Teams are now all confirmed for the Super Six stage stage of the ICC Men’s U19 Cricket World Cup 2024, with Ireland cementing their place after India accounted for the USA in the final Group A first round match today.
The Super Six will involve the top three sides from each of the four round-robin groups, with teams progressing from Groups A and D comprising one Group, and those from B and C the other. Each team carries forward the points and Net Run Rate they earned against fellow Super Six qualifying teams.
The six teams to make it from Groups A and D are:
- India
- Bangladesh
- Ireland
- Pakistan
- New Zealand
- Nepal
The six teams to make it from Groups B and C are:
- South Africa
- England
- West Indies
- Australia
- Sri Lanka
- Zimbabwe
USA, Afghanistan, Namibia and Scotland – the four teams who didn’t make the second stage of the tournament – will contest in play-offs for the last four places.
Super Six format
The teams will play two matches in the Super Six stage against their opponents from the corresponding group who finished in a different position in their group.
The top two sides from the two Super Six groups will then progress to the semi-final stage.
The two semi-finals are scheduled to be held on 6 and 8 February. The final is set to take place on 11 February, with all three knockout games set to take place in Benoni.
Super Six fixtures
30 January
- India v New Zealand in Bloemfontein
- Sri Lanka v West Indies in Kimberley
- Pakistan v Ireland in Potchefstroom
31 January
- Nepal v Bangladesh in Bloemfontein
- Australia v England in Kimberley
- Zimbabwe v South Africa in Potchefstroom
02 February
- India v Nepal in Bloemfontein
- West Indies v Australia in Kimberley
- South Africa v Sri Lanka in Potchefstroom
03 February
- Pakistan v Bangladesh in Benoni
- New Zealand v Ireland in Bloemfontein
- England v Zimbabwe in Potchefstroom