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Topic: counter-intuitive fact from everyday mathematics
Replies: 17   Last Post: Jun 24, 2012 10:11 AM

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Posts: 92
Registered: 6/21/12
Re: counter-intuitive fact from everyday mathematics
Posted: Jun 21, 2012 11:13 AM
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On Thursday, June 21, 2012 10:10:17 AM UTC-4, (unknown) wrote:
> On Thursday, June 21, 2012 12:15:43 PM UTC+1, scattered wrote:
> > On Thursday, June 21, 2012 4:59:47 AM UTC-4, Paul wrote:
> > > There are a few iconic "surprising facts" from elementary mathematics
> > > which are often talked about at high-school level. A few that spring
> > > to mind are the birthday stuff, the let-the-penny-double (or any power-
> > > of-2 equivalent), and the fact that if you shuffle a set there's a
> > > probability of 1-1/e that at least one element in the shuffled set is
> > > in the correct position. (This is referenced as shuffling cards, or
> > > an absent-minded postman etc.)
> > >
> > > I'd like to add my own example to this famous list. I wouldn't be
> > > terribly surprised if someone has already discussed it, but I've
> > > certainly never heard it referenced.
> > >
> > > If you roll a pair of unbiased dice 500 times, the probability that
> > > you never get a double six is less than 1 in 1.3 million.
> > >
> > > I think that 1) To the "person in the street", this is even more
> > > surprising than anything in the standard list. I think most people
> > > would be delighted to spend a dollar on rolling a pair of dice 500
> > > times to be rewarded with a million dollars if they miss 66 each
> > > time. But this would not be good odds.
> > >
> > > 2) This fact has the advantage of being true both theoretically and
> > > practically, with almost no necessity for discussion about the extent
> > > to which real-world practice matches theory.
> > > In that sense, it is a "cleaner" example than anything in the standard
> > > list above. Everything in the standard list has possible real-world
> > > objections. With the birthday stuff, it's a debatable assumption that
> > > births are uniformly distributed throughout the year. With the
> > > shuffling stuff, approximation is used and other assumptions made (the
> > > answer is actually rational, not 1 - 1/e). The power-of-2 stuff often
> > > raises issues of meaning, depending on how large the power of 2. (You
> > > can't literally fold a piece of paper 1000 times. It's unclear what
> > > it means to talk about a sum of money of £10^50 etc.)
> > >
> > >
> > > Did I come up with a decent example? Or is it something that's been
> > > given many times before without me being aware of it?
> > >
> > > Thank You,
> > >
> > > Paul Epstein

> >
> > I fail to see how the result is counterintuitive. If anything, what is counterintuitive is that the probabilty is as high as it is. If a person has any experience at all with games involving dice then they would know that a double 6 is not all that rare. A typical game of Monopoly would feature several double sixes over the course of the game. 500 rolls without a double six would clearly be a run of luck of epic proportions. I can't speak for the mythical "person on the street" but I would imagine that most people wouldn't be surprised that rolling 500 pairs of dice and never getting a double six is somewhat akin to winning a lottery.

> Here's one that's interesting (to me). And I don't actually know the answer.
> People keep doing the 500-dice-roll game until eventually someone wins the game and gets a 66-free 500 roll set.
> What is the probability that this winner got a total number of 6's among all 1000 dice that was less than expected (in other words 166 or less)?
> It's an interesting one because, without bothering to think about it much, the answer is obviously > 50% but I really haven't got much of a clue beyond that. Nothing inside the 60% to 99.999999% range would particularly surprise me.
> Paul Epstein

That is an interesting question. My hunch is that it is closer to the 99% end of the range. Concentrate on the second die in the experiment of 500 rolls. It would be expected to come up 6 roughly 83 times. On average 83/6 = 14 of those times would be when the first die is 1, 14 of those times when the first die is 2, ..., 14 times when the first die is 6. If there are no double sixes, then on average it seems that the second die will be "missing" 14 sixes, with similar hueristic reasoning applying to the first die. Thus I would guess that the expected number of sixes given no double sixes would somewhere in the vicinity of 140-150. But I don't have much intuition about what the variance would be. I'll think for a bit about how to compute the expected number and the probability that you are asking about exactly. A computer simulation would also be fun.

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