08 Mar '17 15:07

I sometimes watch this programme on U.K t.v and I got to thinking about the maths behind it:-

There are 15 players each day.

Each day 3 players get through to the final round and take no further part in the game.

These 3 are replaced on day 2 by three new players making 15 again.

A player is allowed 3 attempts(i.e 3 days) to get to the final round and if he fails to do so is eliminated and replaced.

I reckon that on day four 21 different players have taken part.

The logic here is supposing players 1, 2, and 3 get to the final on day 1 there are 3 new players on day 2.

On day 2 and on day 3 players 1, 2 and 3 get through again allowing 6 new players.

So on day four players 4 to 15 have used up all their attempts and are replaced as well as players 1 to 3.

3+3+15 = 21 different players over 4 days.

There are 15 players each day.

Each day 3 players get through to the final round and take no further part in the game.

These 3 are replaced on day 2 by three new players making 15 again.

A player is allowed 3 attempts(i.e 3 days) to get to the final round and if he fails to do so is eliminated and replaced.

I reckon that on day four 21 different players have taken part.

The logic here is supposing players 1, 2, and 3 get to the final on day 1 there are 3 new players on day 2.

On day 2 and on day 3 players 1, 2 and 3 get through again allowing 6 new players.

So on day four players 4 to 15 have used up all their attempts and are replaced as well as players 1 to 3.

3+3+15 = 21 different players over 4 days.

**The question is does it alter the overall numbers if say player 15 for example gets to the final on his 2nd attempt ? or any other random combination**