A Tunnels and Trolls® play-by-post adventure run by khara_khang
Jax would have taken his gloves off to halve his chances of his symbiont's absorption of one of Jay's spells.
I think you got it backwards? Here's from Taran's and Jax's character sheets:
If the host doesn't wear gloves, there is a 50% chance that these abilities will not activate.
So this says that if Jax doesn't wear gloves, the spells might fail. So it's good he had them on! I think Jax drank too much sea water on the boat ride over? <Tusky grin>
The character sheets are correct, but tarandracon has misinterpreted them.
Wearing the gloves is good for a bad spell (since there is a chance based on your Lv3 IQ SR that the worm will absorb the spell), but wearing gloves is bad for a good spell (since by taking them off, you have at least even odds of the worm not absorbing the good spell, no matter what your SR).
I get it now. Took me a few days, but I finally get it.
With gloves on:
With gloves off:
So in the aforementioned scenario, Jax being the smartly Orc, would take his gloves off, hoping for the 50% chance that the magic absorption would not trigger. If he'd left them on, then the magic absoprtion would automatically try to trigger, and would work on a successful L3SR vs IQ.
So then it becomes a matter of probability. What are the chances of attaining a total of 30 or more given 3d6+20? [Tarandracon really means how often Jax can expect to make the L3SR on IQ. --ed.] And, are they greater or less than 50%? No wonder Jax's brain hurts all the time!
Slightly greater. It averages 30.5 on 3d6+20.
Tarandracon's question boils down to how often 3d6 sums to 10 or more. By my back-of-the-scabbard calculations, this probability is precisely 62.5%.
The context in which tarandracon posted his question was T&T saving rolls, which are defined as 2d6 with "doubles adding and rolling over." This thread got me thinking how much the "doubles add and roll over" rule buys you on average.
The average score of 2d6 (without the doubles bonus) is 7, but what is the average T&T saving roll? The answer may surprise you.
For T&T fandom everywhere, I therefore propose a contest in two parts:
The first person to post at the BFT the answer to Part I (answer must be exact) will win:
The first person to post at the BFT the answer to Part II (proof must convince me) will win:
As usual, I am not online on the weekends, so don't let someone else's submission stop you (unless of course you are sure that it is correct). I will post the answer and proof next week.
Special thanks to hobbit_king for volunteering the prizes and prize descriptions, and for offering to mail the prizes to the winner(s).
Good luck!
-Jax
Been about 4 days or so, so I might as well answer....
What are the chances of attaining a total of 30 or more given 3d6+20? And, are they greater or less than 50%?
Tarandracon's question boils down to how often 3d6 sums to 10 or more. By my back-of-the-scabbard calculations, this probability is precisely 62.5%.
Which is more than 50%, and all that was asked. ;)
The context in which tarandracon posted his question was T&T saving rolls, which are defined as 2d6 with "doubles adding and rolling over."
Was it? The 3d6 thing threw me off for what it was all about.
This thread got me thinking how much the "doubles add and roll over" rule buys you on average.
Not too much, actually. But it's still nice to have.
...what is the average T&T saving roll? The answer may surprise you.
Or maybe not, if math doesn't excite you anyway.
Interesting question.
Ditto.
The first person to post at the BFT the answer to Part I (answer must be exact) will win....
Can I use fuzzy math?
Keen.
The first person to post at the BFT the answer to Part II (proof must convince me) will win....
So it doesn't have to be accurate, just convincing?
I like the "Wizards Experimental Subject" character creation section.
As usual, I am not online on the weekends, so don't let someone else's submission stop you (unless of course you are sure that it is correct).
Well, not a lot of response. People generally shy away from math, I guess.
Special thanks to hobbit_king....
He's a dem fine burrahobbit, a dem fine burrahobbit.
Well, here's what I'm going with, starting with Part II to get to Part I.
There are 36 possible rolls on 2d6. Pretty obvious for that: 6 possibilities per die, 2 dice, and 6x6 is 36. Six possibilities (1+1, 2+2, 3+3, 4+4, 5+5, 6+6) are doubles and are added and rerolled. The rest are not doubles, and end the accounting there.
The thirty all have an average of 7. There's any number of ways to check that out. The easiest way to clearly see this (but most time intensive way) is to write out the 30 non-double possibilities out (1+2, 1+3, 1+4, 1+5, 1+6, 2+1, 2+3, 2+4, 2+5, 2+6, 3+1, 3+2, 3+4, 3+5,3+6, 4+1 ... 6+5) to get the following 30 results: 3,4,5,6,7,3,5,6,7,8,4,5,7,8,9,5,6,7,9,10,6,7,8,9,11,7,8,9,10,11.
Then add them together (getting 210) and divide by the number of possibilities (30) to get your average of 7.
Or go to a few other, much easier ways using all the statistics or discrete mathematics you learned and forgot from high school and college.
At any rate, the 30 average 7. The other 6 doubles are the trick. Those will come out to 2 plus a roll of 2d6, or 4 plus a roll of 2d6, or 6 plus a roll of 2d6, etc., etc., etc. The doubles, not counting the additional 2d6, average a 7: (1+1, 2+2, 3+3, 4+4, 5+5, 6+6 = 2+4+6+8+10+12 = 42/6 = 7). So, the doubles average 7, so it's basically 7 + 2d6 whenever you roll doubles.
Of that second 2d6, 30 of those will average out to 7 (the math is exactly the same as for the first set of 2d6), giving you an average of 14. (The original doubles plus the new 2d6 roll).
The original possibilities that averaged 7 and these few that average 14 makes the overall average (so far) an 8. (There are six times as many 7's as there are 14's, hence the average is an 8.)
But 6 possible results will get you another set of doubles, which would be again added to the 14 average, totaling 21. And that will again follow the progression of adding 7, on average to 30 possibilities, and another set of doubles to the remaining 6.
Well, that average of 21, being a lot less common, only pushes the overall average from an 8 to roughly 8.3: it's 1/36 less common than rolling the original 7, and 1/6 as less common as the 14, so it doesn't carry a lot of weight when averaged out.
Continuing the same mathematical pattern with the even less common 28 from doubles, and you come up with roughly 8.378 as the average.
Continuing the same pattern with the even less-less common 35 from doubles, and you come up with roughly 8.395 as the average.
Continuing the pattern with the even less-less-less common 42 from doubles, and you come up with roughly 8.3991 as the average.
Continuing the pattern with the even less-less-less-less common 49 from doubles, and you come up with roughly 8.3998 as the average.
Clearly, the law of diminishing returns is coming into play.
It's past midnight here in Japan, and I have a 5 am wakeup, so I'm just going to graph it out---a perfectly legitimate mathematical technique!
I'm getting an asymptotic line at 8.4, so that's what I'm saying for the answer, and this whole mess is my reasoning for it.
That's my math answer.
Now, my philosophical/math answer: Well, you could conceivably roll doubles every throw, an infinite number of times, and since each has some value, you'd have infinity. No matter how you want to divide (average) infinity, it's still infinite as all infinities are equal (by infinity mathematics since one number can always be paired with another number in any comparison of infinite series [This only works for countable sequences! --ed.]), so the answer is: the average roll is infinite. ;)
How's that?
We have a winner, but the contest is not over!
Nice approach, eickeric!
Clearly, the law of diminishing returns is coming into play. It's past midnight here in Japan, and I have a 5 am wakeup, so I'm just going to graph it out---a perfectly legitimate mathematical technique!
Legitimate, but unconvincing in this case! Your reasoning lacks the final nail on the coffin lid. You have not convinced me, for example, that the limit is not 8.399995 or 8.40000000001.
I'm getting an asymptotic line at 8.4, so that's what I'm saying for the answer....
The answer is 8.4, and your calculations agree with this fact, but they do not prove it.
You win Part 1! Congratulations! Contact hobbit_king for your prize.
Part II is still open!
In the absence of any more submissions to Part II, this contest now closes with the proof of the following theorem.
The average value of a T&T saving roll is 8.4.
Let y be the average value of 2d6. We will find x, the average value of 2d6 with DARO (doubles add and roll over).
The crucial insight is to recognize the recursion: when doubles are rolled (1/6 of the time), then another T&T SR is effectively rolled and added in. This gives
x = y + x/6.
One can confirm that y = 7, e.g. by summing the 36 possible values of 2d6 and dividing by 36. This gives
x = 7 + x/6,
whose solution is precisely x = 8.4.
Congratulations again to eickeric, who won Part I!
-Jax
P.S: Eickeric's submission for Part II only lacked the summation of the infinite geometric series:
x = 7 + 1/6 * (7 + 1/6 * (7 + 1/6 * (7 + 1/6 * (...)))) = 7 / (1 - 1/6) = 8.4.
P.P.S.: A related statistic (and the subject of tarandracon's initial inquiry, which inspired this contest) is how often Jax and Taran can expect to make the Lv3 SR on IQ required for their symbionts to absorb a spell (assuming that if they are not wearing gloves, they have first beaten even odds of the absorption not occurring in the first place). This is approximately 25.6% of the time for Jax (IQ 20) and approximately 2.4% of the time for Taran (IQ 9). Compare this to precisely 19.4% (Jax) and 0% (Taran) if the T&T SR did not have the "doubles add and roll over" bonus.
Nice approach, eickeric!
It would have been if it had worked. Ah well.
I can't blame you for being unconvinced: I wouldn't have been either. Like I said, it was late, and I forgot all the math I learned from high school and college. Oh well.
The answer is 8.4, and your calculations agree with this fact, but they do not prove it.
Good enough. That's the only prize I needed anyway. ;)
You win Part 1! Congratulations! Contact hobbit_king for your prize.
Hooray!
Part II is still open!
Mebbe I'll revisit it.... Nah.
Congratulations again to eickeric, who won Part I!
Thanks.
P.S: Eickeric's submission for Part II only lacked the summation of the infinite geometric series:
Bummer. That's what I was getting at with:
But getting there isn't making the stop. Math isn't horseshoes. I wouldn't have given it the go either. (Although I still think that infinity thing was pretty good....)
One can apply the technique in the above Proof to show that the average value of 3d6 with TARO (triples adding and rolling over, as is a common house rule for generating Prime Attributes) is exactly 10.8, a subtle improvement over the average 3d6 roll of 10.5.
But we may generalize even further. Let mARO be the generalization of DARO and TARO, such that when all m dice rolled are equal, another roll follows, adding to the total.
The average value x of m s-sided dice with mARO is given by
x = m (s + 1) sm - 1 / (2 (sm - 1 - 1))
This follows from the the formula for the average value y of m s-sided dice:
y = m (s + 1) / 2
and the fact that m equal dice occurs with probability s1 - m.
What about the probability of rolling n or more with mARO? This was the gist of tarandracon's query that started this thread.
With some help from Eric Weisstein's excellent mathworld resource, we can answer this question in full generality.
The probability P(n, m, s) of rolling n or better on m s-sided dice with mARO is 1 if n < m; otherwise it is given by the recursive formula
P0(n, m, s) is the probability of rolling at least n on m s-sided dice, not all equal. P0(n, m, s) is 0 for n > sm; otherwise it is given by
C(n, m, s) is the number of ways to roll n on m s-sided dice. C(n, m, s) is 0 for n < m; otherwise it is given by
For sources on the derivation of C(n, m, s), see mathworld.
The formula for P(n, m, s) splits the contribution between unequal dice and otherwise. P0 sums up all the possible ways to reach n or higher with m s-sided dice, then subtracts off the contributions from m equal dice and divides by the total number of dice configurations. The rest of the formula for P considers the contributions of the s equal-dice rolls with the additional rolls they generate; each of these s rolls occurs with probability s-m.
Let's compute the probability of rolling 18 or more under the TARO rule mentioned above. Here, n = 18, m = 3 and s = 6 in Theorem 2.
Clearly, C(18, 3, 6) = 1, as triple sixes is the only roll that sums to 18, but it is an equal triple, so we subtract it off: P0(18, 3, 6) = 0.
What remains are the contributions of triple ones, twos, threes, fours, fives and sixes with TARO. To hit 18 or more, we need the probability of rolling what's left after the triple. These probabilities are exactly
Since each of these six contributions occurs only 1/216 of the time, we divide the sum of these six probabilities by 216 to get
P(18, 3, 6) = 1016461223 / 52242776064,
or about 0.019456493310286276291.
So while your chances of rolling 18 or higher are still low with TARO, they are over four times better than your chances without: 1 / 216 = 0.004629629....
And consequently, when rolling 3d6 with TARO to generate the seven Prime Attributes and Speed for a T&T character, you have probability 1 - (1 - P(18, 3, 6))7 of generating a character with at least one stat 18 or higher. That's about a 12.85% chance, compared to about 3.20% without TARO.
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