Complex 10 side die game mechanics question

Hello, I couldn't find anything similar to my question so here goes. Sorry of I missed a similar topic.

Here is my setup: the player rolls n 10 sided dice against a target number T (non zero positive integer).
If a 1 is rolled on a die, the die is discarded from the roll. All such dice are considered "bad".
All the other dice (non-1) are considered "good" dice.
Then, each bad die is rerolled. If a 1 is again rolled then that die is considered "cancelling".
Each cancelling die removes one good die from the roll, always the highest one left.
Finally, each remaining good die that scored a 10 adds another die with which it is summed up. The last step can be repeated as long as 10s are rerolled and a die stack can grow to infinity.

Now, each bad die is an automatic failure. Each good die stack that exceeds the target number T is a success. Each die stack that doesn't beat T is diacarded as irrelevant.

The entire roll is a success if there is at least one success die stack left at the end.
Sometimes it will be necessary to know how many successes have been scored.
It is a failure if there are 0 successes.
It is a catastrophic failure if there are 0 successes and there is at least one cancelling die.

What I need to do is produce a probability chart that shows the probabilities of each one of the three possible outcomes and takes into account the two parameters n and T.
I have a basic idea about the component probabilities but I don't know how to get them working together.

Here's what I got so far but I'm not sure if it's worth anything:
A die is bad with a probability of 1/10. It is cancelling with a probability of 1/100.
The probability of a die beating T is (10-T)/10 if it's less than 11. For 11 and geater it's that times 1/10 (since you need to roll a 10 first). I don't know how to generalize that for an arbitrary T. I suspect it has something to do with a converging geometric series but...
I've no idea how to take into account the cancelling of the highest good die.

Please help I'm way in over my head on this one :(

A ten sided die is not a regular polyhedron. It has two sides of five surfaces. I'm sure the "dice setters" would love a chance to throw those around for money.

After reading your detailed description, I still don't know what the objective is. When do you "Win" and when do you "Lose"?

Quote: s2dbaker

A ten sided die is not a regular polyhedron. It has two sides of five surfaces. I'm sure the "dice setters" would love a chance to throw those around for money.

After reading your detailed description, I still don't know what the objective is. When do you "Win" and when do you "Lose"?

Last time I used a 10 sided die was to kill an elf.

We use a D10 in lieu of the spinner in the game of "life". But yeah, the D10 I got from my D&D days, in which I surely would have used D10 to assess damage from one of the spells I would have cast with my +2 wand of ravenclaw or whatever.

Ok but let's assume a "fair"10 sided die then with equal probabilities for all faces 1 to 10 ;)

Theobjective is to get as many successes in the die roll as possible. My problem is to get an expression that would let me compute the probabilities for all the outcomes I outlined given T, n.

Maybe an example for the algorithm above will clear things up.

Let's have a roll of 6 dice against a target nnumber of 5.
n=6
T=5

Say we rolled this: 1, 3, 4, 7, 10, 10

First we reroll the"1" and we get a "1". This means we have to discard one highest good die which is a 10. So we're left with 3, 4, 7, 10.
The 10 lets the player roll an additional die and add it. They roll a 6.
So the roll is 3, 4, 7, 16.

Since T was 6 there are only 2 successes: the 7 and the 16.

So my question is this: how to compute the probability of racking up S successes in an n dice roll against a target number T?

Is this all done with one single toss of all of the n dice at once (barring a 1 or 10 which makes for further die tossing resolution)? Can T be 10 or greater? Why would anyone re-roll the 10 more than the amount of tries needed to beat T?

N Dice are rolled together but the additional rerolls are applied in later steps. Hmm not sure what you mean really :D
Yes T can be anything equal to or greater than 1, but it has to be an integer. So it can easily be set to 27 or 1000000.
That's why you would want to reroll and stack up your 10s as many times as possible.

I'm not good at this.

First, I think the game mechanics are a bit asinine. When target numbers are 10 or less, it doesn't matter if rolling a 10 stacks, since the 10 automatically means you succeeded anyway. And the cancellation effect of rolling a 1 followed by another 1 is too small. Only 1/100 of the time will a cancellation occur.

If you want exact numbers, I can't give them to you for every possibility. But, if you are ok with approximates, the cancellation effect is extremely minor. Thus, the probability of any die being a success is equal to (10-T)/10. Better put, the probability of failure on one die is equal to T/10. The probability of N dice resulting in no successes is equal to (T/10)^N. Since only 1 success is necessary, it's easier to calculate the probability of failure, since all N dice have to fail in order for the roll to be unsuccessful.

Now, here is what I mean about the cancellation effect being minor. I can only think of this with an example. Suppose T=7, and N= 4. Thus, 4 dice are being rolled, with a target of 7, meaning you have to roll 8, 9, 10 on any of the dice to succeed. The formula above would show you that the only way to fail, rolling 7 or less on all 4 dice, is equal to .7*.7*.7*.7 = .2401.

However, there are additional ways to fail the roll. One is to roll 1 success, and one cancellation effect on one of the other dies, and 2 failures on the remaining two dice. Thus = combin(4,1)*.3 * [combin(3,1)*.1*.1] * .7 * .7 = .01764

The other way to fail is to roll two success, and two cancellations. This would be = combin(4,2)*.7*.7*.1*.1*.1*.1 = 0.000294

Thus, the additional ways to fail add up to .01764 + .000294 = .017934

And your total way of failing is equal to .2401 + .017934 = .258034

So, the approximate probability of failing is 24%, while a more accurate probability of failing is 25.8%. So I'd say the approximate is close enough.

Of course, I'm not sure any of my math is done properly. But, in a strictly thought logical sense, the idea of a 1 having to be rolled twice in a row in order to cancel out a good die roll seems like such small potatoes as to not matter much.

For target numbers greater than 10, simply use the same ideas. Figure the probability of success on one die. For a target of 12, a 10 would have to be rolled, followed by another roll of 3 or greater. Thus P(success) = 1/10*8/10 = .08. The failure of one die is equal to 1-P(success), in this case .92. Take the P(failure) to the Nth power to determine the probability of failing with N dice. 0.92^4 = .71639 of no success in 4 dice. Again, the relatively minor consequence of the double 1 roll can be factored in, but it is really minor.

One minor complaint I have of this system, and it is very reminiscent of an RPG that I've read in the past although I can't place it at the moment, is that the target numbers of 9 and 10 have the same probabilities as each other. The why should be fairly obvious. Similarly, the target numbers of 19 and 20 have the same probability as each other. 29 and 30 are equal, etc, etc.

Basically what most folks here are telling you without saying it is that d10 dice cannot be made as "Legal" dice usable in Casino Gaming.
I had a similar ideal years back involving d12 dice. NOPE, cant makem "Legal". There will be some form of mold-mark or injection mark that cannot be removed while keeping the edges sharp.
Dice manufacture is very strict, needless to say, so strict about them edges you can almost cut your finger on them... and they DO chip and sliver.

Not in Nevada. Casinos in NV can get their dice from wherever they want, Toys R Us or your local RPG store included.

its not where, but if they conform to NGL.