Let us label the primary particular person to talk as A, second particular person to talk as B and so forth.

Let’s first have a look at the final case. Number of hats that A can presumably see :

6,7, 8 and 9.

Case 1: A sees 6 or 9 hats
If A sees 6 hats then he’ll know on his first flip itself that he’s carrying 2 hats. This is as a result of it’s on condition that he can solely put on 1 or 2 hats. It can be on condition that whole variety of hats is 8,9 or 10. 6+1 = 7 which isn’t attainable. The solely risk when he sees 6 hats, subsequently, is that he’s carrying 2 hats.

Similarly, if he’s seeing 9 hats then it must be the case that he’s carrying 1 hat.

So, if A calls out 1 or 2 on his first flip, then B,C,D and E will know that they’re carrying a complete of 9 or 6 hats respectively amongst themselves. They will then have the ability to work out their hat rely on their first flip itself.

Case 2: A sees 7 or 8 hats
If A says, “I do not know” on his first flip then it would point out to B,C,D and E that there are a complete of both 7 or 8 hats amongst them 4.

Case 2a : There are literally 7 hats amongst BCDE
In this case, all people amongst B,C,D and E who’s seeing 5 hats as the overall hat rely among the many different 3, will have the ability to guess that he’s carrying 2 hats. This is just because they know that amongst B,C,D and E, there are a complete of seven or 8 hats as a result of A stated, “I do not know” on the primary flip. Now, once they see 5 hats among the many different 3, it turns into apparent to them that they’re carrying 2 hats they usually name it out on their first flip itself. This, 3 of those 4 folks will have the ability to guess their hat rely on their very first flip.

Case 2b: There are literally 8 hats amongst B,C,D and E. In this case, within the first spherical, all people would see 6 hats and they’d say, “I do not know.” This is just because, through the first spherical, when B,C D and E would see 6 hats whole on the opposite 3 folks then they might not be certain in the event that they themselves have 1 or 2 hats on their hats. This is as a result of each B+C+D+E = 7 and B+C+D+E= 8 are attainable.

Now, very importantly discover that 3 various things occur in circumstances 1, 2a and 2b .

In case 1, all people is ready to guess their hat rely within the first spherical.

In case 2a, although A says, “I do not know”, within the first spherical, folks amongst B,C,D and E who’re carrying 2 hats every are capable of guess their hat rely within the first spherical itself.

In case 2b, all people says, “I do not know” within the first spherical.

So, when A says, “I do not know” within the first spherical and but, some persons are capable of guess their hat rely then it’s indicative that the one attainable case is case 2a i.e there are 7 hats amongst B,C,D and E. Thus the one particular person amongst B,C,D,E who’s carrying 1 hat will have the ability to guess his hat rely after he hears anyone amongst B,C,D, E guess their hat rely. A, nevertheless, won’t ever know his hat rely.

Also, when all people says,”I do not know” within the first spherical then it would inform all people that they’re in case 2b and that there are 8 hats amongst B,C, D and E. In the subsequent spherical, B,C, D and E will have the ability to guess their hat rely. A once more will not have the ability to inform his hat rely.

So, the reply is that within the case of 9 whole hats, there are both 7 or 8 hats amongst B,C,D and E. This implies that the group is in case 2a or case 2b. And now we have confirmed above that, in both of those circumstances, all people aside from A will have the ability to guess their hat counts.