c4countdown

Is the target number completely random?! Why do I ask this? Because I don't know the target that haven't a solution.

It is random. CECIL (the generator yoke) gives a random target. All Rachel has to do is push the button. (And work it out of course!)
It does give unsolvable (is that a word?) solutions, but that is quite rare and any that have eluded her are usually solved by the brains of our Countdown Forumites!

Eoin Monaghan wrote:It is random. CECIL (the generator yoke) gives a random target. All Rachel has to do is push the button. (And work it out of course!)
It does give unsolvable (is that a word?) solutions, but that is quite rare and any that have eluded her are usually solved by the brains of our Countdown Forumites!

With alternative solutions provided by Mark K.

Dmitry Goretsky wrote:Is the target number completely random?! Why do I ask this? Because I don't know the target that haven't a solution.

My first game had these numbers for the third numbers game.

2, 2, 50, 75, 100, 25. Target: 720.

721 is as close as you can get, there are some six small solutions where it is impossible to get within a hundred of some solutions.

David O'Donnell wrote:
...there are some six small solutions where it is impossible to get within a hundred of some solutions.

I fear the day when 1, 1, 2, 2, 3, 3 happens.

Jordan F wrote: I fear the day when 1, 1, 2, 2, 3, 3 happens.

I estimate that if 6 small was chosen once per game, this would happen about once every 150 years on average.

No, it's completely pseudo-random.

Ray Folwell wrote:

Jordan F wrote: I fear the day when 1, 1, 2, 2, 3, 3 happens.

I estimate that if 6 small numbers were chosen once per game, this would happen about once every 150 years on average.

The number of different combinations of 6 numbers from 20 is given by 20 ! / (14 ! x 6 !) or 38,760.
Combination 1, 1, 2, 2, 3, 3 is one of them, so the odds of this occurring are 38,759 to 1 against.

This means that if 6 small numbers were chosen once per show, the probability would be 1 in 7,752 in every week. This would work out as once int 172 years, assuming 45 weeks or 225 shows per year.
(Assuming 52 weeks or 260 episodes per year, we get a 'once in 149 years' result, like Ray's.)

The exclamation marks in 20 ! , 14 ! and 6 ! are not expressions of surprise. They are factorials, defined as follows:

1 ! = 1
2 ! = 2 x 1 = 2
3 ! = 3 x 2 x 1 = 6
4 ! = 4 x 3 x 2 x 1 = 24

and so on ....

Just to add to Mark's post, the x isn't him offering his affection in the form of (several!) kisses, but in fact is a symbol meaning 'multiply'. It's a bit like a short-hand for lots of addition, rather than writing out (say) 3 + 3 + 3 + 3 + 3, you can instead just write 3 x 5.

Michael Wallace wrote:Just to add to Mark's post, the x isn't him offering his affection in the form of (several!) kisses, but in fact is a symbol meaning 'multiply'. It's a bit like a short-hand for lots of addition, rather than writing out (say) 3 + 3 + 3 + 3 + 3, you can instead just write 3 x 5.

lol !

Further to my first post on combinations, it must be remembered that 1, 1, 2, 2, 3, 3 can only be made in one way, so the probability of it occurring on a '6 small' game is 1 in 38760.

A group like 1, 1, 2, 2, 3, 4 is easier to achieve, as there are two 3s and two 4s in the small number pack.
Combining these options means that 1, 1, 2, 2, 3, 4 is four times easier to obtain, i.e. a probability of 1 in 9690. This is because we aren't bothered as to which of the two 3s or 4s we're after.

An 'all-different' combination like 1, 2, 4, 5, 7, 8 is similarly 64 times easier, giving an 'easier' probability of 1 in 606.

There are only two of each small number in the pack. Don't know if you knew that, but I think it might require a rejig of the calculations.

JackHurst wrote:There are only two of each small number in the pack. Don't know if you knew that, but I think it might require a rejig of the calculations.

I think the calculations are correct - 2 of each number 1-10 = 20 in total.
It easier to think about if you imagine that one set of small numbers is blue and the other red - then there is only one way to get 1,1,2,2,3,3 i.e. 1R,1B,2R,2B,3R,3B but 4 ways to get 1,1,2,2,3,4 (3R,4R; 3R,4B; 3B,4R; 3B,4B).

Do we have any statistics on how often contestants choose 6 small or any of the other combinations ?

Mark Kudlowski wrote:The number of different combinations of 6 numbers from 20 is given by 20 ! / (14 ! x 6 !) or 38,760.
Combination 1, 1, 2, 2, 3, 3 is one of them, so the odds of this occurring are 38,759 to 1 against.

That's ABSOLUTELY RIGHT! I just recognized it when I read a "colored tile' message.

Dmitry Goretsky wrote:

Mark Kudlowski wrote:The number of different combinations of 6 numbers from 20 is given by 20 ! / (14 ! x 6 !) or 38,760.
Combination 1, 1, 2, 2, 3, 3 is one of them, so the odds of this occurring are 38,759 to 1 against.

You forgot to divide this by (2!)^3! This is because there are two 1s, two 2s and two 3s. So the probability of THIS occuring is 1/(C(20,6)/((2!)^3))=1/4,845

Probability genius my hairy pods.

Dmitry Goretsky wrote:

Mark Kudlowski wrote:The number of different combinations of 6 numbers from 20 is given by 20 ! / (14 ! x 6 !) or 38,760.
Combination 1, 1, 2, 2, 3, 3 is one of them, so the odds of this occurring are 38,759 to 1 against.

You forgot to divide this by (2!)^3! This is because there are two 1s, two 2s and two 3s. So the probability of THIS occuring is 1/(C(20,6)/((2!)^3))=1/4,845

I think you are probability not a probably genius.

Michael Wallace wrote:

Dmitry Goretsky wrote:

Mark Kudlowski wrote:The number of different combinations of 6 numbers from 20 is given by 20 ! / (14 ! x 6 !) or 38,760.
Combination 1, 1, 2, 2, 3, 3 is one of them, so the odds of this occurring are 38,759 to 1 against.

You forgot to divide this by (2!)^3! This is because there are two 1s, two 2s and two 3s. So the probability of THIS occuring is 1/(C(20,6)/((2!)^3))=1/4,845

I think you are probability not a probably genius.

Sorry, I was wrong!

Jon Corby wrote:

Dmitry Goretsky wrote:

Mark Kudlowski wrote:The number of different combinations of 6 numbers from 20 is given by 20 ! / (14 ! x 6 !) or 38,760.
Combination 1, 1, 2, 2, 3, 3 is one of them, so the odds of this occurring are 38,759 to 1 against.

You forgot to divide this by (2!)^3! This is because there are two 1s, two 2s and two 3s. So the probability of THIS occuring is 1/(C(20,6)/((2!)^3))=1/4,845

Probability genius my hairy pods.

Sorry, I was wrong!

Dmitry Goretsky wrote: Sorry, I was wrong!

About being a probability genius? Yeah, we guessed

Jon Corby wrote:

Dmitry Goretsky wrote: Sorry, I was wrong!

About being a probability genius? Yeah, we guessed

No, only about THIS probability! I just recognized it when I read a "colored tile' message. So PM me if you have any problems with probabilities. Genius never makes an error twice

Dmitry Goretsky wrote:Sorry, I was wrong!

Dmitry Goretsky wrote:Sorry, I was wrong!

I like the double post. I presume it's some sort of reference to getting tails in the Sleeping Beauty Problem.

Gavin Chipper wrote:

Dmitry Goretsky wrote:Sorry, I was wrong!

Dmitry Goretsky wrote:Sorry, I was wrong!

I like the double post. I presume it's some sort of reference to getting tails in the Sleeping Beauty Problem.

Maybe the second "I was wrong" was referring to the first one. So maybe he's saying he wasn't wrong in the first place.

Howard Somerset wrote:

Gavin Chipper wrote:

Dmitry Goretsky wrote:Sorry, I was wrong!

Dmitry Goretsky wrote:Sorry, I was wrong!

I like the double post. I presume it's some sort of reference to getting tails in the Sleeping Beauty Problem.

Maybe the second "I was wrong" was referring to the first one. So maybe he's saying he wasn't wrong in the first place.

No, I said "Sorry, I was wrong!" to each of the two users

Dmitry Goretsky wrote:I'm a probability genius

Since it's very rare for things such as this to have a probability of exactly 0 or exactly 1, what do you rate is the probability that you really are a probability genius, in view of the fact that you've admitted to getting things wrong earlier in this thread?

Howard Somerset wrote:

Dmitry Goretsky wrote:I'm a probability genius

Since it's very rare for things such as this to have a probability of exactly 0 or exactly 1, what do you rate is the probability that you really are a probability genius, in view of the fact that you've admitted to getting things wrong earlier in this thread?

0.9

Dmitry Goretsky wrote:

Howard Somerset wrote:

Dmitry Goretsky wrote:I'm a probability genius

Since it's very rare for things such as this to have a probability of exactly 0 or exactly 1, what do you rate is the probability that you really are a probability genius, in view of the fact that you've admitted to getting things wrong earlier in this thread?

0.9

So do I take it that you think that there's a 10% chance that you're wrong when you give an answer of 0.9?

Howard Somerset wrote:Since it's very rare for things such as this to have a probability of exactly 0 or exactly 1,

Meh, I'd've given the probability of that claim being shot down in flames as 1...

Michael Wallace wrote:Just to add to Mark's post, the x isn't him offering his affection in the form of (several!) kisses, but in fact is a symbol meaning 'multiply'. It's a bit like a short-hand for lots of addition, rather than writing out (say) 3 + 3 + 3 + 3 + 3, you can instead just write 3 x 5.

You bastard! I thought I'd pulled Wait, did I just inadvertently multiply your insult?
xxx < Kisses for you Raccoon just for putting me straight, and don't tell CF