Guessing game: what is the wining probability?

Question

Numbers mean Natural Numbers.

The Game: There are two players A,B and ten numbers $1,2,3,4,5,6,7,8,9,10$. A number is being fixed say $7$ (or say $x$) provided A and B does not know about it. Now they individually guess two numbers say $p$ by A and $q$ by B. Now the one who guessed closed to $x$ (here $7$) wins.

Example: A guessed $5$, B guessed $6$ then since $|7-6|=1<2=|7-5|$ so B won.

Problem: Let us vary $x$ (the number fixed before the match), $p$ (A's guessing) and $q$ (B's guessing). Can we find a formula/ function to find the winning probability for a player?

I am unable to proceed! Where should I start? Can someone give me a detailed explanation/solution? Any help will be appreciated.

P.S.: This is neither a textbook problem nor from any competition/exam. This problem was intentionally created by me after watching some guys guessing the age of Bollywood actors on YouTube.

Thanks in advance!

Answer 1

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 $0.43$.

Answer 2

So assuming both chooses uniformly at random and that there is no dependence between the two choices, then by a symmetri argument we would have $$P(\text{A wins}) = P(\text{B wins})$$ and we also know that $$P(\text{A wins})+P(\text{B wins}) + P(\text{Tie}) = 1,$$ so if we know $P(\text{Tie})$, then we also know that $P(\text{A wins})=\frac{1-P(\text{Tie})}{2}$.

Now $P(\text{Tie})$ depends on the value of $x$. So let us first consider a fixed $x$, such as $x=7$.

Given that $A$ chooses a value $i\in \{1,2,3,7\}$, there is exactly one value of $B$ which gives a Tie, namely $B=i$, so $P(Tie | A=i)=0.1$. And if $A$ chooses a value $i\in \{4,5,6,8,9,10\}$, there is two values of $B$, that give a Tie, so $P(Tie | A=i) = 0.2$. Therefore \begin{align*} P(\text{Tie}) &= \sum_{i=1}^{10}P(\text{Tie} |A=i)P(A=i) \\ &= 0.1(4\cdot0.1 + 6\cdot 0.2) = 0.16 \end{align*} and we would get $P(\text{A wins}) = \frac{0.84}{2} = 0.42$

Now we could also consider cases where $x$ is chosen uniformly at random, and we would then compute the probability of ties as

$$P(Tie) = \sum_{i=1}^{10}P(\text{Tie} |x=i)P(x=i)$$ where $P(\text{Tie} |x=i)$ can be calculated with the same reasoning as above. In this case we get $P(Tie)=0.14$ if my calculations are correct.