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Boardgame/dice issue - rolling for 50?

BitstreamBitstream Registered User regular
edited January 2009 in Help / Advice Forum
OK H/A, I have a couple of oddball questions for you:

I'm working on a boardgame, and one of the issues I've run into is that when it comes to dice and statistics I'm fairly boned beyond knowing expected outcomes. I'm hoping someone with more dice experience or better math skills can help me out with two forms of rand50 usage. Forgive my lack of proper terminology, it's been ages since stats class:

1. rand50, 0 through 50 with bell-curve expected outcomes. I can fudge this one a little (5d10 gives me a range of 0-45, and I can adjust accordingly) if necessary, but I'd really rather have an outcome range from 0 to 50 proper. Unfortunately I've never seen a 0-5 d6. Is there one out there, or an even simpler solution I've missed?

2. rand50, 1 through 50 with straight odds of hitting any one outcome. There's always the 50-sided die, or the mysterious 5-sider for the tens column and 10-sider for the ones, but again neither is really feasible.

And a related question - if a d5-ish really is my best option, does anyone have experience with custom dice places? I assume I'd be able to get a d6 (with 0-5) or a d10 (with 1-5 twice over) done up, but I'm worried about costs for prototyping. Plus, I've got enough crap in this game, and telling a company they'll have to get custom dice isn't going to help me sell this thing.

Bitstream on

Posts

  • WillethWilleth Registered User regular
    edited January 2009
    When rolling 5d10 it's impossible to roll lower than 5, surely? Also, a die with a possible roll of 0-50 would have 51 sides.

    I'm sure there's an easier way of doing this, but you can just go through all possible combinations:

    1 1 1 1 1 - 5
    1 1 1 1 2 - 6
    1 1 1 2 1 - 6
    ...
    10 10 10 10 9 - 49
    10 10 10 10 10 - 50

    and then just plot your curve from there. You'll probably find someone out there has done this already. Just thinking about it, it looks like a symmetrical bell curve.

    EDIT: Yeah on re-reading your post it looks like you get all that but I misunderstood because your dice start from 0 (a practice I've never heard of). Not entirely sure why you're looking for an 'outcome range of 0-50 proper' because this is impossible with five ten-siders.

    Willeth on
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  • BitstreamBitstream Registered User regular
    edited January 2009
    Willeth wrote: »
    When rolling 5d10 it's impossible to roll lower than 5, surely?
    Nope, d10s are numbered 0-9.
    Also, a die with a possible roll of 0-50 would have 51 sides.
    Right, but the straight-odds roll I need is 1-50.
    I'm sure there's an easier way of doing this, but you can just go through all possible combinations:

    1 1 1 1 1 - 5
    1 1 1 1 2 - 6
    1 1 1 2 1 - 6
    ...
    10 10 10 10 9 - 49
    10 10 10 10 10 - 50

    and then just plot your curve from there. You'll probably find someone out there has done this already. Just thinking about it, it looks like a symmetrical bell curve.
    Yeah, I have the math of it and any multiple-die roll has a nice symmetrical bell curve (watch someone with actual math skill come in and call me on this :P). Unfortunately I'm having trouble figuring out how to roll to get a 51-outcome bell curve.

    [Edit] - Hmm, I guess I should say all of my d10s are 0-9. and yeah, my 5d10 roll is just the easiest near-solution I could think of.

    Bitstream on
  • KivutarKivutar Registered User regular
    edited January 2009
    Bitstream wrote: »
    Willeth wrote: »
    When rolling 5d10 it's impossible to roll lower than 5, surely?
    Nope, d10s are numbered 0-9.
    Also, a die with a possible roll of 0-50 would have 51 sides.
    Right, but the straight-odds roll I need is 1-50.
    I'm sure there's an easier way of doing this, but you can just go through all possible combinations:

    1 1 1 1 1 - 5
    1 1 1 1 2 - 6
    1 1 1 2 1 - 6
    ...
    10 10 10 10 9 - 49
    10 10 10 10 10 - 50

    and then just plot your curve from there. You'll probably find someone out there has done this already. Just thinking about it, it looks like a symmetrical bell curve.
    Yeah, I have the math of it and any multiple-die roll has a nice symmetrical bell curve (watch someone with actual math skill come in and call me on this :P). Unfortunately I'm having trouble figuring out how to roll to get a 51-outcome bell curve.

    [Edit] - Hmm, I guess I should say all of my d10s are 0-9. and yeah, my 5d10 roll is just the easiest near-solution I could think of.
    o_O
    Seriously?
    Rather than point out all of the various things wrong with that, try this - use a d10 for each decimal place, divide by two, rounding up.

    Kivutar on
  • BitstreamBitstream Registered User regular
    edited January 2009
    Kivutar wrote: »
    o_O
    Seriously?
    Rather than point out all of the various things wrong with that, try this - use a d10 for each decimal place, divide by two, rounding up.

    ...

    *facepalm*

    See, this is the kind of thing a caffeine-and-game-design-addled brain needs spelled out to it. Thanks for making me know I'm an idiot, rather than just feeling like one.

    So that's number two solved. Who's got an equally shaming method for getting the first one? If you need me I'll be crying myself to sleep :P

    Bitstream on
  • DjeetDjeet Registered User regular
    edited January 2009
    For getting a random number 0-5 where there's equal representation what's wrong with (d6 - 1)? Meaning instead of trying to find a custom dice with 0-5 on the face, just have the roller subtract 1 from the result.

    I always assumed "0" on a d10 was "10", making a min roll on 5d10 = 5 (like Willeth was saying).

    Djeet on
  • Cynic JesterCynic Jester Registered User regular
    edited January 2009
    It's been my experience that it's better to adapt your game to a common set of dice as opposed to adapting the dice to your game. Why would you need 0-50 instead of 1-50? Also, why not 1-100?

    Edit: And yeah, in most systems the 0 on a D10 is 10.

    Cynic Jester on
  • PowerpuppiesPowerpuppies drinking coffee in the mountain cabinRegistered User regular
    edited January 2009
    50d2 - 50 would offer a bell curve, wouldn't it?

    Powerpuppies on
    sig.gif
  • DirtyDirtyVagrantDirtyDirtyVagrant Registered User regular
    edited January 2009
    Any die roll offers a bell curve. That's the nature of the dice. It is a creature of averages.

    That said, 25d2 offers a possible value range of 25-50. So it won't work.

    What you really need to do, I think, is take a serious look at this system and decide whether there isn't some other range of values you could conceivably use for whatever it is that you're doing. A sort of "Does it really need to be this way?" kind of thing.

    DirtyDirtyVagrant on
  • MegaMan001MegaMan001 CRNA Rochester, MNRegistered User regular
    edited January 2009
    It sounds like you are making this game on your own, which isn't really a great way going about things. I'm having trouble imagining why you would need 50 separate outcomes for a single event (the dice roll). That is, I can't think of any board game where one event needed 50 possible outcomes and still have the experience entertain me.

    What I am trying to say is this, do you really need 50 separate outcomes or could you trim it down to a much more manageable number utilizing standard dice?

    This doesn't solve your problem on how to roll a D50 - but honestly, that's not really your issue.

    MegaMan001 on
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  • ThomamelasThomamelas Only one man can kill this many Russians. Bring his guitar to me! Registered User regular
    edited January 2009
    Why not just use a d50? While not exactly common, they do exist. Or you could use d100 and divide by 2.

    Thomamelas on
  • DirtyDirtyVagrantDirtyDirtyVagrant Registered User regular
    edited January 2009
    Please take a moment to imagine a board game that came with a 50-sided dice in the box.

    DirtyDirtyVagrant on
  • romanqwertyromanqwerty Registered User regular
    edited January 2009
    For the bell curve could you just use 10d6 -10? That should give a value of 0-50 (10-60) probably far better spread then 50d2.

    romanqwerty on
  • ThomamelasThomamelas Only one man can kill this many Russians. Bring his guitar to me! Registered User regular
    edited January 2009
    Please take a moment to imagine a board game that came with a 50-sided dice in the box.

    "Dude...this fucking rules?" :lol:

    Thomamelas on
  • RookRook Registered User regular
    edited January 2009
    Bitstream wrote: »
    1. rand50, 0 through 50 with bell-curve expected outcomes. I can fudge this one a little (5d10 gives me a range of 0-45, and I can adjust accordingly) if necessary, but I'd really rather have an outcome range from 0 to 50 proper. Unfortunately I've never seen a 0-5 d6. Is there one out there, or an even simpler solution I've missed?

    I think you probably need to think harder about what you're doing. Offering a bell curve like that means that most people who play will never ever see any of the results outside the interquartile range, so essentially you'd be asking them to spend 2-3 mins adding shit up so they can have the same set of 10 or so events/encounters etc for the 50th time.

    What games will tend to do instead is give you a d100 roll to make and fudge odds that way.
    e.g.

    99-90 - Face
    89-85 - Neck
    84-60 - Arms
    59-40 - Chest
    39-34 - Groin
    33-11 - Legs
    10-00 - Feet

    Or say give you sets of tables to roll from say 2D6 and then maybe a further 2D6 roll on each of those table. You could probably have some fun working out the odds

    Rook on
  • BitstreamBitstream Registered User regular
    edited January 2009
    Alright guys, thanks for the input. I may need to rethink this in some way, as the consensus seems to be that I'm overcomplicating things.

    I should have explained the reasoning behind this in the first place - I'm making a game about a presidential campaign, and of course it ends with elections. Because there's an essentially unbreakable 25% base for each side, I figured a 50% spread would be easier to deal with (this is where I can fudge with the 5d10 roll, making the base slightly larger on each side).

    Modern US elections are naturally very nearly 50/50 affairs, with massive upsets being fairly uncommon, so a bell curve heavily favoring the middle quartile as Rook pointed out is actually pretty ideal. There'd be a few regions voting separately to avoid one especially bad roll tanking you, and a couple of variables that would add/subtract from your total.

    Crap, I may be making this too complicated. Looks like I should maybe take this to CF, but any ideas H/A folks have are welcome.

    Bitstream on
  • PowerpuppiesPowerpuppies drinking coffee in the mountain cabinRegistered User regular
    edited January 2009
    Toss 50 coins and count the heads.

    Powerpuppies on
    sig.gif
  • DirtyDirtyVagrantDirtyDirtyVagrant Registered User regular
    edited January 2009
    The odds of flipping 50 coins and coming up with all heads are mathematically insignificant.

    DirtyDirtyVagrant on
  • EggyToastEggyToast Jersey CityRegistered User regular
    edited January 2009
    Bitstream wrote: »
    Modern US elections are naturally very nearly 50/50 affairs, with massive upsets being fairly uncommon,

    I think you're really focusing on the last few elections. If you're thinking of the presidency, elections are historically NOT 50/50, and there have been actually quite a few upsets, given the small sample size (Obama is only 44, after all).

    Thing is, too, if you're worried about probabilities, you probably are overthinking it. The probabilities should suit the mechanics of the game, not vice versa -- after all, you want people to play this game for fun, right? Two games that use probability to great effect are Settlers of Catan and Elasund, both of which use the most probable outcome of 2d6 as a "spoiler" role, and those dice function as a mechanic of the game that fits into other, less probabilistic elements.

    Of course, the vast majority of games do not use a literal die roll as dictating the numbers -- how would rolling 2d6 be less bell-curvy? And, of course, if the end of the game is based entirely on a roll of some dice, that's a mechanic discussion for CF.

    EggyToast on
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  • theclamtheclam Registered User regular
    edited January 2009
    Most people aren't going to want to do much math for a board game.

    Think of doing something like this:

    2d6

    2 = +10 Democrat
    3 = +8 Democrat
    4 = +6 Democrat
    5 = +4 Democrat
    6 = +2 Democrat
    7 = Tiebreak (side with the most money/campaign workers/whatever wins)
    8 = +2 Republican
    9 = +4 Republican
    10 = +6 Republican
    11 = +8 Republican
    12 = +10 Republican

    It's a bell curve, it's simple to remember, and you can make a simple chart for people who forget or are still learning. You could also have each side roll 1d6 and (highest - lowest) will give you the winner and margin of victory.

    Of course, I don't know anything about your game, so do whatever fits, but this may be a helpful example.

    theclam on
    rez_guy.png
  • shutzshutz Registered User regular
    edited January 2009
    To add a certain level of excitement, you could try "dice chaining" (I don't know if that's a real term or not, I just came up with this as it sounds good and seems relevant enough.)

    Let's say, at the end of your game's election campaign, you're rolling for the election. I would assume that whatever the players have done before then has caused a large percentage of the population to decide on one side or the other, leaving a certain percentage of "undecideds".

    Let's say, for your particular game, the Democrats have 38% of the voters "locked", while the Republicans have been able to lock 35%. That means 73% have decided, leaving 27% undecideds.

    At this point, each player rolls 1d6. Now, here comes the "chaining" I mentioned at the beginning: any roll of '6' gives that player another roll. So let's say the republican player rolls a 3, and the democrat player rolls a 6. That player rolls again. He gets another 6! He rolls again, and this time, he gets a 1.

    Now, each player adds his roll to his "locked" percentage. The dice rolls represent the votes of the undecided. So the republicans now have 35+3 = 38%, while the democrats have 38+6+6+1 = 51%.

    The player with the highest percentage wins the election. Any remaining percentage of undecided just means those people didn't vote in the election, or went to have their votes canceled.

    If the two players' percentages added together is higher than 100%, in the case of one player rolling a long chain of 6's for example, that just means that that player was able, at the last minute, to sway some voters that the other side had previously thought "locked", which is something that happens, sometimes.

    Since 1d6 might not seem like enough room to maneuver initially, depending on the locked percentages, you might want to have both players roll more than 1d6. Maybe 2d6 or 3d6. Any 6's give that player an extra roll, as described above.

    The distribution for this will tend to look more like a binomial distribution, but then again, even a binomial distribution can approach a normal distribution, when enough rolls are concerned (plus, if you let players roll multiple dice initially, you're already skewing towards more of a normal distribution.)

    Naturally, none of my suggestions above deal with such complicated stuff as the Electoral College. I have a pretty good excuse for that, though: I'm Canadian. You'd be in deep trouble trying to model our current situation, considering we have a minority government with 3 other parties that each control a healthy chunk of the country (and which tried, before the holidays, to form a coalition government to overthrow the minority government, except that that's probably not going to happen, now, because one of the parties has a new leader...)

    I shudder to think how I would design a game that models the current Canadian election situation...

    shutz on
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    Check out my new blog: http://50wordstories.ca
    Also check out my old game design blog: http://stealmygamedesigns.blogspot.com
  • JohnnyCacheJohnnyCache Starting Defense Place at the tableRegistered User regular
    edited January 2009
    Bitstream wrote: »
    OK H/A, I have a couple of oddball questions for you:

    I'm working on a boardgame, and one of the issues I've run into is that when it comes to dice and statistics I'm fairly boned beyond knowing expected outcomes. I'm hoping someone with more dice experience or better math skills can help me out with two forms of rand50 usage. Forgive my lack of proper terminology, it's been ages since stats class:

    1. rand50, 0 through 50 with bell-curve expected outcomes. I can fudge this one a little (5d10 gives me a range of 0-45, and I can adjust accordingly) if necessary, but I'd really rather have an outcome range from 0 to 50 proper. Unfortunately I've never seen a 0-5 d6. Is there one out there, or an even simpler solution I've missed?

    2. rand50, 1 through 50 with straight odds of hitting any one outcome. There's always the 50-sided die, or the mysterious 5-sider for the tens column and 10-sider for the ones, but again neither is really feasible.

    And a related question - if a d5-ish really is my best option, does anyone have experience with custom dice places? I assume I'd be able to get a d6 (with 0-5) or a d10 (with 1-5 twice over) done up, but I'm worried about costs for prototyping. Plus, I've got enough crap in this game, and telling a company they'll have to get custom dice isn't going to help me sell this thing.



    you seem to be dead set on construction situations where you build an odd-numbered die at some point. The problem with 0-50 is it's a 51 position count and that makes for a really awkward die roll.

    What are you goal odds for the election scenario? Is there any reason to include position 0?

    JohnnyCache on
  • MagicPrimeMagicPrime FiresideWizard Registered User regular
    edited January 2009
    Blue_dice_d5.jpg

    The D5.

    MagicPrime on
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  • BitstreamBitstream Registered User regular
    edited January 2009
    Some really good points here.
    EggyToast wrote:
    I think you're really focusing on the last few elections. If you're thinking of the presidency, elections are historically NOT 50/50, and there have been actually quite a few upsets, given the small sample size (Obama is only 44, after all).

    Actually, that's sort of the point. I'm consciously modeling the game on the post-Nixon era, because that's where things really get ridiculous with electoral shenanigans and media scandals (Nixon once made an opponent cry on national television). Earlier than that and it feels like the game gets too dry and historical.
    theclam wrote:
    Most people aren't going to want to do much math for a board game.

    Think of doing something like this:

    2d6

    2 = +10 Democrat
    3 = +8 Democrat
    4 = +6 Democrat
    5 = +4 Democrat
    6 = +2 Democrat
    7 = Tiebreak (side with the most money/campaign workers/whatever wins)
    8 = +2 Republican
    9 = +4 Republican
    10 = +6 Republican
    11 = +8 Republican
    12 = +10 Republican

    I like this. It's simple, and it makes natural sense. Definitely under consideration.
    shutz wrote: »
    To add a certain level of excitement, you could try "dice chaining" (I don't know if that's a real term or not, I just came up with this as it sounds good and seems relevant enough.)

    I like this too. Something about the idea of a sudden election-day surge for one guy in a given region just fits really well.

    And yeah, I don't envy you Canadians right now. Working on this game has made me realize just how much more effective a two-party system is at conferring "legitimacy" and political capital to actually get shit done, regardless of the other problems it causes.
    you seem to be dead set on construction situations where you build an odd-numbered die at some point. The problem with 0-50 is it's a 51 position count and that makes for a really awkward die roll.

    What are you goal odds for the election scenario? Is there any reason to include position 0?

    This is a good point. The odds of rolling a zero value are so minute that it's most likely not worth this much hair-pulling. I've definitely made things hard for myself for the sake of the numbers, when I should be focusing mon making sure the mechanics are actually fun.

    Thanks for some good ideas, folks, you just made this a lot easier for me.

    This thread should be archived so we can point to something as evidence that a lone hermit making a game without someone to bounce ideas off of isn't always this awesome.

    [Edit] - A wild d5 appears! They're definitely cool, but they're rare and (mass-production-wise) expensive.

    Bitstream on
  • EchoEcho ski-bap ba-dapModerator, Administrator admin
    edited January 2009
    How to roll 1d100:

    Roll 1d10 twice. The multiply the first roll by 10, treating 0 as 0. Add up. Technically this gives a range of 0-99, so add 1 if you want.

    How to roll 1d50, method the first:

    Same as above. Divide by two, round up.

    How to roll 1d50, method the second:

    Roll two d10, divide the first roll by two, round up, multiply by 10. Add up.

    I'm assuming you want it simple and not require a fistful of rolls.

    Echo on
  • EggyToastEggyToast Jersey CityRegistered User regular
    edited January 2009
    It's also less fun, but it's much easier to have a deck of cards if you want to have random probabilities. Cards labeled 1-100 would give you "even odds" for any number. But you could also have 10 "50's", 8 "40's" and so on.

    Actually, if you wanted it to be more like a real election, you could have smaller percentages on the cards, and they represent a constituency. So you would draw a card and add 3% to your "base."

    You could also modify the mechanics so that you would have to give up some percentages to gain others, with some cards being face-up so people know what they're losing. And by having an area where you shuffle and then place cards face up, each game would be different -- but the strategies could be similar. Say 5% is up for grabs in California but you'll lose the 2% cards on George and North Carolina. But if California only gets a 2% card, then that would change the strategy.

    But yeah, there are a couple games where you can actually get a deck of cards to replace the die rolls. They're actually more statistically "perfect" because in a deck of cards, if the deck isn't shuffled during play, each number will appear at least once. In some games you could roll 2d6 and never get a 12, but it's destined to show up based on its statistical probability in a deck of cards. You can also roll 5 12s in a row, too. Just something to think about.

    EggyToast on
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  • shutzshutz Registered User regular
    edited January 2009
    Bitstream wrote: »
    shutz wrote: »
    To add a certain level of excitement, you could try "dice chaining" (I don't know if that's a real term or not, I just came up with this as it sounds good and seems relevant enough.)

    I like this too. Something about the idea of a sudden election-day surge for one guy in a given region just fits really well.

    The basic idea behind my suggestion is, don't waste too much time trying to have your dice probabilities precisely model a real election, as that will make for a really boring game. Always remember, the main quality of a game is that it must be fun. Realism must always be secondary to the fun aspect, otherwise, all you get is a realistic, but boring or tedious game, and nobody wants to play such a game (well, maybe hardcore wargamers, but they're a weird bunch.)

    Also, someone suggested using cards instead of dice. That's a pretty good idea, too. Or you can go for a hybrid solution, where dice rolls tell you how many cards to draw, or card draws tell you how many dice to roll. Just figure out what the general distribution of the rolls you decide on is, and then check to see if the numbers will make sense with the rest of your game, but don't try to be exact, just make sure the extremes can still lead to usable (even if ridiculously improbable) results.
    Bitstream wrote: »
    And yeah, I don't envy you Canadians right now. Working on this game has made me realize just how much more effective a two-party system is at conferring "legitimacy" and political capital to actually get shit done, regardless of the other problems it causes.

    We're in a weird situation, yes, but I still don't envy the American system either, as you're stuck with only 2 parties, and their ideologies are gradually calcifying into archetypes, since they never really have any other new ideologies to fight against. All their strategies have to do with how fight back against the opposing ideology. Who do you vote for if don't agree with the trickle-down economics and religious fundamentalism of the Republicans, but also don't believe in the potential wastefulness of bleeding-heart socialism that the democrats keep pushing? Who do you vote for, when you're a centrist?

    That's one of the reasons why Obama makes so many people hopeful: he did his best to rise above the petty party-based ideologies, to try and appeal to everyone. But my bet is, unless he makes a huge difference in all facets of your country, within his first 4 years, things are going to go back to the two opposing ideologies duking it out with the same strategies as before.

    shutz on
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    Check out my new blog: http://50wordstories.ca
    Also check out my old game design blog: http://stealmygamedesigns.blogspot.com
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