Adventuring: A Dice Game

[Endovior]

Here's some rules for a fantasy-themed dice game I just thought up. Feedback is welcome.

Adventuring:
A dice game. The objective is to earn money. You do this by rolling 4 six-sided 'Hero' dice (each of which must be distinguishable from the others... typically they are different colors): a Cleric, a Fighter, a Rogue, and a Wizard. The house rolls another six-sided die called a Demon. To earn a lot of money, you wish to roll your Heroes on the same number; which is NOT the same number as the Demon. You earn an amount of gp equal to the number each Hero rolls. If two heroes match, you get double the amount show amount. If three match, you get triple the amount shown. If all heros match, you get five times the amount shown! However, if a Hero die matches the Demon die, the Hero dies... and you lose an amount equal to five times what you would have just made with that hero! If multiple heroes match the Demon, then you lose 5 times the amount that they would have made! You may not roll a dead hero unless you opt to raise him; this costs 5 Gp times the amount shown on his die... and you can only do this when your balance is positive. If you roll while your balance is negative, you instantly lose a number of points equal to 10% of your current negative score, rounded up to the nearest cp. You may leave the table (gaining or paying your current balance) at any time that all your heroes are alive... and you MUST do so when all of your heroes are dead (of course, that does not prevent you from playing again...). Fortunately, each Hero die has a special power.

Cleric: Healing: If the Cleric's roll is equal to the roll of a dead Hero, he may instantly raise that hero for only twice the amount shown, as opposed to five times that amount. He can even do this while the balance is negative!
Fighter: Courage: Fighters put forth a heroic effort in combat; when they die, you only lose an amount equal to four times what you would win (this also applies when other party members die with the fighter!)
Rogue: Stealth: Rogues do better on their own then most; they earn one extra gold when their roll does not match any other Hero.
Wizard: Magic: A Wizard's magic doubles the amount you gain when his roll matches that of another Hero, a doubling which STACKS with normal increases! However, if he dies, you lose twice as much as normal, too!

 

[D.M. Punks]

Nice. Mechanics are a bit confusing at first but with enough games, they become second nature (played against myself a handful of times). I could easily see this being played in-game at an tavern in your favorite fantasy game.

 

[theRogueRooster]

I haven't run the numbers yet, but at first glance this seems pretty neat. As D.M. Punks said, I can easily see this as a tavern game in a campaign world. In fact, consider it yoinked with just a few changes. I think it might work well in that role since it's quick and easy.

I'll have to think about it some more to make sure the probabilities work out agreeably, but I like what you've got. So far the only change I'd make for my own use is to remove the ability to raise a character (but keep the cleric ability). I would want to eventually force the current player to pass the dice and to keep a rich player from hoarding it forever. But that's just for my variant "multiplayer, pass the dice around the table" version, not your "casino-style, player versus the House" rules.

One question I have is, are you rolling the Demon die each time you roll the Hero dice? You must be, or otherwise the Cleric ability makes no sense, since the Cleric would be dead if he rolled the same number as a dead hero (which would be the same as the Demon die).

-tRR

 

[Endovior]

Yes. Each round, all surviving hero dice are rolled against the demon.

That being said, by all means twink the probabilities... I just made it up myself, and in my test experience, there's about an even chance of winning lots or losing lots... it tends to balance on the Wizard.

 

[ssampier]

a. Interesting. I haven't quite figured out the house advantage. But it seems to me the house has an advantage since the probability of getting the same number on 2d4 is 1/16 (1/4*1/4). The odds of getting the same number on 3d4 is 1/64 (1/4*1/4*1/4). On 4d4 the probability is 1/256 (1/4^4).

b. The odds of not getting the demon die on any die: (Math wizes please correct my math):

1/3

1. x(1) not receiving the demon die number=3/24 (3/4 * 1/6)
2. x(2) "
3. x(3) "
4. x(4) "

3/24^4=1/3

The odds of getting the demon die on any one hero die is 2/3.

 

[orsal]

(Math wizes please correct my math):

Will do.

a. Interesting. I haven't quite figured out the house advantage. But it seems to me the house has an advantage since the probability of getting the same number on 2d4 is 1/16 (1/4*1/4).

Nope. The first die can be anything; the second one has a probability of 1/4 of matching it. Or, if you prefer, there are four possible combinations (out of the 16) that match, so the probability is 4/16=1/4.

The odds of getting the same number on 3d4 is 1/64 (1/4*1/4*1/4).

The first die can be anything, the other two must match it, so probability is (1/4)^2=1/16.

On 4d4 the probability is 1/256 (1/4^4).

(1/4)^3=1/64, by same reasoning as above.

b. The odds of not getting the demon die on any die: (Math wizes please correct my math):

1/3

Way wrong.

1. x(1) not receiving the demon die number=3/24 (3/4 * 1/6)

Same mistake as before -- this is the probability that the demon die will get a 1 (or 2 or 3 or 4, but the number must be fixed in advance) and the hero die something different.

P(demon die gets 1 or 2 or 3 or 4)=4/6=2/3
P(hero die gets different, given demon die gets 1 or 2 or 3 or 4)=3/4
P(demon die gets 5 or 6)=2/6=1/3
P(hero die gets different, given demon die gets 5 or 6)=1

Altogether: P(hero die different from demon die)=(2/3)(3/4)+(1/3)(1)=5/6.

Or equivalently, whatever the hero die gets, the demon die has a probability 1/6 of matching it, or 5/6 of getting something else. Simpler that way, but the final calculation won't be -- we'll need to do it thinking first of the demon die (see below).

3/24^4=1/3

Umm.. (1/8)^4=1/4096. I'm not sure where you got that 1/3 from, but it's horribly off. Besides, even if your 3/24=1/8 figure were correct, you can only multiply probabilities to get the probability of all events happening if the probabilities are independent. In this case they aren't -- if you know hero die #1 was a match, you know the demon die got at most 4, making it more likely that the other hero dice match.

Here's how we can calculate the probability that none of the hero dice match the demon die:

P(demon die gets at most 4)=2/3
P(no hero dice match it, given demon die has at most 4)=(3/4)^4
P(demon die gets 5 or 6)=1/3
P(no hero dice match it, given demon die has 5 or 6)=1

Altogether, P(no hero dice match the demon die)
= (2/3)(3/4)^4+(1/3)
=(27/128)+(1/3)
=209/384, or approximately 0.544.

 

[ssampier]

I appreciate the correction. I am confused because all dice are cast at the same time. The first die is not the comparison; each die has a 1/4 independent chance of obtaining that same number.

The second part, I tried to correct with Excel, but I think I messed up with my Excel formulas.

 

[Arkhandus]

They're 6-sided dice, not d4s, y'know....... Not sure if that changes the probability curves or anything, but then, I hate doing complicated math, and statistics/probability math is definitely not my strong point.....

Anyway.......Fighter's ability should probably be changed to losing only 3 times the amount, it's not very significant as-is I think, and the rogue's ability is definitely least useful.....maybe they should earn 2 or 3 gold more instead of just 1. Or perhaps when the rogue dies, you lose no gold from the rogue, because they tend to sneak ahead of the rest of the group?

Hrm. Mayhap I will try this game out later if I can find someone to try it with.

 

[ssampier]

I must of missed that. I thought the heroes were d4s and the demon die was a d6.

Danke.

 

[ssampier]

In any case, I forgot the AND/Or rule. I read some statistics forums online and I think I understand the difference now.

in this case: 1 - (5/6)^n.