# Thread: Expected number of questions to win a game

1. ## Expected number of questions to win a game

Let, a person is taking part in a quiz competition. For each questions, there are 3 answers, and for each correct answer he gets 1 point.
When he gets 5 points, he wins the game.
If he gives 2 consecutive wrong answers, then his points resets to zero (i.e. if his score is now 4 and he gives 2 wrong answers, then his score resets to 0).

My question is, on an average how much questions he needs to answer to win the game?

2. ## Re: Expected number of questions to win a game

Originally Posted by concept
Let, a person is taking part in a quiz competition. For each questions, there are 3 answers, and for each correct answer he gets 1 point. When he gets 5 points, he wins the game.
If he gives 2 consecutive wrong answers, then his points resets to zero (i.e. if his score is now 4 and he gives 2 wrong answers, then his score resets to 0). On an average how much questions he needs to answer to win the game?
This seems to be quite complicated. The difficulty will in modeling the outcomes.

This person can win on the $5^{th},~6^{th},~\cdots~10^{th}$ game.
If the person gets to play the $10^{th}$ and does not win, then the score must reset.

Now that is not to say that the score does not reset before the tenth game.
In fact, it could reset after the second game.

Have you worked with conditional expectation?

3. ## Re: Expected number of questions to win a game

Yes, I know about conditional probability, but I can not figure out how to solve this problem

4. ## Re: Expected number of questions to win a game

Originally Posted by concept
Yes, I know about conditional probability, but I can not figure out how to solve this problem
No it is not conditional probability, it is conditional expectation.

5. ## Re: Expected number of questions to win a game

Originally Posted by Plato
No it is not conditional probability, it is conditional expectation.
I did not work with conditional expectation. I searched in "Probability and Statistics- walpole" but could not find. Then found in wikipedia, but could not understand properly.
Could you please give hints how to solve the problem using conditional expectation or without using conditional expectation?

Thank you

6. ## Re: Expected number of questions to win a game

Originally Posted by concept
I did not work with conditional expectation. I searched in "Probability and Statistics- walpole" but could not find. Then found in wikipedia, but could not understand properly. Could you please give hints how to solve the problem using conditional expectation or without using conditional expectation?

In that case I suppose you are not to use it.
Do you understand what I have said about the setup?

Now I assume the person is simply doing random guessing.
The probability that the person wins on the eight game is $\binom{5}{3}{\left( {\frac{2}{3}} \right)^3}{\left( {\frac{1}{3}} \right)^5}$.

The reasoning behind this is that the fifth win is gotten on the eighth game requires that there be four wins and three losses none of which are together.

Now $\binom{5}{k-5}{\left( {\frac{2}{3}} \right)^{k-5}}{\left( {\frac{1}{3}} \right)^5}$ is the probability if winning on the $k^{th}$ game, where $k=5,6,7,8,9,10$.

The points reset to 0 for the eleventh game. And the next chance to win is on the fifteenth game.
All of that assumes no-resets.

The computations are quite tedious. I am not prepared to do more.

7. ## Re: Expected number of questions to win a game

Originally Posted by concept
Let, a person is taking part in a quiz competition. For each questions, there are 3 answers, and for each correct answer he gets 1 point.
When he gets 5 points, he wins the game.
If he gives 2 consecutive wrong answers, then his points resets to zero (i.e. if his score is now 4 and he gives 2 wrong answers, then his score resets to 0).

My question is, on an average how much questions he needs to answer to win the game?
Is the person randomly guessing answers? If yes, suppose we start from the beginning or, from a "reset to zero".

Let n = the number of questions answered to win, l = the # of losses, w = the # of wins

n = l + w, and as mentioned $n \geq 5$. The probability of winning in 5 questions is simple (1/3)^5. Note that the winning string of letters must consist of the substrings W and LW (any combinations of these does not result in a "Reset"). e.g. n = 6 = LW W LW W (as just one possibility). Now, P(W) = 1/3 and P(LW)=2/9, and we can find the probability of winning after n questions by:

$P(n=5) = {5 \choose 0} \left(\frac{1}{3}\right)^5 \left(\frac{2}{9}\right)^0 = 1/243$
$P(n=6) = {5 \choose 1} \left(\frac{1}{3}\right)^4 \left(\frac{2}{9}\right)^1 = 10/729$ i.e. LW W W W W
$P(n=7) = {5 \choose 2} \left(\frac{1}{3}\right)^3 \left(\frac{2}{9}\right)^2 = etc.$ i.e. LW LW W W W
$P(n=8) = {5 \choose 3} \left(\frac{1}{3}\right)^2 \left(\frac{2}{9}\right)^3$ i.e. LW LW LW W W
$P(n=9) = {5 \choose 4} \left(\frac{1}{3}\right)^1 \left(\frac{2}{9}\right)^4$ i.e. LW LW LW LW W
$P(n=10) = {5 \choose 5} \left(\frac{1}{3}\right)^0 \left(\frac{2}{9}\right)^5$ i.e. LW LW LW LW LW

The one thing I don't have time to help you with is the conditional probability prior to having won (according to the above possibilities). So, you must calculate the probability of seeing a string that combines LW's and W's and then LL at the end (so, in this case, you can't have more than 4 wins, or else you've won already!). This part is quite complicated, because you could have a large number of "resets to zero", and these particular strings would be in various sizes. Seems like it would take less time to do this via a computer program to play this game thousands of times to estimate the answer.

8. ## Re: Expected number of questions to win a game

Originally Posted by majamin
$P(n=7) = {5 \choose 2} \left(\frac{1}{3}\right)^3 \left(\frac{2}{9}\right)^2 = etc.$ i.e. LW LW W W W
$P(n=8) = {5 \choose 3} \left(\frac{1}{3}\right)^2 \left(\frac{2}{9}\right)^3$ i.e. LW LW LW W W

There is a problem with those calculations without further assumptions.
They are correct with no resets.
But $n=7$ is possible with a reset on $n=2$.

Thus $P(n=7) = {5 \choose 2} \left(\frac{1}{3}\right)^3 \left(\frac{2}{9}\right)^2 +\left(\frac{2}{3}\right)^2P(n=5)$
That is no resets plus one reset.

Now for $n=8$ we could have a reset on $n=2\text{ or }n=3$ or none at all.

9. ## Re: Expected number of questions to win a game

Originally Posted by Plato
There is a problem with those calculations without further assumptions.
They are correct with no resets.
But $n=7$ is possible with a reset on $n=2$.

Thus $P(n=7) = {5 \choose 2} \left(\frac{1}{3}\right)^3 \left(\frac{2}{9}\right)^2 +\left(\frac{2}{3}\right)^2P(n=5)$
That is no resets plus one reset.

Now for $n=8$ we could have a reset on $n=2\text{ or }n=3$ or none at all.
Yes, that was my intent, to only calculate a "winning" string. I agree with your addition to the problem, and I stand by my original comment that it would be difficult to include all possibilities (especially for large n) without the aid of some kind of computer assisted code. My hunch is that a "nice", explicit solution is out of reach.