X = number of dice I have
Y = number of dice you have

Vague issues: if I have three sixes and you have two, will I roll all my three sixes against your two in a tie reroll, or only two? If we tie in both sixes and fives, will we roll only the sixes first, or both at once? Is intensity calculated for both sides separately, or is it summed together? I'll assume the former in all cases.

General analysis: At the beginning of the process the dice scatter uniformly to all possible values. Any tie rerolls will also scatter the dice they reroll uniformly. What this means is that at any given value at any point in the reroll scheme the proportion of my dice to your dice, X/Y, will be the same, as any rolled and rerolled die pools will have preserved the proportion.

As any rerolls take the dice from the current maximal value and scatter them over all values, the overall effect is to increase the number of dice in the lower portions of the scale. Meanwhile, the probability of the scheme continuing depends on the number of dice still available in the higher values, because the rolling stops when only a single player has the highest value. Overall, when considering the final profile of the die pool, this causes a curve leaning towards the lower end of the scale, the strength of the lean depending on the number of dice in the smaller of the players' pools.

Because I'm lazy today and should be working on my own game, I won't do much analysis. Instead, I'll just throw out some rough numbers that should work well enough. No guarantees.

Ben wrote:
1) Given two die pools of arbitrary size (around 3-20 dice per pool), what are the odds of victory?

In case of no ties: I win with a X/(X+Y) chance, because the highest die might be any of the X+Y rolled dice.
In case of a tie: Because our die values scatter uniformly, at a given value (including the tying value) we will have an expected X/6 and Y/6 dice. When these are rolled, the probability of my winning among the rerolls is equal to the above case. Furthermore, as the probability is equal to the probability of me winning among the remaining originals, it doesn't matter whether the new dice are higher than the remaining originals; the probability of me winning is X/(X+Y) regardless.

2) What is the distribution of the intensity?

Generally, each rolled or rerolled die will increase intensity by 1/6. The overall intensity will be divided by the proportion of X/Y, because all rolled and rerolled pools will have that proportion. It's perhaps simpler to calculate the overall number of rolled and rerolled dice than otherwise. For that purpose let's say X+Y=Z. Rough distribution follows:
initial roll: average of A=Z/6 dice rerolled
1st reroll: average of B=Z/36+Z/6*(5/6)^A dice rerolled
2nd reroll: average of C=Z/36*(5/6)^A+Z/6*(4/6)^B dice rerolled
3nd reroll: average of D=Z/36*(4/6)^B+Z/6*(3/6)^C dice rerolled
... and so on, the symmetry is clear.
The rough assumption here, made to simplify the calculations, is that both X and Y are high enough to cause ties at nearly all values of the initial roll but low enough to justify cutting off the iteration (around 5-20 dice is fine). At low numbers a better approximation is gained by dropping more rerolls, skewing closer to the base value of Z/6 (which you'd get without the reroll mechanic). At higher numbers of dice you should approximate towards the geometric curve, upwards (with X and Y both arbitrarily high, the rerolls will follow a linear increase of z/6 per reroll, z being number of dice rerolled last). What it all boils down to is the following rough table of intensity distribution:
Z    Intensity
12    2+1/3+1/18+0.25+0.04...=[around]= 2.7 (difference of .7 from the base)
18    3+1/2+1.74+0.29+1.33+0.22+0.75...=[around]= 7.83 (difference of almost 5 from the base)
24    4+2/3+1.93+0.32+1.4+0.23+1+0.21...=[around]= 10 (difference of about 6 from the base)
30    5+5/6+2+0.33+1.48+0.25+1.25+0.21+1,1+0.18...=[around]= 13 (difference of about 8 from the base)
More values can be calculated, but as the approximation seems to carry the theory, we can posit a careful conclusion: the intensity at low numbers of dice is practically identical to Z/6, but rises strongly from the linear base on a geometrical curve with higher numbers. The divergence is caused by the increasing number of ties; the number of ties increases as a linear function of the number of dice, while the rerolls remove dice from further ties only at a asymptotic pace. Thus the proportion of tied dice to the overall number rises when the number of dice increases.

To get the intensities for separate sides from the above iterations, divide the given value with X/Z or Y/Z.

3) What are the odds of a calamity?

A calamity happens when all dice get the result of '1'. With low numbers of dice (insignificant ties) the probability approaches 1-(5/6)^Z (where Z is as above). However, the probability does not actually change significantly even with ties: assuming we have z non-'1' dice and we roll another one in the hopes of causing a reroll and, perhaps, a calamity, what happens? At best (let's say it's one of my dice against z of yours, distributed over all values) we can hope to reroll 1+z/6 dice, of which 1/6+z/36 will end up ones. Thus we'd have to have z>30 to actually lessen the expected number of non-'1' dice.

The reason for the above phenomenon is that with an increasing number of dice it's also increasingly unlikely to reroll all dice into ones before a highest value emerges. Thus the odds of a calamity do not get a significant boost over the baseline when the number of dice increases. There is a slight boost, that's true, but the overall trend is strongly downwards. For practical play I wouldn't expect a calamity with more than five dice.

4) If more than two pools are rolled against each other, how do things look?

The probability of ties increases, and thus there is a slight intensity/calamity increase with large pools. I doubt whether the effect will be significant with any number of dice. The winner is still determined by the ratios of pool sizes.

The extreme case would be a big number of one-die pools rolling against each other. Assuming no second-place ties rerolled, the situation would look very much like the two-player version, with perhaps one or two more reroll rounds.

5) What is the average number of rerolls before final success or failure?

Taking a look at my intensity calculations, above, we'll notice that adding six dice total to the rolls seems to add one reroll, more or less. This is the case with equal-sized pools. I'd deduct one reroll for each full six dice difference between pools, as well, as an educated guess. So with two 15 die pools I'll give you five rerolls average, while 24 against 6 gives you about two. This is a rule of thumb, but should be good enough.

1) For arbitrary (3-20ish) pools, what are the odds of success and failure?

This is for the unopposed case. It's simple, because you fail if and only if you didn't roll any sixes. With X dice the probability for that is (5/6)^X. The probability for success is the inverse. Some numbers:
Dice Failure
1    83%
2    69%
3    58%
4    48%
5    40%
6    33%
7    28%
8    23%
9    19%
10    16%
11    13%
12    11%
... and so on

2) What is the average intensity of a success?  A failure?

Because there is no rerolls, the average intensity is always near X/6, where X is the number of dice rolled. With small numbers of dice success or failure affect this somewhat:

1 die: success is always 0 intensity, failure is average of 1/5 intensity.
2 dice: success average is 1/6 intensity, failure average is 2/5 intensity.
3 dice: success average is 2/6 intensity, failure average is 3/5 intensity.
X dice: success average is (X-1)/6 intensity, failure average is X/5 intensity.
... So failure averages are always a little bit stronger than success, but overall the differences are small.

Ben wrote:
3) Given the previous case of me rolling 3 6s versus your single 6, if all rolled dice turn up 1s, that's a calamity.  The rest of the dice on the table don't matter anymore (except for ones, which add intensity.)  Therefore, I think that they would be more common, right?  (Not common, just more common than you mention.)  The idea is hopefully to get something close to 1/50 for a "strange result" effect.

So let me see if I get this: after the initial roll, if any single reroll comes up all ones, that's a calamity? Only the dice in the current reroll matter for that? That changes things quite a bit. Remember that the profile of the rerolls will always be the same with reasonably large initial pool sizes: we'll get some dice tied for value '6', we'll reroll them, then we either roll even less dice from '6' or a little bit more from '5', so on until something happens. Also, remember that with any roll with more than 4 dice the probability of calamity is less than .08%, so those are meaningless for us. What this means is that we actually have a locally homogenous system here: regardless of the initial number of dice in the roll, we will end up with a reroll of perhaps 2-4 dice from value '6' sooner or later, after which the situation works out the same regardless of how many dice are "waiting" in fives or so. A quick calculation tells me that the "descent profile" we're talking about gives us roughly... .5+2.7+somestuff...=3.5% chance of a calamity per roll, regardless of value of Z.

If one side has less dice than the other, the probability of a calamity is less than that, because the probability of a reroll is less. Furthermore, the chance lessens linearly, because the rerolls are cut off before the crucial last, small rolls (the pools for both sides diminish independently in the reroll process), if there's less dice on one side. Taking this into consideration a rule of thumb tells me that you'll get 3.5*X/Y percent, if X<Y. So with Y=20 and X=10, for instance, I'll give you a 1.75% chance for a calamity. With Y=20 and X=5 I'll give .9%

Furthermore, there's the case of small die pools to begin with (<6 dice in both pools). For those I'll give .05*Z/2 percent. So with Z=12 I'll give .3%, but with Z=6 only .15%, based on the probability of there being even one reroll to begin with. (When Z goes to ridiculously small numbers like Z=3, you'll again get probs. like 2% or so, from the initial roll rolling all ones.)

Overall I'd say you have a surprisingly good prob. of getting exactly what you want here. If one side is stronger than the other a calamity is not as likely as otherwise, but even in the best case it's only 3.5%, which shouldn't be too much. And even with pretty large differences in pool size the chance stays over 1% for a long while. Pretty promising. The only problem are small pools (4<Z<12), with which you're unlikely to get a calamity at all.

You seem to indicate that it the average intensity of a 15-15 roll is the same as a 29-1 roll.  Clearly, this is not the case (the 15-15 roll has a much higher chance of producing a tie).  How strong is the effect?  Is your chart based on an assumption of an even die pool size?

Yep, difference in pool size indeed lowers the incidence of ties. My intensity chart there assumes that both sides have a "large number" of dice, large enough to cause those rerolls in the first place. I'd say that if the difference is not over 1:2 and the smaller pool has over six dice and Z<50, it works well enough. In a sense that simplified iteration I do there works only if we assume that a reroll will happen each time, and the roll won't be resolved without the rerolls lowering the number of dice. There's also some minor terms of the calculation I dropped which would become significant with a large number of dice, but we won't care about those.

If handling smaller pools (<6) on both sides, then you can assume that the intensity will be Z/6+0.5, or 1-2, because any reroll with more than one die per side is so unlikely as to not affect the expected value for our purposes. On the other hand, when having a small pool (with x dice, let's say, x<6) only on one side, the intensity will be roughly Z/6+(x/6)*B, where B is from the earlier approximation. This is significantly closer to Z/6 than the case with even pool size. Practically, here's my calculations for a six-die pool against an increasing pool the size of Y:
Y    Intensity
6    [around]= 2
12    3+1/2+1.74... =[around]= 5.24 (difference of over 2 from the base)
18    4+.66+1.93... =[around]= 6.59 (difference of over 2 and a half from the base)
24    5+.83+2.00... =[around]= 7.83 (difference of almost 3 from the base)
As you can see, the intensity is still a little bit superlinear, but significantly less than with equal pool sizes. This is because with a small number of dice on one side it's excessively unlikely to get two rerolls. I think the real numbers are even about 10% lower than these, but I'm not going to make the calculations to confirm. Let somebody with a suitable math program do some Monte Carlo simulation for that.