First result progression analysis

[Sagy]

Does anyone have a statistical analysis of the progression rates in WC (UEFA) based on the first game results.

I'm particularly interested in the progression rates in the following two cases (I know the answer for the third case ;)):
Group X Group Y
A 3 E 3
B 3 F 1
C 0 G 1
D 0 H 0Basically, Groups A & B in 2006 WC.

Thanks,

Sagy

[numerista]

I know the answer for the third case ;)

Assuming all results (W/L/T) are equally likely, here's what I get for a team in the third case (1-1-1-1):

P(finish first) = 0.173
P(two-way tie for first) = 0.136
P(second) = 0.086
P(four-way tie) = 0.037
P(two-way tie for second) = 0.173
P(third) = 0.086
P(two-way tie for third) = 0.136
P(fourth) = 0.173

The reason I mention this is to point out that in this case, a team has a better than 1-in-3 chance (34.6%) of its position being meaningfully affected by a tie-breaker.

I'll try to put up the other scenarios later ...

[numerista]

Scenario: 3-3-0-0

Team w/ 3 Pts.

P(finish first) = 0.3704
P(two-way tie for first) = 0.1605
P(second) = 0.1605
P(three-way tie for first) = 0.0247
P(four-way tie) = 0.0123
P(two-way tie for second) = 0.1235
P(three-way tie for second) = 0.0123
P(third) = 0.0741
P(two-way tie for third) = 0.0370
P(fourth) = 0.0247

Team w/ 0 Pts.

P(finish first) = 0.0247
P(two-way tie for first) = 0.0370
P(second) = 0.0741
P(three-way tie for first) = 0.0123
P(four-way tie) = 0.0123
P(two-way tie for second) = 0.1235
P(three-way tie for second) = 0.0247
P(third) = 0.1605
P(two-way tie for third) = 0.1605
P(fourth) = 0.3704

[numerista]

Scenario: 3-1-1-0

Team w/ 3 Pts.

P(finish first) = 0.4938
P(two-way tie for first) = 0.0741
P(second) = 0.1605
P(three-way tie for first) = 0.0370
P(two-way tie for second) = 0.0741
P(three-way tie for second) = 0
P(third) = 0.0988
P(two-way tie for third) = 0.0247
P(fourth) = 0.0370

Team w/ 1 Pt.

P(finish first) = 0.1481
P(two-way tie for first) = 0.0741
P(second) = 0.1728
P(three-way tie for first) = 0.0247
P(two-way tie for second) = 0.1111
P(three-way tie for second) = 0.0123
P(third) = 0.1852
P(two-way tie for third) = 0.0741
P(fourth) = 0.1975

Team w/ 0 Pts.

P(finish first) = 0.0494
P(two-way tie for first) = 0.0247
P(second) = 0.1358
P(three-way tie for first) = 0.0247
P(two-way tie for second) = 0.0741
P(three-way tie for second) = 0.0123
P(third) = 0.1728
P(two-way tie for third) = 0.0741
P(fourth) = 0.4321

Disclaimer: all results come from a script I wrote this morning ... apologies in advance for any mistakes.

[Sagy]

numerista,

Thank you very much.

So far I only looked at 2002 and 1998 WC actual results.

There were 8 type X groups (4 in each WC) and only 1 of the 16 first game winners (Costa Rica in 2002) failed to advance.
There were 7 type y groups (3 in 1998, 4 in 2002), 2 of the 7 first game winners (2002 Argentina and Russia) failed to advance.


It would be interesting to know if a full analysis will bring the results closer to the statistical expectations or does the assumption that all results have equal likelihood virtually guaranties the actual to be meaningfully different from the expectation?

[numerista]

It would be interesting to know if a full analysis will bring the results closer to the statistical expectations or does the assumption that all results have equal likelihood virtually guaranties the actual to be meaningfully different from the expectation?

Because the simulation is an extreme case, where the teams with 3 points are no better than the ones with 0, it's only likely to be close to the truth when all four group members are similar in quality.

Because groups in the European Cup Finals tend to have more parity, I tried the simulation out on them. Results are below; unless noted, all ties are two-way for second place.

Nuts and Bolts
2004
type x - 1 - (1 of 2 teams advanced; 0-1 in tiebreakers)
type y - 3 - (3 of 3; 1-0 a tie-breaker, a three-way tie for first)

2000
type x - 2 - (3 of 4)
type y - 2 - (1 of 2; 0-1)

1996
type x - 1 - (1 of 2; 0-1)
type y - 2 - (2 of 2)
type z - 1

1992 (2 pts a win from here back)
type y - 1 (1 of 1)
type z - 1

1988
type x - 1 (1 of 2)
type y - 1 (0 of 1)

1984 (start of present group format)
type x - 1 (1 of 2)
type z - 1

Totals for European Cup Finals, 1984-2004

Type X - 12 teams
7 advance w/o tiebreakers (predicted 8.29)
2 tie-breakers needed (predicted 2.07)
3 eliminated w/o tiebreakers (predicted 1.64)

Type Y - 9 teams
6 advance w/o tiebreakers (predicted 6.07)
2 tie-breakers needed, 1-1 record (predicted 1.00)
1 eliminated w/o tiebreakers (predicted 1.93)

Surprisingly (to me, at least), in this case the simulation comes close to reality.