Which of the following statements about the relative sizes of the balls is NOT NECESSARILY true?

Ball	H1	H2	H3	H4
B1	P
B2	P
B3	X
B4	P
B5	P
B6	P

From clue (3), every ball except B3 pings on H1, so the H1 column is $P$ for B1, B2, B4, B5, B6 and $X$ for B3.

---

Each ping means $ball \le H1$, and B3's miss means $B3 > H1$.

Since $B1, B2, B4, B5, B6 \le H1$ while $B3 > H1$, B3 is larger than every other ball.

So B3 is the largest ball.

Ball	H1	H2	H3	H4
B1	P	X
B2	P	P
B3	X	X
B4	P	X
B5	P	X
B6	P	X

From clue (4), only B2 pings on H2, so the H2 column is $P$ for B2 and $X$ for B1, B3, B4, B5, B6.

---

Here $B2 \le H2$ while every other ball is $> H2$.

So B2 is smaller than every other ball.

So B2 is the smallest ball.

Ball	H1	H2	H3	H4
B1	P	X	X
B2	P	P
B3	X	X
B4	P	X	P
B5	P	X
B6	P	X

From clue (2), B4 pings on H3 but B1 does not, so H3 is $P$ for B4 and $X$ for B1.

---

This gives $B4 \le H3$ and $B1 > H3$, so

$B4 \le H3 < B1$.

Therefore $B4 < B1$.

Ball	H1	H2	H3	H4
B1	P	X	X	P
B2	P	P
B3	X	X
B4	P	X	P
B5	P	X		X
B6	P	X		P

From clue (1), B1 and B6 ping on H4 but B5 does not, so H4 is $P$ for B1 and B6 and $X$ for B5.

---

This gives $B1 \le H4$, $B6 \le H4$ and $B5 > H4$, so

$B1 \le H4 < B5$

$B6 \le H4 < B5$.

Therefore $B1 < B5$ and $B6 < B5$.

Ball	H1	H2	H3	H4
B1	P	X	X	P
B2	P	P
B3	X	X	X	X
B4	P	X	P
B5	P	X		X
B6	P	X		P

The hoops are now ordered using the marks already placed:

$H2 < H1$, since B1 misses H2 but pings H1: $H2 < B1 \le H1$.

$H4 < H1$, since B5 misses H4 but pings H1: $H4 < B5 \le H1$.

$H3 < H4$, since B1 misses H3 but pings H4: $H3 < B1 \le H4$.

$H2 < H3$, since B4 misses H2 but pings H3: $H2 < B4 \le H3$.

Combining these gives $H2 < H3 < H4 < H1$.

---

H1 is the largest hoop, and $B3 > H1$, so B3 is larger than every hoop.

B3 therefore misses every hoop, so its H3 and H4 cells are $X$.

B3 makes $0$ pings in total.

Ball	H1	H2	H3	H4
B1	P	X	X	P
B2	P	P	P	P
B3	X	X	X	X
B4	P	X	P
B5	P	X		X
B6	P	X		P

B2 is the smallest ball with $B2 \le H2$, and from the hoop order $H2 < H3 < H4$.

So $B2 < H3$ and $B2 < H4$.

---

B2 therefore pings H3 and H4, so both cells are $P$.

B2 pings all four hoops.

Ball	H1	H2	H3	H4
B1	P	X	X	P
B2	P	P	P	P
B3	X	X	X	X
B4	P	X	P	P
B5	P	X		X
B6	P	X		P

B4 pings H3, so $B4 \le H3$, and from the hoop order $H3 < H4$.

So $B4 < H4$, meaning B4 pings H4 and that cell is $P$.

Ball	H1	H2	H3	H4
B1	P	X	X	P
B2	P	P	P	P
B3	X	X	X	X
B4	P	X	P	P
B5	P	X	X	X
B6	P	X		P

B5 misses H4, so $B5 > H4$, and from the hoop order $H3 < H4$.

So $B5 > H4 > H3$, meaning B5 misses H3 and that cell is $X$.

Ball	H1	H2	H3	H4
B1	P	X	X	P
B2	P	P	P	P
B3	X	X	X	X
B4	P	X	P	P
B5	P	X	X	X
B6	P	X	P/X	P

For B6 on H3: B6 pings H4 so $B6 \le H4$, and B6 misses H2 so $B6 > H2$.

Thus $H2 < B6 \le H4$, with H3 lying in between, and no clue fixes whether $B6 \le H3$ or $B6 > H3$.

So B6 on H3 is undetermined, marked $P/X$.

---

The grid is complete; the only undetermined cell is B6 on H3.

Hoop order: $H2 < H3 < H4 < H1$.

Ball facts: B2 is smallest, B3 is largest, with $B4 < B1 < B5$ and $B6 < B5$; the place of B6 relative to B4 and B1 is not fixed.

Pings per hoop: $H1 = 5$, $H2 = 1$, $H3 = 2$ or $3$, $H4 = 4$, giving a total of $12$ or $13$ pings.

---

Established orderings: B2 is smallest, B3 is largest, with $B4 < B1 < B5$ (and $B6 < B5$).

Option 1, $B1 < B5 < B3$: $B1 < B5$ holds and $B5 < B3$ holds, so necessarily true.

Option 2, $B4 < B5 < B3$: $B4 < B1 < B5$ gives $B4 < B5$, and $B5 < B3$, so necessarily true.

Option 4, $B2 < B1 < B5$: B2 is smallest and $B1 < B5$, so necessarily true.

Option 3, $B1 < B6 < B3$: $B6 < B3$ holds, but $B1 < B6$ is not forced. We only know $H2 < B6 \le H4$ and $H3 < B1 \le H4$, so B6 could be smaller than B1. This is NOT necessarily true.

Answer: option 3.