**Cinerati**, about the probability distribution of a different sort of "exploding" dice: Saving throws in

*Tunnels and Trolls*. I have previously written about Exploding D10, and this is a closely related problem. I'll let Christian explain it ...

I am getting ready to do a few posts on Tunnels and Trolls. Specifically, I will be proposing an alternate combat resolution system -- one based on an existing system within the game. I want to play around with the concept of using T&T "Saving Throws" as the basis for all mechanical resolutions in the game.

In T&T the saving throws are resolved by comparing a statistic to a difficulty # and using the result as the basis for a 2d6 roll. The actual equation is [15 + (level of challenge x 5)] - Statistic = Target Number. So a character with a Luck of 12 attempting a 1st level challenge would need a result of (15+5-12=8) eight or better to succeed on the attempt. Figuring out the probabilities on a basic 2d6 roll is a simple affair, but in T&T a player who rolls doubles keeps the result, re-rolls and adds the result to the prior sum until the player no longer rolls doubles. It's an open ended doubles system. What would be basic equation be to determine probabilities in this case. Logically, given that the chance of doubles is 1/6, it seems that the probabilities would be the same as a normal open ended d6 roll for each die (which I believe produces an average of 4.3 per die), but I'd like an equation I can use to determine the game balance as characters advance etc. I'd like to use the base probability of an "average" character of a given level attempting a task as the basis for my new system.

If you could be of assistance, it would be greatly appreciated.

A friend of mine had noted that asking me questions like this is like teasing a small child with a shiny toy, holding it just above my reach; You just

*know*I'm going to jump up and try to get it - and I did. ;-)

As I said, this is very similar to how the probability for Exploding D10 work in the

**MechWarrior 3rd edition RPG**. With D10X you count the roll if is in the range 1-9, and if it is a "10" you count the 10

*and roll again*(that the explosion). Rinse and repeat until done. In

**T&T**we have a 2d6 roll that explodes if a tie is rolled on the two dice instead of the highest value on either one.

As with D10X, this breaks down into

The probability of rolling a ties on 2d6 is 1/6, so the geometric series starts of like this:

0 ties with probability = 5/6 = 0.8333

1 tie with probability = (1/6)*(5/6) = 5/36 = 0.1389

2 ties with probability = (1/6)*(1/6)*(5/6) = 5/216 =0.02315

3 ties with probability = (1/6)*(1/6)*(1/6)*(5/6) = 5/1296 = 0.003858

... and so on out to infinity. The average number of ties in a series is (1/6)*1/(5/6) = 1/5 = 0.2, which is nice it you only want to know the average 2d6X roll (which is 8.4, btw), but we need the entire probability distribution.

Now there are six different ways to roll a ties, and they have values of 2,4,6,8,10,12, all with 1/6 probability, which is just the same as 2 times the value of a 1d6 roll. I will note this as 2*

**1**d6. Combining this with the geometric series, we get this:

a

*"0"*with probability = 5/6 = 0.8333

a

*2**roll with probability = (1/6)*(5/6) = 5/36 =0.1389

**1**d6a

*2**roll with probability = (1/6)*(1/6)*(5/6) = 5/216 = 0.02315

**2**d6a

*2**roll with probability = (1/6)*(1/6)*(1/6)*(5/6) = 5/1296 = 0.003858

**3**d6... and so on out to infinity.

NOTE: The probability of zero ties is really 0.8333 (the first bar); I chopped off the graph to show the what ties actually add in, which really isn't very much. I calculated the probabilities out to 6 consecutive ties, which is only a little bit of overkill. It could go on forever, but unless you are infinitely lucky (and have a

*lot*of spare time on your hands) your streak of ties will eventually come to an end, where you add the sum of the the final not-tied dice to your previous total. That probability is a modification of the usual discrete triangular 2d6 distribution, and it looks like this (2d6-ties):

The final step to pull this all together is a bit messy, so pardon my hand-waving over the details (if you really want the nitty-gritty, look in the accompanying spreadsheet). You multiply each probability in the geometric series with each the probabilities in the 2d6-ties distribution and sum up the corresponding values of the roll, then tally up the total probability for all combinations with the same sum. It looks like this:

This looks a lot like the 2d6 distribution with a long "tail" tacked onto the right hand side. Here's the cumulative probability:

Finally, a table of the rolls and cumulative probabilities, though if you really want to numbers it's probably easier to grab them from the spreadsheet. Again, the table theoretically should go to infinity, but in practice you will only rarely see any rolls greater than 20.

I'm curious to see what Christian does with this, and I'll link to his post(s) on the topic when he has them up.

Have fun rolling the dice!

[Edit: Typo corrected, added decimal probabilities.]

[Edit2: Added link to spreadsheet, which I swear I did once already.]