Countdown probability problem
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- Martin Gardner
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Countdown probability problem
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.
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.
Last edited by Martin Gardner on Thu Mar 05, 2009 8:58 pm, edited 1 time in total.
If you cut a gandiseeg in half, do you get two gandiseegs or two halves of a gandiseeg?
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Re: Countdown probability problem
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...
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...
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Re: Countdown probability problem
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.
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Re: Countdown probability problem
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.Simon Myers wrote: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...
If you cut a gandiseeg in half, do you get two gandiseegs or two halves of a gandiseeg?
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Re: Countdown probability problem
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 wrote: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.
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Re: Countdown probability problem
I took logs of each side.Michael Wallace wrote: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 wrote: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.
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.
If you cut a gandiseeg in half, do you get two gandiseegs or two halves of a gandiseeg?
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Re: Countdown probability problem
Or just a^b = c => a = c^(1/b)...
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Re: Countdown probability problem
Hmm, so why the fcuk didn't they teach us that instead?Michael Wallace wrote:Or just a^b = c => a = c^(1/b)...
If you cut a gandiseeg in half, do you get two gandiseegs or two halves of a gandiseeg?
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Re: Countdown probability problem
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.Martin Gardner wrote:Hmm, so why the fcuk didn't they teach us that instead?Michael Wallace wrote:Or just a^b = c => a = c^(1/b)...
Last edited by Michael Wallace on Thu Mar 05, 2009 10:59 pm, edited 1 time in total.
Re: Countdown probability problem
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.
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Re: Countdown probability problem
They taught you to use logs for doing a slightly different question.Martin Gardner wrote:Hmm, so why the fcuk didn't they teach us that instead?Michael Wallace wrote:Or just a^b = c => a = c^(1/b)...
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.
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Re: Countdown probability problem
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.David Roe wrote: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.
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Re: Countdown probability problem
Logs is currently C2 (ie. AS level).
16/10/2007 - Episode 4460
Dinos Sfyris 76 - 78 Dorian Lidell
Proof that even idiots can get well and truly mainwheeled.
Dinos Sfyris 76 - 78 Dorian Lidell
Proof that even idiots can get well and truly mainwheeled.
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Re: Countdown probability problem
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.Martin Gardner wrote: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.
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Re: Countdown probability problem
Or from the discussion in the spoiler thread!Martin Gardner wrote:I just came up with this one at random.
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Re: Countdown probability problem
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.Neil Zussman wrote: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.Martin Gardner wrote: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.