8/15/2023 0 Comments Rachel riley maths![]() Through the ultra-scientific methods of trying different numbers in my calculator until I got the right answer, I find that (100% - 73% 8 ) is about 92% - more than the 90% we were trying to find. ![]() There are three maths challenges every show, so the chance per show of only watching solvable challenges is already 90% 3 ~= 73%. If we make some more reasonable ones, we can actually calculate the amount of time it takes for 90% of people to have seen at least one unsolvable episode. I've made some fairly big, and very wrong assumptions for the sake of clarity. The chance of that happening is 100% - the number I gave above, which is basically 100% (because the number I gave is so small). We actually wanted to find the chance of watching one of the 8.8% games where the challenge can't be solved at all. That's the chance that we only watch episodes where everything is solvable. Working out that is generally much easier for the general case - it's just ( 100% - 90% N ), where N here is the total number of games we end up watching.Īssuming there are 52 episodes each year, and that a person watches ever week for their entire 70 year life, we end up with 90% 3640 ~= 2.8×10 -165 % (that is 0.0000.(×165)28% - very very small). That means that we can also find the correct result by working out the chance that all games are solvable, and subtracting that from 100%. We can actually work this out more simply - the chance that in N games we end up watching at least one unsolvable game is actually the chance that we don't only watch solvable games. Our chances of seeing it happen have almost doubled by watching two episodes. (90% × 10%) - the the case that only the second is unsolvable.(10% × 90%) - in the case that only the first is unsolvable.(10% × 10%) - in the case that they're both unsolvable.In the second week, we also have a 10% chance of seeing an unsolvable game, but the chance that we see at least one unsolvable game in both weeks is the sum of: The first week, we have 10% of seeing an unsolvable game. ![]() Let's round the numbers to 90%/10%, just for the sake of maths, and assume we watch one episode of countdown every week, without fail (and that there's only one number round each week). The chance is actually quite high, if we take into account all the games we watch.
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