Use conditional probability.
The youngest person can sit anywhere at the table, so his position doesn't matter. The second person must sit next to the first, so
Let A be the event that the people at the table are in ascending order; L be the probability that the second youngest is seated to the left of the first person and R be the probability that he's on the Right.
If the second youngest is not seated next to the first, all bets are off, so we'll use this as an event to condition on in order to make the problem "smaller."
So p(A) = p(A,L) + p(A,R)
= p(A|L)p(L) + p(A|R)p(R)
P(R) = p(L) = 1/4
Since the youngest person is already seated, there are only 4 places the next in line can sit.
Now, p(A|L) = probability(the third youngest sits next to the second, the fourth next to the third and the fifth in the last seat)
Again we could apply conditional probability
= p(3 next to 2 | L) p(4 next to 3| L, 3 next to 2) p(5 in last seat | 4 next to 3, 3 next to 2 , L)
p(3 next to 2 | L) = 1/3 since 2 seats are taken there are 3 places 3 can sit and only 1 of them is next to 2
p(4 next to 3| L, 3 next to 2) = 1/2
p(5 in last seat | 4 next to 3, 3 next to 2 , L)=1/1=1
So, p(A|L) = 1/3* 1/2* 1= 1/6
Since the problem is the same in both directions,
By symmetry, p(A| R) = 1/6
So, subbing everything in,
P(A) = 1/4* 1/6. + 1/4*1/6