Minor dice rolling/ probability question

[Ulzgoroth]

Originally Posted by RyanW

Okay, I've double checked, and it is indeed:

5n/6^(n+1)

If there are n rerolls, the (n+1)th roll is the final roll. The probability of that in the simple exploding d6 case is 5/6^(n+1). The number of rerolls is the infinite sum of (number of rerolls times probability of that many rerolls) or 5n/6^(n+1).

Oh, yes, now I see.

[talonthehand]

Originally Posted by Anthony

Another way of looking at this: this is a finite state machine with 4 terminating conditions and 6 states. This means the mean number of dice rolled is 1.5, so the average will be (average ignoring rerolls) * 1.5, or since there are 6 values and the average is division by 6, it's just (sum ignoring rerolls) / 4 (and in a more general case, it's (sum ignoring rerolls) / (total terminating conditions).

The sum ignoring rerolls is 0 + 2 + 3 + 4 + 5 + 5 = 19. 19/4 = 4.75.

For giggles I went ahead and made a java program to do this kind of roll 100000 times then average it, and yeah, it came to 4.75.

[RyanW]

Originally Posted by talonthehand

For giggles I went ahead and made a java program to do this kind of roll 100000 times then average it, and yeah, it came to 4.75.

It's always nice to have empirical evidence to back up the theory. And multiple models that come to the same conclusion.

[Celjabba]

One detail that may or may not have its importance : it is impossible to roll a '6' (or an 11) with the perk : you can roll 2,3,4,5,7,8,9,10,12, ...
I don't know the system so no idea if it may matter ?
I sort of remember a game, can't put a name on it right now, where a similar advantage was actually a problem : you couldn't roll some desirable results.

[hal]

There are many game systems out there like that in which a target number might be 2 higher than the number of sides to the die itself.

Take for instance, needing to roll a 1+ on your target, and you get to reroll 6's and treat it as adding +5 to the next die roll.

What are the odds of rolling a 7?

Technically, you have a 1 in 6 chance of rolling a 6. Then, because of +5, the next roll becomes:

1+5 = 6 (or 1 in 6 times 1 in 6) or a 1 in 36 chance
2+5 = 7 (ditto)
3+5 = 8 (ditto)
4+5
5+5
6+6 (enters into a new sequence

New sequence:
1+10 = 11
2+10 = 12
3+10 = 13
4+10 = 14
5+10 = 15
6+10 = enter new sequence



You would need to add up all of the probabilities to get a given number to see what the statistical odds of rolling a given number or higher are by adding all of the numbers equal to a given target or less, or a given target or a given target and higher (depending on the rule design).

What if you had a rule that said "Roll a 6 sided, and on rolls of 6, add 1d6-1 to the next roll - but any 6 keeps rolling?

It gets "interesting" ;) That's largely why I don't like rolling pools of dice against a target number, and re-rolling any die that gets the highest roll possible and adding until you don't get the highest die roll. The statistical odds for rolling a specific target number jump around a fair bit. But that's just me. ;)

[Jasonft]

In the reboot of TORG that this is from increments of five are very important. In the dice rolling question I posted, you are adding your total to your base weapon strength and comparing it to the toughness/ armor/ etc of the target.

Each step of five gives you additional points of Shock(stun) and number of Wounds(HP) inflicted. In certain builds the advantage(Perk) in question would very much seem to be worth it.

Thanks for the help. Math lost me completely right about the time the teacher started adding in multiple letters.