The '99 Aus-SA Semi-final: A tie no matter how you slice it
Today I read about a new system to be put in place for resolving ties in decider matches. The system has already been adopted for the ongoing Asia Cup, in case the final ends in a tie. Bowl-outs have been used in T20 games a few times, but this new idea seems a tad more systematic and less random. Instead of 6 balls being bowled at the stumps with no batsman, there will be batsmen facing the deliveries. The side with most runs will presumably win.
Sounds good. But what if runs scored even in that one over are the same for each side?
This got me thinking about the most famous tie in ODI history - the 1999 World Cup semi-final between Australia and South Africa. My friend RajK has always insisted that the outcome was unfair to South Africa and the match should have been replayed. Of course, he will say anything to make SA look good. He is their only remaining fan. So I wondered, suppose this system had been in place back then. And the one-over-a-side eliminator had also resulted in a tie (in terms of runs scored, as well as sixes and fours hit). How would the result have looked? Of course, both teams scoring the same number of runs in the one over eliminator is a huge assumption, considering South Africa is involved. Given their penchant for choking, they would have probably lost it in that over itself. But let us assume a tie nonetheless.
The general impression I had looking back was that South Africa, riding on Klusener's strokeplay and Gibbs' early cameo, had a more six-four-full innings. So I dug up the scorecard. Well, it turns out that both teams hit 3 sixes each in their innings. Fair enough, I thought. Not too many sixes are usually hit in low-scoring matches, so the likelihood of those two small numbers being equal was reasonably high. But surely the number of fours would give us a result?
It does not! Both sides hit exactly 17 fours each in their innings! What are the odds?
Under the rules for the new tie-breaker, each team will nominate three batsmen and one bowler. Each side bats one over, with the innings being declared closed if it loses two wickets.
Sounds good. But what if runs scored even in that one over are the same for each side?
If the teams finish tied on runs scored in that one over, the side with the higher number of sixes in its full innings and in the one-over eliminator will be declared the winner. If the teams are still tied, the one with the higher number of fours in both innings will win.
This got me thinking about the most famous tie in ODI history - the 1999 World Cup semi-final between Australia and South Africa. My friend RajK has always insisted that the outcome was unfair to South Africa and the match should have been replayed. Of course, he will say anything to make SA look good. He is their only remaining fan. So I wondered, suppose this system had been in place back then. And the one-over-a-side eliminator had also resulted in a tie (in terms of runs scored, as well as sixes and fours hit). How would the result have looked? Of course, both teams scoring the same number of runs in the one over eliminator is a huge assumption, considering South Africa is involved. Given their penchant for choking, they would have probably lost it in that over itself. But let us assume a tie nonetheless.
The general impression I had looking back was that South Africa, riding on Klusener's strokeplay and Gibbs' early cameo, had a more six-four-full innings. So I dug up the scorecard. Well, it turns out that both teams hit 3 sixes each in their innings. Fair enough, I thought. Not too many sixes are usually hit in low-scoring matches, so the likelihood of those two small numbers being equal was reasonably high. But surely the number of fours would give us a result?
It does not! Both sides hit exactly 17 fours each in their innings! What are the odds?
