In which we attempt to determine FC Bayern’s chance of overtaking Hannover for the final Champions League spot, and come away moderately optimistic. Put down your beer bongs and your crack pipes and pay attention, ‘cause we’re diving into the deep end.
With 5 games left, both Bayern and Hannover have an opportunity to earn as many as 15 (and as few as 0) points. Looking back on both clubs’ records over the entire 2010-2011 season across all competitions, FC Bayern has 25 wins, 10 losses and 7 draws, while Hannover has picked up 17 wins, 11 losses and 2 draws. Using this as our (admittedly rough) template, Bayern has a 59.5 percent chance to win each game, a 23.8 percent chance to lose each game, and a 16.6 chance to draw. Hannover, meanwhile, comes into each game with a 56.6 percent chance to win, a 36.6 percent chance to lose, and a 6.6 percent chance to draw.
“Wait!” you may say. “If Bayern has a greater chance to win and smaller chance to lose, then how is Hannover sitting 1 point ahead of us in the standings?” But keep in mind this data tracks both teams ACROSS ALL COMPETITIONS - aside from the horrendous crash-outs at the end, Bayern had a decent record in both the Champions League and the Pokal, while Hannover lost in the first round of the Pokal and did not play in any European competition.
A warning: this next part is incredibly math-heavy. If you want to jump in later, after the hard work has been done, meet me after the series starts (like this: * * * *) about two pages down.
Now, here’s where we break out the calculators: with 5 games left and a possible win, loss, or draw in each, there are 237 unique combinations of wins, losses and draws that each team could run into over the last month of the season. Note that (for now) we’re looking at each team individually, not what BOTH teams will do on a given weekend. This number includes all combinations in different orders. Specifically, it works out like this:
NOTE that this section has been edited. Reader Qian Liv pointed out a miscalculation in the original post - you'll see where he's right a few paragraphs down.
Possible Combinations in which …
All five results are the same: 3
- 5 straight wins
- 5 straight losses
- 5 straight draws
(clearly, for ^ these, there is only one possible order for each)
Only two types of results are present: 90
- 4 wins + 1 draw (5)
- 3 wins + 2 draws (10)
- 2 wins + 3 draws (10)
- 1 wins + 4 draws (5)
- 4 losses + 1 draw (5)
- 3 losses + 2 draws (10)
- 2 losses + 3 draws (10)
- 1 loss + 4 draws (5)
- 4 wins + 1 loss (5)
- 3 wins + 2 losses (10)
- 2 wins +3 losses (10)
- 1 win + 4 losses (5)
Under each possible combination, the number in parentheses tells us the possible total number of different sequences in which that combination could take place. This is important for determining the likelihood that each combination might happen, but not important when it comes to the number of points earned in the standings.
All three types of results are present (W, L, and D), with three of one type: 54
- 1 win + 1 loss + 3 draws (18)
- 1 win + 1 draw + 3 losses (18)
- 1 loss + 1 draw + 3 wins (18)
Again, for each, the number in parentheses means there are 18 sequences in which this combination could arise. This means that each is much more likely to occur than 5 straight results of a single type, even though there would be no difference in the standings between W-W-W-L-D and L-D-W-W-W. My thanks go out to my accountants Tim and Timothy for their help with this part.
* NOTE that this ^ paragraph is incorrect - there are actually 20 of each combination, not 18, as Qian Liv has correctly noted. Therefore, the number of total possible combinations (the "denominator" increases by 6. This won't have a huge effect on the relative likelihood of winning, because it would change the odds equally for both teams, but we thank Qian Liv for the help *
where all three types of results are present (W, L, D), with two of two types and one of the third: 90
- 2 wins + 2 losses + 1 draw (30)
- 2 wins + 2 draws + 1 loss (30)
- 2 draw +2 losses + 1 win (30)
I actually have all these possible sequences written out in a Word document - e-mail me at email@example.com if you want to see it. Though I couldn't imagine why.
Now, we know the theoretical likelihood of a win, loss, or draw for each game with respect to both FC Bayern and Hannover. Therefore, we can apply these probabilities to all the possible combinations, as follows:
|W-L-D combo||Out of 237 possible combinations, how many lead to this sequence REGARDLESS OF ORDER ?||Points derived therefrom||Likelihood for Bayern||Likelihood for Hannover|
|237 possible combinations of W, L and D||100.13||97.72|
Under the first column, a win is represented by a 3, a draw by a 1, and a loss by a 0, with each result represented by the number of points it would lead to. The second column indicates the relative likelihood that this combination would come up over 5 games if we counted all times it would come up, regardless of sequence. The third column details how many points total would result (basically, just adding up the 5 numbers in column 1), and the fourth and fifth show how likely each team is to hit upon that result in light of their proven record of wins losses and draws. Again, note the update above and the fact that the denominators would actually be 243, not 237. Thanks.
In order to calculate columns 4 and 5, we substitute the chance of a win, loss, or draw, as a factor of one, for each team (0.595 - 0.238 - 0.166 for Bayern, and 0.566 - 0.366 - 0.066 for Hannover) for each result in the first column. Then we multiple those 5 numbers, then multiply that result by the number of possible sequences in which the combination could arise.
To take an easy example - Bayern has a 59.5 percent chance of winning each game. So, their chance of winning five straight is 0.595 ^ 5, or .07457 as a factor of one, or 7.457 percent. Then, we multiply that number by 1 (only 1 possible sequence results in 5 straight wins) to get 7.457%.
To take a slightly more complicated example, move down three rows. Hannover has a 56.6 percent chance of winning each game, and a 6.6 percent chance of drawing. Therefore, their chance of winning then next 4 and then drawing the last game is (0.566 ^4) x .066 as a factor of one, or about 0.6773 percent. But, there are 5 possible sequences of 4 wins + 1 draw (all of which count for the same number of points in the standings), so we multiply that number by 5, to get about a 3.386 percent chance that they will win 4 and draw 1 in any order.
The number on the very bottom of the two right-most columns show that the totals don’t add up to exactly 100%, because of rounding after 3 or 4 decimal places. Still, I think 100.13 and 97.72 percent accuracy is pretty good.
* * * *
If you jumped down to this spot to avoid the heavy math, welcome aboard. You didn’t miss much. We are now going to take our previous chart and combine possible result combinations in which the point totals are the same. E.G., we calculated the chance that each team would pick up 2 wins + 0 draws + 3 losses, and the chance that each team would pick up 1 win + 3 draws + 1 loss. But since both of those would result in 6 points for the standings, we combine the likelihood that each will happen to get the total chance that each team, in light of its propensity to win, lose, and draw games, will pick up 6 points. The resultant chart looks like this:
|Points||% chance Bayern will end up with this many points||% chance Hannover will end up with this many points|
Again, the figures at the bottom of the right-most columns show that our percentages don’t add up to exactly 100, but pretty damn close. Alert students will recognize this as the appropriate time to break out the dual-axis chart:
The height of each bar (mit Bayern im die Rotten, selbstverständlich) represents the percentage chance that each team will pick up that number of points over the final 5 games. Two things stand out:
- The chance of either team getting 14 points is zero - no combination of wins, losses, and draws over the last 5 would give a team exactly 14 points.
- Hannover’s aversion to draws skews the graph a bit - I.E., they have a larger chance of getting 9 (which requires 3 wins but no draws) then they do of getting 7, 8, 10, or 11 (all of which would require at least 1 draw).
If this graph is to be believed (which of course it isn’t, this is all just time-wasting foolishness because we still have 4 days until the next game and no one wants to hear about Dimitar Berbatov) Bayern stands a pretty decent chance of passing Hannover. Our likelihood of winning all 5 is almost 2% higher than theirs, and our likelihood of winning 4 and drawing 1 is about 7% higher than theirs. They have a slightly better chance of getting 12 points (winning 4 + losing 1 outright), due to their aversion to draws, but we make that up with a bigger chance to get 11 points.
Put another way, FC Bayern’s chance to end up with one of the 4 highest possible point totals (11, 12, 13, or 15 points) is about 38.557 percent, while Hannover’s chance of securing one of those point totals is just south of 29 percent. Since we’re all looking for a reason for optimism, this is a pretty good place to start.
In all seriousness, I have started to get over the disappointment of Saturday, and I do still like our chances of finishing in the top three. Hannover is a good club, but they’re on the road for two straight against tough opponents, including a Thursday night game at Freiburg next week. Now that we’ve finally shook ourselves of LvG, we have no more excuses - we need sharp passing, tough defending, and we need to create chances every time we go forward.
We’ll be back later this week with a look ahead to this weekend’s Leverkusen game, which - quick preview - is very, very important. I promise this one will have pictures. Thanks for reading.