Numbers Game Solved With One Number, Is It Possible?

[Gav Grant]

Hey fellow Countdowners, I'm new to the forum so go easy on me as I pose my first question regarding the now infamous "Numbers Game." Sitting here on the nightshift, watching re-runs of some of the classic Countdown episodes from years gone by on my shiny new mp4 player, as you do, and forgive me if this has been asked before but if, after choosing your six numbers and Carol presses her magic button, the random number lands on one of the selected numbers. Would it be fair to assume that one could therefore solve the sum with one number?

The mind boggles, oh well only another 6.5hrs to go

Kind Regardz,

Gavin

[Dan Vanniasingham]

The target generated is in the region of 100-999, so yes, it's possible for the target to be 100 with a 100 in the selection.

Unless it's actually 101-999 and my mind is playing tricks on me...

[Joseph Bolas]

Forgive me if this has been asked before but if, after choosing your six numbers and Carol presses her magic button, the random number lands on one of the selected numbers. Would it be fair to assume that one could therefore solve the sum with one number?

Welcome to the forum Gavin.

I think in the earlier years, you could've had 100 as a target, but I do believe that CECIL can now only generate numbers from 101-999.

The last time 101 actually came out in a numbers round was on the 4th September and there was a 100 and a 1 there, so I think you will find that the answer to your question is that a numbers game can only be solved with a minimum of two numbers, never one.

[Charlie Reams]

CECIL hasn't set 100 as the target in any of the 2061 numbers games available on the wiki, so statistically we can be about 90% sure that 100 is no longer a valid target.

[Martin Gardner]

100 has definitely been used before, in a Tony Vick game available on the Countdown Page, so it definitely was possible at one point.

[Damian E]

Cecil.........(pronounced SeeSill), is programmed to select ANY 3-figure number, so 100 is legitimate.

[Joseph Bolas]

100 has definitely been used before, in a Tony Vick game available on the Countdown Page, so it definitely was possible at one point.

I have had a look on the Coundown page but I can't find this game. Have you got a link to this game please?

[Charlie Reams]

I have had a look on the Coundown page but I can't find this game. Have you got a link to this game please?

I can't find it either, the nearest I could find was this: http://www.thecountdownpage.com/gotw-001001.htm

[Joseph Bolas]

If you go back to the older days of Calendar Countdown, when they used HENRY, the first wheel had a '0' on it and therefore I believe that you could've had '010', '025', '050', '075' and '100' etc as targets too.

[Paul Howe]

If a target of 101 was unattainable and there was a 100 in the selection the 100 could count as a single number "solution". Whether such a numbers game exists is unclear, I just spent a few minutes trying to make one and was unsuccessful.

Edit: have just read the thread and realised 100 is a legitimate target. I think the numbers generator on Countgen only does 101-999 unless Charlie changed it, to me that seems like a better target range.

[Charlie Reams]

Yep, CountGen does 101-999 only, although as you rightly say, it's still potentially worthwhile to declare anything as low as 91.

I have to say I think someone has changed CECIL without informing Damian, because 100 has not come up in any of the last 2484 numbers games, which makes it 93.7% likely that CECIL can't generate it. (Perhaps one of our resident statonerds can confirm my back-of-the-envelope calculation.)

[Damian E]

Charlie, i think the idea is, you change the programming on Apterous to incorporate the possibility of a 100 target, then i come out a week or two later and say it was a mistake and that 101 is actually the lowest. It's called messing you about.

Seriously, i do think that 100 is the lowest, but if its not been 'chosen' by Cecil in 2484 numbers games, it doesn't mean that much. If you were to list all the targets from 100 -999 that HAVE come out in that time, there must be dozens that have not been picked surely?

Why don't we have a small compeitition?

Pick a random number from 100 (or 101) to 999, and see who's comes out first, starting from tomorrow's show. I'll donate a good Countdown prize to the winner.

I'll take a guess at 774.

[Kirk Bevins]

Why don't we have a small compeitition?

Pick a random number from 100 (or 101) to 999, and see who's comes out first, starting from tomorrow's show. I'll donate a good Countdown prize to the winner.

I'll take a guess at 774.

Haha.

[Kirk Bevins]

If a target of 101 was unattainable and there was a 100 in the selection the 100 could count as a single number "solution". Whether such a numbers game exists is unclear, I just spent a few minutes trying to make one and was unsuccessful.

Well if we have a 100 in the selection and 101 to be impossible, we need the 5 small numbers to not be consecutive (since if you have say 2, 4, 6, 7, 9, then you can do 7-6 to make the 1). This means the only two plausible selections are: 100, 2, 4, 6, 8, 10 and 100, 1, 3, 5, 7, 9. Taking the first, it's clear 1 can be made by 6/(4+2) and taking the second it's clear that the 1 can be made by just, well, using the 1. So in a 1 large selection, it will be always possible to get 101 if there was a 100 in the selection sadly.

For four large, you will have 100, 50, 75, 25 and so 101 will always be possible too (100 + (50+25)/75).

For 2 large and 3 large, it's a bit trickier. An imaginary prize for anybody who can give me a selection with a 100 in with a target of 101 which is impossible to obtain.

[Charlie Reams]

Seriously, i do think that 100 is the lowest, but if its not been 'chosen' by Cecil in 2484 numbers games, it doesn't mean that much. If you were to list all the targets from 100 -999 that HAVE come out in that time, there must be dozens that have not been picked surely?

You're right, there are a handful of other targets that haven't come up (including 999) but it does seem odd that 100 is one of them.

Why don't we have a small compeitition?

Pick a random number from 100 (or 101) to 999, and see who's comes out first, starting from tomorrow's show. I'll donate a good Countdown prize to the winner.

I'll take a guess at 774.

I'll go for 774. Gut feeling.

[Kirk Bevins]

You're right, there are a handful of other targets that haven't come up (including 999) but it does seem odd that 100 is one of them.

I'm confused. http://www.c4countdown.co.uk/viewtopic. ... =999#p7025

[Charlie Reams]

You're right, there are a handful of other targets that haven't come up (including 999) but it does seem odd that 100 is one of them.

I'm confused. http://www.c4countdown.co.uk/viewtopic. ... =999#p7025

I only considered games which are on the wiki. That one isn't.

For 2 large and 3 large, it's a bit trickier. An imaginary prize for anybody who can give me a selection with a 100 in with a target of 101 which is impossible to obtain.

There are none. Do I still get a prize?

[Kirk Bevins]

For 2 large and 3 large, it's a bit trickier. An imaginary prize for anybody who can give me a selection with a 100 in with a target of 101 which is impossible to obtain.

There are none. Do I still get a prize?

No. Prove it.

[Charlie Reams]

No. Prove it.

Okay, proof for 2 large:

Say the selection is 100, N, a, b, c, d, where N is the other large number and a, b, c and d are the 4 small numbers. Rearrange the selection so that the small numbers are in ascending order, so we can assume that a is the smallest and d the largest. Clearly a must be at least 2 or we can do 100+1 regardless of the other numbers. As you said, the numbers can't be the equal or consecutive as that would give us the 1 trivially, so b must be at least 4. Likewise c must be at least 6 and d at least 8. Working the other way, d can't be more than 10 (it's a small number), so c can't be more than 8, b can't be more than 6 and a can't be more than 4. So we now have quite a reduced set of cases to consider:

Assume a = 2. b must be one of 4, 5 or 6, but in each of these cases then any of the three values of c (6, 7 or 8) allow us to easily generate the 1 without even considering d and N. So a can't be 2.
Assume a = 3. b can't be 4, and if b is 5 or 6 then again we can easily get the 1 just using a, b and c. So a can't be 3.
Assume a = 4. b can't be 4 or 5, so it must be 6, hence c can't be 6 or 7 so it must be 8, and hence d can't be 8 or 9 so it must be 10. So we have the selection 100, N, 4, 6, 8, 10, and 100+(4+6)/10 gives us the 101. So a can't be 4.

But we already said a would have to be no greater than 4, so that's all the cases, and it's impossible to find a selection where 101 can't be made.

It's interesting that we never have to consider cases for N.

The argument for 3 large is left as an exercise for the reader (it's the same but more tedious.)

[Martin Gardner]

I have had a look on the Coundown page but I can't find this game. Have you got a link to this game please?

I can't find it either, the nearest I could find was this: http://www.thecountdownpage.com/gotw-001001.htm

I can't remember how I found it but I did, if Mike Brown could read this I'm sure he could find it in a flash.

[Martin Gardner]

Ok found it, this seems to settle the argument: Link

The final numbers game, pity he didn't go for 4 large.

[Damian E]

Ok found it, this seems to settle the argument: Link

The final numbers game, pity he didn't go for 4 large.

You see Charlie, I told you

Well researched, Martin.

Am surprised though that nobody has yet spotted that Junaid Mubeen has been on before................

Cue the mass protests from Joseph Bolas. Oh how i love life.

[Charlie Reams]

Ok found it, this seems to settle the argument: Link

The final numbers game, pity he didn't go for 4 large.

You see Charlie, I told you

That was 12 years ago! Like I said above, I think it has been changed at some point.

Well researched, Martin.

It was actually me that found the page.

Am surprised though that nobody has yet spotted that Junaid Mubeen has been on before.:

Oh yes. Almost exactly ten years ago. Given that he can't be much over 20 now, he must've been pretty young back then. Seems fair enough to give him another shot, but I'm looking forward to the inevitable voices of dissent.

[Damian E]

I'll find out for definite and let you know.

Its not that crucial anyway is it, but even so its good to put it to rest. How did you research the 100 game if Martin posted it?

[Charlie Reams]

I'll find out for definite and let you know.

Its not that crucial anyway is it, but even so its good to put it to rest. How did you research the 100 game if Martin posted it?

Not at all, but it would be good to put it to rest. Martin and I were talking about it last night on MSN. He remembered that the large number was 50, although he misremembered the contestant being Tony Vick, but I eventually tracked it down with the awesome power of Google.

[Chris Corby]

Am surprised though that nobody has yet spotted that Junaid Mubeen has been on before................

I think he may well be the singing love child of host Des. I am sure I can remember "Nothing Compares 2 U" by Junaid O' Connor.............

[Gavin Chipper]

Yep, CountGen does 101-999 only, although as you rightly say, it's still potentially worthwhile to declare anything as low as 91.

I have to say I think someone has changed CECIL without informing Damian, because 100 has not come up in any of the last 2484 numbers games, which makes it 93.7% likely that CECIL can't generate it. (Perhaps one of our resident statonerds can confirm my back-of-the-envelope calculation.)

100-999 is 900 numbers so if 100 is available then the chances of it not appearing in 2484 consecutive games is (899/900)^2484 = 0.063. So doing 1 minus that and multiplying by 100 you get your 93.7%. Not significant at the 5% level even! However, the way you've worded it seems to be a bit of a misuse of statistics. All we can really say is that in a situation where 100 is possible, then over the course of 2484 games there is a 93.7% chance that it will come up. You wouldn't say after three heads in a row that there is a 7/8 chance (or 3/4 using a two-tailed (as opposed to two headed) test I suppose) that it's a doubled-headed coin.

But anyway, it's not that statistically freaky so I see no reason to doubt Damian on this one. Oh, Damian - yes, I see.

If a target of 101 was unattainable and there was a 100 in the selection the 100 could count as a single number "solution". Whether such a numbers game exists is unclear, I just spent a few minutes trying to make one and was unsuccessful.

Obviously this and the following discussion may have assumed perfect contestants, but 100 could still be offered as a single number "solution" if the contestant simply failed to see how to get 101.

[Paul Howe]

If a target of 101 was unattainable and there was a 100 in the selection the 100 could count as a single number "solution". Whether such a numbers game exists is unclear, I just spent a few minutes trying to make one and was unsuccessful.

Obviously this and the following discussion may have assumed perfect contestants, but 100 could still be offered as a single number "solution" if the contestant simply failed to see how to get 101.

I meant that a solution is any method that gets you as close as possible to the target, not just anything the contestants happen to offer.

[Damian E]

Excellent explanation, Gavin.
I wouldn't have put it better myself.


I just wouldn't.

[Peter Dunwoody]

Slightly unrelated to whether the target is attainable with one number, but yes I have seen a game where 100 has been the target. It was a CoC final (either 7 or 8, I can't remember which), however the large number in the selection was 50.

Is it possible that maybe if 100 was in the selection and was also generated as the target that the sequence would be reshot? Which might explain it so rarely appearing...

[Ian Volante]

Cecil.........(pronounced SeeSill), is programmed to select ANY 3-figure number, so 100 is legitimate.

But Richard always pronounced it Cecil! Is this humour? I never can tell.

[Johnny Canuck]

http://wiki.apterous.org/Episode_1504

[Soph K]

If, for example, the target was 100 and one of the large numbers was 100, could the method just be 100 or would you have to do, say, 9-8=1 and 4-3=1 and 100+1-1=100? If you know what I mean...?

[Rhys Benjamin]

If, for example, the target was 100 and one of the large numbers was 100, could the method just be 100 or would you have to do, say, 9-8=1 and 4-3=1 and 100+1-1=100? If you know what I mean...?

RW:
I'll give you 10 points anyway, but can you find another way?"