c4countdown

Did DECANTERS really come out backwards, I wasn't paying attention.

Joseph.

Yep it did.

Yes. And I started thinking about permutations and the odds of it happening, when I realized I've wondered the same thing. Making me think it's happened before.

metalized_deionizer wrote:Yes. And I started thinking about permutations and the odds of it happening, when I realized I've wondered the same thing. Making me think it's happened before.

What are the actual odds of that happening? How would you work that out exactly?

JBolas wrote:

metalized_deionizer wrote:Yes. And I started thinking about permutations and the odds of it happening, when I realized I've wondered the same thing. Making me think it's happened before.

What are the actual odds of that happening? How would you work that out exactly?

I'll set the problem. Given the letters have been selected, find the probability the words DECANTERS is spelt backwards.
There is only one solution SRETNACED, out of 9!/2! (9 letters, but there are 2 Es) = 1/181440.

It gets more complicated when you try to work out the odds of the word SRETNACED coming out in that order on the show because you have to factor in the tile distributions.

Janowsky wrote:It gets more complicated when you try to work out the odds of the word SRETNACED coming out in that order on the show because you have to factor in the tile distributions.

Also, many contestants choose their letters in blocks of vowels and consonants, making it unlikely that anything sensible is going to emerge.

Conor wrote:

JBolas wrote:

metalized_deionizer wrote:Yes. And I started thinking about permutations and the odds of it happening, when I realized I've wondered the same thing. Making me think it's happened before.

What are the actual odds of that happening? How would you work that out exactly?

I'll set the problem. Given the letters have been selected, find the probability the words DECANTERS is spelt backwards.
There is only one solution SRETNACED, out of 9!/2! (9 letters, but there are 2 Es) = 1/181440.

Blimey Conor, how old are you again?! *bows down"

juj wrote: Blimey Conor, how old are you again?! *bows down"

16. But it's not a very difficult problem once you've simplified it to that, it's just some arrangements maths.

I agree that it's not that hard but in fairness, an equivalent problem recently proved too difficult for K. Bevins BSc Maths (Hons).

Hmm well it's impressive to me, I may once have known stuff like this when I was doing A-level maths - just for the short while I needed to know it so as to pass the exam - but I've certainly forgotten every last iota of it now...

Charlie Reams wrote:I agree that it's not that hard but in fairness, an equivalent problem recently proved too difficult for K. Bevins BSc Maths (Hons).

Indeed this problem proved too complicated for me. I have no understanding of arrangements or combinations. I know how to calculate factorials and know how to calculate something such as 7C2 but giving me a wordy problem and asking how many different arrangements, I wouldn't know where to start. Nobody has been able to explain it to me - which really p***es me off.