Countdown probability problem

[Martin Gardner]

I just came up with this one at random. You'll have difficulty with this if you don't have maths up to AS-Level (like me) or you're just naturally good at maths (like me 10 years ago).

Player A is playing Countdown. The probability of player A finding the max for all types of round; letters, numbers and conundrums alike is x, which is somewhat obviously a value between 0 and 1. What value does x have to take so that player A maxes exactly half of his games?

What is the value of x for nine rounds?

Edit: What about 15 rounds?

Excuse me if the wording is not perfect, I'm not used to setting this sort of puzzle.

[Simon Myers]

Only spent a minute or two thinking about this, and I'm presuming you mean "what is the lowest such x?"

In that case I suppose it's x^(number of rounds) > 0.5

So that for a nine round game, about 0.926 so that 0.926^9 = 0.50609. Or put into words, one would have to max ~93% of rounds to max around half of nine round games. I think that's right...

[Michael Wallace]

I presume you mean "What value of x means that a player would expect to max half of his games" (you've used the word 'statistically' in an odd way). And yeah, Simon has pretty much answered it - your probability of maxing a game is just x^(number of rounds) = p, say, and then that's just the proportion of your games you'd expect to max.

[Martin Gardner]

Only spent a minute or two thinking about this, and I'm presuming you mean "what is the lowest such x?"

In that case I suppose it's x^(number of rounds) > 0.5

So that for a nine round game, about 0.926 so that 0.926^9 = 0.50609. Or put into words, one would have to max ~93% of rounds to max around half of nine round games. I think that's right...

Yes if you say exactly half of his games. And yes, but how do you work it out? There is a nice little formula that works it out perfectly.

[Michael Wallace]

Yes if you say exactly half of his games. And yes, but how do you work it out? There is a nice little formula that works it out perfectly.

A 'nice little formula'? To solve a^b = c? I fail to see what more needs to be done than just rearranging it like any other equation.

[Martin Gardner]

Yes if you say exactly half of his games. And yes, but how do you work it out? There is a nice little formula that works it out perfectly.

A 'nice little formula'? To solve a^b = c? I fail to see what more needs to be done than just rearranging it like any other equation.

I took logs of each side.

Edit: having said that, I might as well do it now.

There's a nice little log rule that says log a^b = b * log a. This now turns b, the power, into a coefficient. So:

9(log x) = log 0.5

Rearrange:

log x = (log 0.5)/9

Then you do 10^(log x) and that gives you 0.925874712

To check this, do 0.925874712^9 = 0.5. Perfecto.

Obviously you just replace the nine with however many rounds you want, and that gives you a definitive answer. So 15 is 0.954841603. In other words you need to go from 92.6% of rounds maxed to 95.5% in order to keep up the rate of maxing one game out of two when moving from 9 rounds to 15.

[Michael Wallace]

Or just a^b = c => a = c^(1/b)...

[Martin Gardner]

Or just a^b = c => a = c^(1/b)...

Hmm, so why the fcuk didn't they teach us that instead?

[Michael Wallace]

Or just a^b = c => a = c^(1/b)...

Hmm, so why the fcuk didn't they teach us that instead?

When did you do GCSE maths? That was on the syllabus when I did mine (2000), and I'm pretty sure it's still on it now.

[David Roe]

If you're over 45, you couldn't work out the 15th root of anything unless you had access to a mainframe computer. If you're under 45, the powers that be didn't realise pocket calculators had been invented, so carried on teaching logs.

[Howard Somerset]

Or just a^b = c => a = c^(1/b)...

Hmm, so why the fcuk didn't they teach us that instead?

They taught you to use logs for doing a slightly different question.

In your case you are trying to solve for x the equation x^b = c, when you know b and c. As has already been pointed out the solution is simply x = c^(1/b).

It is when it's the power that you're trying to find that taking logs is an appropriate method. For example trying to solve for x the equation a^x = b, when you know a and b. In this case, by taking logs you get
log (a^x) = log b
x log a = log b
and hence x = (log b)/(log a)

To answer a question raised in this thread, the technique is now in the AS level maths syllabus, in module C2 for most courses.

[Howard Somerset]

If you're over 45, you couldn't work out the 15th root of anything unless you had access to a mainframe computer. If you're under 45, the powers that be didn't realise pocket calculators had been invented, so carried on teaching logs.

If you're over a slightly older age, when mainframe computers were a thing of the future, you would work out the 15th root of something using well thumbed log tables, as I used to do at school.

[Kai Laddiman]

Logs is currently C2 (ie. AS level).

[Neil Zussman]

In other words you need to go from 92.6% of rounds maxed to 95.5% in order to keep up the rate of maxing one game out of two when moving from 9 rounds to 15.

It's surprising (to me at least) that you don't have to be *that* much better to max 15 than 9. Although I suppose once you're at a 92.6% level, it would be hard to improve much more.

[Gavin Chipper]

I just came up with this one at random.

Or from the discussion in the spoiler thread!

[Charlie Reams]

In other words you need to go from 92.6% of rounds maxed to 95.5% in order to keep up the rate of maxing one game out of two when moving from 9 rounds to 15.

It's surprising (to me at least) that you don't have to be *that* much better to max 15 than 9. Although I suppose once you're at a 92.6% level, it would be hard to improve much more.

Yeah, that's quite a significant improvement; you need to go from 7.4% missed maxes to 4.5%, which is quite a big change. Even then the underlying assumption is basically wrong: the 15-round format has a higher proportion of letters games than the 9-round format, and on average it's harder to max letters games than numbers game. And that's even before you talk about the added pressure in a 15-rounder as you approach the max game simply because the game is longer, and other such psychological factors.