Comparison of Randomly Selected Numbers

It should be clarified that the generated numbers are independent of each other and that every number has probability $0$ to be generated. Otherwise, there is not enough information.

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Given those properties, it follows that $P(a>b)=P(b>a)$ , as the events are independent of each other (it is equally likely that the first number is greater than the second in comparison to the second being greater than the first). Since $P(a>b)+P(a<b)=1$ , we find $P(a>b)=\frac{1}{2}$ .

As said in the problem, it is not safe to assume that the likelihood of 4 or 6 appearing as a digit in a random number is equal. Therefore, we will have to write a more general expression for these probabilities.

Let $P(4)$ be the probability of 4 appearing as a digit, and $P(6)$ that of 6. Now let

$P(4)=x$ and $P(6)=1-x$

The next thing to do is to "imagine" the possible values of $a$ and $b$ and compare them.

Of course, whenever you compare two positive numbers, you start by comparing the value of their leftmost digits. If their values are equal, you move on the next leftmost digits, and so on.

This method of comparing undoubtedly applies even to numbers like those generated by Perpetua.

Given that Perpetua can only generate a number between 1 and 10, the only possible ones digits of $a$ and $b$ are obviously 4 and 6. Now it is possible that the ones digit of $a$ is 4 while $b$ 's is 6 and vice versa. It is also possible that $a$ and $b$ have the same ones digit. Taking note of this, we calculate the probabilities for each of these outcomes, which we can name $P(a_{ones} ∩ b_{ones})$ . Therefore,

$P(4 ∩ 4)=x^2$

$P(4 ∩ 6)=x(1-x)$

$P(6 ∩ 4)=x(1-x)$

$P(6 ∩ 6)=(1-x)^2$

If the ones digit of $a$ is greater than that of $b$ , $a$ is greater than $b$ , regardless of what the other digits of either $a$ or $b$ are. The same applies to $b$ . So, what have we gotten so far? Looking at the probability expressions above, we know now that the probability of $a$ being greater than $b$ is at least $x(1-x)$ . Now let's write a temporary expression for $P(a>b)$ . If we only consider this case, $P(a>b)=x(1-x)$ .

Next we consider the case that $a$ and $b$ 's ones digits are both equal. The comparing will now be done on $a$ and $b$ 's corresponding tenths digits. Again, the possible values of these digits will be considered and then calculated like we did with the ones digits. However, in computing the probabilities, the fact that the ones digits of both $a$ and $b$ are equal must be taken into account. If we can tell that $a≠b$ by looking at their tenths digits, $P(a>b)=x(1-x)+((x^2+(1-x)^2)[x(1-x)]$ . This expression takes into account both the possibility that $a$ and $b$ 's ones digits are different (which contributes to the first term) and the possibility that $a$ and $b$ 's ones digits are equal while their tenths digits are different . Still, this is not the whole story. You can extend this chain of reasoning to the hundredths, thousandths, and so on digits, and it still wouldn't be enough. If you were to consider ALL the possibilities, you would have to go as far as considering the possibility of $a$ and $b$ 's "rightmost" digits being different while all the digits to their left being equal. This means going practically infinitely far off into the decimal sequences to compare $a$ and $b$ . Remember that Perpetua can tell if $a$ and $b$ are equal or not no matter how close their values are. Considering ALL these possibilities, you get a "nice" expression for the sum of probabilities.

$P(a>b)=x(1-x)+(x^2+(1-x)^2)[x(1-x)]+(x^2+(1-x)^2)^2[x(1-x)]+(x^2+(1-x)^2)^3[x(1-x)]+⋯$

This sum is a definitely an infinite geometric series with common first term equal to $x(1-x$ and common ratio equal to $(x^2+(1-x)^2)$ . Since its common ratio is less than 1, this geometric series is converging , and can therefore be written as $\frac{x(1-x)}{1-(x^2+(1-x)^2)}$ . This simplifies to $\frac{1}{2}$ .