We know that in the first innings the predicted score and in the second innings the predicted winning percentage will be calculated in the WASP.

But there is only one formula for calculating both score and percentage.

V(b,w) =r(b,w) +p(b,w) V(b+1,w+1) +(1-p(b,w)))V(b+1,w)

How is this one formula enough to get two results?

This question is not related to the previous one. I’m not asking about logical explanation here; I just want to know, what is the change that they do in the formula to get the percentage for the second innings?

Providing an example is a great thing for this type of question.

  • If someone explain the formula in your previous question (s)he will also explain how both results are calculated.
    – Himanshu
    Commented Jan 29, 2014 at 7:10
  • Like these two questions. Commented Jan 29, 2014 at 7:21

1 Answer 1


If you are looking at the complete explanation of that rule I came across nice blog that clears all the doubts regarding working of WASP.

To illustrate, in the first innings model to calculate the expected additional runs when a given number of balls and wickets remain, we could just average the additional runs scored in all matches when that situation arose. This would work fine for situations that have arisen a lot such as 1 wicket down after 10 overs, or 5 wickets down after 40 overs, etc.), but for rare situations like 5 wickets down after 10 overs or 1 wicket down after 40 it would be problematic, partly because of a lack of precision when sample sizes are small but more importantly because those rare situations will be overpopulated with games where there was a mismatch in skills between the two teams. Instead, what we do is estimate the expected runs and the probability of a wicket falling on the next ball only. Let V(b,w) be the expected additional runs for the rest of the innings when b (legitimate) balls have been bowled and w wickets have been lost, and let r(b,w) and p(b,w) be, respectively, the estimated expected runs and the probability of a wicket on the next ball in that situation. We can then write

V(b,w) =r(b,w) +p(b,w) V(b+1,w+1) +(1-p(b,w)))V(b+1,w)

Since V(b*,w)=0 where b* equals the maximum number of legitimate deliveries allowed in the innings (300 in a 50 over game), we can solve the model backwards. This means that the estimates for V(b,w) in rare situations depends only slightly on the estimated runs and probability of a wicket on that ball, and mostly on the values of V(b+1,w) and V(b+1,w+1), which will be mostly determined by thick data points. The second innings model is a bit more complicated, but uses essentially the same logic.

WASP Working in Details

Hope this helps.

  • 1
    While this link may answer the question, it is better to include the essential parts of the answer here and provide the link for reference. Link-only answers can become invalid if the linked page changes.
    – Himanshu
    Commented Feb 13, 2014 at 11:29
  • -1 Even after you were told twice to quote the source content and summarise it, you didn't. This is pure plagiarism.
    – Himanshu
    Commented Feb 18, 2014 at 4:17
  • 2
    This isn't that helpful anyway; repeating the definition doesn't make it any easier to understand. Commented Feb 18, 2014 at 5:52

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