Archive for the 'Gaming' Category

End of an era

November 6th, 2009 — Wordman

During and after college, I invested a whole lot of energy into the setting and game of Shadowrun. About all I have to show for it is a large shelf of books that hasn’t really been touched since I moved into my house six or so years ago.

So, I’m selling it all. (And also, some separate fanzines.)

Well, I am keeping a few things. I’ll keep my limited edition hardbacks from third and fourth edition (and the new 20th Anniversary limited edition, if it ever ships), hardbacks from 1st and 2nd edition, an extra copy of the greatest gaming aid in the history of man, and the Denver boxed set. And memories, I guess. And PDFs.

Popularity: 1% [?]

Roleplaying industry predictions

June 19th, 2008 — Wordman

If you follow gaming at all, you know that the fourth edition of Dungeons & Dragons has been released. As I suspected, the rules are, essentially, the rules that govern a video game captured in book form. As I did not suspect, the result is actually pretty good. It’s still very crunchy (i.e. rules-heavy), but it is pretty well designed crunch, with a lot of design focus given to keeping what is fun and ditching what is not.

page imagesThe effort is helped quite a bit by some interesting layout choices. In particular, their use of white comes as welcome change from their 3E design and somehow looks modern and slick. After years of role-playing products designed with ink on most of the page, usually with some kind of pale or gray image as the background, the open style used in 4E might change the way a lot of books get designed. I remember noticing the use of white in the gorgeous Ptolus and wondering why more RPG books didn’t use it. Fourth Edition’s color and font choices (mostly a family called Mentor) are also an interesting break from their past and, I think, well selected.

A few days ago, Wizards of the Coast finally unveiled their new Gaming System License (GSL). Not so good. It’s many shortcomings are being debated in forums all over now, but a thread on ENWorld is particularly notable, as it includes several publishers of 3E supplements using the Open Gaming License (OGL). Posts from a user named Orcus are worth reading, in particular, because he is a lawyer as well as a game publisher (Clark Peterson from Necromancer Games). Even though Necromancer’s page currently claims certain products will be ported to 4E, this forum indicates that at least some of the books mentioned (something called the Tome of Horrors, in particular) will not be, now that the details of the license have been unveiled.

One of the main issues is what some are calling the “poison pill” clauses (even though it isn’t really a classical poison pill). Essentially, it makes converting a “product line” (whatever that means) from the OGL to the new GSL a one-way process, and contains language that essentially would put a publisher’s future into the hands of Wizards of the Coast. I’m not a lawyer, but from how I read the GSL (PDF here), it seems to me that publishers would be fools to sign it as it is presently written, particularly if they already have created OGL content. It also pretty much shuts down any fan-based computer tools, like character trackers, reference tools and the like (though there will supposedly be a “fansite” related policy released later).

So, unless the license is changed, you can bet that not many publishers are going to fully embrace it, though many will probably make a few books for it. I’m guessing that most of the following will occur:

  • Well established product lines (that is, those who could claim a measure of brand recognition and loyalty) will continue to publish these lines under the OGL. They will eventually be forced to remove the d20 logo from them, but it will probably not matter.
  • Well established companies, if they publish for 4E at all, will do so with entirely new product lines. They will be able to leverage their name, but not their brands.
  • A few new companies will arise that make only 4E products, but will focus more on adventures than anything else. At least one of these companies will, in fact, be owned by another established company that is still publishing OGL-only material.
  • After realizing that the GSL won’t let it follow its current plans for 4E integration, Pathfinder will remain 3.5, and will become something of a flagship product for those still using the OGL, mutating into the semi-official mechanism by which the 3.5 rule set evolves.
  • Because of this, friction will be created between Wizards and Paizo, Pathfinder‘s publisher. Since Paizo has a very close relationship with Wizards, the result will be that Pathfinder will be sold before the year is over.
  • More than half of newly created companies that enter into the D&D related publishing business will publish under 3E rules using the OGL.
  • The amount of shelf space given to 4E products in gaming stores will not exceed that given to 3E products. Ever. In Borders and other large book stores, however, you won’t be able to find 3E products at all.
  • Numerous fan sites will emerge that convert third party OGL content into 4E. They will be constantly under threat from Wizards of the Coast, who will pay a lot of legal fees to continually fight them as they shift around.
  • Wizards will become more vocal about copyright infringement on p2p networks.

I’d love to see some other open source game take off in the wake of GSL backlash. Unfortunately, if this happens, it is likely to be Pathfinder, rather than a better system, such as FATE (also available under the OGL license) or Wushu Open (released under the Creative Commons). Even more unfortunately, there aren’t that many other viable alternatives. There are certainly a number of free RPGs out there, but few of them are open source.

Popularity: 6% [?]

Inside joke

May 19th, 2008 — Wordman

Only seven people in the world will actually understand this, much less think it is as funny as I do. And only half of them are probably reading this. But hey, my blog, my rules.

In doing some spring cleaning, I came across this scrap of paper in one of the many piles in my office:

Tanador note

Good times.

UPDATE: not long after posting this, I got a “mysterious” text page from one of the seven people saying “There is no rash.” Trust me, that was hilarious.

Popularity: 6% [?]

Open letter to White Wolf

April 15th, 2008 — Wordman

To: White Wolf

The advantage of electronic books is that they are easier to store, searchable and, until now, cheaper.

As you know, electronic versions of your two recent releases (Yu-Shan and Scroll of Kings) are listed for $18, nearly $5 more than books with equivalent page counts released just months ago. That’s a price increase of almost 50% and marks the first time I can remember the electronic version of one of your books costs more than the print version. While retail for the print version is $25, Amazon sells it for $17. (They also continue to sell the “books with equivalent page counts” mentioned above for $17.)

As someone who has legally purchased electronic copies of nearly all of your First and Second Edition Exalted titles, I find this, of course, extremely irritating. But, more to the point, if this price change is here to stay (which I hope it doesn’t), then I will now be much more demanding of features in these electronic books that, until now, I’ve been giving you a pass on not providing. In particular, for the additional $5 for a bunch of electrons, I now expect and demand…

  • …reduced security. At the very least, I should be permitted to edit and save my own bookmarks and have the ability to add margin notes and save them. At best, eliminate it entirely. (Yes, I do know how to strip it off, but I’d prefer not to have to.)
  • …free updated versions of all affected files whenever you make corrections or errata to existing books. (Other companies, much smaller than you, do this already, by the way.)
  • …the person producing the PDF to spend time to make sure the file size is small and the page render times fast. Many of your books (particularly the White and Black Treatises) have exceedingly long draw times. (A good test here is to keep clicking on the “next page” button. If you do this quickly and the majority of the pages barely render before you click the next one, it’s to slow.)

Or, you could, you know, put your prices back down to a reasonable level.

I learned a while ago that I follow the following pattern when buying gaming books, even if I can’t explain exactly why: if the PDF costs around a third of the cost of the printed version, I buy both the printed version and the PDF. If the PDF costs around half the cost of the printed version, I buy the PDF only. If the PDF costs more than half of the printed version, I buy neither.

Update: I thought posted this a while ago, but it looks like I didn’t. In the interim, White Wolf released a new “fatsplat” book for the same price as other flatsplats when they were first offered. Older flatsplats are now $16, so it looks like White Wolf might be pricing at a premium when the book is initially released, then reducing prices later. I think this practice really, really sucks, and has made me take another big step toward abandoning Exalted entirely. In a much better move, they also, for the first time, reissued a title with corrections as a free upgrade. While I welcome this development, I must note that it is much less compelling when the bookmarks in the new version are much, much worse than those in the original. Given how easy it is to automatically generate bookmarks in programs like InDesign, this is disgusting.

Popularity: 5% [?]

Fitting tribute

March 5th, 2008 — Wordman

Does anyone else think we should start a fund to build a mausoleum for E. Gary Gygax (who died Monday), based on the layout of the Tomb of Horrors? I mean, building a gelatinous cube and the sphere of annihilation would be a challenge, but it would be so worth it. Plus, Lake Geneva could use a tourist attraction. So what if it would kill 98% of the people who went into it.

map
Image thanks to Mad Irishman, and is probably © Wizards of the Coast

Update: It also appears that EGG’s death is causing a number of people to break out their old, original D&D books to play “tribute” games. I suspect that people might wind up seeing how much better the original was in some ways (worse in others, of course). For example, people who have never played a D&D game that didn’t involve feats may be in for a surprise. It would be cool if this brought more people to the idea that simple rules sets are often better, and bail out on all these systems that are built around micro-rules, such as feats and charms and spells and so on, that really exist as a driver to sell more books.

For those itching for some original AD&D goodness, but who don’t have the books and can’t find them in a torrent somewhere (*cough*), there is an alternative. The Old School Reference & Index Compilation (OSRIC) is an open license version of what are basically the original AD&D rules, legally sanitized to remove “artistic representation” and trademarks owned by TSR Wizards of the Coast. Happy hunting.

Popularity: 6% [?]

Chance of Reign

January 31st, 2008 — Wordman

Based on an opinion voiced in an Exalted forum, I picked up the hardback of Reign, a game self-published by Greg Stolze. Stolze is the co-author of the incredibly good Unknown Armies and wrote the best “how to run a game” chapter that I’ve ever read (published in the otherwise unremarkable Gamma World Game Master’s Guide), so I was excited to see how he tackled the epic fantasy genre. Mechanically, he does so with something called the One Roll Engine. (If you want to sign up to a site with a not-very-helpful user interface, you can download a version of the basic system for free.) This engine makes use of dice in a way I haven’t seen a game utilize before, and I figured I’d blog about the probability of the system because a) I didn’t immediately know how to calculate it, b) a post on how to do such probability calculations will hopefully be useful to fellow gamers and c) I wanted an excuse to play with the statistical and graphing features of Numbers.

Reign uses pools of ten-sided dice (d10s), where results are determined by looking at how many dice match each other. It is a bit like poker, where a roll might generate, say, a three-of-a-kind, a pair, and some singles. Each of these matches is called a “set”, and the number of dice in the set is called the “width” of the set. So, for example, if you roll five dice and get 4, 4, 4, 2, 8, you have one set of width three. For many rolls in Reign, the actual number being matched, called the “height” of the set (4 in the previous example), doesn’t particularly matter. For other rolls, only the height matters. Still others rely on both the height and the width. Usually only a single set in a roll matters, but there are some mechanics that make use of the other sets as well. In this way, the rules in Reign pull a lot of meaning out of a single roll mechanic, and the probabilities employed are interesting but not obvious. I’ve done some (very) minor probability documentation on some other systems (an invented five-suit card game and the “excellencies” of Exalted, for example), but this is a bit more tricky.

Simple calculations

The first basic observation to make is that if you have a pool with more than 10 dice in it, you will always get at least one set. If you have n+1 items that can only have n distinct values, at least one of those values will come up more than once. Reign is well aware of this, and has rules to match. Consequently, pools only exceed 10 dice in very extraordinary circumstances; however, these circumstances, though rare, do exist in the game, so this post will consider results with up to 15 die pools.

Secondly, for any given roll, the odds of achieving exactly a specific height are identical to achieving any other exact height. That is, odds of rolling two tens using 5d10 are exactly the same as rolling two nines. This means that, were you to build a chart where rolling height x means one thing and height y means something else, you have equal chances of getting either result. Thus, if you want one result more than the others, you need to assign that result to more than one height on the chart. Reign makes use of this idea at least once in the rules (as a hit-location table). More often, however, height is used in rules where the height must be at least that number or higher. This makes reaching a certain height harder as the number increases. A height of 10, for example, can only be reached if the dice conspired to built a set containing 10s, but a height of nine is reached by those same sets plus sets containing nines, making it more likely. As a result, when this post considers height, it will do so in the sense of reaching at least that height.

Tools

Because Reign deals with sets, the key technique in calculating probability is combinatorics. In particular, the concept of a combination is crucial. This looks scary, but you probably had to do some of this in high school math class. The basic idea is that you have some number of items in your hand, and you have to select a certain (smaller) number of those items. The combination calculation measures how many different ways you could make such as selection, where order does not matter. For example, say you have four cubes in your hand, one red (r), one blue (b), one green(g), one violet (v). You are instructed to select two cubes, where the order doesn’t matter. You can do this, it turns out, six different ways: rb, rg, rv, bg, bv, gv. This would be notated as C(4,2), read “four choose two”. Generically, C(n,k), or “n choose k“. You may also have vague memories of a permutation, which is the same idea as combination except that order does matter. For example, rb and br are considered the same combination, but different permutations. A permutation of two items from a set of four is notated as P(4,2). Calculating both combinations and permutations makes use of the factorial operator. This is a fairly simple idea to wrap your head around. A number like “4 factorial”, notated “4!” just means to multiply 4 × 3 × 2 × 1. So n! just means n × n−1 × n−2 × … × 3 × 2 × 1. Combinations and permutations are calculated as follows:

C n = ( n ) = n!   P(n,r) = n!
k k k! (nk)! (nr)!

Fortunately, before you run screaming from this, most spreadsheets (and more advanced calculators) have functions for combinatorics. Excel, for example, uses combin(n,k). It also has permut(n,r), fact(n) and power(x,y), which are also useful. (One note here: Numbers lacks a permut(n,r) method, which is extremely irritating.)

Stepping-stone calculations

Calculating odds of rolls like this is somewhat similar to calculating odds in poker: you figure out how many possible ways there are to get a certain result, and you divide by the total number of all results. This gives the percent chance of achieving that result. Calculating the total possible results for a die roll is easy: start with a single die and figure out how many results this die can yield. Unless you are using very strange dice, this will equal the number of faces on the die. For each die you add to the pool, multiply by the number of ways that die can come up. So, if you roll a d20, a d12 and a d6, the total possible outcomes are 20 × 12 × 6. Since Reign uses pools of n d10s, for any given roll the total possible outcomes are 10n.

Calculating the ways to get a given result is usually more difficult. One very trivial thing to figure out: what are the odds of all of the dice of a roll matching? In this case, all of the dice have to match, and there are ten possible values this match can have. Thus, no matter how many dice are rolled, there are only ten possible rolls that result in all of them matching. This means the probability is always 10 ÷ 10n = 1 ÷ 10n−1 = 101−n.

Another basic building block is also easy to calculate: when rolling n dice, what is the chance that you will get no set at all? To figure this out, you need to know how many ways you can roll the dice such that none match any of the others. So, try to actually construct such a roll, rolling one die at a time. The first die could come up as anything, so it would have 10 possible outcomes. The second die could be anything but the value of the first, so would have nine possible outcomes. The third could have eight possible values and so on. This sequence is sort of like 10! (that is 10 × 9 × 8 × 7 × 6 × 5 × 4 × 3 × 2 × 1) with the last few terms hacked off based on how many dice you had. If you had six dice, for example, you’d need to hack off the 4 × 3 × 2 × 1 part. Note that this is 4! and that “hacking off” the terms would be done by taking 10! ÷ 4!. Note that this is exactly the formula of P(10,6), above. (This is not a coincidence, but rather exactly what permutation is.) Knowing the chance of no match, you can take 1 minus this number to figure out the chance of at least one match. The outcome looks like this (note that this matches with page 57 of the main Reign rulebook):

Dice Rolls without match Possible rolls Chance of no match Chance of at least one match
1 P(10,1) = 10 10 100.000% 0.00%
2 P(10,2) = 90 100 90.000% 10.00%
3 P(10,3) = 720 1,000 72.000% 28.00%
4 P(10,4) = 5,040 10,000 50.400% 49.60%
5 P(10,5) = 30,240 100,000 30.240% 69.76%
6 P(10,6) = 151,200 1,000,000 15.120% 84.88%
7 P(10,7) = 604,800 10,000,000 6.048% 93.95%
8 P(10,8) = 1,814,400 100,000,000 1.814% 98.19%
9 P(10,9) = 3,628,800 1,000,000,000 0.363% 99.64%
10 P(10,10) = 3,628,800 10,000,000,000 0.036% 99.96%
11 0 100,000,000,000 0.000% 100.00%

Getting more detailed results requires more complexity. It happens that one of the tools needed to proceed requires figuring out the odds of rolling exactly one set with a given width, and all the rest of the dice not matching at all. By itself, this number is more of an intermediate result, but a general formula for figuring it out will help calculating more important results. We’ll call this formula “exact” and we do it like this:

  1. Let n be the pool size, i.e. the number of dice.
  2. Let d be the number of sides. In Reign, d is always 10.
  3. Let w be the desired width.
  4. If you were to hand pick a set of the desired width from your pool, you would need to select the height of the set you are going to build, and a number of dice. The number of possible ways to do this is d × C(n,w).
  5. The dice that are not part of the set have dnw possible values, but some of these would either make the current set larger, or could combine to make sets of their own. We want to make sure that all of the remaining dice not only don’t match the set set, they don’t match each other, either. Since we used w dice to make the set, that leaves nw dice. Using ten-sided dice, the first of these dice would have nine possible ways of not matching the set. The next would only have eight possibilities, as it should match neither the set nor the previous die. The next die (if there is one), would have seven possibilities and so on. If you look at the pattern, you can see that this winds up being equal to P(d−1,nw).
  6. Taking these all together, that makes d × C(n,w) × P(d−1,nw) possible ways to roll n d-sided dice such that they generate exactly one set of exactly width w.
  7. In Reign, the formula is Exact(n,w) = 10 × C(n,w) × P(9,nw)

For reference, this formula gives the following values:

Width Dice
2 3 4 5 6 7 8 9 10 11 12 13 14 15
2 10 270 4,320 50,400 453,600 3,175,200 16,934,400 65,318,400 163,296,000 199,584,000
3 10 360 7,200 100,800 1,058,400 8,467,200 50,803,200 217,728,000 598,752,000 798,336,000
4 10 450 10,800 176,400 2,116,800 19,051,200 127,008,000 598,752,000 1,796,256,000 2,594,592,000
5 10 540 15,120 282,240 3,810,240 38,102,400 279,417,600 1,437,004,800 4,670,265,600 7,264,857,600
6 10 630 20,160 423,360 6,350,400 69,854,400 558,835,200 3,113,510,400 10,897,286,400 18,162,144,000
7 10 720 25,920 604,800 9,979,200 119,750,400 1,037,836,800 6,227,020,800 23,351,328,000
8 10 810 32,400 831,600 14,968,800 194,594,400 1,816,214,400 11,675,664,000
9 10 900 39,600 1,108,800 21,621,600 302,702,400 3,027,024,000
10 10 990 47,520 1,441,440 30,270,240 454,053,600
11 10 1,080 56,160 1,834,560 41,277,600
12 10 1,170 65,520 2,293,200
13 10 1,260 75,600
14 10 1,350
15 10

Calculating width odds

Armed with this information we can start answering real questions about the probability of specific results. Unfortunately, there is no one formula that provides a generic answer. Each roll of n dice acts a bit like a poker hand of n single-suited cards or, more closely, a roll in a game of poker dice (which is essentially a simplified version of Yahtzee). You have to look at the various possible results, and count them independently. Unfortunately, you have to do this separately for each value of n. We’ll show details of n=7, since it is large enough for interesting things to occur. With seven dice, you have 107 possible rolls, with the following possibilities:

  • No match (0x). This we calculated above: P(10,7)
  • Single pair (2x). This we calculated above: Exact(7,2)
  • Two pair (2x,2x). There are C(10,2) ways to select the heights of the pairs. The first pair can be selected C(7,2) ways, the second C(5,2). The remaining three die can come up P(8,3) different ways without matching the two pair or creating new pair of their own (that is, 8 × 7 × 6). So, the total number of exactly two pair rolls with seven dice is C(10,2) × C(7,2) × C(5,2) × P(8,3). Note that this suggests a general formula for the number of exactly two pair results for any number of dice n: C(10,2) × C(n,2) × C(n−2,2) × P(8,n−4), with the realization that this is only possible for some values of n (i.e. if n is less than 4, you can’t actually build two pair, and P(8,n−4) becomes undefined as well).
  • Three pair (2x,2x,2x). Similar to two pair, using this general formula: C(10,3) × C(n,2) × C(n−2,2) × C(n−4,2) × P(7,n−6).
  • Single triple (3x). This we calculated above: Exact(7,3)
  • Triple and a pair (3x,2x). Ten ways to select the height of the triple, the dice for which can be chosen C(7,3) ways. That leaves nine possibilities for the height of the pair, the dice of which can be selected C(4,2) ways. For the remaining two dice, there are 8 possible values for the height for the first and seven for the second (otherwise they match either each other or one of the other sets). Note that this is P(8,2). The total is then 10 × C(7,3) × 9 × C(4,2) × P(8,2). Again, a general formula suggests itself: 10 × C(n,3) × 9 × C(n−3,2) × P(8,n−5)
  • Triple and two pair (3x,2x,2x). Ten ways to select the height of the triple, the dice for which can be chosen C(7,3) ways. There are C(9,2) ways to select the heights of the pairs. The first pair can be selected C(4,2) ways, the second C(2,2). The remaining three die can come up 7 different ways without matching the other sets (note that this is P(7,1)). 10 × C(7,3) × C(9,2) × C(4,2) × C(2,2). For more than seven dice, there would dice left over that could not match the value of any of the sets or each other, so permutations of the seven remaining possible values would be used: 10 × C(n,3) × C(9,2) × C(n−3,2) × C(n−5,2) × P(7,n−7).
  • Two triples. Calculated much like two pair, with a general formula of C(10,2) × C(n,3) × C(n−3,3) × P(8,n−6).
  • Four of a kind (4x). This we calculated above: Exact(7,4)
  • Four of a kind and a pair (4x,2x). Similar to a triple and a pair: 10 × C(n,4) × 9 × C(n−4,2) × P(8,n−6)
  • Five of a kind (5x). This we calculated above: Exact(7,5)
  • Five of a kind and a pair (5x,2x). Similar to a triple and a pair: 10 × C(n,5) × 9 × C(n−5,2) × P(8,n−7)
  • Six of a kind (6x). This we calculated above: Exact(7,6)
  • Seven of a kind (7x). This we calculated above: Exact(7,7)

You can do similar analysis of all the other lengths, and you end up with the table below. Each row in the table shows a possible set combination that may be rolled, and the general formula to figure out how many possible outcomes will generate that set combination. Naturally, not all combinations are possible for a given number of dice. Try as you might, you’ll never get a width three set from two dice. Each cell in the table shows two different bits of information for given result for a given number of dice. The first is the numeric result of that number. The second, in bold, is the percentage chance of that result when rolling that many dice. Because the table gets obnoxious as the number of dice increases, this only shows results for two to 10 dice, but you should be able to use techniques shown above to extend it to more if you need to. (I also wrote some really horrid Python code to double-check these numbers by counting the sets made by every possible roll from 2 to 10 dice. It all matches.)

Result General Formula Dice Rolled
2 3 4 5 6 7 8 9 10
0x P(10,n) 90
90.0%
720
72.0%
5,040
50.4%
30,240
30.24%
151,200
15.120%
604,800
6.0480%
1,814,400
1.81440%
3,628,800
0.362880%
3,628,800
0.0362880%
2x 10 × C(n,2) × P(9,n−2) 10
10.0%
270
27.0%
4,320
43.2%
50,400
50.40%
453,600
45.360%
3,175,200
31.7520%
16,934,400
16.93440%
65,318,400
6.531840%
163,296,000
1.6329600%
3x 10 × C(n,3) × P(9,n−3) 10
1.0%
360
3.6%
7,200
7.20%
100,800
10.080%
1,058,400
10.5840%
8,467,200
8.46720%
50,803,200
5.080320%
217,728,000
2.1772800%
4x 10 × C(n,4) × P(9,n−4) 10
0.1%
450
0.45%
10,800
1.080%
176,400
1.7640%
2,116,800
2.11680%
19,051,200
1.905120%
127,008,000
1.2700800%
5x 10 × C(n,5) × P(9,n−5) 10
0.01%
540
0.054%
15,120
0.1512%
282,240
0.28224%
3,810,240
0.381024%
38,102,400
0.3810240%
6x 10 × C(n,6) × P(9,n−6) 10
0.001%
630
0.0063%
20,160
0.02016%
423,360
0.042336%
6,350,400
0.0635040%
7x 10 × C(n,7) × P(9,n−7) 10
0.0001%
720
0.00072%
25,920
0.002592%
604,800
0.0060480%
8x 10 × C(n,8) × P(9,n−8) 10
0.00001%
810
0.000081%
32,400
0.0003240%
9x 10 × C(n,9) × P(9,n−9) 10
0.000001%
900
0.0000090%
10x 10 × C(n,10) × P(9,n−10) 10
0.0000001%
2x,2x C(10,2) × C(n,2) × C(n−2,2) × P(8,n−4) 270
2.7%
10,800
10.80%
226,800
22.680%
3,175,200
31.7520%
31,752,000
31.75200%
228,614,400
22.861440%
1,143,072,000
11.4307200%
3x,2x 10 × C(n,3) × 9 × C(n−3,2) × P(8,n−5) 900
0.90%
43,200
4.320%
1,058,400
10.5840%
16,934,400
16.93440%
190,512,000
19.051200%
1,524,096,000
15.2409600%
3x,3x C(10,2) × C(n,3) × C(n−3,3) × P(8,n−6) 900
0.090%
50,400
0.5040%
1,411,200
1.41120%
25,401,600
2.540160%
317,520,000
3.1752000%
4x,2x 10 × C(n,4) × 9 × C(n−4,2) × P(8,n−6) 1,350
0.135%
75,600
0.7560%
2,116,800
2.11680%
38,102,400
3.810240%
476,280,000
4.7628000%
4x,3x 10 × C(n,4) × 9 × C(n−4,3) × P(8,n−7) 3,150
0.0315%
201,600
0.20160%
6,350,400
0.635040%
127,008,000
1.2700800%
4x,4x C(10,2) × C(n,4) × C(n−4,4) × P(8,n−8) 3,150
0.00315%
226,800
0.022680%
7,938,000
0.0793800%
5x,2x 10 × C(n,5) × 9 × C(n−5,2) × P(8,n−7) 1,890
0.0189%
120,960
0.12096%
3,810,240
0.381024%
76,204,800
0.7620480%
5x,3x 10 × C(n,5) × 9 × C(n−5,3) × P(8,n−8) 5,040
0.00504%
362,880
0.036288%
12,700,800
0.1270080%
5x,4x 10 × C(n,5) × 9 × C(n−5,4) × P(8,n−9) 11,340
0.001134%
907,200
0.0090720%
5x,5x C(10,2) × C(n,5) × C(n−5,5) × P(8,n−10) 11,340
0.0001134%
6x,2x 10 × C(n,6) × 9 × C(n−6,2) × P(8,n−8) 2,520
0.00252%
181,440
0.018144%
6,350,400
0.0635040%
6x,3x 10 × C(n,6) × 9 × C(n−6,3) × P(8,n−9) 7,560
0.000756%
604,800
0.0060480%
6x,4x 10 × C(n,6) × 9 × C(n−6,4) × P(8,n−10) 18,900
0.0001890%
7x,2x 10 × C(n,7) × 9 × C(n−7,2) × P(8,n−9) 3,240
0.000324%
259,200
0.0025920%
7x,3x 10 × C(n,7) × 9 × C(n−7,3) × P(8,n−10) 10,800
0.0001080%
8x,2x 10 × C(n,8) × 9 × C(n−8,2) × P(8,n−10) 4,050
0.0000405%
2x,2x,2x C(10,3) × C(n,2) × C(n−2,2) ×
C(n−4,2) × P(7,n−6)
10,800
1.080%
529,200
5.2920%
12,700,800
12.70080%
190,512,000
19.051200%
1,905,120,000
19.0512000%
3x,2x,2x C(10,2) × C(n,2) × C(n−2,2) ×
8 × C(n−4,3) × P(7,n−7)
75,600
0.7560%
4,233,600
4.23360%
114,307,200
11.430720%
1,905,120,000
19.0512000%
3x,3x,2x C(10,2) × C(n,3) × C(n−3,3) ×
8 × C(n−6,2) × P(7,n−8)
201,600
0.20160%
12,700,800
1.270080%
381,024,000
3.8102400%
3x,3x,3x C(10,3) × C(n,3) × C(n−3,3) ×
C(n−6,6) × P(7,n−9)
201,600
0.020160%
14,112,000
0.1411200%
4x,2x,2x C(10,2) × C(n,2) × C(n−2,2) ×
8 × C(n−4,4) × P(7,n−8)
151,200
0.15120%
9,525,600
0.952560%
285,768,000
2.8576800%
4x,3x,2x 10 × C(n,4) × 9 × C(n−4,3) ×
8 × C(n−7,3) × P(7,n−9)
907,200
0.090720%
63,504,000
0.6350400%
4x,3x,3x C(10,2) × C(n,3) × C(n−3,3) ×
8 × C(n−6,4) × P(7,n−10)
1,134,000
0.0113400%
4x,4x,2x C(10,2) × C(n,4) × C(n−4,4) ×
8 × C(n−8,2) × P(7,n−10)
1,134,000
0.0113400%
5x,2x,2x C(10,2) × C(n,2) × C(n−2,2) ×
8 × C(n−4,5) × P(7,n−9)
272,160
0.027216%
19,051,200
0.1905120%
5x,3x,2x 10 × C(n,5) × 9 × C(n−5,3) ×
8 × C(n−8,3) × P(7,n−10)
1,814,400
0.0181440%
6x,2x,2x C(10,2) × C(n,2) × C(n−2,2) ×
8 × C(n−4,6) × P(7,n−10)
453,600
0.0045360%
2x,2x,2x,2x C(10,4) × C(n,2) × C(n−2,2) ×
C(n−4,2) × C(n−6,2) × P(6,n−8)
529,200
0.52920%
28,576,800
2.857680%
714,420,000
7.1442000%
3x,2x,2x,2x C(10,3) × C(n,2) × C(n−2,2) ×
C(n−4,2) × 7 × C(n−6,3) × P(6,n−9)
6,350,400
0.635040%
381,024,000
3.8102400%
3x,3x,2x,2x C(10,2) × C(n,3) × C(n−3,3) ×
C(8,2) × C(n−6,2) × C(n−8,2) × P(6,n−10)
31,752,000
0.3175200%
4x,2x,2x,2x C(10,3) × C(n,2) × C(n−2,2) ×
C(n−4,2) × 7 × C(n−6,4) × P(6,n−10)
15,876,000
0.1587600%
2x,2x,2x,2x,2x C(10,5) × C(n,2) × C(n−2,2) ×
C(n−4,2) × C(n−6,2) × C(n−8,2) × P(4,n−10)
28,576,800
0.2857680%

Calculating height

As mentioned above, calculating odds for height is only interesting when you need a roll of at least a given height. Such calculation is surprisingly cumbersome, even if you mostly leave width out of it. In order for width to “count”, you must have at least one match, and this is what makes things hard to figure. It is honestly easier just to write code to count all possible outcomes, but if you want to figure it out do the following:

  1. Let n be the pool size, i.e. the number of dice.
  2. Let d be the number of sides. In Reign, d is always 10.
  3. Let h be the minimum height needed. Naturally, 0<hd.
  4. You now need to break the problem into sub-problems. So let m take on values from 2 to n. For each value of m, figure out the following:
    1. Figure out how many rolls of n dice of d sides generate exactly m dice that have values of h or greater. This can be done with a function X(n,d,m,h) = C(n,m) × (d-(h-1))m × (h-1)(n-m). Note that when h=1 and m=n, the final term in that function is 00, which is technically undefined, so is omitted in that case.
    2. Figure out how many possible ways the n dice of d sides can come up. We did this above: dn.
    3. Find the probability of X(n,d,m,h). This is just the value of the function, divided by the total number of possible rolls. So ProbXm = X(n,d,m,h) ÷ dn.
    4. Now, given the m dice that came up h or greater, figure out the number of possible ways those dice can match. We do this much the same way we figured out the chance of at least one match, above. This time, though, we have only d-(h-1) possible values. So the possible ways to have no match with these dice is P(d-(h-1),m). The number of possible ways these dice can come up is ÷ (d-(h-1))m. So Y(d,m,h) = (d-(h-1))m − P(d-(h-1),m).
    5. Figure ProbYm = Y(d,m,h) ÷ (d-(h-1))m.
    6. Remember Probm = ProbXm × ProbYm.
  5. Now add up all values of Probm, and you have the probability of rolling at least height h on n dice of d sides.

This looks like this, again with the total number of rolls that meet a given height, followed by the percentage chance:

Height Dice Rolled
2 3 4 5 6 7 8 9 10
1 10
10.0%
280
28.0%
4,960
49.60%
69,760
69.760%
848,800
84.8800%
9,395,200
93.95200%
98,185,600
98.185600%
996,371,200
99.6371200%
9,996,371,200
99.96371200%
2 9
9.0%
252
25.2%
4,491
44.91%
63,954
63.954%
792,225
79.2225%
8,952,624
89.52624%
95,403,447
95.403447%
982,427,886
98.2427886%
9,941,058,909
99.41058909%
3 8
8.0%
224
22.4%
4,016
40.16%
57,888
57.888%
729,600
72.9600%
8,413,056
84.13056%
91,453,568
91.453568%
957,516,032
95.7516032%
9,803,680,256
98.03680256%
4 7
7.0%
196
19.6%
3,535
35.35%
51,562
51.562%
660,835
66.0835%
7,771,456
77.71456%
86,191,903
86.191903%
918,898,750
91.8898750%
9,546,080,131
95.46080131%
5 6
6.0%
168
16.8%
3,048
30.48%
44,976
44.976%
585,840
58.5840%
7,022,784
70.22784%
79,471,872
79.471872%
863,672,832
86.3672832%
9,124,660,224
91.24660224%
6 5
5.0%
140
14.0%
2,555
25.55%
38,130
38.130%
504,525
50.4525%
6,162,000
61.62000%
71,144,375
71.144375%
788,768,750
78.8768750%
8,490,265,625
84.90265625%
7 4
4.0%
112
11.2%
2,056
20.56%
31,024
31.024%
416,800
41.6800%
5,184,064
51.84064%
61,057,792
61.057792%
690,950,656
69.0950656%
7,588,071,424
75.88071424%
8 3
3.0%
84
8.4%
1,551
15.51%
23,658
23.658%
322,575
32.2575%
4,083,936
40.83936%
49,057,983
49.057983%
566,816,382
56.6816382%
6,357,469,311
63.57469311%
9 2
2.0%
56
5.6%
1,040
10.40%
16,032
16.032%
221,760
22.1760%
2,856,576
28.56576%
34,988,288
34.988288%
412,797,440
41.2797440%
4,731,954,176
47.31954176%
10 1
1.0%
28
2.8%
523
5.23%
8,146
8.146%
114,265
11.4265%
1,496,944
14.96944%
18,689,527
18.689527%
225,159,022
22.5159022%
2,639,010,709
26.39010709%

Popularity: 7% [?]

iCrushYourHead

November 1st, 2007 — Wordman

When a coworker was recently comparing experiences with the iPod Touch, he mentioned how grateful he was to the Kids in the Hall, as emulating them had trained his finger muscles to use the device’s zoom feature. For the uninitiated, zooming out on the iPod Touch involves touching two fingers to the screen and spreading them apart, while zooming in involves moving your fingers together. This is basically the same motion used in a series of Kids sketches involving a semi-crazy guy saying “I’m crushing your head”, while making a similar motion from a distance, like this attempt to crush the head of a statue of Buddha:

Crushing Buddha's head

All this head crushing and its relation to to the iPod Touch UI got me thinking about the perfect game for the device: iCrushYourHead. The idea would be that pictures of people would randomly drift across the screen at various speeds, and using the “pinching” UI gesture, you would have to crush their heads as they passed. The crushing would be animated with a cheesy accordion-fold type effect. As the game progressed, they would drift by faster and faster, and more would be on screen at once. Naturally, you’d need to be able to import photos of people who deserve head crushing.

The same coworker, hearing this idea, suggested another game, this one for kids. It is a port of those kids books where the page is divided into three sections, one for the head, one for the body and one for the legs, where the kid will mix and match various parts to make fun combinations. It would make use of another cool UI trick on the touch, the sliding scroll gesture where you can kind of “throw” a section of a screen and it will scroll with a sort of natural deceleration. It’s hard to explain if you haven’t used the iPod Touch, but it is very natural. In the case of this game, it would act a bit like the wheels of a slot machine, that you could spin independently with varying degrees of force. The number of choices for the body parts could also be vastly larger than a physical book would allow.

Feel free to send me royalty checks if you build these games.

Popularity: 10% [?]