Dice statistics help - chase odds?

[RangerWickett]

I need to figure out if I'm making a chase mechanic too easy.

The idea I have in mind is that each side rolls d20+speed, and whoever rolls higher either gets one success or negates one of their opponents' successes. Repeat each 'round,' which might be a minute or an hour.

So you basically have a scale that goes:

Pursuer 3 successes -- catch-up
Pursuer 2
Pursuer 1
Tied
Quarry 1
Quarry 2
Quarry 3 successes - get-away

Now, I'm not a natural at statistics, so I was hoping folks might help me out. If people have the same bonus, what are the odds of either side winning, and how long will it take for a winner to occur on average? How does it change as speeds diverge?

Obviously the chase could go forever, except the quarry can eventually reach a safe haven. So how many rounds of running do I want to make it take for a fair, challenging race?

Or is my plan badly designed? Would you recommend something else?

 

[Morrus]

Hmm. I'm not statistician either. If each has the same score (so a 50% chance of winning the contest each round), on average isn't the chase going to continue for ages? I guess we're looking for the probability of three wins, which are 50%x50%x505=12.5%.

Erm. I dunno what I now do with that 12.5%. Or even if it's useful.

 

[MatthewJHanson]

My recommendation is to just sit down and try it. Get two different colored d20s, assign them to a runner, and roll through a few chases on your own.

 

[Nagol]

Here the first 10 turn probabilities for a equal chance chase:

	1	2	3	4	5	6	7	8	9	10	
W3	0.00%	0.00%	10.72%	12.32%	19.74%	21.57%	26.75%	28.50%	32.17%	33.72%
W2	0.00%	22.56%	3.38%	15.61%	3.85%	10.91%	3.68%	7.73%	3.26%	5.54%
W1	47.50%	4.75%	32.51%	6.45%	22.57%	6.61%	15.88%	6.05%	11.31%	5.21%
T	5.00%	45.38%	6.78%	31.22%	7.69%	21.82%	7.37%	15.45%	6.52%	11.07%
L1	47.50%	4.75%	32.51%	6.45%	22.57%	6.61%	15.88%	6.05%	11.31%	5.21%
L2	0.00%	22.56%	3.38%	15.61%	3.85%	10.91%	3.68%	7.73%	3.26%	5.54%
L3	0.00%	0.00%	10.72%	12.32%	19.74%	21.57%	26.75%	28.50%	32.17%	33.72%

So it'll end half the time by turn 7 and 2/3s of the time by turn 10. Even at turn 10 though more than one in ten chases are tied.

This model is sensitive to bonuses. Heres the result with the pursuer having a +1 bonus:

	1	2	3	4	5	6	7	8	9	10
W3	0.00%	0.00%	14.47%	16.53%	26.47%	28.80%	35.69%	37.91%	42.74%	44.69%
W2	0.00%	27.56%	3.93%	18.93%	4.44%	13.13%	4.22%	9.21%	3.71%	6.53%
W1	52.50%	4.99%	35.70%	6.74%	24.60%	6.86%	17.16%	6.23%	12.11%	5.33%
T	4.75%	45.11%	6.41%	30.83%	7.23%	21.38%	6.88%	15.00%	6.04%	10.64%
L1	42.75%	4.06%	29.07%	5.49%	20.03%	5.59%	13.97%	5.08%	9.86%	4.34%
L2	0.00%	18.28%	2.60%	12.55%	2.94%	8.70%	2.80%	6.11%	2.46%	4.33%
L3	0.00%	0.00%	7.81%	8.93%	14.29%	15.55%	19.27%	20.47%	23.08%	24.13%

and a +2 bonus:

	1	2	3	4	5	6	7	8	9	10
W3	0.00%	0.00%	18.76%	21.30%	33.85%	36.64%	45.12%	47.71%	53.49%	55.71%
W2	0.00%	32.78%	4.42%	21.93%	4.87%	14.80%	4.52%	10.10%	3.87%	6.96%
W1	57.25%	5.15%	37.96%	6.79%	25.47%	6.74%	17.29%	5.97%	11.86%	4.98%
T	4.50%	44.00%	5.92%	29.30%	6.51%	19.78%	6.05%	13.50%	5.17%	9.31%
L1	38.25%	3.44%	25.36%	4.54%	17.02%	4.50%	11.55%	3.99%	7.92%	3.33%
L2	0.00%	14.63%	1.98%	9.79%	2.18%	6.61%	2.02%	4.51%	1.73%	3.11%
L3	0.00%	0.00%	5.60%	6.35%	10.10%	10.93%	13.46%	14.23%	15.95%	16.61%

 

[Octangula]

Slight, necro, but:

Assuming that the odds are equal to move in each direction, there's a very simple rule for the average duration of random walks. You multiply together the distances in both directions. Here, it's 3 each way, so 3*3=9.

This does assume that one side wins the roll each time. In reality, there is a 5% chance that the result will be tied, which will lengthen it slightly. This is probably a shortcut that's mathematically incorrect, but my mind wants to say 9.45

I'm not sure how the walk is affected by p not being 0.5; I'm looking into this now, mostly to satisfy my own curiousity.

Edit: My gut feeling is that with p being something other than 0.5, then it makes it more likely that the walk (and therefore, in your case, the chase) will end faster. If you understand why birthdays not being equally distributed across the year make the birthday paradox more likely to happen, then my thinking is that this applies here as well.

 

[Flatus Maximus]

Slight, necro, but:

Assuming that the odds are equal to move in each direction, there's a very simple rule for the average duration of random walks. You multiply together the distances in both directions. Here, it's 3 each way, so 3*3=9.

This does assume that one side wins the roll each time. In reality, there is a 5% chance that the result will be tied, which will lengthen it slightly. This is probably a shortcut that's mathematically incorrect, but my mind wants to say 9.45

I'm not sure how the walk is affected by p not being 0.5; I'm looking into this now, mostly to satisfy my own curiousity.

Edit: My gut feeling is that with p being something other than 0.5, then it makes it more likely that the walk (and therefore, in your case, the chase) will end faster. If you understand why birthdays not being equally distributed across the year make the birthday paradox more likely to happen, then my thinking is that this applies here as well.

Ooh, Markov Chains -- my favorite!

Indeed, each side is equally likely to win (assuming they start tied); and if equal rolls are not allowed, then the "expected time to absorption" into one of the winning states is 9 turns. If we allow for ties, but otherwise moving "up," "down," or "staying put" are equally likely with probability 1/3, then the expected number of turns is 13.5. (Assuming we start tied.)

Actually, the process cannot go on forever. The game will end in finite time with probability one!

I cannot answer the question about how things change as the speeds diverge because I'm not sure what you mean.

I can show the math if anyone is interested....

 

[Flatus Maximus]

I just realized how silly it was to consider the 1/3 probability case for each direction. As was suggested, the correct calculation is with 1/20 for a tied result. In this case, the expected time to absorption is 180/19, which is approximately 9.47 -- nice intuition, Octangula!

 

[Cerebral Paladin]

Yeah, the farther it is from even, the shorter the average length. Take the end case of 100% chance of success: the average (and uniform) time to the end of the walk will be 3. Next think about a high chance of success: if you have a 90% chance of success, better than 70% of the cases will resolve in three rounds, way more than the 21% at even. Then glance at the calculated values from Nagol, and you can see that even at just more than half (a +1 bonus), you have meaningfully more resolved games by round 9 than at even odds, and the remaining games are less likely to be at the midpoint of the walk. So the more you move away from random, the faster the walk resolves, with the average bounded at 9 rounds (for random, plus a little bit for ties) and 3 rounds for guaranteed success. But it will also be nonlinear in the time in between.

The flip side is that (except when success is guaranteed) it's just an average. If the odds are 50-50, a meaningful chunk of the time it will go much longer than average. So if you want to bound it at a reasonable time length, you need to either impose a cap or make sure it's way off from 50-50.

 

[Flatus Maximus]

The flip side is that (except when success is guaranteed) it's just an average. If the odds are 50-50, a meaningful chunk of the time it will go much longer than average. So if you want to bound it at a reasonable time length, you need to either impose a cap or make sure it's way off from 50-50.

I calculated a variance (in the "fair" case) of about 53, so one should expect the majority of games to take less than, say, 15-20 rounds.

 

[Umbran]

Actually, the process cannot go on forever. The game will end in finite time with probability one!

You should, in this scenario, be able to calculate a non-zero probability of it not having resolved for any arbitrarily large number of steps, which (IIRC) should mean it sums to one only at infinity, not in finite time.

Mind you, that's academic - the chances of it going on for times a mathematician would call "infinite" are not a practical concern. Whether you're rolling dice for 15 minutes to resolve the chase is more the question.