Entertainment For Lively Minds
Take a Chance on Me...
Posted by Fraser M on 6 April 2011 - 11:45am.
Not an ABBA thread - any experts on probability?
This morning, I took part in an online BBC experiment about risk. If you fancy doing it yourself, the link is: https://www.bbc.co.uk/labuk/experiments/risk/
If you're planning on taking part yourself, don't read the comments, as I give one of the answers away.
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SPOILERS AHOY
I answered 3. They claim the answer is 6.
Now, I understand that when a die is rolled, the probability of it coming up with any given number is 1 in 6.
But my feeling is that if you have a large group of children all throwing a fair die, some will roll a six first time, some second, some third, some forth, some fifth, some sixth and so on but with a sufficiently large group, that number should even out to roughly equal numbers getting a six on each of the rolls, ergo, it should appear somewhere about the middle, hence I chose 3.
Possibly the confounding factor is in the fact that there's no requirement that they need roll a six within the first six rolls, but can you help me understand this?
Hopefully this is actually quite easy, and not as brain-hurty as the Monty Hall Problem...
3
Was my answer too. At first I put 6, but the use of the word 'average' in the question threw me and made me reconsider. There are a few of the questions where the wording seems designed to throw you a bit. That's probably also part of the test, seeing if you just go with your gut or think about things.
Just done it
And that's exactly the one I supposedly got wrong too, also answering three (or four, if it had been an option - the midpoint between 1 and 6, anyway).
Then I thought about it again.
One of every group of six children would be expected to throw any given number - including one - on the first throw. So (assuming the sample size was large enough, as you say), there would be even distribution of those throwing each of the six possible outcomes. On a graph, a straight line. So one sixth of the kids would throw each of the numbers - again, including one -on their first/second/third/etc. throw.
Now what was the question again? "On average, how many throws do they take to throw a one?" Well, half of them took one, two or three throws, and the other half took four, five or six. So the average is whatever number is bang in the middle: 3.5.
Conclusion: I still don't understand why that answer is "wrong". If the odds of something happening are 1 in 6, that doesn't mean you'll take six attempts to do it. It means that it will definitely happen if six attempts are made. But it's just as likely to happen on the first attempt as it is on the sixth - dice have no memory - so surely the correct answer is the mean value between the two extremes. Innit? Or is it?
Where's Joe "Maths Boy" Rivers when we need him?
Where am I?
See below.
No, it doesn't. In this case, every single event (dice roll) is independent. If you think about it, each number has a 1/6 chance of appearing when you roll a dice. By your logic, if you rolled 6 times, you'd be guaranteed to get one of each number, which isn't the case. So, to roll a 1 here, it's quite conceivable you'd need more than 6 throws.
Again, not quite. If it happens on the first attempt, game over. For it to happen on the third attempt, it also has to have NOT happened on both the first and second attempts. As an isolated event, the third roll is just as likely to be a 1, but you've got to get to the third roll before you get there. There's a probability of 1 you'll make it to the first roll, but only a probability of 25/36 you'll make it to the third roll.
A single photon of light peeps through the low'ring cloud
So, what you're saying is that one-in-six isn't the certainty, but the average. It could take quite a few more than six throws, yes, I can see that. I think.
But I still don't get how you can quantify the top end of how many throws it'd take for 100% of the kids to have thrown a 1.
And this is where I get even more confused. Shouldn't the chance of throwing a number x on the n'th throw be one in six by one in six, i.e. 36:1? In that case, why isn't the right answer the median between 1 and 36, i.e. 18 throws?
(Yes, I know. Pitiful. But I was in the arts stream me.)
Attempts
Isn't another option that it takes more than six attempts?
(I haven't done the test, btw)
Just redone the quiz and
here are all the choices:
1
2
3
6
12
Given Joe's explanations, which are indeed truly head-hurty (that Wikipedia article whistles over my bonce), does these potential responses affect which one should pick?
Oh, and I realised I discussed the children throwing a 'six' in my attempt to explain my thinking in my first post in the comments, when it should have been about them throwing a 'one' (not that it makes any substantial difference). Apologies.
I'm not 100% sure
but I'll give it a bash.
The odds of you getting a 1 on your n-th throw (and not before) is:
((5/6)^(n-1))*(1/6)
That is, you throw "not a 1" (the odds of which are 5/6) on every throw until the n-th one; the odds of which are 1/6.
It can take you more than 6 goes to get a 1; the odds of you throwing a 1 first time might well be 1/6, but the odds of you doing it second time aren't: they're 5/6 * 1/6 = 5/36.
Here is where my theory starts to fall down...
If you multiply number of throws (n) by the odds of that number of throws being the number you require until getting a 1, then you see that n=6 gives you the highest value (actually, it's the same with n=5). However, I can't see why the answer is definitively 6 and not 5, and I'm not sure why you'd multiply the formula by n. Clearly, my degree is going to good use here.
Can I just say that I'm almost certain there's a much better - and less complicated - explanation for this.
Incidentally, I love The Monty Hall Problem; it's one of my very favourite things in all of maths.
EDIT: Dredging up some old memories has made me realise it's a Binomial distribution. Scroll down to the section on Probability Mass Function. The problem you describe is a special case where k=1 (and n=number of throws, p=5/6, i.e. the probability of not throwing a 1). Doesn't explain why the answer is 6 rather than 5 though!
I fear probability.
It's one of the least naturally intuitive things in the world. We seem wired to want outcomes to be related to each other even when they're not: there's a big part of our brains that thinks the roll of one die could affect the result of the next, completely different die, and it's quite hard to turn that off.
It's like the bit of us that goes FUCKING PENCIL when it falls off the table: we seem to want to invest everything with motives and intentions.
Also explains why we still play the lottery, despite it being significantly more likely that you will die on the way to the shop to buy the ticket than win the jackpot.
I'm not sure of the maths here
But somebody once said to me that if you live in London you have better odds on shagging the Queen than you do on winning the Lottery.
Windsor
And she gives you a scratchcard afterwards, so it's win-win for us Londoners.
I'm not a statistician, but..
Imagine an infinite number of kids, each with a die they keep rolling. They stop when they roll a one. After six throws, a goodly proportion will have rolled a one. After twelve throws, almost all will have rolled one. And so on, but will continue to be some kids rolling away, waiting for a one to come up. The graph tails away into infinity, skewing the average to six throws rather than five.
Or something.
That was a very interesting test.
Apparently I'm goodish at gauging risk, but am deeply unethical. Sounds about right.
Surely
the right combination for a con man!
Have you considered a career change?
Do you need to?
Another experiment on that site
Which might be of interest to the average Word person.
https://www.bbc.co.uk/labuk/experiments/musicality/
I've (sort of) got it
Assume that you'll never take more than 50 throws to roll a 1. The chance of it happening on your 50th throw are around 1/1000) anyway.
For each number, 1 to 50, apply the formula, ((5/6)^(n-1))*(1/6) and then multiply each of these numbers by 100. This is the equivalent of repeating the test 100 times, and how many of those times you'd roll a 1 on your n-th go. You'll find you roll a 1 first go approx. 17 times, second go approx. 14 times, third go approx. 12 times, and so it continues.
Then you need to find the average of these numbers, which is ((17*1)+(14*2)+(12*3)+...)/100. If you do this using Excel - as I have - you get an answer around the 5.99 mark. As probabilities for n>50 are so minuscule, you can assume (making a bit of a leap of faith) that if n was infinity, the average would be exactly 6.
Yes, counter-intuitive and yes, not the best explanation, but I think - with a little push in the right direction from Lenny and Archie - I got there.
Thanks Joe!
That's a long way past my intuitive grasp of probability, and though I can understand what you're saying, I fear that faced with a similar question I would get it wrong again!
However, I will decide to change which door I picked now you've shown me the goat. Thanks.
Nobody else
posting results?
My biggest risk area is - Social: Would you attempt a stand-up comedy routine, or confront a stranger in a crowd?
Everything else was low, apart from ethical.
My risk knowledge score was 81%
My risk numeracy is 66% - which is about 65% more than my 1st year university stats exam results
On the wheely thing, I chose the risky on 66% of the time up to the rigged one, then it dropped to 33%
Am I any wiser?
Nope, but it was fun!
Ok, here goes
...my risk knowledge was 73% and my risk numeracy 86%. My weakness was ethical. I WOULD sleep with a married woman. She just wouldn't sleep with me.
Oh, did it mean someone ELSE's wife? Oops.
On the music quiz
...I guess I'm an enthusiastic amateur.
Big fat zero for musical ability. "You scored low for perception, which suggests that you don’t have a strong ear for music. You are likely to find it challenging to keep a beat or sing in tune, and you may also have difficulty in distinguishing between different instruments or musical genres. Your scores in the musical tests will give you a good idea of where your specific strengths and weaknesses lie."
Not sure how that fits with my high score on social risk taking. And propensity for alcohol.
Oh dear.