Rolling successes with rerolls, additional rolls, adds, and finding the avarage outcome

Okay, I know the basic math behind the system, but this get's over my head.

I have five ten sided dice, or 5d10. Each die that rolls 4 or more, is counted as a success (thus, an avarage of 0.7 successes per dice, or 3.5 successes for 5d10). After thus rolling, I am allowed to add 2 to the result of any one single dice. On avarage, there are thus 1.5 non-successes, eg 3 or lower, rolls. Of those rolls, two thirds (results of 2 and 3, but not 1), can be modified with this +2 to turn one success - so out of the 1.5 fails, 2/3 can be saved - resulting in 0.5 fail, and 4.5 success on avarage. Am I correct this far?

Now, it gets a bit more complicated. I get to reroll up to two dice that failed, if I want to - but I have to reroll them both at the same time if I choose two, so I can't reroll one dice, see if it fails, and then reroll it again. So only one reroll, of zero, one, or two dice on my choice. At this point I think I get lost, since I assume that the answer would be 0.5 (the fails) * 0.7 (the chance of success) = 0.35 success, leaving me at an avarage of 4.85 successes per roll with the above mentioned modifiers.

However.... If I roll one or more natural 10's, I get to add 2d10 to my pool and roll them as well. And this is where I get lost.
How do I calculate the chance of getting a natural 10 on five dices, provided that I will reroll 1, 2 or 0 dice depending on the amount of successes ( I never reroll a success)? If I have 0 or 1 failures, should I still reroll two dice (even if one or both of them are successes) in an effort to gather more success on avarege?

To make this even more complicated - The +2 to one dice is added after -everything else-. The +2 dice are rolled as soon as I get a natural 10, so this can occur either before, or after the reroll. (A dice is natural 10 as long there's no substraction or addition to it - rerolls still count as natural rolls).

Questions:
Provided that I never reroll any successes and always reroll as many failures (failures after counting in the +2 for one of them) I can, how many successes I get on avarage?
Provided that I have rolled 4 successes but no natural 10's on the roll, should I reroll only the failure, or should I also reroll one of the successes in order to maximize avarage successes (in an effort to get the natural 10, that is).
Provided that I have rolled 5 successes but no natural 10's on the roll, should I reroll: zero, one, or two of the successes in effort to get that natural 10 and maximize avarage successes? Or am I better off not rerolling anything?
What is the chance of rolling 5 or more successes with the above mentioned conditions by rerolling only failures? What is the optimal strategy to get -at least- 5 successes?

Thanks in advance!

- Gamer

Quote: Gamer

Okay, I know the basic math behind the system, but this get's over my head.

I have five ten sided dice, or 5d10. Each die that rolls 4 or more, is counted as a success (thus, an avarage of 0.7 successes per dice, or 3.5 successes for 5d10). After thus rolling, I am allowed to add 2 to the result of any one single dice. On avarage, there are thus 1.5 non-successes, eg 3 or lower, rolls. Of those rolls, two thirds (results of 2 and 3, but not 1), can be modified with this +2 to turn one success - so out of the 1.5 fails, 2/3 can be saved - resulting in 0.5 fail, and 4.5 success on avarage. Am I correct this far?

No. Although you can save 2/3 of all of the failures, if you have more than one failure, you can only save one of them. In the following table, p(successes) is the binomial distribution of getting n-successes where the probability of success is 0.7. p(save) is the probability that at least one of the failures is a 2 or a 3, or 1 - the probability that all of the failures are a 1. Expected saves is p(successes)*p(save). In the lower right hand cell, you see that you will save only 0.67232 of them on average.

Successes	p(successes)	p(save)	expected saves
0	0.00243	0.995884773662551	0.00242
1	0.02835	0.987654320987654	0.028
2	0.1323	0.962962962962963	0.1274
3	0.3087	0.888888888888889	0.2744
4	0.36015	0.666666666666667	0.2401
5	0.16807	0	0
		Total	0.67232

When you get two additional rolls from a natural ten, can you apply the +2 to those rolls? If so, then this increases the likelyhood that you have a failure that you can save.

Yeah. You add the +2 to one single dice after the reroll and after the possible +2 dice from a natural 10.

So: Roll 5 dice. Check for a 10 - if there's at least one ten, roll to additional dice. After that, decide which dice to reroll, rerolling 2. After that, add +2 to a single dice.
Alternatively: If there are initially no 10's, reroll 0, 1 or 2 dice. Did a 10 come up? If no, add +2 to a single dice. If yes, roll 2 additional dice, and then add +2 to any single dice.

Quote:

In the lower right hand cell, you see that you will save only 0.67232 of them on average.

Does this mean that I can "save" 0,67232 failures per roll on avarage? In effect, that (without taking the +2 dice to account), expected avarage is 3.5+0.67232=4,17232 successes per roll?

EDIT: Or, that out of an avarage 1.5 failures per roll, I can save 0,67232 of them? Meaning, 1,00848?

Thanks for the chart by the way!

It means that you can save 0.67232 overall, so your expected average is 3.5+0.67232 = 4.17232.

My other thoughts about this are as follows:

Re-roll as many failures as you can, but re-roll 1's first, since a 2 or a 3 can be saved if necessary.

If you have 5 successes, but no perfect 10, re-roll two of them. On average, you will end up with 2.08 successes instead of the 2 that you decided to re-roll. If you only re-roll one of the dice, then you will end up with 1.076 instead of the 1 that you re-rolled.

I think that this applies to 4 successes without a perfect 10 also. Re-roll the failure and another die, even if the failure can be saved. You will end up with 2.08 successes instead of 1 or 2 that you had before.

Cool, thanks! But how did you arrive on that 2.08?
Rolling 2 dice: 1.4 successes, and 0.2 perfect 10's on avarage. perfect 10= +2 dice = 1.4 more successes on avarage, 0.2*1.4=0.28.
So if I understand the math correctly ( I might not), then rerolling 2 dice results on 1.4+0.28 successes on avarage=1.68 successes, + 0.67232 for the "save" = an avarage of 2,35 successes?
Or should I only count the expected saves of 0, 1 and 2 dice (since at this stage, I'm only rerolling 2 dice) for a total of 0.00242+0.028+0.1274=0.15782, for a total of 1.4+0.28+0.15782=1.83782?
Do I have the math all wrong, or am I just missing something?
(most likely both)

Roll 2 dice:
You will have 1.4 successes without "saving" any.
You will get one or more perfect tens 0.19 of the time {the probability of rolling at least one ten is 1 minus the probability of rolling all 2-9s = 1-(0.9)^2 }.
From the additional rolls, you will get 0.19*1.4 additional successes = 0.266.
Because you haven't used your "save" yet, you can use this. On average, you will be able to save 0.414 rolls. This takes into account getting the additional rolls for a ten also. I don't have a pretty write up for this, and my spreadsheet wouldn't even make much sense. But, to calculate, you must consider every possible outcome. For instance:

Roll 0 successes/0 tens: you rolled 2 dice total, so the probability that you can save one of the two is 1-(1/3)^2
Roll 1 success/0 tens: the probability that you can save the remaining roll is 1-(1/3)^1
Roll 1 success/1 ten: you get two more dice, so now, you have to consider the outcome of the other two dice, and your total successes can be 1, 2, or 3. With each of these, you need to calculate the probability that you can save one of the failures.
Roll 2 successes/0 tens: p(saving) = 0, because everything is a winner already
Roll 2 successes/1 ten: must consider the outcome of the other two dice, and your total successes can be 2, 3, or 4. With each of these, calculate the probability that you can save one of the failures.

All together, you get 1.4 + 0.266 + 0.414 = 2.08