circuit finals tentative first round matchups


Every group is so stacked. Good lord.


looks like it will be fun :slightly_smiling: Josh Sharpe isn’t on the list?


Add one more to the list, I just confirmed today.


@PinballNarcissist final list is out, maybe time to update your match up?


There’s only 39 on that list. I received an email from @PAPA_Doug saying there was a late cancellation and they are looking to fill that position. Considering that I am tied for 58th, looks like they are going pretty deep into the standings. Unfortunately I won’t be going. Good luck to whomever gets that last spot!


Updated. Looks like I’m in for a great time for round 1 :slightly_smiling:

If I have any matchups wrong let me know. The order PAPA shows on thier list for the ties is different than what it says it should be on the site. (Riel/Hansen and Nayman/Zeltmann)


Haha, not only are you in a killer group but you guys are gonna get the chum of leftover picks. Better to be ranked 40th and play good games rather than 5-8 and play EBD/ Party Groan


Can’t wait to see what Jim decides to pick… :persevere:

I added current rankings and I’ll total them up in a few


Gagno and @BMU’s groups are both sub 200 in the world rankings. Just kinda guessed that #40 would be somewhere around 100 in the world rankings.


Vegas odds on Jim’s picks

Party Bone


Farty Zone?


I think we’ll be playing 4 games per round. With 3 games, 6 points was considered enough in general to pass the round. I guess now it would be 8 points?



8 would be an average of 2nd place and a good benchmark. But … Here’s the situation:

10 points is a guarantee to the next round. Fewer is no guarantee.

9 points guarantees at worst a two-way playoff for the second spot [10-9-9-0].

It’s possible to have a 4-way playoff at 7-7-7-7.

It’s possible to get in via a playoff with as few as 4 points [16-4-4-4].

As always, it depends upon how dominant the top player in the group is: the more points they take, the fewer of the remaining points are needed for the next-best player to move on.

It’ll be interesting to see if a 4-way tie occurs!


Time to compute the probability of a 4-way tie :slight_smile:


SWAG says 1 in 24^2 = 1/576 … to lazy to test that right now. Your turn?


I got the same thing (assuming players are equally likely to end up in each place). I can type it up if anyone is curious.


Ok. I was curious if there was any way it could happen without having everyone finish each place once. But: everyone needs to win once and finish third once (to make an odd number), and then you’re locked.

I’m not entirely sure why it’s 24^2 but there’s probably some mathy proof of it.

“So you’re saying there’s a chance…”


Hey guys: show your work…Bonus credit if you include some far-out mnemonic device using woodland creatures and combinations of quasi-rhyming words to create a mathematical formula.


In order for a four way tie to occur each player would have to have a 1st, 2nd, 3rd, and 4th place finish. No other result in four games will give 7 points. Since each player has to have one of each place the results of the first game aren’t important to the calculations. To make the explanation a bit simpler let’s assume that in game one P1 was 1st, P2 was 2nd, P3 was 3rd, and P4 was 4th.

The probability of a 4-way tie is simply the number of outcomes that result 4-way tie divided by the total number of possible outcomes. Each match has 24 possible outcomes. For the three remaining matches there are 24^3=13824 possible outcomes. P1 needs a 2nd, 3rd, and a 4th in these matches. There are six ways that this can occur. The next step is to figure out how many possible ways there are for this to result in all players tying.

Let’s assume that P1 finished in 2nd in game two, 3rd in game three, and 4th in game four. That leaves three possibilities for how P2 can finish (341, 413, 143). If P2 finishes 341, then P3 has to finish 412 and P4 has to finish 123. P2 finishing 413 also leaves only one way P3 and P4 can finish (P3:142, P4:321). If P2 finishes 143, then there are two different ways that P3 and P4 can finish (P3: 412, P4: 321 or P3:421, P4:312). This gives us four possible ways for there to be a 4-way tie for players 2-4 for each of the original six possibilities of how P1 finished. The result is there are 4*6=24 ways for there to be a 4-way ties out of 24^3 total possibilities for the final three games. This gives us a 1/576 probability of a 4-way tie.


The PAPA site is now showing Eric Fisher as 40th seed; rank 233rd. That should complete the groups. Looks like my group would be me, Josh Henderson, Johnny Modica, and Bowen. If so, we’d have the narrowest rank range, the highest average rank, the widest age range, and three great guys I’m going to enjoy competing with! Ok, so we get last choice of games. Whatever, we’ll just try to have fun.