# What are the average and median scores for attributes in Tunnels and Trolls?

Attributes for characters in Tunnels and Trolls are rolled as follows (in newish editions; jargon term is TARO):

1. Roll 3d6.
2. If all the dice match (you rolled a triple), roll another 3d6.
3. Continue with step 2 until you no longer get triples.
4. Sum all the dice.

(The process is for each attribute; there is no interaction between them.)

Example roll 1: Dice show 3, 6, 6. The rolls do not match so we sum them to get 15.

Example roll 2: Dice give 3, 3, 3. Since the dice do match, we roll another 3d6 and happen to get 2, 2, 2. The result is still a triple so we roll another 3d6 and happen to get 2, 6, 4. The dice are no longer a triple so we sum everything for a result of 27.

What is the average score? What about median?

This can be approximated, to arbitrary depth, with a variety of methods. It is reasonable to truncate after N time 3d6 rolls - even if they are all triples - as probability of any continuing combination (and therefore its effect on statistics) becomes very low. Technically we can perform that approximation because the probabilities multiply by a fraction less than one, whilst the result of getting lucky only adds a limited amount.

I have calculated the results of doing this for up to 8 iterations (i.e. up to 7 triple results, plus an 8th one where we ignore whether it is another triple). This is a reasonable approximation, for example the highest possible result is an attribute of 144, but that occurs so rarely that it contributes just 3 in 10^17 to the mean.

Here is a summary of the results:

• Mean 10.800 (compared to 10.5 for unmodified 3d6)

• Median 11 (compared to 10.5 for unmodified 3d6)

Interesting things only start to happen when you look at top percentiles:

• The 95th percentile is a score of 16 compared with 15 unmodified

• The 99th percentile is a score of 24 compared with 17 unmodified

• The 99.9th percentile (1 in 1000 rolls) is a score of 32, compared to 18 unmodified.

• The 99.99th percentile (1 in 10,000 rolls) is a score of 41.

So this game mechanic is very much like ordinary 3d6 except for rare lucky exceptions where there can be a large boost.

As a bonus, here's a quick graph to compare T&T version (in red) with plain 3d6 (in yellow):

You can see some of the multiples of 3 have a small amount of probability shaved off, which is balanced by higher probabilities later on (most noticeable at "triple value" + 10|11) and a long low probability tail.

The expected value of a standard 3d6 roll is 21/2 and with TARO the expectation is 10.8, which is 10 + 4/5.

I present a non-rigorous argument below. The rigorous argument with the same idea is too long for an answer here; instead, please see Expected characteristic in Tunnels & Trolls character creation, with generalizations (T. Brander) in the journal Mathematics for applications, Vol. 7, No. 2 (2018).

We have a 1/36 chance of getting triples, which (by linearity of expectation; see details below) increases the average by another average of 3d6 multiplied by 1/36; that is, we get $$\ 21/2 + 21/2 \cdot 1/36\$$.

But we may get triples again; there is a further 1/36 chance of this happening and if it does happen, then on average we add 21/2. Continuing like, the expectation is $$\mathbb{E}(\text{TARO}) = \frac{21}{2} + \frac{21}{2} \cdot \frac{1}{36} + \frac{21}{2} \cdot \left( \frac{1}{36} \right)^2 + \ldots.$$

This is a geometric series, so $$\mathbb{E}(\text{TARO}) = \frac{21}{2}\cdot \sum_{j=0}^\infty \left(\frac{1}{36}\right)^j = \frac{21}{2}\cdot \frac{1}{1-1/36}.$$

Let us simplify the expression: $$\mathbb{E}(\text{TARO}) = \frac{21}{2}\cdot \frac{1}{1-1/36} = \frac{21}{2}\cdot \frac{1}{35/36} = \frac{21}{2}\cdot \frac{36}{35} = 21 \cdot \frac{18}{35} = 3 \cdot \frac{18}{5} = \frac{54}{5} = 10 + \frac{4}{5}.$$

The median of the distribution is 11. The median for standard 3d6 roll is {10, 11}. Since DARO increases the results, the median may only increase (or stay the same). If the original roll was (3, 3, 3), DARO increases it to at least 13; this is enough to guarantee that the median for DARO must be at least 11.

To get an upper bound for the median, we can assume that every triple we originally roll gives us the final result of infinity. This moves a probability mass of 6/216 away to the infinity. The probability of rolling exactly 11 is 27/216, so moving a mass of 6/216 can't move the median beyond that. Hence, the median is at most 11.

Since the median is at least 11 and at most 11, it must be 11.

Appendix

With the same calculation one can figure out the average of rolling n dice with s sides each, and rerolling and adding as long as all of the newly rolled dice match. The average is $$(1-s^{1-n})^{-1}n(s+1)/2.$$

Plugging in three for n and six for s gives the above result. The average with DARO (doubles add roll over, the resolution mechanic in the game), ignoring automatic failures, is consequently 8.4 or 42/5, significantly larger than the average of 7 with a typical 2d6 roll.

• This doesn't work, because the 1/36 probability of continuing is not independent of the previous result, it only happens on very specific results that sum to 3,6,9,12,15,18. Your calculation of expected result would apply to a roll of 3d6, plus a separate "decision" roll of 2d6, where if you get a 12, you roll and add another 3d6. – Neil Slater Jan 12 '17 at 8:19
• @NeilSlater Expectation is linear even when the events are not independent; my intuition tells that this works, but, as mentioned, I haven't done the rigorous calculation. For example, the expectation of 6d6 and d6 multiplied by 6 is exactly the same. – Tommi Jan 12 '17 at 16:30
• @NeilSlater Also, we got the same result, which would be quite a coincidence otherwise. – Tommi Jan 12 '17 at 16:31
• No your maths is incorrect. It happens to be close in this example, because the adjustment due to re-rolls is slight. The linearity of adding expectations is true when you are able to add up independent variables. But the value of the 3d6 is not independent from the addition of the next 3d6. As acounter-example, consider if the rule was to re-roll and add for any result over 11. In that case the median value would still be 10.5, and the expected value would not match your formula either. – Neil Slater Jan 12 '17 at 16:51
• @NeilSlater The correct answer turns out to be precisely 10+4/5. See the updated answer. – Tommi Jan 14 '19 at 10:05

A simple approximation... The mean of 3d6 is 10.5
the odds of open ending are 1/36 (6 of 216)

the 1st recursion adds 10.5/1 * 1/36
= 10.5/36
= 3.5/12=7/24

the mean thus should be
10 1/2 + 7/24
= 10 19/24

a second recursion is
10 19/24 *(7/24 & 1/36)
= 10 18/24 + 7/(24 * 36)
= 10 19/24+ 7/864
= 10 (674+7)/864
= 10 681/864
= 10 227/288

But note some hiccups... 3 has 0 chance; 6 is rarer than normal in 3d6. So any probability table showing a 3 for a TARO roll is dead wrong.

Edit: Having sat down with my second favorite spreadsheet, I worked out to 2 recursions...
(magnitude houses delimited by thin spaces, decimal by full stop)

$$\begin{array}{|r|rr|} \hline \text{N} & \text{P(N)/10 077 696} & \text{Percent}\\ \hline \text{3} & \text{0} & \text{0.000 000%} \\ \text{4} & \text{139 968} & \text{1.388 889%} \\ \text{5} & \text{279 936} & \text{2.777 778%} \\ \text{6} & \text{419 904} & \text{4.166 667%} \\ \hline \text{7} & \text{700 488} & \text{6.950 874%} \\ \text{8} & \text{981 072} & \text{9.735 082%} \\ \text{9} & \text{1 121 689} & \text{11.130 411%} \\ \hline \text{10} & \text{1 263 603} & \text{12.538 610%} \\ \text{11} & \text{1 265 550} & \text{12.557 930%} \\ \text{12} & \text{1 126 884} & \text{11.181 961%} \\ \hline \text{13} & \text{989 517} & \text{9.818 881%} \\ \text{14} & \text{711 537} & \text{7.060 513%} \\ \text{15} & \text{432 264} & \text{4.289 314%} \\ \hline \text{16} & \text{294 258} & \text{2.919 894%} \\ \text{17} & \text{154 959} & \text{1.537 643%} \\ \text{18} & \text{14 365} & \text{0.142 543%} \\ \hline \text{19} & \text{15 684} & \text{0.155 631%} \\ \text{20} & \text{15 708} & \text{0.155 869%} \\ \text{21} & \text{14 436} & \text{0.143 247%} \\ \hline \text{22} & \text{15 756} & \text{0.156 345%} \\ \text{23} & \text{15 780} & \text{0.156 583%} \\ \text{24} & \text{14 508} & \text{0.143 961%} \\ \hline \text{25} & \text{15 180} & \text{0.150 630%} \\ \text{26} & \text{14 556} & \text{0.144 438%} \\ \text{27} & \text{12 634} & \text{0.125 366%} \\ \hline \text{28} & \text{12 006} & \text{0.119 134%} \\ \text{29} & \text{10 080} & \text{0.100 023%} \\ \text{30} & \text{7 500} & \text{0.0744 22%} \\ \hline \text{31} & \text{6 210} & \text{0.061 621%} \\ \text{32} & \text{4 266} & \text{0.042 331%} \\ \text{33} & \text{2 316} & \text{0.022 981%} \\ \hline \text{34} & \text{1 656} & \text{0.0164 32%} \\ \text{35} & \text{990} & \text{0.009 824%} \\ \text{36} & \text{322} & \text{0.003 195%} \\ \hline \text{37} & \text{300} & \text{0.002 977%} \\ \text{38} & \text{276} & \text{0.002 739%} \\ \text{39} & \text{252} & \text{0.002 501%} \\ \hline \text{40} & \text{228} & \text{0.002 262%} \\ \text{41} & \text{204} & \text{0.002 024%} \\ \text{42} & \text{180} & \text{0.001 786%} \\ \hline \text{43} & \text{156} & \text{0.001 548%} \\ \text{44} & \text{132} & \text{0.001 310%} \\ \text{45} & \text{109} & \text{0.001 082%} \\ \hline \text{46} & \text{87} & \text{0.000 863%} \\ \text{47} & \text{66} & \text{0.000 655%} \\ \text{48} & \text{48} & \text{0.000 476%} \\ \hline \text{49} & \text{33} & \text{0.000 327%} \\ \text{50} & \text{21} & \text{0.000 208%} \\ \text{51} & \text{12} & \text{0.000 119%} \\ \hline \text{52} & \text{6} & \text{0.000 060%} \\ \text{53} & \text{3} & \text{0.000 030%} \\ \text{54} & \text{1} & \text{0.000 010%} \\ \hline \end{array}$$

Statistically, long right tail, Mean 10.7998, mode 11, range 4-54, midline of range [28, 29], median result 10.

Nobody seems to have posted an AnyDice program to calculate this yet, so let me do that:

function: taro ROLL:s {
if 1@ROLL = 3@ROLL {
result: ROLL + 3d6
} else {
result: ROLL
}
}
output 3d6 named "normal 3d6"
output [taro 3d6] named "3d6 explode triples"

function: explode ROLL:s as DIE:d up to N:n times if all equal {
if N > 0 & 1@ROLL = (#ROLL)@ROLL {
result: ROLL + [explode DIE as DIE up to N-1 times if all equal]
} else {
result: ROLL
}
}
function: explode DIE:d if all equal {
result: [explode DIE as DIE up to 2 times if all equal]
}
output 3d6 named "normal 3d6"
output [explode 3d6 if all equal] named "3d6 explode triples"


The first program above only rerolls at most once, while the second one rerolls up to twice and can be easily modified to reroll any number of times. (The second program also works correctly as written with any number of dice of any size, not just 3d6.) In practice, it makes no difference — the probability of getting two (or more) rerolls on the same roll is one in $$\36^2 = 1296\$$, i.e. less than $$\0.1\%\$$.

In other words, as the other answers note, the distribution is basically the same as that of a normal 3d6 roll, except for a small ($$\1\:/\:36 \approx 2.8\%\$$) chance of a reroll making it look approximately like a 6d6 instead and a vanishingly small chance of even higher results: