Pat Cummins insists there’s enough support for him and Mitchell Starc in Australia’s bowling stocks, despite the pair taking almost 60 per cent of the team’s wickets at the World Cup.
Starc and Cummins have each taken nine wickets in this year’s tournament, with the latter claiming figures of 3-33 to break through Pakistan on Wednesday.
The next highest wicket-takers are Adam Zampa and Marcus Stoinis with four apiece, while Nathan Coulter-Nile and Kane Richardson have two and Aaron Finch one.
The rest of Australia’s bowlers are also significantly more expensive than the frontline duo, with Starc (5.38 economy rate) and Cummins (4.40) the only bowlers going at less than a run a ball.
But Cummins said it was only natural, with wickets easier to come by with the new ball and the opportunity to be more attacking.
“I think we do (have enough support),” Cummins said.
“The last two or three years I have bowled mainly the middle overs and taking wickets is so much harder than in that first 10.
“Richo (Kane Richardson) and Coults (Nathan Coulter-Nile) have played a bit of cricket but still not heaps. I think they’re always going to get better.
“That role in the middle, it’s always so hard to take wickets. And that’s where someone like Zamps has been great.”
Cummins has enjoyed an impressive start to the World Cup, after former quick Brett Lee labelled him the kind of player who can win the tournament for Australia.
He has bowled the first over in the past two matches against India and Pakistan, rewarding coach Justin Langer with a wicket early on Wednesday.
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After playing just two games in the last World Cup, the Penrith paceman is now leading the 2019 edition for most dot balls bowled – 112 from 230 deliveries.
“That’s probably been the big improvement in my game the last couple of years, just the ability to hold a length and make it tough to get runs,” he said.
“If you give away easy singles than 300 becomes a quite risk-free score sometimes if a team bats well.
“Cutting down the singles is huge, it means you can get hit for a boundary and might not have that big over of 10 or 11. It still might only be a five or a six.”