c4countdown

Obviously based on Peter's thread about numbers... 11,900,000-odd numbers games is a lot. How many letters games are there? More? Or less?
What's your hunch? My guess is less. Or actually maybe more, thinking about all the duplicated letters in the piles.

And who's going to work it out...? (state your letter distribution when doing so please!)

Can't be bothered to do a proper calculation, but randomly picking 9 letters and dividing by 9! to remove duplicate selections (ignoring duplicate letters within selections) gives just over 15,000,000. Refining this to pick valid numbers of consonants and vowels brings it down to around 3,500,000.

Thomas Cappleman wrote:Can't be bothered to do a proper calculation, but randomly picking 9 letters and dividing by 9! to remove duplicate selections (ignoring duplicate letters within selections) gives just over 15,000,000. Refining this to pick valid numbers of consonants and vowels brings it down to around 3,500,000.

That doesn't seem right, are you sure? Dividing by 9! seems wrong (there's only one AAAAAAAAA but you're saying there's 9! of them, aren't you?) If I have an alphabet of 3 letters, and I'm choosing a selection of 3 letters, the possibles are AAA,AAB,AAC,ABB,ABC,ACC,BBB,BBC,BCC,CCC. 10 selections. Your calculation would give (3^3)/3! = 4.5

Jon Corby wrote:

Thomas Cappleman wrote:Can't be bothered to do a proper calculation, but randomly picking 9 letters and dividing by 9! to remove duplicate selections (ignoring duplicate letters within selections) gives just over 15,000,000. Refining this to pick valid numbers of consonants and vowels brings it down to around 3,500,000.

That doesn't seem right, are you sure? Dividing by 9! seems wrong (there's only one AAAAAAAAA but you're saying there's 9! of them, aren't you?) If I have an alphabet of 3 letters, and I'm choosing a selection of 3 letters, the possibles are AAA,AAB,AAC,ABB,ABC,ACC,BBB,BBC,BCC,CCC. 10 selections. Your calculation would give (3^3)/3! = 4.5

Yeah, I was deliberately ignoring that for simplicity. I suspect that this error will be greater than the fact that I'm duplicating some consonants illegally for my second estimate, so looking probably around 5-6 million.

Your answer seems hideously inaccurate. You're assuming that there are 9! repetitions for each selection, but that only applies when all the letters are different. Even with a selection of 3 letters from an alphabet of 3 your calculation gives only 45% of the true answer! With bigger figures that will surely be amplified! I think you've massively understated the answer.

The formula for finding the number of combinations of r letters chosen from n different kinds of letter is (n+r-1)! / r!(n-1)! (where repetition is allowed and the order doesn't matter)

The number of ways of dealing 3 vowels from 5 different kinds of vowels is 7!/(3!*4!) = 35
dealing 4 vowels = 70
dealing 5 vowels = 126

The number of ways of dealing 4 consonants from 21 different kinds of consonants is 24!/(4!*20!) = 10626
dealing 5 consonants = 53130
dealing 6 consonants = 230230

So pairing these up for the three different kinds of letters rounds:
Number of possible 4C 5V letters rounds= 10626*126 = 1,338,876
Number of possible 5C 4V letters rounds= 53130*70 = 3,719,100
Number of possible 6C 3V letters rounds= 230230*35 = 8,058,050

Adding these up gives 13,116,026. I think.

A bit of reading around leads me to learning about binomial(n+k-1,k) of choosing (with replacement) k objects from n.
So 3 vowels is binomial(7,3) * binomial(26,6) = 8058050
4 vowels is binomial(8,4) * binomial(25,5) = 3719100
5 vowels is binomial(9,5) * binomial(24,4) = 1338876

which added together gives us 13,116,026.

So that's what it would be if there were at least 6 of every consonant and at least 5 of every vowel in their respective piles (i.e. that count includes stuff like UUUZZZZZZ, which can't happen.)

I'm not sure now how to count those though, and remove them.

And I'm still unclear as to whether we will end up with MORE or FEWER possible letters games than numbers games. Isn't this exciting?

Edit: Ha, brilliant Robert. Nice to have that confirmed anyway! Any ideas on how we deal with the actual distribution (if we can find it), and do you have a hunch for the difference it will make to the figure?

The other thing of course is that even with the correct distribution of letters, I believe in reality it's still not a completely random selection because somebody sifts through to make sure there aren't too many of the same letter in a row and that sort of thing. But we should ignore that for this exercise.

Is this open to pure guesswork? If it is, my guess is 6,031,769.

http://www.c4countdown.co.uk/viewtopic. ... 82#p140469 is probably our best shot at a decent consonant distribution.
That's 1 JKQXYZ, 2 BFHVW, 3 C, 4 GMP, 5 L and 6+ DNRST.

As for vowels, it doesn't matter as there's going to be at least 5 of each, so Rob's numbers are fine.

Thomas Carey wrote:http://www.c4countdown.co.uk/viewtopic. ... 82#p140469 is probably our best shot at a decent consonant distribution.
That's 1 JKQXYZ, 2 BFHVW, 3 C, 4 GMP, 5 L and 6+ DNRST.

As for vowels, it doesn't matter as there's going to be at least 5 of each, so Rob's numbers are fine.

Cool, cheers. Erm... now how do we absorb this information into the calculations?

JimBentley wrote:Is this open to pure guesswork? If it is, my guess is 6,031,769.

Hmm. I'm saying 6,031,770.

Used this to work out how to determine the number of possible consonant combinations: http://mathforum.org/library/drmath/view/56197.html. It turns out the way to choose n consonants is to take the coefficient of x^n in the expansion of (1+x)^6*(1+x+x^2)^5*(1+x+x^2+x^3)*(1+x+x^2+x^3+x^4)^3*(1+x+x^2+x^3+x^4+x^5)*(1+x+x^2+x^3+x^4+x^5+x^6)^5. The first term is because there are 6 letters with 1 tile, the second is 5 letters with 2 tiles, etc. This gives 9149 ways of getting 4 tiles, 41670 for 5, and 161415 for 6. Combining with the ways of choosing vowels gives a final answer of 9,719,199.

Thomas Cappleman wrote:Used this to work out how to determine the number of possible consonant combinations: http://mathforum.org/library/drmath/view/56197.html. It turns out the way to choose n consonants is to take the coefficient of x^n in the expansion of (1+x)^6*(1+x+x^2)^5*(1+x+x^2+x^3)*(1+x+x^2+x^3+x^4)^3*(1+x+x^2+x^3+x^4+x^5)*(1+x+x^2+x^3+x^4+x^5+x^6)^5. The first term is because there are 6 letters with 1 tile, the second is 5 letters with 2 tiles, etc. This gives 9149 ways of getting 4 tiles, 41670 for 5, and 161415 for 6. Combining with the ways of choosing vowels gives a final answer of 9,719,199.

I just wrote a program to work it out a different way (nothing to do with coefficients of polynomials) and got the same answer.

Great stuff guys (he says, abandoning the brute-forcer he was in the middle of writing).


So there are fewer letters games than numbers games. WHAT A DAY THIS HAS BEEN.

I love how this forum generally has taken off recently.

So based on this, and using an estimate of the number of valid conundrumable words (probably 10,000 or so), we should be able to work out how many different possible shows there can be.

Johnny Canuck wrote:So based on this, and using an estimate of the number of valid conundrumable words (probably 10,000 or so), we should be able to work out how many different possible shows there can be.

That will get more complicated again, as each letters round will remove a load of other possibilities for the later rounds (as the letters tiles get used up).

How many possible contestants are there?

Gavin Chipper wrote:How many possible contestants are there?

The population of the world is 7 billion, so 6,999,999,999 unless you somehow persuade Matt Bayfield to apply.

Graeme Cole wrote:

Gavin Chipper wrote:How many possible contestants are there?

The population of the world is 7 billion, so 6,999,999,999 unless you somehow persuade Matt Bayfield to apply.

6,999,999,998 until I graduate. Unless I flunk out. Then we're back up to 6,999,999,999.

Graeme Cole wrote:

Gavin Chipper wrote:How many possible contestants are there?

The population of the world is 7 billion, so 6,999,999,999 unless you somehow persuade Matt Bayfield to apply.

Can we have 'likes' back etc.? Seriously Matt though, apply now.

Graeme Cole wrote:

Gavin Chipper wrote:How many possible contestants are there?

The population of the world is 7 billion, so 6,999,999,999 unless you somehow persuade Matt Bayfield to apply.

Gavin Chipper wrote:

JimBentley wrote:Is this open to pure guesswork? If it is, my guess is 6,031,769.

Hmm. I'm saying 6,031,770.

Well done Gev, you were closer.

Jon Corby wrote:

Gavin Chipper wrote:

JimBentley wrote:Is this open to pure guesswork? If it is, my guess is 6,031,769.

Hmm. I'm saying 6,031,770.

Well done Gev, you were closer.

Yeah, but he won't get anywhere on Manic Miner with his guess.

JimBentley wrote:

Graeme Cole wrote:

Gavin Chipper wrote:How many possible contestants are there?

The population of the world is 7 billion, so 6,999,999,999 unless you somehow persuade Matt Bayfield to apply.

Can we have 'likes' back etc.? Seriously Matt though, apply now.

Kind of you to suggest it, Jim, but Graeme's right. I'm sure the last sort of applicant the Countdown team want is yet another university-educated white male ex-Scrabbler, especially one who has also played extensively on apterous.