It's not possible to solve the game in all cases with fewer than 7 guesses, because after 6 guesses you have at least 95 unguessed numbers and 6 bits of information and $2^6<94.$
But actually 7 is not achievable, since at most 26 primes can appear in any given block of 101. Thus the first guess, in the worst case, will have at least 74 remaining numbers, and so at least 8 guesses are required since $2^6<68$.
The goal at each step is to find numbers that split the remaining numbers into two sets of roughly equal size. Here are the beginnings of a strategy for 9. Ask about 3. If the number is 3 you're done; if n + 3 is prime it should not be hard to guess the number in 8 more tries.
Otherwise 74 numbers remain: 1, 5, 6, 7, 9, 11, 12, 13, 15, 17, 18, 19, 21, 22, 23, 24, 25, 27, 29, 30, 31, 32, 33, 35, 36, 37, 39, 41, 42, 43, 45, 46, 47, 48, 49, 51, 52, 53, 54,
55, 57, 59, 60, 61, 62, 63, 65, 66, 67, 69, 71, 72, 73, 74, 75, 77, 78, 79, 81, 82, 83, 84, 85, 87, 88, 89, 90, 91, 92, 93, 95, 96, 97, 99. Now ask about 10. If n+10 is prime it should be easy to find the number in 7 more tries.
Otherwise 50 numbers remain: 5, 6, 11, 12, 15, 17, 18, 22, 23, 24, 25, 29, 30, 32, 35, 36, 39, 41, 42, 45, 46, 47, 48, 52, 53, 54, 55, 59, 60, 62, 65, 66, 67, 71, 72, 74, 75, 77, 78, 81, 82, 83, 84, 85, 88, 89, 90, 92, 95, 96. Now ask about 12. If n is 12 you're done; if n+12 is prime you have 6 tries to find 16 numbers.
Now 33 numbers remain: 6, 15, 18, 22, 23, 24, 30, 32, 36, 39, 42, 45, 46, 48, 52, 53, 54, 60, 62, 65, 66, 72, 74, 75, 78, 81, 82, 83, 84, 88, 90, 92, 96. Now ask about 1. If n+1 is prime there are 15 possibilities and you have 5 tries.
Now you have just 18: 15, 23, 24, 32, 39, 45, 48, 53, 54, 62, 65, 74, 75, 81, 83, 84, 90, 92. This is a bit tricky, but guess 1139 (or 1148) to split the remainder into two sets of 9.
Either way you go at least one of the sets has 5 members (4 would be possible only if you picked one of the 18).
This is a most intriguing problem and took many, many, many thinking hours to solve, I even wrote a script to better understand it.
It is possible for the sinners (players) to win every time. The number of players playing is irrelevant and the number of numbers they choose per turn is irrelevant as long as the players get to choose more numbers than the devil. In other words:
Theorem. Let $n,m,k\in\mathbb{N}$ such that $n>m$, then a $$n\text{-to-}m\text{ bias game & with }k\text{ players playing,}$$ has a winning solution for the players.
To show how this solution works, lets first demonstrate the $3$-to-$2$ bias scheme with $3$ players playing. Let $N=10^{2101}$, and $P=\{1,\ldots,10^{2101}\}$ be a large set.
Player 1 has only one job, pick one number in each of the sets $\{3n+1,3n+2,3n+3\}$; this can easily be done since the devil can only pick up to two numbers at a time.
So, Player 2 is now presented with $N'=10^{2101}/3$ sets, $P'=\{A_1,\ldots, A_{N'}\}$. Let $$P'_1=\{A_1,\ldots,A_{1000}\} \\ P'_2=\{A_{1001},\ldots,A_{2000}\} \\ \vdots \\ P'_{10^{2101}/3000}=\{A_{10^{2101}/3-999},\ldots,A_{N'}\}$$
such that each set, $A_i$, is size three and in each set, $A_i$, Player 1 has chosen at least one number. Now the devil chooses two numbers, in say $P_i'$ and $P_j'$, whatever the numbers delete the sets that contain those numbers from $P_i'$ and $P_j'$. Now pick two numbers from two different sets remaining in $P'_i$ and $P'_j$; for your third choice pick a number from a set, different than the first two number's sets, from either $P'_i$ or $P'_j$ whichever of the two has least number of sets that you have chosen numbers from. For some fixed $i$, $P'_i$ will eventually have about 80% of it's sets deleted (because the devil will have chosen from those sets), however, $P'_i$ will now consist of about 200 sets that each contain a number that Player 2 has picked. We continue in this fashion till $P'_i$ contains about $$1000\cdot (.2)^3 \approx 8$$ sets that Player 2 will have chosen all the numbers in and thus Player 1 and Player 2 will be sharing a number in one of the sets in $P'_i$.
Eventually Player 3 will be presented, $R=\{R_1,\ldots,R_{10^{2101}/3000}\}$, with $$R_1=\{1,\ldots,3000\} \\ R_2=\{3001,\ldots,6000\} \\ \vdots $$ a collection of about $10^{2101}/3000$ sets and the only knowledge that Player 3 has, is that Player 1 and Player 2 share an element in each $R_i$. The devil goes first and picks two numbers; whatever numbers the devil picks, delete those sets from $R$. For example, suppose the devil picks $1$, then delete the entire set $R_1=\{1,\ldots,3000\}$ from $R$. Now from the remaining sets (which the devil hasn't touched yet) choose three numbers, one from each of three sets in $R$. Phase 2: The devil will now pick two more numbers from two more sets, maybe from sets you have chosen from, maybe from sets that you haven't, whatever the case, delete those sets from $R$ and again choose from the remaining sets (which the devil hasn't touched) three more numbers from sets that you haven't already chosen. Since you always get one extra move than the devil, then when all sets in $R$ have been exhausted, you have thrown out at most ruffly 80% of the sets (since the devil chose from them); however the remaining sets will all have one number in them that you have chosen and no numbers that the devil has chosen. We define the remaining sets as $R'$. Again, $R'$ is about 20% the size of $R$. We now repeat the process on $R'$ and arrive at $R''$, a set 20% the size of $R'$ but now every set in $R''$ has two elements that you have chosen and the devil has chosen none of numbers in those sets. With this pattern, a calculation of $$\frac{10^{2101}}{3000}\cdot (.2)^{3000} \approx 4$$ shows that there will be about $4$ sets remaining that Player 3 will have chosen all the numbers in and the devil hasn't chosen any. Thus, all three players will have chosen at least one number in common.
You can see that large numbers are required to ensure that a winning solution is reached. Let us suppose for a moment that a fourth player joined our group of three players, now how to we proceed to ensure that they still win?
The whole process above, ensures that the first three players will be sharing a number in the set $\{1,\ldots,10^{2101}\}$. We will repeat this process with another set $\{10^{2101}+1,\ldots,2\cdot 10^{2101}\}$. Player 3 will have two sets to juggle a $A$ and $B$, which each contain $10^{2101}/3000$ sets. The devil may play entirely in $A$ or play entirely in $B$ or may choose to play in each set. Player 3 simply needs to make sure that $A$ and $B$ lose sets at an equal rate. If Player 3 focuses all their attention on $A$ and the devil keeps playing in $B$, then Player 3 will end up with 0% useful sets in $B$. To keep this from happening Player 3 should play in $A$ if the devil plays in $A$ and should play in $B$ in the devil plays in $B$. If the devil chooses one number in $A$ and one number in $B$, then Player 3 should choose one number in $A$ and one number in $B$ and then use their third choice on the set that has (currently) the least number of 'good sets'.
This will give us two sets, $\{1\dots,10^{2101}\}$ and $\{10^{2101}+1,\ldots,2\cdot 10^{2101}\}$, in which the first three players share a number in those sets. How many sets will we need to accomplish like this till Player 4 has enough sets? Well we want $N$ sets such that: $$N\cdot (.2)^{10^{703}}>1.$$ So, a &%#@ing &*#@ number of sets.
But it is possible and then Player 4 will be presented with $N$ sets that each have $10^{2101}$ numbers such that in each set there is at least one number that the first three players share. And so, Player 4 can continue in the same fashion that Player 3 did.
If anyone knows of an easier way to describe my method, please let the community know.
Best Answer
As Leander Tilsted Kristensen notes, the pertinent probability to compute is the probability of a tie; by symmetry, each player's probability of winning is half the remaining, complementary probability.
Assuming $x$, $p$, and $q$ are chosen uniformly at random among the numbers $1$ to $10$, there are $1000$ different outcomes, all equally likely, and we need only count the number that result in ties. There are two ways a tie can happen: If $p=q$, or if $x=(p+q)/2$ (i.e., if $|x-p|=|x-q|$). The former happens in $100$ different ways; the latter happens in only $50$ different ways, namely when $p$ and $q$ have the same parity, so that $(p+q)/2$ is an integer. So our initial count is $100+50$. However this doublecounts the $10$ cases when $p=q=x$, so the final count for the number of ways a tie can occur is
$$100+50-10=140$$
and thus the probability of a tie is $140/1000=0.14$, giving each player a probability of winning of $0.43$.