,106,IF(B1<=8,107,108))),IF(A1=105,105,IF(A1=109,109,0)))))))))"

That gives you one roll. To simulate the next roll, add two more cells:

• D1: "Randbetween (1,10)" (or take B1 and copy it to D1)
• E1: Take C1 and copy it to E1.

And so on, for how many rolls you want.

After simulating the desired number of rolls, add three cells to record whether you ended up at F, ended up at A or failed to arrive anywhere. After, say, ten rolls, your ending point will be in U1. You will thus put a 1 in V1 if U1 is 105 (you ended up at F), a 1 in W1 if U1 is 109 (you ended up at A) and a 1 in X1 if U1 is anything else (you failed to arrive at F or A). You can do this automatically with more IF/THEN formulas:

• V1: "=If(U1=105,1,0)"
• W1: "=If(U1=109,1,0)"
• X1: "=if(V1+W1=0,1,0)"

You now have one row in Excel simulating ten (or how ever many) rolls through Chart A. This of course doesn't give you the odds of anything. It simply tells you what happened one time.

Now, pull that row down, say 1000 times. You may have to pull A1 down separately to get 102 in each cell. Otherwise B1 may become 103, C1 may become 104 and so on. Obviously, you don't want that.

Now you have 1000 rolls. And it only took you a few seconds.

Finally, add three SUM formulas at the top or the bottom (or anywhere else, really). It's probably better to add them at the top, thus instead of your sets of rolls going from rows 1 to 1000, you can make them go from rows 5 to 1004 or whatever. You want the sum of F's, the sum of A's and the sum of no results. Assuming you've pulled the whole set down so as to start at row 5, you then have:

• V1: "=SUM(V5:V1004)"
• W1: "=SUM(W5:W1004)"
• X1: "=SUM(X5:X1004)"

Let's try it for the first set of 1000 rolls. I get:

F (success): 503
A (accident): 32
No Result: 465

What about a second set of 1000 rolls? As you probably know, just by changing what is contained in one cell--say a dummy cell somewhere--every random result (all 10,000 of them) is automatically re-rolled by Excel.  So, now I get:

F (success): 558

A (accident): 33

No Result: 409

Here's a third set:

F (success): 528

A (accident): 28

No Result: 444

We have discovered a few things:

1. After ten rolls, while we do not know precisely what the chance of success is, it looks like it's 50% to 55%.
2. The chance of an accident is much much lower--only around 3%.
3. Roughly 40% to 45% of the time, you get nowhere.

I find that interesting. It's certainly not obvious from the chart. Among other things I would have thought the chance of an accident would be higher.

Of course it presumably would be higher if the character had a low intelligence. On the other hand it would probably drop do almost negligible if the character had a high intelligence.

I hope some of you find this useful and helpful. Though again, I imagine a few of you have already done this on your own. Despite its seeming complexity, it actually doesn't take very long to set up, at least if you're familiar with some of the simpler Excel formulas. Of course, writing the formula for Charts B or C would take somewhat longer.

If you can believe it, I did write the formula for Chart C and then I lost it (or maybe it's some anonymous "Workbook1" in my documents). But I did save the results. After ten rolls on Chart C the numbers are:

F (success): 90

A (accident): 78

No Result: 832

Yeah, there's an almost 10% chance you'll figure out how to operate that permanent cybernetic installation after only a few hours. On the other hand, there's also a close to 10% chance you'll blow yourself up (or whatever) in the same amount of time. 

Is that how you thought it would turn out?

Please let me know if you have any questions or if I made any mistakes on the above. I think it's all correct, but it's not very difficult to make mistakes with so many formulas. 

NEXT POST: An oddity that these results reveal about Chart A...    

Zylarthen Test Dungeon: 2nd Level

This is the second and probably last of the Test Dungeon levels, as I've learned a few things I want to incorporate in a larger randomly generated megadungeon. I was partly inspired by Barrowmaze, which was, according to the author, in large part generated randomly.

I think random generation is great. The apparent patterns as well as odd results act as 'imagination engines' to help fill things out. And obviously things would need to be filled out, not only to add tricks and traps but also the architectural and other details that would make the setting really come alive. I hope the Boars make a reappearance in the megadungeon, as well as of course the Sea Horses, but we'll just have to see.

Here are the rooms:

1. Monster: Pixie (1) AC 9, HD 1+2, hp 8, AT 1-6 plus various magical attacks and defenses MV 9/18, AL Neutrality LN Pixie, Common, Kobold (fluent), Hobgoblin, Leprechaun.
2. Empty
3. Empty
4. Empty
5. Empty
6. Treasure: 6,000 cp, Magic Shield, map to 1 unguarded gem (10 sp value), 64 miles away.
7. Treasure: 2000 cp, 2 jewelry (500, 30 sp value), Magic Shield.
8. Empty
9. Monster: Sea Horses, Medium (7) AC 7, HD 2+1, hp 8,8,6,11,10,13,6, AT 1-6, MV 0/18 Treasure: 11,000 cp.
10. Empty
11. Empty
12. Empty
13. Monster & Treasure: Orcs (2) AC 6, HD 1, hp 5,2, AT 1-6, MV 9, AL Chaotic LN Orcish, 400 gp, 3 jewelry (120, 70, 80 sp value).
14. Empty
15. Empty
16. Empty
17. Empty
18. Treasure: 200 gp, 6 jewelry (130, 850, 700, 40, 350, 90 sp value).
19. Empty
20. Empty
21. Empty
22. Empty
23. Monster: Cyborgs (2) AC 9, HD 3, hp 16,10, AT 1-6 plus assimilation, MV 12, AL Neutrality LN simple Common plus long range ESP.
24. Empty
25. Empty
26. Empty
27. Empty
28. Empty
29. Monster: Boars (5) AC 7, HD 3, hp 13,10,14,12,12, AT 1-6, MV 15.
30. Empty
31. Monster & Treasure: Poisonous Snakes (17) AC 9, HD 1 hp, AT 1 hp plus poison, MV 3, 400 gp, 8 gems (10 sp value each), 8 jewelry (5 sp value each).
32. Empty
33. Empty
34. Monster: Giant Ants (4) AC 3, HD 1/2, hp 1,2,3,3 AT 1-6 plus paralyze, MV 18.
35. Empty
36. Treasure: 350 gp, 2 gems (50, 5 sp value).
37. Treasure: 1 gem (10 sp value).
38. Monster: Ogre (1) AC 5, HD 4+1, hp 17, AT 3-8, MV 9, AL Neutrality LN Ogre, Merman.
39. Empty
40. Treasure: Cursed Sword -2, Map to 500 gp and 18 gems (10 sp value each), 3 miles away, unguarded.
41. Empty
42. Empty
43. Monster & Treasure: Giant Ants (8) AC 3, HD 1/2, hp 1,3,3,2,1,1,3,3 AT 1-6 plus paralyze, MV 18, 70 gems (10 sp value each).
44. Monster: Lizard Men (3) AC 5, HD 2+1, hp 9,10,4 AT 1-6, MV 6/12, AL Neutrality LN Lizard Man.
45. Empty
46. Empty
47. Empty
48. Monster: Boars (2) AC 7, HD 3, hp 17,8 AT 1-6, MV 15.
49. Empty
50. Empty

Notes:

1. I used the random generation guidelines from the four Seven Voyages of Zylarthen booklets as summarized in this post, covering the 1st level of Zylarthen's First Dungeon.
2. I wanted to create a random number of rooms--41-60--but I rolled a 10, giving me exactly 50 again.
3. This time I included the monster stats, using the format of B1: In Search of the Unknown. For the intelligent monsters, I randomly generated languages for them (LN), using the procedure described in Zylarthen, Vol. 4, pp. 32-36. If I rolled the same language twice, I assumed that meant fluency.
4. For monsters, on any roll of 1-2 out of 6, I randomly picked a previously generated monster. Presumably monster types would cluster.
5. Of course I rolled Boars again. Twice.
6. We now also have Sea Horses.
7. Does that have anything to do with the fact that the Ogre speaks Merman?
8. Treasure hordes were adjusted to reflect the guidelines in Vol. 4, p. 4. Basically I multiplied all results except those for cp's and magic items by 20%. There's still almost certainly too much.
9. This level makes it clear that if you use the treasure assignment scheme as given or close to as given--1 in 4 rooms have monsters, 50% of monsters have treasure and 1 in 6 unoccupied rooms have treasure (which roughly tracks the OD&D recommendations), then you may very well end up with more treasure not guarded by monsters than guarded by them. Even though much if not most of the unguarded treasure will be by assumption trapped or hidden, this still may not be a desirable ration for some.
10. I love the map to that one 10 sp value gem, 64 miles away (unguarded of course).
11. Also, getting a magic shield twice was interesting. I initially thought the number and power of magic items would have to be adjusted downward if using the random generation scheme previously outlined. Now I'm wondering whether they might need to be adjusted upward.
12. Overall, the 2nd level perhaps looks easy. But I think this is deceptive. Remember, the world of Zylarthen can be quite harsh. If you go down to zero hit points you're probably dead, maimed or very hurt.
13. I made the Snakes poisonous (you don't have to), but that turns these C level creatures into potentially the most deadly encounter on the level.
14. It is true that the 2nd level is lonely compared with the 1st. Instead of the teaming low-level humanoid melting pot of Bandits, Vikings, Martians, Half-Elves and all the rest, you get only 1 Pixie, 2 Orcs and 3 Lizard Men. Presumably however the Wandering Monster Table for the 2nd Level would feature many of the 1st Level humanoids.
15. And yes, all of them are wearing clothes. Don't even start with that.
16. Okay, maybe not the Lizard Men.

Randomly Generated Dungeon for Seven Voyages of Zylarthen

Or rather, the 1st level of one. There are 50 rooms:

1. Empty
2. Empty
3. Empty
4. Empty
5. Monster: 4 Bats
6. Empty
7. Monster: 2 Black Knights
8. Empty
9. Empty
10. Empty
11. Monster: 4 Skeletons
12. Empty
13. Empty
14. Empty
15. Empty
16. Empty
17. Empty
18. Empty
19. Empty
20. Monster & Treasure: 3 Elves, 4000 SP, 100 GP, 1 Gem (10 SP value)
21. Monster & Treasure: 2 Snatchers, Dagger +1, +2 vs. Goblins and Kobolds, Scroll of 1 Spell: Pass-Wall, Mail +4 (carried)
22. Treasure: 9000 CP, 1 Gem (10 SP value)
23. Monster: 3 Crocodiles
24. Empty
25. Empty
26. Empty
27. Empty
28. Empty
29. Empty
30. Monster: 6 Bandits, 60 CP (carried)
31. Empty
32. Empty
33. Treasure: 1 Jewelry (600 SP value)
34. Empty
35. Empty
36. Empty
37. Empty
38. Empty
39. Monster: 2 Giant Dragonflies
40. Monster & Treasure: 6 Vikings, 36 SP (carried), 1100 SP
41. Monster: 2 Gargoyles
42. Monster: 9 Red Martians, 90 pi coins (carried)
43. Treasure: 500 SP
44. Treasure: Scroll of 7 Spells: Animate Objects, Contact Higher Plane, Magic Jar, Telekinesis, Wall of Stone (2), Animate Dead, 1 Map to treasure horde of 1300 GP (97 miles away)
45. Empty
46. Empty
47. Monster & Treasure: 1 Boar, 800 CP
48. Monster: 2 Berserkers, 8 SP (carried)
49. Empty
50. Monster: 3 Half-Elves

Notes:
1. I used the guidelines on p. 4 of Vol. 4, The Campaign, figuring that 1 in 4 rooms would contain monsters, 50% of monsters would have treasure and 1 in 6 unoccupied rooms would have treasure.
2. Monsters were generated from the tables on pp. 9-10.
3. Magic Items and spells were generated from the tables in Vol. 3, Book of Magic.
4. For treasure hordes, I used the treasure class tables on p. 38 of Vol. 4. But I multiplied the results by 10% (11000 SP became 1100 SP, and so on) on any roll of 1-5 out of 6. Perhaps I should have let the CP numbers stand just to give the player-characters more to carry. 
5. For unguarded treasure or monsters without a treasure class, I randomly chose from treasure classes 1-3.
6. Amazingly, the total treasure value (if you include the horde referenced by that map) comes out to approximately 15,000 SP--well within the guidelines on p. 4.
7. I didn't cheat (not once). I suppose the scroll of 7 5th level spells in Room 44 might be adjusted, as well as (maybe) the Mail +4. Then again, the party might need some of those spells for a few of the monsters (1 magic dagger against 2 Gargoyles is tough).
8. Many of the monsters seem easy for 1st level characters, but obviously a few of the monsters are very powerful with serious TPK potential. This is a good dungeon level to teach players how to pick their battles (if they can) and when to hide or run away. Similarly, making friends (there are obviously a number of potential allies) or at the least not making enemies will also be paramount. You don't want those 9 Red Martians (all presumably armed with Radium Pistols) to have a problem with you.
9. I have no idea how the Boar got in there.

Of Orcs and Probability

In the last post, I addressed what in the comments section I called a "category error"—wrongly imputing the existence in the fantasy world of a unit of measurement in the game mechanic that existed only in our world. There are no hit points in the fantasy world, nor are there abilities such as Strength or Constitution, nor +1 modifiers to magic swords, nor players, GM's or potato chips. But there are genes (probably), creatures that are strong or have a sturdy constitution, magic swords of varying utility (some of which talk back to you), player-characters (as opposed to players), gods (sort of like GM's but easier to sweet talk) and almost certainly potatoes. Mixing categories, or mixing up worlds leads to all sorts of errors. Constructing a theory of evolution in the fantasy world based on hit points is a bit like basing it on potato chips.

But I want to leave that point for now. For the fun of it, let's assume that hit points do have a real existence in the fantasy world, like genes perhaps. And thus we can now go along with constructing a kind of evolutionary theory that might show how a community of, say, Orcs has its low hit point members gradually (or not so gradually) weeded out through fighting. There's no question that they would be weeded out eventually, of course, but it might be fun and interesting to track how the process would occur.

The initial set-up was specified in this blog post. You start with 100 one hit die Orcs with an average spread of 1-8 hit points—12.5 would have 1 hit point, 12.5 would have 2 hit points, and so on. They fight three rounds against a similar group of Orcs. Each Orc has an Armor Class of 6 and carries a spear that does 1-6 points of damage. At the end of this, how many Orcs would survive? Perhaps more interestingly, what proportion of each hit point group of Orcs would survive? And how would this change the overall spread of the different hit point groups?

The claim was made that:

From the final count, if we presume that any of these humanoids we're meeting have been in only 3 rounds of combat, only 1 in 100 humanoids should have 1 hit point.  Only 13% should have 3 or less. Nearly half, 45%, should have either 7 or 8 hit points.  More than three quarters, 77%, should have 5 or more.

This is because:

Trying to get my math right here.  The numbers are based on the chances of being hit once plus the chance of being hit twice and the chance of being hit three times, multiplied by the chance of any of those hits killing the humanoid.  The humanoid's attacks are not considered - only the chance of a humanoid with an armor class of 6 surviving three spear attacks during a given combat.

Actually, he didn’t get his math right.

That’s okay, we all make mistakes, and I owe a debt to that blogger for giving me an interesting puzzle in probability to occupy myself for an hour or so after the kids had gone to sleep.

The easiest way of seeing part of it is to focus on the weakest group—the 1 hit point Orcs. All they need is one successful hit against them and they go down. You don’t need to figure out the chances of, say, one hit versus two hits occurring, or how much expected damage might or might not be done with each strike. Rather, it’s quite simple: One hit and you’re dead.

What are the odds that you’ll be hit? Well, flip it—what are the odd’s you won’t be hit? Every round there’s a 60% chance you won’t be hit (1-12 is a miss, 13-20 is a hit). Thus, the odds for surviving three rounds are 60% x 60% x 60%, or roughly 22%. That roughly tracks this blog post’s leading graphic, above. 4 Orcs would reduced to 1.  12.5 Orcs would be reduced to 2.7. However, asking how many 1 hit point Orcs would remain out of 100 survivors has to take into account that many of the other stronger Orcs would also have fallen. As we shall see later, it comes to about 50%. So out of 100 surviving Orcs (imagine you started with 200), there would be 2.7 x 2, or roughly five 1 hit point guys still standing. So the proportion of the wimpiest would have been reduced from 12.5% to 5%. It’s a tough world. But not quite as tough or tough so quickly on the 1 hit point Orcs as was originally claimed. The answer after three rounds is not 1 in 100 but more like 5 in 100.

Quick digression: the original blogger pegged the to hit chances at 35% not 40%. Every early edition of D&D that I’ve seen, from the original 1974 version to the 1e Players Handbook, to Moldvay/Cook has one hit die monsters hitting AC 6 opponents on a 13. So why 35%? Most likely, the blogger made the same mistake that I sometimes make in my head: 13 to hit means a probability to hit of (20-13)/20, right? Wrong. Don’t forget to count the 13. The correct formula is 20 minus the highest roll to miss (12) not the lowest roll to hit (13). Or if you prefer, you can also just add 1 to the numerator. But pegging the chances at 40% rather than 35% actually kills the weaker Orcs quicker. If the chances to hit had only been 35%, then 27% or 3.4 (as opposed to 2.7) would have survived.

Moving on to computing the survival odds for all Orcs, I’m not going to explicitly go through the whole thing, but here’s a sketch:

1. The first thing to do is break down the odds for any Orc in a three round battle being hit 0 times vs. 1 time vs. 2 times vs. 3 times. You can represent it like this:

Permutation	1st round	2nd round	3rd round	# of hits	Probability
1	miss (60%)	miss (60%)	miss (60%)	0	21.6%
2	miss (60%)	miss (60%)	hit (40%)	1	14.4%
3	miss (60%)	hit (40%)	miss (60%)	1	14.4%
4	miss (60%)	hit (40%)	hit (40%)	2	9.6%
5	hit (40%)	miss (60%)	miss (60%)	1	14.4%
6	hit (40%)	miss (60%)	hit (40%)	2	9.6%
7	hit (40%)	hit (40%)	miss (60%)	2	9.6%
8	hit (40%)	hit (40%)	hit (40%)	3	6.4%

This gives totals of:

Chance of 0 hits	21.6%
Chance of 1 hit	43.2%
Chance of 2 hits	28.8%
Chance of 3 hits	6.4%

So, now, without doing any more calculations, we can also see why 7 and 8 hit point Orcs make out so well in the original blogger’s example. Since the maximum damage is 6 hits, an Orc with 7 or 8 hit points must be hit at least two times to be killed (and even then there’s a good chance he won’t be killed). But there’s only about a one-third chance this will happen. Why, in the original example, all the Orcs are fighting with spears as opposed to doing standard damage of 1-8, or using some of the better weapons assigned to Orcs in, say, the Monster Manual is a good question. But it shows up an interesting and almost paradoxical pattern. The less effective the weapons the worse the wimpier Orcs will fare relative to their 7 or 8 hit point comrades. The evolutionary process would happen quicker if Orc armies wielded daggers. Conversely, if the Orcs were all wielding two-handed swords or halberds, the proportions of surviving wimpy Orcs versus surviving strong Orcs would be less pronounced. Even doing the calculations with weapons that did 1-8 points of damage (as opposed to those spears doing only 1-6) would smooth things out on the survival curve (as opposed to the discontinuous break in the original example that separates the 7 hit point and 8 hit point Orcs out from the rest).

2. Now compute the expected damage chances for 1 hit, 2 hits and 3 hits. Here, we’re actually in familiar territory as we’re simply calculating the odds for achieving various totals using 1d6, 2d6 and 3d6. Many OD&D players almost carry those odds around in their heads.

3. Next multiply the two together in all the possible cases. I used an Excel spreadsheet, and again I won’t go though the details, but the final expected survival numbers after 3 rounds of battle are given on this table.

1 hp	2 hp	3 hp	4 hp	5 hp	6 hp	7 hp	8 hp
2.7	3.6	4.6	5.7	6.9	8.2	9.6	10.2

That gives the expected number of survivors as 51.5 out of a starting group of 100.

That’s actually an interesting number. It shows that for one hit die or 1st levelish creatures with moderate armor, if you (as a low level character) fight them for three rounds, you are likely to have reduced their numbers by about half. That’s another morale break point, I think. Low level OD&D combats shouldn’t last very long.

We can rewrite the results using percentages (by dividing the results by 51.5%). Drop the %’s and you have the number of Orcs in each hit point category out of 100 surviving ones:

1 hp	2 hp	3 hp	4 hp	5 hp	6 hp	7 hp	8 hp
5.2%	7.0%	8.9%	11.1%	13.4%	15.9%	18.6%	19.8%

These numbers are closer to those of the original blogger at the high ends (though not at the low ends), though they are still not quite as pronounced.  So replace

Only 13% should have 3 or less. Nearly half, 45%, should have either 7 or 8 hit points.  More than three quarters, 77%, should have 5 or more.

With

Only 21% should have 3 or less. Over a third, 38%, should have either 7 or 8 hit points.  More than two thirds, 68%, should have 5 or more.

Keep in mind, though, that raising the to hit chances to 40% helps the original blogger’s case. The numbers would be even more off if we had stayed with 35%.

Finally, you can rerun the numbers using the new proportions—5.2, 7.0 etc. vs. 12.5, 12.5, etc.—to find results if the surviving Orcs decide to fight additional three round battles. Sure enough, if you fight enough three round battles—3 actually—the wimpy 1 hit point Orcs will be reduced to that magic 1 in 100 number.

  

# Battles	1 hp	2 hp	3 hp	4 hp	5 hp	6 hp	7 hp	8 hp
1	5.2%	7.0%	8.9%	11.1%	13.4%	15.9%	18.6%	19.8%
2	1.9%	3.4%	5.5%	8.4%	12.4%	17.5%	23.9%	27.0%
3	0.6%	1.5%	3.1%	5.8%	10.4%	17.4%	27.9%	33.5%

But these numbers are quite different from those originally claimed:

hp	surviving after three battles
1	1 in 11,248 (!!! -ed.)
2	1 in 252
3	1 in 46
4	1 in 15
5	1 in 6
6	1 in 3
7	4 in 7
8	2 in 3

Again, see here.

Extra credit: instead of computing probability formulas, you can simulate Orc battles using the RANDBETWEEN and IF functions of Excel.

For the first to hit roll on a d20 it’s a1=RANDBETWEEN (1,20), then to compute damage you go b1=IF(a1>=13, RANDBETWEEN (1,6), 0). Do that three times and then add the “damage cells” to determine whether you have a kill. If f2 is the number of starting hit points, then the formula is =IF((b2+d2+e2)>=f2, 1, 0) where 1 is a kill and 0 is a survival. Of course starting hit points can be determined by =RANDBETWEEN(1,8). Or one can simply start out with 12 or 13 in each category. For successive battles featuring survivors, you can use a more complicated RANDBETWEEN or even RAND function using the new proportions.

If anyone has ever done something like this in Excel, you know that once you have the formulas set up, you just have to touch a random non-used cell (or fill it in) to simulate another battle and thus get a different result. It takes less than a second.

Probability can be weird. On only my fifth battle I had a situation where the 1 hit point Orcs actually survived in greater numbers than the 8 hit point Orcs.

More power to them!

Final note: I could of course have made all sorts or errors. If any readers would like to try their hand at identifying them, I would of course be interested and not offended. But please, no four-letter words, Alexis.