GameBanshee Forums

I know there a few math types that come to Gamebanshee on a regular basis, and it is by far not my strong point so I appeal to you, oh mathmeticians of GB.

I'm working on a game that uses a series of dice rolls to simulate difficulty, and I am trying to find the odds of success.

What I need to know is this:
What are the odds of a success when using this die sequence?

First roll a d4, whatever number you result in from that roll write down. You then need to roll that number or higher on a d6. Then roll the result or higher on a d8, d12, and d20 as before.

If possible, I would like to know the odds of success for the entire series and the odds of success if you omit the d4; the d4 and d6; the d4, d6, and d8; and omitting the d4, d6, d8, and d12.

I'm not sure if anyone knows how to do any of this, but I thought I would give it a shot. TIA to anyone who can help.

Hmmmm - it has been a long time since I've done any statistics of importance, but I can't actually remember any type of generel statistics that covers your case, because of the alternating value of the possiblity and the influence between the tries.
So it is a bit more complex then I've delt with - back when I did statistics
So I can't help you calculate the specifics you need.

However - if nobody else can help either - then you could simply start calculating the various outcome.
Like 1/4 chance with a d4 and 1/6 chance with a d6 and so on, and if you have a specific number of 1/4 you know how bit a possiblity (number of times 1/6) you have with a d6.

Besides - what do you need to know the excat odds of success for?

I knew it was a pretty tough situation.. LOTS of variables here. I'm thinking the only person that can get firm numbers is a statistician.

The only reason I want the real numbers is because I want to see if it is possible to do(I know it's possible actually, I've done it), and how hard is it. It needs to be fairly hard for the application I need it for.. Also, I want it to get easier with every die that is removed from the sequence. I'm hoping that it will get easier by about the same amount with each die taken away. I'm sort of wondering if anyone is this kind of statistical wizard. I'm terrible with it.

I still fail to see what you need to the probabilities for if you need to code a program/application. I can understand the curiosity of finding out a mathematical formula can be written, but if you need to code some computer thingy - then I can't see the usage for that.

I'd personally not hardcode probability chances into a program if there is any way to avoid it.

EDIT:
Oh - and if you simply want the numbers, then it is "easy" to simply code a program that will calculate all the probabilities for you anyways
Instead of trying to make a general forumla

I don't need it for any computer application at all.. I am making a tabletop game and I just want to check to see if the system is sound. I'm pretty bad with programming to.. I have a real problem with numbers.

That's a really difficult system to work out mate. Why do you want something so complex?

There is an RPG which uses a system like that (and it is the only one I've found which uses my d30 as well), I think it is Requiem or Lost Souls. Basically, the more difficult the task, the larger the dice you roll, trying to get under a target number. So a real easy task, with a TN of 5, you get to roll a d6. For a real difficult task, you roll a d30.

Cheers,
TGHO

[QUOTE=jopperm2]I don't need it for any computer application at all.. I am making a tabletop game and I just want to check to see if the system is sound. I'm pretty bad with programming to.. I have a real problem with numbers. [/QUOTE]

Well - my advice would then be to simply manuallly calculate the possiblity for every possible outcome - it is a rather limited number still.

How about a guestemation formula?

There is a chance of 1/4th on each of the dice on the d4 (well duh), so a average should be 2.5. The average on a d6 is 3.5.

So, in order to beat the 2.5 on a d6 you'll need to throw at least a 3. So 3,4,5 and 6 are positive numbers. That equals 66.67 of the dice.

So a average chance to beat a d4 with a d6 would be 66%.


Now I have 2 thoughts:

1- Problem is that we now have to deal with higher numbers, as lower numbers no longer count.

To beat a d6 with a d8, statistically chances have increaced that it in the top 4 region (3,4,5,6). That leaves the d8 with 7 and 8 plus the chance that it's higher than the average of 3,4,5 and 6 = 4.5. So a minimal of 5 is what we have here. 5,6,7 and 8 is 50% of 8. So chances have shrunk... that means I have to calculate the rest too...

For the d10, average of the d8 is 6.5 so a minimal roll of 7 is needed. 7,8,9 and 10 are positive. That's 40%.

And then the d12. Average of the d10 would be 7,8,9,10 -> 8.5 so the d12 needs a 9. 9,10,11 and 12 are positive, that's 33%.

If the rolls are done after eachtother we caluculate that to actually make the final roll, the d12, chances are 66%*50%*40%*33%= 4.4%
So there's a 4,4% chance to actually succesfully reach and win the d12, according to this little guestemation.



2- Another possebility would be to keep going with the average of each dice. Then it's a 66% chance on each roll.

Rolling them after one another trying to beat each last roll, then that would be 66% (d6)-> 43% (d8) -> 29% (d10) -> 19% (d12)
(so 66% on 66% on 66% etc.) So there's a 19% chance to actually succesfully reach and win the d12, according to this little guestemation.


Quite the difference between the two formulas.
I think the first one is more accurate.


I'm no mathaticawizard but if they are both not correct then please work it out. I find it interesting.

I think I agree on your first one, Rob

I made a worst case scenario estimate and if you rolled the highest number on every die the chance of succeeding with the d20 was like 2/5%, pretty hard. I'm not sure if I can tweak it or not.. I want something that is relatively hard but not impossible at its hardest.

Perhaps you can fail a roll, but still need to roll the next rolls.
In the end you as GM improvise what not-making one or more rolls would mean. Or you make rules for it.
Not makeing one roll - little effect
etc.
Not makeing any off them - critical miss

Possibly.. I would have to rethink how that would effect the system. It's designed to simulate processes where one step depends on how well you accomplished the previous step.

This is a formula dependant on 5 variables, and as such I don't see immediately what the chances are. I'll go ask some higher-ups for help.

edit: There are 46080 combinations. Go tiger.

Groundbreaking statistics every day here at Gamebanshee..

Thanks for the effort, hopefully someday we'll know the answer.

I guess you probably have lost interest in this by now, but my bed is currently occupied by an agressive animal, and I have nothing else to do...

There are 24 possible combinations you can roll a D6 and a D4, and in 18 of those combinations the D6 will be higher. That means the chance of success is 75%.

There are 192 combinations you can roll a D6, D4 and D8. The D8 will be higher than the D6 and the D6 will be higher than the D4 in 88 of those I think. The Reason is that in one case all 8 results on a D8 will result in sucess, in two cases 7 results will be sucess etc. A table will look like this:

D4|D6 6_5_4_3_2_1
1 ____ 3+4+5+6+7+8
2 ____ 3+4+5+6+7
3 ____ 3+4+5+6
4 ____ 3+4+5

The number in the table is the number of successful results on the D8 for a certain result on the D6 and D4. It all adds up to 88, wich means 88/192, or about 46% of the results are successful so far.

When it comes to the D12 I am not sure how to make this visible, but basically you can imagine a diagram like the above, only with 3 dimensions instead of two. The number of total combinations is by now 2304 so its gets hard to do calculation without having an effective way of doing it, but after laborating a bit i think the number of successful combinations are 499. That would mean about 22% chance.

Im afraid the D20 makes too many combinations for using the previous method though.

I am not good at statistics, but im quite sure I've got the first two right. The third I think is correct too, but I wouldnt bet anything expensive on it. To check I made a small program on my calculator that tries to solve the problem empirically. (Dont have any compiler of any kind on this computer ) The result was as follows:

first check: 78%
first and second: 45%
first, second and third: 27%
all checks: 18%

So the result seems resonable. The calculator is very slow though, so this is only about 700 test, wich is not very much.

Actually, I am still interested and thanks for the effort. Someone PMed me a while back with the result and I forgot to post it. I'll edit in the PM too. THanks a bunch Dottie and MR Waesel!

[QUOTE=Mr.Waesel]Here's a link to a thread where I asked your question.[/QUOTE]