Callum Todd wrote: ↑Fri Jun 20, 2025 6:42 am
Philip A wrote: ↑Tue Jun 17, 2025 12:44 pm
If you removed ten small numbers each from 1 to 10, so that each small number appears once only instead of twice (and therefore 14 numbers instead of 24: 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 25, 50, 75, 100), how much per cent of all standard numbers games would be impossible to score any points at all?
Wow, great question. My intuition says it will be less than under the current system. Let's see how wrong I am.
Okay, so I had a go at trying to find the answer to this as a bit of computer programming practise. Before having any faith in the accuracy of my results please consider these three key details:
1. This is Ask Graeme but I am not Graeme.
2. I have no experience, skill, or training in either computer programming or mathematics.
3. The following is the products of some mathematics I programmed a computer to do.
With that out of the way, I gather that there would be 3,003 possible selections of the Countdown numbers game if there were only one of each small number in the deck:
45 possible 4 large selections
480 possible 3 large selections
1,260 possible 2 large selections
1,008 possible 1 large selections
210 possible 6 small selections
As there are 900 possible numbers targets (100-999 inclusive), this means there would be 3,003 * 900 =
2,702,700 possible distinct numbers rounds under this system. Of them, only
127 are impossible to score from. This makes
0.0047%. These 127 impossible rounds are spread across just 5 selections:
1 2 3 4 5 6 has
95 impossible targets: 821-829, 851-853, 875-889, 911-949, and 971-999.
1 2 3 5 6 7 has
15 impossible targets: 983-997
1 2 3 4 5 7 has
7 impossible targets: 991-997
1 2 3 4 6 7 has
7 impossible targets: 907-913
1 2 3 4 5 8 has
3 impossible targets: 851-853
Thanks to some wizardry from Maus and Corby on
an old thread I found on c4c, they've already done the work on the existing Countdown game with two of each small number in the deck:
Thomas Carey wrote: ↑Wed Jan 06, 2016 11:45 pm
55+840+3690+5808+2850=
13,243 numbers selections.
Times the targets and you have 13,243*899=
11,905,457 of whatever I was working out.
Jon Corby wrote: ↑Mon Jan 11, 2016 2:57 pm
Here is the breakdown by type, and how far away you can get:
Code: Select all
Zero away 1 away 2 away 3 away 4 away 5 away 6 away 7 away 8 away 9 away 10 away 11+ away
6S 1963726 (76.64%) 353472 65417 26954 15232 9859 7357 5780 4589 3945 3326 102493
1L 4966076 (95.11%) 220295 21730 5813 2527 1272 713 470 339 264 217 1676
2L 3192103 (96.23%) 112258 8487 1880 865 435 313 207 144 114 78 426
3L 693131 (91.79%) 53875 4577 1332 724 461 296 206 139 87 64 268
4L 43710 (88.40%) 4661 556 179 112 75 42 28 22 18 16 26
However, since their excellent work we have discovered that 100 is a valid numbers target which changes things ever so slightly. Specifically it adds another 13,243 possible rounds (one for each selection) to give us
11,918,700 possible rounds. I haven't done the hard maths on this but I am going to run with the assumption that 1 1 2 2 3 3 is the only selection for which it is impossible to get within 10 of 100, so that adds just one to the total of impossible (11+ away) games, giving us a new total of
104,890, or
0.88%
So, in conclusion: Yes, having only one of each small number in the deck would significantly cut the percentage of rounds it would be impossible to score on, from 0.88 to 0.0047.