Question 442709
When drawing your first number, you have a 1/49 chance of drawing any number.

If you draw a 1: You have a 1/48 chance of drawing a number consecutive to it. 

If you draw a 49: You have a 1/48 chance of drawing a number consecutive to it.

If you draw anything else: You have a 1/24 chance of drawing a number consecutive to it.

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2nd draw:

If you have drawn a 1, and then a 2, you have a 1/47 chance of drawing a number consecutive to it.

If you have drawn a 49, and then a 48, you have a 1/47 chance of drawing a number consecutive to it.

If you have draw something other than that, then you have a 4/47 chance of drawing a number consecutive to it.

For instance you have drawn 11,17

10, 12,  16, 18 are all candidates.

We also excluded ... so 

4/44 = 1/11 chance.
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On our third draw:

probability of 1,2,3 =  (1/49)(1/48)(1/47)
probability of 49,48,47 = (1/49)(1/48)(1/47)
Anything else: 6/46 = (3/23)


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We now have one run.

probability of 1,2,3 and then no run = (1/49)(1/48)(1/47)(45/46)(39/45)(35/44)
Same probability with 49,48,47 and then no run

Our anything else probability is: (1/49)(1/24)(4/47)(3/23)(37/45)(33/44)

probability(B) = (61465/10068347520)
probability(C) = (61465/10068347520)
probability(A) = (14652/2517086880)

P(A or B or C) = 2* (61465/100068347520) + 14652/2517086880)

= 2* (61465/100068347520) + (14652/2517086880) 


= (18682943161 / 2650260149893440)

Hopefully... lol