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Group Dice Rolling Probability - Multiple Choice

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Sat, 07/09/2016 - 01:08
Rimmsolin
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For this scenario, imagine four players rolling 4 d6's each. If each needs to contribute a "1" result to a common pool to win as a group, what is the probability? What if there are two or three players? What if only one player is rolling trying to get at least one "1"?

I've driven myself mad (and a little sad I didn't take statistics and probability in school.) I've done the research. I've used www.anydice.com. I'm attempting to understand the math. I'm sure it's obvious to those that have been taught. For those of us reinventing math from scratch, it's a little tougher. :)

Here's the likely outcomes I see. If you have the answer and can explain WHY not just WHAT the answer is, I'd much appreciate it. (Be gentle...)

A) www.AnyDice.com - http://anydice.com/program/8cd9
This shows 48% of a d6 array results in a 0, hence any result that has at least one "1" or more is 52%. Any additional players is multiplied such as 52% or 0.52 * 0.52.
1P = 52%
2P = 27%
3P = 14%
4P = 7%

B) Each die has a 1/6 chance to be the winning result, or 16.67%. To have at least one die a "1" with four dice is: 1/6 + 1/6 + 1/6 + 1/6 = 4/6 or 67%. Additional players are multiplied 0.67 * 0.67.
1P = 67%
2P = 45%
3P = 30%
4P = 20%

C) Four dice have a single chance to be all "1's" out of 6*6*6*6 or 1296 total chances. Or 1/6 * 1/6 *1/6 * 1/6 Or 1/1296. Three dice have a 1/6 * 1/6 * 1/6 chance, or a 1/216, or 6/1296. Two dice have a 1/6 * 1/6 chance, or 1/36, or 36/1296. And one dice has a 1/6 chance, or 216/1296. Adding up the possible winners is 1 + 6 + 36 + 216 = 259. 259/1296 = 20%. Each player has the same chance to contribute at least one "1" die at 20%. So For all players to add in, it's 20%*20%, etc. or...
1P = 20%
2P = 4%
3P = 1%
4P = <1/10%

I think I'm close on C, but it seems super low chance practically looking at it. Is it option A, B, C or a D?

-Adam

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Sat, 07/09/2016 - 02:13
#1
czarcastic
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I believe option A is

I believe option A is correct. There is an 83.33% chance that any d6 will NOT be a 1. As you are rolling 4 dice, the calculation for each player will be .8333 to the 4th power, or 48.23%.
So roughly 52% chance for a player to roll a 1.
Then that is to the power of x, where x= the number of players.

TBH, I almost jumped on B, but extrapolated to 6d6 having a 100% chance, and we all know that isn't true.
Then I remembered reading somewhere that for these types of probability problems, you have to calculate the chance that the target DOESN'T show, rather than the chance it will.

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Sat, 07/09/2016 - 05:28
#2
X3M
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Option A starts out

Option A starts out correctly. 4D{1,0,0,0,0,0} is correct to use.
But then you talk about multiple players.

What are they doing?

Working together?
2P = 8D6{1,0,0,0,0,0}
3P = 12D6{1,0,0,0,0,0}
4P = 16D6{1,0,0,0,0,0}

Or are they against each other?
(Sorry, this is a bit more complicated to use any dice for)

That is very important to calculate the right answer. But your 2P, 3P and 4P are probably wrong.

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Sat, 07/09/2016 - 12:59
#3
czarcastic
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If I understood correctly,

If I understood correctly, each player is a separate set testing for individual success. Group success requires all players to individually succeed. Also, it's assumed that players cannot give each other their extra successes (separate sets).
With 2 players, there are 4 possible results:
A and B succeed = Group success
A fail, B succeed = Group fail
A succeed, B fail = Group fail
A and B fail = Group fail
So the group only succeeds 27% of the time (.52x.52)

If they can share successes, then it just increases both the trial pool and the successes required. I think this works out to the same 52% chance, but haven't run the math.

Unless for some reason you want things to get exponentially harder for the players the more there are, I suggest sharing successess.

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Sat, 07/09/2016 - 13:44
#4
Zag24
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Here's how to think about

Here's how to think about this sort of problem.

Problem 1, rolling 4d6, what are the odds of getting at least one 1?

Whenever the question are the odds of getting at least 1 of something, turn the question upside down: What are odds of rolling 4 dice and getting not any 1s (or getting 2-6 on all 4 dice)? When you have to get "all" of something, you just multiply. So the odds of getting not-1 on one die is 5/6, so the odds of getting not-1 on four dice is 5/6 x 5/6 x 5/6 x 5/6 = (5/6) ^ 4 = 625/1296.

So, if the odds of rolling 4 dice and getting all not-1 is 625/1296, then the odds of getting at least one 1 is just 1 minus that, or 671/1296.

The chance of 4 people all doing that in one try is just that ^ 4.

| # players | odds of all getting at least one 1 with 4 dice |
|---------|----------------------------------------------|
| 1 | 671/1296 (call this p) ~= 51.8% |
| 2 | p ^ 2 ~= 26.8% |
| 3 | p ^ 3 ~= 13.9% |
| 4 | p ^ 4 ~= 7.2% |

Baah. I can't seem to get the table feature to work.

Anyway, it looks like your first results were correct. Now you know why. :)

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Sat, 07/09/2016 - 13:53
#5
X3M
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I get it now. Each player

I get it now. Each player needs to provide at least 1 success. Then A is completely correct. Didn't read correctly, sorry.

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Mon, 07/11/2016 - 18:02
#6
Rimmsolin
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Ok, so we're agreed A is

Ok, so we're agreed A is correct.

1P = 52%
2P = 27%
3P = 14%
4P = 7%

Yes, each player is a set and has to contribute one success. Think of it as a STAR die face, instead of a number if that helps. The number is irrelevant. A player may spend this special side, at will so the other faces or uses do not matter.

Allowing sharing of successes would make it easier. At least X successes, where X is number of players, from any dice roll would be something like 15-35% or so for 2P-4P.

What would be ideal is a system that scaled with number of players with the same (or nearly so) odds instead of the variance above so that the game mechanic can be balanced.

Essentially this mechanic is for a super power that can be called in, when each player gives up a die to contribute to the party ability. Something like 10-20% occurrence is probably about right. Each player will have 4d6 that are identical as above. A unique 5th die can have any combination of symbols.

If the number of players always goes up, and the percentage of successes is always under 100%, then it would seem to be escalating difficulty to have all successes. Does that make sense? Any suggestions?

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Mon, 07/11/2016 - 19:28
#7
let-off studios
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Sharing Successes

I can think of a few ideas off the top of my head. To "cover" for a player who did not roll any stars/successes:
OPTION 1 - One player can use two additional stars to count for another player's success. If they have three additional stars to contribute (that is, they rolled all stars on their 4d6) they can cover their own needed success plus that of two other players. Results as follows:
---> Rolling 1 star = 1 success
---> Rolling 3 stars = 2 successes
---> Rolling 4 stars = 3 successes

OPTION 2 - Players can contribute to a pool, with the same ratios as above. This seems to me the easiest to accomplish.

OPTION 3 - A player can "freeze" a die out of their own pool for one round to contribute a success for another player. On the subsequent round, they'll have a pool of only 3d6 to work with. Their "frozen" die comes back the following round.

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Mon, 07/11/2016 - 20:33
#8
Rimmsolin
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I really like the OPTION 3,

I really like the OPTION 3, freezing a die with one small change. Players can keep the dice frozen and take it back at will.

So in a 4 player game, players A, B, C, and D will get 4 dice.

During round one combat, A and B each contribute a STAR result to the pool.
During round two, player C contributes a STAR result to the pool.
During round three, player D does not get a STAR, but has the choice to blow a one-time ability to re-roll in the hopes of getting a STAR, or the team may need to abort and take their dice back.

The interesting choice is opportunity cost. Players A, B, and C lost one of the main action dice in the hopes to pull off the super ability. The pressure would be on Player D to make it happen. Also this gives a great risk/reward team dynamic.

This mitigates the rolling all sets on one turn, but doesn't seem to change the odds for one turn. How would one figure the odds between turns? For 4 players is it 0.52 * 0.52 * 0.52 * 0.52 = 0.07. Then for the 2nd round with only 2 players left it is only 0.52 * 0.52 = 0.27? Essentially the odds get better based on remaining players. Something feels off on this. I think addition needs to be applied, from what I know. :)

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Mon, 07/11/2016 - 22:18
#9
let-off studios
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One Time Ability

If you're relying so much on slim probabilities and a one-shot ability also relies on chance, then the players will likely feel hopeless a lot more often. Particularly when so many dice from the group are devoted to this super ability and there's still no guarantee when the one-shot ability is used.

I think testing would bear this out, but from how you describe it here, it sounds to me that you need other ways for players to directly and deliberately control the dice to give them a snowball's chance of success. Otherwise it hinges on complete randomness with only a single outcome that means success, as opposed to games that offer opportunities for even the worst rolls to be useful for something.

I enjoy dice games as much as the next guy, probably more, but I'll enjoy it more if there's something I can do with whatever roll comes up.

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Tue, 07/12/2016 - 00:45
#10
Rimmsolin
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Yeah, I agree on the

Yeah, I agree on the randomness. I suppose I could have detailed the system a little more... The good news is this is not output randomness, but rather input randomness. Meaning a player rolls the dice and decides where best to spend them. No basic actions require a dice roll.

The dice are a boosting mechanic for attacks, heals, etc, not a measure of success in of themselves. Most of a players abilities can be used at a base level--for instance a crossbow attack deals 3 DMG. However using one or more of the four dice can provide +1 target, _2 DMG, or even turn the bolts into flaming bolts to inflict BURNING when damaging the target.

So as described here, spending these boost dice to power a super, party ability takes away the small boosts in the short term, to queue up one powerful event such as Healing Rain group heal, or Bombardment encounter wide AOE in the long term. It's up to the player to decide when to use the ability after queuing it up for maximum effect. Combat runs several rounds and boss fights even longer so a properly timed super can turn a fight.

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