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Players of a certain TRPG have characters with 6 ability scores, each ability score ranging from 3-18. One method of generating those is by rolling 4d6 drop lowest. That means four six-faced-dice are rolled, and the three highest results are added.

What's the probability that, given 5 players, one player will have a highest ability score equal or lower than the lowest ability score of another player?

The related question here shows how to get the distribution of 4d3 drop lowest, but how do I get from there to an answer of my question above?

A good answer would explain the result in a way that a statistics novice can follow.

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  • While this is not a *statistical* answer, it's worth noting that http://anydice.com/ provides a calculator for precisely this purpose. – Sycorax Aug 05 '14 at 13:01
  • An answer that contains a script for anydice that shows me the answer to my questions would be accepted (if it was the best or only answer, of course) – Mala Aug 05 '14 at 13:06
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    More general version of the question [1](http://stats.stackexchange.com/q/32309/4485). Almost the same question [2](http://stats.stackexchange.com/q/30585/4485). – Affine Aug 05 '14 at 13:10
  • @Affine The answers to those are not understandable for a novice. I am asking here because I would like a nice explanation that people who are not otherwise into statistics can follow (and that I can link to). I will update the question to clarify that. – Mala Aug 05 '14 at 13:16
  • While these should be possible to do analytically, I'd be inclined to answer via simulation. However, beware of the issue of calculating probabilities for events specified post hoc -- i.e specified after they occur -- as if they'd been specified before the fact. (see the wheelbarrow-full-of-dice example [here](http://stats.stackexchange.com/questions/67721/is-there-a-law-that-says-if-you-do-enough-trials-rare-things-happen/67723#67723)) – Glen_b Aug 05 '14 at 13:17
  • Although I am sympathetic to the question, ultimately it consists of a sequence of steps each of which has been very well explained and illustrated elsewhere on this site. I recognize that assembling such a sequence in itself might constitute an original answer, but I am concerned that we could eventually accumulate a large number of very closely related questions about dice (and coins, etc) that offer nothing new except details of the (common) arithmetic needed to solve each separate one. However, if this post were edited to focus on the *last* question it could be of lasting interest. – whuber Aug 05 '14 at 13:36
  • @whuber The question linked to as duplicate does explain how to get the distribution of 4d6k3, but it does not explain what I asked. Maybe it is clear to an expert how to get from that answer to the answer of my question; but an expert would likely not need to ask such a question here. – Mala Aug 05 '14 at 13:55
  • What we are looking for from you, Mala, is to focus this set of questions into a single one. You can accomplish this by researching the answers to the preliminary parts and emphasizing the step at which you are getting stuck and need help. It's not a matter of communicating with experts or not, but of [formulating an answerable question.](http://stats.stackexchange.com/help/how-to-ask) – whuber Aug 05 '14 at 15:48
  • I split the question in two different ones, both referencing the original question. I am not sure what else I can say except for that just having the distribution doesn't show me the answer... – Mala Aug 05 '14 at 18:19
  • There's an article that covers the difference in odds between 3d6 and 4d6-drop-the-lowest at http://anydice.com/articles/4d6-drop-lowest/ –  Sep 04 '15 at 21:08

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Unlike in answering your other question, I'm just going to do this with Monte Carlo sampling. It wouldn't be super-hard to solve it out explicitly, but it's late. :p If that's what you're looking for let me know and maybe I'll do it out this weekend.

We can get an explicit pmf for $X_{ij}$ by brute force, as before. Now that we have that, let's just roll up 1,000,000 parties real quick, each with 5 players, each with 6 ability scores. (Computers are nice.)

Using scipy and the score_pmf variable from last time:

from scipy import stats
ability = stats.rv_discrete(
    name="ability", values=(np.arange(score_pmf.size), score_pmf))
samps = ability.rvs(size=(1000000, 5, 6))
min_max_scores = samps.max(axis=2).min(axis=1)
max_min_scores = samps.min(axis=2).max(axis=1)
print('{:%}'.format(np.mean(max_min_scores > min_max_scores)))

printed out 1.002000%, 0.969900%, 0.974200% for me when I repeated it three times. So "about 1%" seems like a good estimate.

If you want to be a little more rigorous, this corresponds to it happening 10,020 + 9,699 + 9,742 = 29,461 times in 3,000,000 trials, which according to an Agresti-Coull binomial confidence interval means that we can be 99% sure it happens between 0.967% and 0.997% of the time.

This is the easiest way to answer any question about dice rolling like this, though it doesn't provide any statistical insight or generalization.

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