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Posers and Puzzles

Posers and Puzzles

  1. Subscriber venda
    Dave
    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.
    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
  2. Standard member wolfgang59
    Infidel
    08 Mar '17 17:29
    Originally posted by venda
    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 h ...[text shortened]... player 15 for example gets to the final on his 2nd attempt ? or any other random combination
    [/b]
    Yes.
    Players who do not make the final are there for 3 days.
    Players who do make the final may have played 1 or 2 or 3 days.

    Consider a new series beginning Monday.
    3 finalists leave and are replaced.
    The 3 replacements could make the final.
    Same again next day.
    After Wednesdays game 21 players have taken part and are eliminated.

    But if the none of the replacements make any final after Wednesdays game
    (all finalists are from original 15) only 15 have been eliminated. 6 go on to
    Thursdays game.
  3. Subscriber venda
    Dave
    08 Mar '17 18:51
    Originally posted by wolfgang59
    Yes.
    Players who do not make the final are there for 3 days.
    Players who do make the final may have played 1 or 2 or 3 days.

    Consider a new series beginning Monday.
    3 finalists leave and are replaced.
    The 3 replacements could make the final.
    Same again next day.
    After Wednesdays game 21 players have taken part and are eliminated.

    But if the no ...[text shortened]...
    (all finalists are from original 15) only 15 have been eliminated. 6 go on to
    Thursdays game.
    As you say, after Wednesday's game 21 player's have been eliminated.
    Numerically that's the 15 original players and the 6 replacements from Monday and Tuesday, so following that line of thought Thursdays player's would be 15 new player's.
    However as you rightly say one or more of Tuesday's and/or Wednesday's "new" player's could have one or two games left for Thursday.
    I suppose someone has put it all into a formula and derived the min/max number of players required for "x" number of games but it's a bit beyond my talents I'm afraid