[Math] Guessing a number between 1 and 100 in 7 guesses or less.

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This question was asked before but I have a lot of trouble understanding the answers given.

Why is it that a number selected at random between 1 and 100 can be determined in 7 or less guesses by always guessing the number in the middle of the remaining values, given that you're told whether your previous guess was too high or low?

Here's a link to the earlier thread: Guessing number between 1-100 always can always be guessed in 7 guess. Why?

I wonder if someone could explain this in greater detail?

Best Answer

The answer is that $2^7=128>100$.

To be more precise:

Say that you start by guessing $50$. Either you're right, in which case you're done, or you're wrong; if you're wrong, then you know that the number is either in $[1,49]$ or $[51,100]$, so that there are only $49$ or $50$ possibilities left.

Now, assuming that you aren't already done, you have a collection of $49$ or $50$ possibilities left; guess the middle one. That is, if you are on $[1,49]$, guess (say) $25$; if you're on $[51,100]$, then guess (say) $75$. If you got it right: great. If not, then you find out whether it should be higher or lower; in particular, if you previously knew it was in $[1,49]$, then you either now know it is in $[1,24]$ or you now know it is in $[26, 49]$. If you previously knew it was in $[51,100]$, then either you know that it is in $[51,74]$ or that it is in $[76,100]$.

In any case, you have either already guess right, or you have narrowed down the possibilities to one of $[1,24]$, $[26,49]$, $[51,74]$, or $[76,100]$. In any one of these cases, there are at most $25$ possibilities left, and it must be one of them.

Continue in this way: cut your current interval of possibilities in half by a guess, so that you are either right or you can discard roughly half of the possibilities based on the announcement of "higher" or "lower".

By continuing this process, in the third step you narrow it down to at most $12$ possibilities; in the fourth, to at most $6$; in the fifth, to at most $3$; in the sixth, to at most $1$; and voila! In your seventh guess, assuming that you're unlucky enough to have not guessed it yet, there's only one number that could possibly be it.

Related Solutions

[Math] Guessing number between 1-100 always can always be guessed in 7 guess. Why

Hint: $2^7=128>100$. That is, you can simply guess in the middle and bisect until you get to the solution.

In general, just find the smallest power of $2$ greater than the max number in your set. For $1000$, you should be able to get it in $10$ guesses, because $2^{10} = 1024$.

Probability – Average Number of Guesses to Guess Number Between 1 and 1000?

Let us take the more ideal case of guessing integers between $1$ and $2^{10}-1=1111111111_2$ writing in base $2$ from now on without explicitly writing subscript $2$ each time. Then the algorithm for the guesses will be like this:

The initial guess is $1\overline 0$ where $\overline 0$ means fill up with zeros until we have a ten digit number
Write the $n$-th guess as $g_n=x_n1\overline 0$ where $x_n$ is a binary string of length $(n-1)$
If the game responds by "higher", the next guess is $g_{n+1}=x_n11\overline 0$
If the game responds by "lower", the next guess is $g_{n+1}=x_n01\overline 0$
If the game responds by "correct" we are done after $n$ guesses

So we see that the game terminates whenever there are only zeros remaining right of $x_n1$. Then note that there are $2^{n-1}$ binary strings $x_n$ of length $(n-1)$. So $2^{n-1}$ numbers are guessed after exactly $n$ guesses. So for this setup the expected number of guesses will be, returning to writing in base $10$: $$ \frac{1}{2^{10}-1}\sum_{n=1}^{10}n\cdot 2^{n-1}=\frac{9217}{1023}\approx9.00978 $$ So the average case here is very close to the worst case of $10$ guesses. This could seem odd at first, but odd is also the clue here ... Half of the time it takes $10$ guesses to reveal that the randomly picked number $x$ is in fact odd. The first $9$ guesses fills up with zeros which in base $2$ means "divisible by $2$". One quarter of the time $x$ is divisible by $2^2=4$ forcing the algorithm to make $9$ guesses before succes. And so on.

I wrote the following code in Python:

L=[1,1000]
G={}
for n in range(1,11):
    i=0
    while i<len(L)-1:
        guess = int(0.5*(L[i]+L[i+1]+1))
        if guess not in L:
            L = L[:i+1]+[guess]+L[i+1:]
            i+=1
            G[guess] = n
        i+=1
S=0
for g in G:
    S+=G[g]
print (10+S+10)

which responded with $8987$, so the average for the $1$ to $1000$ case, assuming that we always make the guess of the rounded average of the limits of the interval we know that $x$ belongs to, must be $8.987$ guesses.

Best Answer

Related Solutions

[Math] Guessing number between 1-100 always can always be guessed in 7 guess. Why

Probability – Average Number of Guesses to Guess Number Between 1 and 1000?

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