The serial number of a $100$ dollar bill contains $8$ digits. Person A wins if she can guess two of these digits (she has two trials), if she doesn't guess two of these, person B wins.
Are my calculations correct?
P(guess is correct)$=1/10 $ Since the events are independent you can just add the probabilities and you end up with P(Person A wins)$=20$%
Am I missing something?
Thank you.
On
The nature of the guessing scheme is not clearly specified. I will illustrate with a few plausible interpretations. We assume in all cases that the digits of the serial number are independent and identically distributed discrete uniform on $\{0, 1, \ldots, 9\}$.
Scenario 1. Person A is allowed to select two distinct integers $m, n$ from $0$ to $9$ inclusive. If for each $m, n$ selected, there exists at least one digit in the serial number that matches, then Person A wins.
Example: The serial number is $12345678$. Person A guesses $m = 0$, $n = 1$. Then she loses because $0$ is not among any of the digits in the number.
Probability of winning: In this case, it does not matter which digits Person A selects. We may assume without loss of generality that the selection is $m = 0$, $n = 1$. Then Person A loses if none of the digits are $0$ or $1$. This outcome occurs with probability $$\left(\frac{8}{10}\right)^8 = \frac{65536}{390625} \approx 0.167772,$$ hence Person A wins with probability $$1 - \frac{65536}{390625} = \frac{325089}{390625} \approx 0.832228.$$
Scenario 2. Person A guesses each digit of the serial number in order, and for each digit, she is told whether the guess is correct or incorrect. She is then allowed to guess again. If, after her second guess, she identifies at least two correct digits, she wins.
Example: If the serial number is $12345678$, and her guess is $31415926$, then she is told that the $5$ is correct and the others are not correct. Then she guesses $98765432$. Note that since she knows the $5$ is correct, she keeps that guess the same. But she loses because none of the other digits in her second guess match.
Probability of winning: Here, the situation is different because the position of the digit must also match. Person A loses if the number of successes is $1$ or $0$. Under optimal playing strategy, Person A chooses two different digits for each digit's guess, unless the first guess is successful; therefore, the probability that she guesses correctly at least once for a single digit is $$\frac{1}{10} + \frac{9}{10}\cdot\frac{1}{9} = \frac{1}{5}.$$ Thus the number of total number of successes in $8$ digits is a binomial random variable $X$ with $p = 1/5$ and the probability this variable is at least $2$ is $$\Pr[X \ge 2] = 1 - \Pr[X \le 1] = 1 - \binom{8}{0}(1/5)^0(4/5)^8 - \binom{8}{1}(1/5)^1(4/5)^7 = \frac{194017}{390625} \approx 0.496684.$$
Scenario 3. Person A must guess the correct digit in the correct position, and must do this for two distinct digits in the serial number. She is allowed two guesses for a given position, thus four guesses in total.
Example: The serial number is $12345678$. Person A guesses that the first digit is $7$, which is wrong. She then guesses the first digit is $1$, which is correct. Then she guesses the last digit is $0$, which is wrong. Her second guess for the last digit is $5$, which again is wrong.
Probability of winning: Again without loss of generality, we may assume she chooses guesses for the first two digits. Then the probability of both correct guesses with two trials each is simply $\frac{1}{5} \cdot \frac{1}{5} = \frac{1}{25}$ per the computation in Scenario 2.
As you can see, I have constructed three interpretations of your question, all valid, all with different answers. Which one is intended?
As noticed the guessing scheme is not completely specified. In the following solution I’m assuming that $2$ independent serial numbers are specified by $A$ which wins if at least $2$ digits are guessed in at least one the $2$ independent trials.
For each trial, the probability for $A$ to loose are as follows
then by $P_{win}=1-(P_{loose})^2$ we obtain