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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.