Puerto Rico seat position... something really ridiculous is going on.

[Rex Goodheart]

So, OK, a bit of introductory material first....

When I play Puerto Rico, which has been quite a lot, I play four player, with very very rare exception. Four player is the absolute best way to play Puerto Rico, no question.

I've also done quite a bit of extensive data analysis with the game, tracking well over a thousand games to determine best seat position, best building purchases, etc, etc.

With regard to seat position what I have found, conclusively, even with the "Balanced" variation of the game, the seat positions from best to worst are 3,4,1,2. I've written about this here and on other sites.

Knowing this, I have also lobbied on THIS site for the need to auction opening seat positions for games like Puerto Rico where such are so important to relative success, and where such is the norm for real life tournaments. Don't get me wrong... you certainly CAN with good play win with seat 2, but you're starting at a distinct disadvantage and other good players will make sure that your chances from that position are much smaller.

And on we go...

So, after taking a hiatus because I got frustrated that I was being unfairly assigned seat 2 much too often I missed it all and decided to return....

Well, in my 14 games back here are the seat positions I've been assigned at the opening:

2, 2, 1, 2, 1, 1, 2, 2, 2, 1, 1, 1, 2, 1

This is pretty comical, actually. With true random selection, the mathematical odds of not drawing either a third or fourth seat in 14 games is, get this, 0.0000000000127645085

That probability is so absurdly low that I'm led to suspect that there's something seriously wrong with the "random" selection process for Puerto Rico.... at the very least, the selection algorithm should be reexamined.

But, more importantly, I think it illustrates just how unfair the ELO process is WITHOUT an opening seat auction. I consider myself pretty decent at the four player game, with a current ELO of 424 despite the seat assignments.... and it's only a game so I don't lose any sleep over any of this.... but my goodness. With a few seat 3 assignments who knows how high that would be! LOL.

So, again, here's a call for an opening seat auction mechanism for Puerto Rico and similar games.

Thanks for your attention.

[sourisdudesert]

This is pretty comical, actually. With true random selection, the mathematical odds of not drawing either a third or fourth seat in 14 games is, get this, 0.0000000000127645085

I may be wrong but the odd is ( 1/2 )^14 = 0,000061035.

So it happens 4781629 more that you think

[Rex Goodheart]

I think you're incorrect.

The calculation would be ((1/2) * (1/3))^14

Any way, either computation = extremely low.

[sourisdudesert]

Sorry, I don't understand your formula.

You are playing 4 players, so you have a probability of 0.5 to pick the seat 1 or the seat 2.

Having seat 1 or seat 2 over the latest 14 games has a probability of 0.5 ^ 14.

The probability can be extremely low, however you played 5151 PR games. I'm not sure of the following formula, but 0.5 ^ 14 * 5151 = 31%, so if I'm right there was a 1/3 possibility to have such an event. So it is not so uncommon.

[Rex Goodheart]

Well, I admit I was overthinking it.

You're correct.... the chances of not getting a 3rd or 4th seat is 50%. The chance of not getting a 3rd or 4th seat within 14 games is .5^14 = 0.0000610351563, correct.

But here is where you're underthinking this:

The chance of this happening exactly once in 5151 games is 31%, but the chance of this happening in my last 14 games is still 0.0000610351563.

[dschingis27]

Actually it is very complicated to calculate the probabilty of at least one streak of 14 in 5151 games.
https://math.stackexchange.com/question ... of-streaks

I guessed some people might be interested.

[Rex Goodheart]

Thanks for the link. My interest in such was piqued by the statement that the chance of it happening within 5151 games is 31%. I took that at face value, but the computation that yielded that number seems too simple for the situation. I'll definitely be reading your link, thanks.

At any rate, if we all were playing regularly in person, sitting next to each other, we would not allow to happen what has happened to me (which now has grown to 15 games). At some point we'd all say "enough is enough", "this has gotten absurdly unfair", "this is no longer a true test" , and we'd make up a substitute algorithm that insures that each player get a chance from each seat within a reasonable period of time.

[Rex Goodheart]

I just recalculated....

I believe that the chance of a 14 game failure streak (as previously described) never happening within 5151 games is 0.730805299, as calculated below:

We've already established that the probability of 14 failures within a 14 game stretch is .0000610351563.

Therefore:

Game # Probability that the Game Just Completed a 14 game failure streak:
1 0
2 0
3 0
4 0
5 0
6 0
7 0
8 0
9 0
10 0
11 0
12 0
13 0
14 .0000610351563
15 .0000610351563
16 .0000610351563
.
.
.
5151 .0000610351563


By the same token:

Game # Probability that the Game Just Completed DID NOT make 14 game failure streak:
1 1
2 1
3 1
4 1
5 1
6 1
7 1
8 1
9 1
10 1
11 1
12 1
13 1
14 1 - .0000610351563
15 1 - .0000610351563
16 1 - .0000610351563
.
.
.
5151 1 - .0000610351563

Therefore, the probability that such a failure streak NEVER happens within 5151 games is:

1^13 * (1 - .0000610351563)^5138 = 0.730805299

Which means that the probability of it happening AT LEAST ONCE is 1 - 0.730805299 = 0.269194701

I'm fairly confident in that particular computation, but the number of times a 14 game failure streak can be expected within 5151 games is a separate, much more complicated endeavor. Hmmmm..... I wonder what that might be. My own expectation is NEAR ZERO times. Going to try to figure it out.

[Rex Goodheart]

OK, at this point I'm entertaining myself, I know.

But I left off my last post wondering how many times I can expect a 14 game losing streak within 5151 games. I found the answer (wasn't as complicated as I thought)!

I'll start by illustrating a very simple problem: if we flip a coin 5 times how many times should we expect heads? The answer is 2.5 times, of course, as follows:

Flip # Probability
1 .5
2 .5
3 .5
4 .5
5 .5
------
2.5

We arrive at our answer by summing up the probability of each roll.

Now, for our 5151 game problem. As already established, the probability of any one game completing a 14 game losing streak is 0000610351563

Therefore

Game# Probability

1 0
2 0
3 0
4 0
5 0
6 0
7 0
8 0
9 0
10 0
11 0
12 0
13 0
14 .0000610351563
15 .0000610351563
16 .0000610351563
.
.
.
5151 .0000610351563
--------------------
0.313598633

That sum means that for every 5151 set of games we can expect a losing streak of 14 games less than one third of once.

What does this all tell me? Well, if we count the number of 1's, 2's, 3's, and 4's within those 5151 games, and if those are spread rather evenly, that I've just been unlucky, nothing more.

But my greater point is that it shouldn't matter either way: awful luck in drawing seats should somehow be mitigated.



Love the site!

[sprockitz]

you are lacking use of some conditional probability with your assumptions.

Take a 3 coin flip event as an example, what are the odds of at least 2 tails in a row?

By the model above you would say it is

1. 0
2. 0.25
3. 0.25

total expected of 0.5 and probability of it happening in a given set as 1 - (1-0.25)^2 = 7/16

In reality there are only 8 possibilities here. Of those 8 TTT, TTH, HTT are the only 3/8 where it occurs, which is 6/16 and not 7/16. The expectation is 4/8 as it occurs twice in the TTT case and once each in the other 2 cases.

So expectation is much easier to compute than the probability of a streak occurring at least once.

I'm just freewheeling after midnight some mathematical thought here, but feel pretty good about my hunch here...

But we know a few things...we know the probability of extending a streak is 0.5 for each step, and each time the streak is extended it will count as another streak of 14. So given a streak hits 14, the streak hits 15 half the time, 16 1/4 of the time...which works out to an average of 2 (it is your basic infinite sum 1 + 1/2 + 1/2^2 + 1/2^3...).

So if the expected value is .31 and we know that an expected streak group length is 2, that means the number of independent streaks is half of that .31 or .155.

So .155 independent streaks occur...but that doesn't mean there is a 15.5% chance in any given 5151 games. But since our groups are now independent we can approximate this. An easy first order approximation says if something occurs once in n% of groups it'll occur at least twice in n^2% of those groups. And this would mean to one order approximation, which is decent for smaller n, we are looking at .155*(1-.155) = .130 or a 13.0% chance of a streak of 14 occurring at least once in a given group of 5151 games.

This is an approximation obviously and also ignores the beginning and end limits of the data in that some streaks would be ended only because the data ended, but these should be a small enough percentage given the precision we are looking to achieve.