Guess 1-9 in three tries

[bonanova]

With 3 yes-no questions you can determine which of eight items, numbered 1 through 8, was chosen at random by a friend.

Suppose there are nine objects numbered 1-9, and you may create multiple replicas of any or all of them. Can you now determine the number on an object chosen at random from the expanded set of objects, with an average of 3 yes-no questions?
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_{Vidi, vici, veni.}

[Trojan Horse]

Seems to be too easy. Let me know if I missed a trap:

Make three extra replicas of the 9. So you'll have four 9's, and one of everything else.

First question: is the number a 9? If the answer is yes, you've done it in one question. Otherwise, you use three more questions to pick out the number from the remaining 8 possibilities; you'll use a total of four questions.

There are 12 objects, and 4 of them have 9's. So you have a 1/3 chance of needing 1 question, and a 2/3 chance of needing 4 questions. Overall, you'll need 3 questions on average.

In fact, by increasing the number of 9's, you can make the average as close to 1 as you want (but you can't actually reach 1).

[bonanova]

No trap.

In fact for the case of one 1, two 2's, three 3's ... nine 9's you can get it with an average of exactly 3 guesses and no fewer. Should be no problem to see how. That's the answer I thought would be found; yours is much simpler.

So - if with 9 we can get the average arbitrarily close to 1 question, we should be able to go past 9 for 3 questions. Is there an upper limit?
_________________
_{Vidi, vici, veni.}

[ralphmerridew]

No upper limit. With objects numbered 1 through 1+2^k, create m=r*2^k copies of the 1.
First ask "Is the number 1?", then use standard methods.
Probabity m/(m+2^k) will take 1 questions.
Probability (2^k)/(m+2^k) will take 1+k questions.

On average, (m + (1+k)*2^k)/(m+2^k) questions needed
== (r*2^k + (1+k)*2^k)/(r*2^k + 2^k)
== (r+k+1)/(r+1)
== 1 + k/(r+1)

Which can be made arbitrarily close to 1.