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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?

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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):

enter image description here

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.

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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.

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    \$\begingroup\$ 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. \$\endgroup\$ – Neil Slater Jan 12 '17 at 8:19
  • \$\begingroup\$ @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. \$\endgroup\$ – Thanuir Jan 12 '17 at 16:30
  • \$\begingroup\$ @NeilSlater Also, we got the same result, which would be quite a coincidence otherwise. \$\endgroup\$ – Thanuir Jan 12 '17 at 16:31
  • \$\begingroup\$ 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. \$\endgroup\$ – Neil Slater Jan 12 '17 at 16:51
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    \$\begingroup\$ @NeilSlater The correct answer turns out to be precisely 10+4/5. See the updated answer. \$\endgroup\$ – Thanuir Jan 14 at 10:05

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