Re: [Phys-L] From a Math Prof (physics BS major) at my institution ( math challenge)

[Bill Nettles <bnettles@uu.edu>]

You have to calculate the probability that none of the 5 numbers are consecutive, then subtract that from 1 to get probability of at least 1 couplet.

Choose a number. The probability that the next choice is not consecutive is 32/34 and the next is 31/33  and 30/32  and 29/31, so the probability that first is not consecutive with 2nd, 3rd, 4th or 5th is 77.5%.  

 Now you have find the probability that the 3rd, 4th and 5th are not consecutive with the 2nd. There are 33 numbers left and two of them are consecutive, so we have 31/33 *30/32 * 29/31. That's 82.4%

For the 3rd not to be consecutive with the 4th and 5th, we have 32 numbers, two of the consecutive with the 3rd, so 30/32*29/31. That's 87.7%

For the 4th not consecutive with the 5th,we have 31 numbers and 2 are consecutive (I'm ignoring end effects. Don't know how important they are) so we have 29/31. That's 93.5%

All of these have to happen so multiply those probabilities : 52.4% probability of no consecutive numbers, 100%-52.4% = 47.6% probability of at least one couplet.

At least that's my analysis, (and I've been through it enough times that it's my paradigm :)   )...  [<--drooling on my shirt] If I have deceived myself, please show me how.

> -----Original Message-----
> From: Phys-l [mailto:phys-l-bounces@phys-l.org] On Behalf Of brian
> whatcott
> Sent: Thursday, February 27, 2014 11:07 PM
> To: Phys-L@Phys-L.org
> Subject: Re: [Phys-L] From a Math Prof (physics BS major) at my institution (
> math challenge)
>
> On 2/18/2014 9:46 AM, Rauber, Joel wrote:
> > The second list was the random list.  As noted, one cannot prove which one
> was the random list, you can only make a probabilistic guess.
> > /snip/the Math Prof. told me that the probability of consecutive numbers
> appearing on a truly random list is 48%, much higher than most people would
> guess.
> >
> I attempted to confirm the expectation of at least two consecutive numbers
> in a 21,5 set, like this:
>
> There are 34 couplets of two consecutive numbers in the range 1..35 The
> combinations of any 2 numbers in 35 are 595 It follows that the chance of at
> least one consecutive couplet  in a row is 34 in 595 and the chance of no
> consecutive couplet in a row is  561 in 595.
> It follows that the chance of no consecutive couplet in 21 rows is 0.29, and
> the chance of at least one consecutive couplet in a
>   21 row set is  71%
>
> Unfortunately, this does not agree with Joel's input which was 48%, so I
> expect that I slipped. But where?
>
>
> Brian Whatcott    Altus OK
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