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



Nice exposition! You threw down the gantlet on end effects, so let's how far i can go wrong on this one: you said "Choose a number. The probability that the next choice is not consecutive is 32/34"
There are two numbers for the initial pick which are special - the end cases of 1 and 35. For these values, there are 33 possible non consecutive choices from 34 possibilities for the second pick of the quintet, but for the other 33 starter values, there are as stated 32/34 chances of a non-consecutive value for the second number.
That would put the probability of finding the first two numbers in a quintet are not consecutive as 33/34 X 2/35 + 32/34 X 33/35 = .05546 + 0.88739 = 94.285% rather than 94.118% your figure - a rather small change!
/suspect shortcut next!/
lets suppose that the probabilities are under estimated in the same ratio for all four joint probabilities (??) - this ratio is 10018/10000, so i am estimating
77.5% X 1.0018^4 = 78.05%
82.4% X 1.0018^3 = 82.85%
87.7% X 1.0018^2 = 88.02%
93.5% X 1.0018 = 93.67%
The product of these possibilities is 53.3% leaving 46.7% as the chance of one or more consecutive pairs ...
Needless to say, this value is not supported by your histogram count.

Brian Whatcott Altus OK

On 2/28/2014 5:50 PM, Bill Nettles wrote:
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.