4. A person rolls a die, adding up the numbers rolled each time, and stops if the total exceeds 20. Then, when he stops, the number he is most likely to have rolled is $\qquad$
4.21.

Consider the value obtained after the dice roll just before the one that exceeds 20 is $x$.
If $x=15$, then only rolling a 6 can result in 21;
If $x=16$, then rolling a 5 or 6 can result in 21 or 22, with each number having a probability of $\frac{1}{2}$;

If $x=17$, then rolling a 4, 5, or 6 can result in 21, 22, or 23, with each number having a probability of $\frac{1}{3}$;

If $x=18$, then rolling a 3, 4, 5, or 6 can result in 21, 22, 23, or 24, with each number having a probability of $\frac{1}{4}$;

If $x=19$, then rolling a 2, 3, 4, 5, or 6 can result in 21, 22, 23, 24, or 25, with each number having a probability of $\frac{1}{5}$;

If $x=20$, then rolling a 1, 2, 3, 4, 5, or 6 can result in 21, 22, 23, 24, 25, or 26, with each number having a probability of $\frac{1}{6}$.

Therefore, the probability of getting 21 is greater than the probability of getting any other number. Hence, 21 is the most likely total sum to be obtained.