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...