"odds of rolling successes on decreasing quality d6s" Topic

I know the topic title is strange…
I came up with a weird idea I might use for an heroic system, it's called Quality Vs Quantity

To perform an action in this system, you need to roll a certain number of successes, or to roll more successes than your foe. If you roll 1d6, success is automatic. If you roll 2d6, you succeed on any die which rolls a 2+.
If you roll 3d6, you succeed on 3+.
if you roll 4d6, you succeed on 4+.
5d6, 5+
6d6, 6.
It's very easy to remember. The higher number of dice you roll, the more potential successes you might score, but the lower your chances of doing so.
Is there any game like this? Do you think it would work?
What are the odds of rolling 1, 2, 3, 4, 5 and 6 successes?
Thanks for any help
Andrea

ganeshagames.blogspot.com

Not sure I understand what you are getting at, but:

"If you roll 2d6, you succeed on any die which rolls a 2+."

By this do you mean if either of the die shows a 2+, there is success, and the only way to fail is to roll snakeeyes?

yes, if you roll one die, you don't fail, but you score only one automatic success
if you roll 2 you fail only on snake eyes and you can get two successes
if you roll 3 dice you fail on a 1 or 2 and you get maximum of 3 successes
and so on

I told you it's weird.:-)

Ok, a success is a die which shows the number you need to roll.. I hope I'm explaining it better now.

So if you roll 3 dice, any dice which comes up 3,4,5 or 6 is counted as a success.
At the end of the contest, the player who rolled more successes wins, and the number of successes rolled is a gauge of effect achieved (in hand to hand, for example, it could be the number of wounds you inflict)

Seems like the "sweet spot" would be about 3 dice. I think everybody would just take that, go for the reasonably sure 2-3 successes. Who would roll 6 dice and try for 6 successes?

Gonk,
sometimes you might need to be more than safe, and need more successes. That could be built in the rules later.

It's a Bell distribution, with both 3 dice and 4 dice having the same expected return. Of course, you could get more with 4, so that is probably the best number to roll all the time.

I would think the trick would be: how do you keep the players from knowing how many dice the other guy will roll? (Assuming an opposed roll of some kind)

jizbrand,
they keep dice hidden in the hand. Use small d6 only or find players with big hands :-)

oooh, a variant: Lesser types (henchmen) have to roll MORE dice. Making it MORE likely to do nothing.

If I were explaining it I'd say you need a 1+ on every die, with a DRM for every die after the first one.

Any reason you'd want to do it this way? Seems like you'd be layering a whole "dice management" game on top of your mechanisms. Could really slow down play as players decide 4 or 5? 4 or 5? How about 2?….

I don't know if I'm gonna use this mechanic or for what, I was thinking for swashbuckling, pulp or any other period where you are supposed to attempt over the top actions.

My Song of Blades and Heroes rules use a system where players have to decide whether they'll roll one,two or three dice on every activation roll (basically once per figure) and it doesn't slow the game too much (actually critics say it's the best part of the game). But it's a skirmish system where players use 5-15 figs maximum, and in addition you can make group activations.

ganeshagames.blogspot.com

You need to look up bimial distributions and calculate accordingly:

link

Or do some "porr man's statistics" and build yourself out some "simple" success matrices.

For example, for 2d6, there are 26 possibilities, so you create them:

11 12 13 14 15 16
21 22 23 24 25 26
31 32 33 34 35 36
41 42 43 44 45 46
51 52 53 54 55 56
61 62 63 64 65 66

Then calculate the percantageges:
0 Success = 1/36 = 3%
1 Success = 10/36 = 28%
2 Success = 25/36 = 69%

Etc.

Expected successes
1D6 = 1*6/6 = 6/6
2D6 = 2*5/6 = 10/6
3D6 = 3*4/6 = 12/6
4D6 = 4*3/6 = 12/6
5D6 = 5*2/6 = 10/6
6D6 = 6*1/6 = 6/6

I played around with something like this but couldn't get it to do what I wanted. Your design needs probably differ so it might work for you.

The extremes are pretty bad. 1D6 for example looks pretty poor compared to 2D6 and 3D6.
2D6 has only a 1/36 chance of not getting at least 1 success.
3D6 has only a 1/27 chance of not getting at least 1 success.

Similarly 6D6 gives up a lot for that 1 in 46,656 chance of getting 6 successes.

Hi all,

Stick with 4+ to succeed for arguments sake. That is a 3/6 chance to succeed on one die. If you roll 2 dice and want to succeed with just one, your chances are as follows.

Success*NotSuccess+NotSuccess*Success

Basically you add up all the permutations of successes multiplied by chance of failure.

This is an easy example. The chance to roll 1 success is as follows.

3/6*3/6+3/6*3/6=9/36+9/36=18/36=.5 or 50% chance of rolling exactly 1 success.

If you want to know 2 successes, then both dice must be success.

Formula is Success*Success

3/6*3/6=9/36=.25 or 25%

So the chances of 1 or two successes would equal 25%+50% or a total of 75% chance of scoring 2 successes.

This example is simple since there are only 2 dice. It gets much more complicate very quickly.

Hope that helps,

John

Not warped enough.

Make the size of the die used increase as well:

2D3
3D4
4D5
5D6

Then you build the possibility of a truely horrendous number of successes. 99D100 rolls anyone?