The probability of obtaining a numbered ticket (six digits)…

The probability of obtaining a numbered ticket (six digits) in which the sum of the first three digits equals the sum of the last three digits is $0.05525$. What is the probability that among two tickets there is such a ticket, if the two tickets $1^{\circ}$ have consecutive numbers; $2^{\circ}$ are independent of each other?

Step 1: The lucky tickets

Ticket numbers run from $000000$ to $999999$. Exactly $55\,252$ of them are "lucky", so $p=0.055252\approx0.05525$.

Step 2: $2^{\circ}$ — independent tickets

\[ P(A_1+A_2)=1-(1-p)^2=2p-p^2\approx0.1075. \]

Step 3: $1^{\circ}$ — consecutive numbers cannot both be lucky

Write the number as $N=1000a+b$ ($a$ = first three digits, $b$ = last three). A number and its digit sum have the same remainder mod $9$, so

\[ S_{\text{first}}-S_{\text{last}}\equiv a-b \pmod 9. \]

For a lucky ticket this difference is $0$. Going to $N+1$:

• if $b\lt999$: $b$ becomes $b+1$, and $a-b$ changes by $-1$;
• if $b=999$: $a$ becomes $a+1$ and $b$ becomes $0$, and $a-b$ changes by $1000\equiv1\pmod9$.

In both cases $a-b$ changes by $\pm1\not\equiv0\pmod 9$, so $N$ and $N+1$ cannot both be lucky. (A computer check over all $999\,999$ pairs confirms: no such pair exists.)

Step 4: $1^{\circ}$ — the answer

The events are mutually exclusive, so

\[ P(A_1+A_2)=2p\approx0.1105. \]

Step 5: The answers

\[ \boxed{1^{\circ}\ \ 2p\approx0.1105, \qquad 2^{\circ}\ \ 2p-p^2\approx0.1075.} \]

Consecutive tickets are slightly better: since they can never both be lucky, there is no "wasted" overlap.