Mark,

>If 2 are selected then the probability is:
>
>2/n * 1/[n-1]

This is obviously not right. If the possibility of one to be selected of n is
1/n (obvious), then the possibility to be one of 2, who selected, should increase (not decrease, as in your formula).
so, it should be 1/n + 1/[n-1], as I understand.



>In our State Lottery, you have to pick 6 out of 6 numbers from a total of 50 numbers, that probability is:
>
>6/50 * 5/49 *4/48 * 3/47 * 2/46 * 1/45 or about 16,000,000 to 1 for a payout of only $4,000,000 or less.

Ok, it's probably right.

What about other kind of lottery, there each number could be from 0 to 9,
say 000000
999999
any number between them.
The probability is 10 in -6, as I understand, so greater, than in State Lottery, which also seems to be non-obvious. Or am I wrong?

>
>I am sure someone can provide the simple formula, like x = n!/[[y-1]**[n-1]] -- which this example is not correct. It's been about 15 years since I had Differential Equations.
>
>>So, I may hope? :) BTW, have you already been the Member of the day? I don't remember...
>>
>>So, the probability may be calcualted as: (right?) in each particular day for each such person:
>>prob=1/337+1/336, if all these people had access in the last 20 days in each day.
>>
>>Let's count for simplicity 2/200 or 1/100.
>>
>>This feature started from Feb. 19th, say, 20 days.
>>How will be prob calculated in this case? I never studied theory of probability seriously :(
>>
>>TIA