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 08-21-2017, 01:26 PM #1 Jasonft   Join Date: Aug 2008 Minor dice rolling/ probability question Minor dice rolling/ probability question (and I'm just OCD enough this is kinda bugging me). I post it here because I know there are people here a lot better at math than I am. Rolling a standard six sided die: average result is 3.5 Rolling a standard six sided die where a '6' gives you +5 to your total and you roll again (adding the results together) until you roll something other than a '6': I think the result is 4 or just a hair more. Not sure of exact number. My question involves the following - Rolling six sided die as per the second example, but adding in the ability to re-roll any '1' (not adding it to the total) until you roll something besides a '1'. If you still add the rolls together as you do in the second example, and re-roll any '1' without adding to the total, what is the average result for the die roll? Not a GURPS question, I know. This actually comes out of the rebooted TORG: Eternity and relates to how valuable an advantage (they call them Perks) is in the game. Any help would be appreciated.
 08-21-2017, 01:46 PM #2 Ulzgoroth   Join Date: Jul 2008 Re: Minor dice rolling/ probability question Clarification required about the rerolls on 6. You described the rerolls as ending if you rolled something other than a 6 - did you mean that the rerolls are exactly like the original roll, or are they exempt from the 'reroll 1s' rule? Assuming the former, the expected value of a die D is: (D + 2 + 3 + 4 + 5 + (5+D))/6 So D*4/6 = 19/6 D = 19/4 = 4.75 With only the exploding 6s it would be: D = (1 + 2 + 3 + 4 + 5 + (5+D))/6 D*5/6 = 20/6 D = 4 With rerolling 1s but no exploding 6s it would be: D = (D + 2 + 3 + 4 + 5 + 6)/6 which is actually the same as exploding 6s only, so D = 4. __________________ I don't know any 3e, so there is no chance that I am talking about 3e rules by accident.
 08-21-2017, 02:11 PM #3 Jasonft   Join Date: Aug 2008 Re: Minor dice rolling/ probability question Re-rolls normally (using option #2 in my example) get re-rolled again only on a '6'. Total is added together of all die rolls in the chain with each '6' adding +5 to the total. All dice in the chain of re-rolls would follow this rule. My question involved the average result if re-rolling a '1' is also forced (due to the Perk/ Advantage being involved). A '6' is still counted as triggering a re-roll in addition to adding +5 to the total. A '1' triggers a re-roll but does not add to the total. All dice in the chain of re-rolls trigger re-rolls on both '1' and '6' with the differences mentioned. Any roll other than '1' or '6' stops the chain of re-rolls for that individual d6 and the rolls are added together to get a total. Situations in game (TORG as declared) might result in one or more d6 being rolled each according to the rules specified and the total for all dice (including re-rolls) being applied to the action. I wanted to know how much the Perk for re-rolling '1' changed the average result per die. Last edited by Jasonft; 08-21-2017 at 02:21 PM.
 08-23-2017, 09:07 PM #4 dripton   Join Date: Mar 2008 Location: Northern Virginia, USA Re: Minor dice rolling/ probability question If 1s are just rerolled, then you're basically rolling a 5-sided die, labeled 2-6. Where the 6 is actually 5 + the result of another die roll. Anyway, I think a closed-form solution for this is hard, so I just wrote some code to estimate it. And got about 4.75 on average.
08-23-2017, 11:20 PM   #5
Ulzgoroth

Join Date: Jul 2008
Re: Minor dice rolling/ probability question

Quote:
 Originally Posted by dripton If 1s are just rerolled, then you're basically rolling a 5-sided die, labeled 2-6. Where the 6 is actually 5 + the result of another die roll.
I hadn't looked at the 1 reroll that way, but that's quite true.
Quote:
 Originally Posted by dripton Anyway, I think a closed-form solution for this is hard, so I just wrote some code to estimate it. And got about 4.75 on average.
A closed form solution is rather easy, and I presented it above. Same answer as you got.
__________________
I don't know any 3e, so there is no chance that I am talking about 3e rules by accident.

 08-23-2017, 11:20 PM #6 RyanW ☣     Join Date: Sep 2004 Location: Southeast NC Re: Minor dice rolling/ probability question Assuming my math is right: The final roll of an "exploding" d6 is effectively a d5 (if it rolls a six, it isn't the last die) which will have an average value of 3. The number of expected 6s rolled is the infinite sum of 5n/6^(n+1), which is 0.2. If each six adds 5 to the total, that's just 0.2*5 = 1. The total average is just 3+1 = 4. --------- If all 1s are rerolled, the number of expected 6s becomes the infinite sum of 4n/5^(n+1), or 0.25. The last die would average 3.5 (it's now effectively d4+1). So 0.25*5+3.5 = 4.75. --------- If only an initial 1 is rerolled, it gets just a little bit trickier, but brute force spreadsheet approximations still work just fine (you just have to have the first line use a different formula from the rest). I get 0.24 rerolls for a total average of 4.6. __________________ RyanW To quote George Washington's famous farewell speech: "Later, losers."
08-23-2017, 11:36 PM   #7
Ulzgoroth

Join Date: Jul 2008
Re: Minor dice rolling/ probability question

Quote:
 Originally Posted by RyanW Assuming my math is right: The final roll of an "exploding" d6 is effectively a d5 (if it rolls a six, it isn't the last die) which will have an average value of 3. The number of expected 6s rolled is the infinite sum of 5n/6^(n+1), which is 0.2. If each six adds 5 to the total, that's just 0.2*5 = 1. The total average is just 3+1 = 4. --------- If all 1s are rerolled, the number of expected 6s becomes the infinite sum of 4n/5^(n+1), or 0.25. The last die would average 3.5 (it's now effectively d4+1). So 0.25*5+3.5 = 4.75. ---------
I'm confused about your sums - shouldn't they be n*(1/6)^(n+1) and n*(1/5)^(n+1)? Cool way to do it though.
Quote:
 Originally Posted by RyanW If only an initial 1 is rerolled, it gets just a little bit trickier, but brute force spreadsheet approximations still work just fine (you just have to have the first line use a different formula from the rest). I get 0.24 rerolls for a total average of 4.6.
Pretty easy to handle with my approach - especially when you've already determined that the value of a die roll with exploding 6s but no rerolled 1s is 4.

d = ((4) + 2 + 3 + 4 + 5 + (5+4))/6 = 27/6 = 4.5
__________________
I don't know any 3e, so there is no chance that I am talking about 3e rules by accident.

08-23-2017, 11:49 PM   #8
RyanW

Join Date: Sep 2004
Location: Southeast NC
Re: Minor dice rolling/ probability question

Quote:
 Originally Posted by Ulzgoroth I'm confused about your sums - shouldn't they be n*(1/6)^(n+1) and n*(1/5)^(n+1)? Cool way to do it though.
I may have transcribed something wrong in writing that post. I'll check it in the morning, as I've already put the computer I was working it out on to bed.
__________________
RyanW
To quote George Washington's famous farewell speech: "Later, losers."

 08-24-2017, 10:25 AM #9 Anthony   Join Date: Feb 2005 Location: Berkeley, CA Re: Minor dice rolling/ probability question 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. __________________ My GURPS site and Blog.
 08-24-2017, 10:38 AM #10 RyanW ☣     Join Date: Sep 2004 Location: Southeast NC Re: Minor dice rolling/ probability question 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). __________________ RyanW To quote George Washington's famous farewell speech: "Later, losers."

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