# Thread: One for the mathematicians amongst us

1. ## One for the mathematicians amongst us

This is beginning to really infuriate me and I am not smart enough to work it out

A, B and C play for the same club – they all come to the final game of the season having taken 23 wickets for 50 so an average of 2.17

In that last game A takes 7-40 (ave 5.71), B 2-25 (ave 12.50) and C 1-22 (ave 22.00) so obviously A must end up with the best bowling average for the season, or so I would have thought, but in fact

A has 30 for 90
B has 25 for 75
C has 24 for 72

So all still have the same average of 3.00!!!!!

If someone can explain the mathematical principles to me, preferably in words of no more than two syllables, I would be a lot happier, and the person who told me this last week, and then confessed he couldn't explain it, would be able to walk past my office again without fear of being struck by flying objects

2. I understand how this comes about but I am struggling to find the words to represent it.

In general, because their overall average is considerably lower than their averages in the last game, the more that happens in their last game will have more effect on their average. So, because A took seven wickets at a greater average but C only took one wicket in the last game it evens itself out.

Its an interesting idea, one which I hopefully helped to understand.

3. They all had 23 wickets for 50 runs so to make a forumla for what is required to average 3 let:

r= runs conceded in the last innings
w= the number of wickets

3 = (50+r)/(23+w)

rearranging to isolate r we get the number of runs that can be conceded given a bag of w wickets in the final game:

r = 3*w + 19

so the first 19 runs takes their totals to, lets say, 23 wickets for 69 runs = ave 3 then;

A takes 7/21 @ 3
B takes 2/6 @ 3
C takes 1/3 @ 3 and they all end up with the same average

4. I'm no mathematician. But it seems to me that there are two tricks here which make the result seem counter-intuitive, particularly to a cricket fan.

1. The figures are deliberately designed to trick cricket lovers. Treat them with care. It may help to multiply the runs conceded and averages by 10 to get a more meaningful picture for a cricket fan. Remember that, given their previous figures, each of the bowlers performed extremely badly in their last innings. Including A, who although he took 7 wickets, haemorrhaged 40 runs, which in the context of his season is masses. It's the equivalent of Bowler D going into the last game averaging 21.7 and conceding 400 runs in that innings. But as a cricket lover rather than a mathematician, it's very difficult to get your head around the idea that taking 7-40 can significantly worsen your average.

2. It's also possible to be misled by the bowlers' averages in the last game alone. These need to be treated with extreme care. However if you bear in mind point 1 above, it becomes possible to digest them properly. Bowler A's average of 5.71 in that game was very poor (again, think of Bowler D averaging 57.1). What makes it worse is that not only did he concede a whopping 5.71 runs per wicket, but he did that 7 times over by taking 7 wickets. Bowler C, however, had an average that was positively disastrous - conceding 22 runs per wicket (think 220 runs per wicket) - but fortunately he only did that once because he only took one wicket.

5. Thanks gents - I'm not going to go as far as to say I now understand but at least I can see the point .................................................. ...... sort of!

6. Originally Posted by zaremba
The figures are deliberately designed to trick cricket lovers. ....although he took 7 wickets, haemorrhaged 40 runs, which in the context of his season is masses. ....it's very difficult to get your head around the idea that taking 7-40 can significantly worsen your average.

7. I wouldn't claim to understand it well enough to explain it, but it works because each bowlers share of the pools wickets remains identical to his share of the pools runs, though crucially the shares of the pool are no longer equal.

What happens the week after when the bloke who takes 7fer has to buy another jug is described in the famous "barstool economics" parable used to discuss tax policies

And there is a "real world" cricket effect - you see this situation with Net Run Rate calculations sometimes - particularly in ICC comps with minnows in. And in our own now killed off FPT - if you had a match rained off against someone you would expect to beat (Ireland, say, sorry Worcester) you not only miss out on the point, but also suffer that everyone else benefits in the net run rate calcualtion from hammering them. You can see this in the example if you now remove one of the earlier better performances from each, say 3 for 6... now the averages diverge, as you would expect.

Not sure if that helps!

8. I don't know if this is so obvious that it isn't worth mentioning, but.....

With 1-22 and 2-25, the latter has one extra wicket and 3 extra runs, so an average of 3 per extra wicket

With 2-25 and 7-40, the latter has 5 extra wickets and 15 extra runs, so an average of 3 per extra wicket

So basically, in order for any of the 3 players' average to "blow out" from 2.17 to 3, they would have to take 1-22. At this point Bowler A has stopped bowling with his total average at 3, whereas the other two have continued on taking a wicket for each 3 runs.

9. To make the same point in a more realistic example, let's say 3 bowlers had taken 19 wickets each for 361 at an average of 19.

Next week bowler A takes 1-39, bowler B takes 2-59, and bowler C takes 7-159. They all end up with the same average of 20.

It's simply a case that bowler A has allowed his average to increase to 20 by doing substantially worse than normal, but has then ended his spell. The other two might have also allowed their averages to increase to 20, but have then started to take each additional wicket for the cost of a further 20 runs.

10. Yes, and this is why even though average is effectively a combination of ER and S/R, it does make sense to pay attention to the two separately as well.

11. Originally Posted by fredfertang
This is beginning to really infuriate me and I am not smart enough to work it out

A, B and C play for the same club – they all come to the final game of the season having taken 23 wickets for 50 so an average of 2.17

In that last game A takes 7-40 (ave 5.71), B 2-25 (ave 12.50) and C 1-22 (ave 22.00) so obviously A must end up with the best bowling average for the season, or so I would have thought, but in fact

A has 30 for 90
B has 25 for 75
C has 24 for 72

So all still have the same average of 3.00!!!!!

If someone can explain the mathematical principles to me, preferably in words of no more than two syllables, I would be a lot happier, and the person who told me this last week, and then confessed he couldn't explain it, would be able to walk past my office again without fear of being struck by flying objects
Here is something that might help: Game is not a unit of time, wicket is a unit of time.

Here is a chart of the three bowlers. In essence, what has happened is that bowler A is getting worse, but at a very slow rate. Bowler B is getting worse, but at a much faster rate. Bowler C just completely went from Malcom Marshall to Mohammad Sami in one innings. So while the average is the same, Bowler B and Bowler C haven't had as much time as bowler A, otherwise their averages would be much worse.

Basically, all three are getting worse, and obviously all three will have an average higher than 3.00 at some point if they keep going at this rate. However (and this seems obvious), they all have to 'cross' 3.00 to get above 3.00. The difference is that bowler A will take the longest to cross 3.00 while B will get to 3.00 and above faster, and C will get to 3.00 the fastest. So it's not a matter of all of them having 3.00 average, it's a matter of them all going past 3.00 at varying rates. What the scenario did was to simply give the wickets it takes for them to get past that number, but that we already knew. You can't average 4 from 2, without averaging 3 in between (well you can, if one wicket takes you there, but you still 'crossed' that number at some point during the innings).

The graph below shows how long it took them to get to that 3.00. Yes, it was one game, but think of time in terms of number of wickets. A 'game' is an arbitrary notion when it comes to averages. The idea of number of games played does not factor into what your average is. Only total runs and total wickets do, so your co-worker gave you a cutoff point which was fake, and did not say anything about the nature of their averages. If we look at a moving average in terms of wickets as a unit of time, you get the graph below, and you can see how all 3 got there at different 'times'.

Remember, we are effectively using wickets as a unit of time, so for proper comparison, that needs to be standardized. If we extrapolate their averages to 30 wickets, here is what the graph would look like, and as you can see, bowler A is miles better at that point. He has more than twice better average than C.

I hope this was beneficial .

12. Originally Posted by silentstriker
Here is something that might help: Game is not a unit of time, wicket is a unit of time.

Here is a chart of the three bowlers. In essese, what has happened is that bowler A is getting worse, but at a very slow rate. Bowler B is getting worse, but at a much faster rate. Bowler C just completely went from Malcom Marshall to Mohammad Sami in one innings. So while the average is the same, Bowler B and Bowler C hasn't had as much time as bowler A, otherwise their averages would be much worse.

Basically, all three are getting worse, and obviously all three will have an average higher than 3.00 at some point if they keep going at this rate. However (and this seems obvious), they all have to 'cross' 3.00 to get above 3.00. The difference is that bowler A will take the longest to cross 3.00 while B will get to 3.00 and above faster, and C will get to 3.00 the fastest. So it's not a matter of all of them having 3.00 average, it's a matter of them all going past 3.00 at varying rates. What the scenario did was to simply give the wickets it takes for them to get past that number, but that we already knew. You can't average 4 from 2, without averaging 3 in between (well you can, if one wicket takes you there, but you still 'crossed' that number at some point during the innings).

The graph below shows how long it took them to get to that 3.00. Yes, it was one game, but think of time in terms of number of wickets. A 'game' is an arbitrary notion when it comes to averages. The idea of number of games played does not factor into what your average is. Only total runs and total wickets do, so your co-worker gave you a cutoff point which was fake, and did not say anything about the nature of their averages. If we look at a moving average in terms of wickets as a unit of time, you get the graph below, and you can see how all 3 got there at different 'times'.

Remember, we are effectively using wickets as a unit of time, so for proper comparison, that needs to be standardized. If we extrapolate their averages to 30 wickets, here is what the graph would look like, and as you can see, bowler A is miles better at that point. He has more than twice better average than C.

I hope this was beneficial .
What do you really think?

13. Whoa, SS posting in a maths thread!?

14. I will be coming with my math notebook in our next CW meet.

15. Thierry Henry pretty much captures the basic idea but another example should help clarify the concept.

Suppose there are three batsmen all with the same average just before the end of their careers. Let's say it's the same average Bradman had before his last innings.

A has just one innings left. He needs 4 runs to average 100 like Bradman. (averaging 4 over that one innings)
B has two innings left. He will need 104 runs . (average 52 over those two innings)
C has three innings left. He will need 204. (average 68 over those three innings)

All three batsmen start off with the same average and end up averaging 100 but have different averages in their remaining careers. This is because these are of different length so those who play longer have to do more to maintain that 100 average. Same with the bowling where those who bowl more have to do more to maintain that average.

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