08212017, 01:26 PM  #1 
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 reroll 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 reroll 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. 
08212017, 01:46 PM  #2 
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.
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08212017, 02:11 PM  #3 
Join Date: Aug 2008

Re: Minor dice rolling/ probability question
Rerolls normally (using option #2 in my example) get rerolled 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 rerolls would follow this rule.
My question involved the average result if rerolling a '1' is also forced (due to the Perk/ Advantage being involved). A '6' is still counted as triggering a reroll in addition to adding +5 to the total. A '1' triggers a reroll but does not add to the total. All dice in the chain of rerolls trigger rerolls on both '1' and '6' with the differences mentioned. Any roll other than '1' or '6' stops the chain of rerolls 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 rerolls) being applied to the action. I wanted to know how much the Perk for rerolling '1' changed the average result per die. Last edited by Jasonft; 08212017 at 02:21 PM. 
08232017, 09:07 PM  #4 
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 5sided die, labeled 26. Where the 6 is actually 5 + the result of another die roll.
Anyway, I think a closedform solution for this is hard, so I just wrote some code to estimate it. And got about 4.75 on average. 
08232017, 11:20 PM  #5  
Join Date: Jul 2008

Re: Minor dice rolling/ probability question
Quote:
A closed form solution is rather easy, and I presented it above. Same answer as you got.
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I don't know any 3e, so there is no chance that I am talking about 3e rules by accident. 

08232017, 11:20 PM  #6 
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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.
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RyanW To quote George Washington's famous farewell speech: "Later, losers." 
08232017, 11:36 PM  #7  
Join Date: Jul 2008

Re: Minor dice rolling/ probability question
Quote:
Quote:
d = ((4) + 2 + 3 + 4 + 5 + (5+4))/6 = 27/6 = 4.5
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I don't know any 3e, so there is no chance that I am talking about 3e rules by accident. 

08232017, 11:49 PM  #8 
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Join Date: Sep 2004
Location: Southeast NC

Re: Minor dice rolling/ probability question
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.
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RyanW To quote George Washington's famous farewell speech: "Later, losers." 
08242017, 10:25 AM  #9 
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. 
08242017, 10:38 AM  #10 
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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).
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RyanW To quote George Washington's famous farewell speech: "Later, losers." 
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