# Math Help - Probability question

1. ## Probability question

Need help with this problem.

A round of a card game consists of shuffling a pack of five cards (numbered 1,2,3,4,5), turning over the top card and writing down its number. This card is then returned to the pack for another round. What is the total number of rounds, on average, needed to obtain two different numbers?

2. Originally Posted by student999
A round of a card game consists of shuffling a pack of five cards (numbered 1,2,3,4,5), turning over the top card and writing down its number. This card is then returned to the pack for another round. What is the total number of rounds, on average, needed to obtain two different numbers?
Suppose that $X$ is the round on which the a match occurs, then $X = 2,3,4,5,6$.
$P\left( {X = j} \right) = \left( {\frac{{j - 1}}{5}} \right)\left( {\frac{{5!}}
{{\left( {6 - j} \right)!\left( {5^{j - 1} } \right)}}} \right)$
.

Now calculate the expected value: $E(X) = \sum\limits_{j = 2}^6 {j \cdot P\left( {X = j} \right)}$

3. Originally Posted by student999
Need help with this problem.

A round of a card game consists of shuffling a pack of five cards (numbered 1,2,3,4,5), turning over the top card and writing down its number. This card is then returned to the pack for another round. What is the total number of rounds, on average, needed to obtain two different numbers?
There is no limit on the number of rounds required; for example, you could draw the number 3 20 times in a row. Not likely, but it could happen.

Let's say the card drawn on round 1 is X.

What is the probability of success on round 2? You have to draw anything but an X. This happens with probability $4/5$.

What is the probability of success on round 3? You have to draw X on round 2 and then something else. This happens with probability $(1/5) (4/5)$.

Success on round 4? You have to draw X on rounds 2 and 3 and then something else. This happens with probability $(1/5)^2 (4/5)$.

Success on round $n$? You have to draw X on rounds 2 through $n-1$ and then something else, with probability $(1/5)^{n-2} (4/5)$.

By now you may have recognized that this is just like a geometric probability distribution except that we are starting at 2 instead of 1. So the mean is $1 + 1/(4/5) = \boxed{9/4}$.

I'm guessing that you have already seen the geometric probability distribution:
Geometric distribution - Wikipedia, the free encyclopedia
If not, we can work through how you compute the expected value, but it requires summing an infinite series.

jw