Expected number of questions to win a game

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In summary: In your case for the states from 1 to 9 there is at least one emerging path that conducts to the adsorbing state 10, so that they are all transient states... Regarding 'other approaches' what I can say is that the problem can't [probably] be solved with 'conventional' statistic approaches... Kind regards $\chi$ $\sigma$In summary, the person in the quiz competition has to answer an average of 15 questions to win the game, assuming that there is a 1/3 probability of answering each question correctly and that 2 consecutive wrong answers reset the score to 0. This can be calculated using Markov chain analysis, where the person starts at one of the 9 transient
  • #1
reg_concept
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Let, a person is taking part in a quiz competition.
For each question in the quiz, 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 4 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?
Someone please help.
 
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  • #2
concept said:
Let, a person is taking part in a quiz competition.
For each question in the quiz, 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?
Someone please help.

An important detail is missing: for each question what is the probability p of correct answer?... if p=1 then the expected number of answers is 5... of course...

Kind regards

$\chi$ $\sigma$
 
  • #3
chisigma said:
An important detail is missing: for each question what is the probability p of correct answer?... if p=1 then the expected number of answers is 5... of course...

Kind regards

$\chi$ $\sigma$

Since there are 3 choices per question, so probability of correct question is 1/3
 
  • #4
Never mind, we can indicate with p the probability of correct answer and q=1-p the probability of wrong answer at each step. The problem is a Markov chain type that uses the following state transition diagram... http://d2yq0g4vt6ipuo.cloudfront.net/c7/8e/i68062919._szw565h2600_.jpg

The probability transition matrix is...

$\displaystyle P = \left | \begin{matrix} q & p & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & q & p & 0 & 0 & 0 & 0 & 0 & 0 \\ q & 0 & 0 & p & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & q & p & 0 & 0 & 0 & 0 \\ q & 0 & 0 & 0 & 0 & p & 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 0 & 0 & 0 & q & p & 0 & 0 \\ q & 0 & 0 & 0 & 0 & 0 & 0 & p & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & q & p \\ q & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & p \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 1 \end{matrix} \right| $ (1)

The presence of the final state 10 makes the Markov chain an Adsorbing Markov chain and, because the state 10 is the only adsorbing state we can compute the fundamental matrix N as...

$\displaystyle N = \sum_{k=0}^{\infty} Q^{k} = (I_{9}-Q)^{-1}$ (2) ... where $I_{9}$ is the 9 x 9 unity matrix and Q is the 9 x 9 matrix obtained from P deleting the last row and the last column...

$\displaystyle Q = \left | \begin{matrix} q & p & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & q & p & 0 & 0 & 0 & 0 & 0 \\ q & 0 & 0 & p & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & q & p & 0 & 0 & 0 \\ q & 0 & 0 & 0 & 0 & p & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & q & p & 0 \\ q & 0 & 0 & 0 & 0 & 0 & 0 & p & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & q \\ q & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \end{matrix} \right| $ (3)

Now the expected number of steps is the vector...

$\displaystyle \overrightarrow{t} = N \cdot \overrightarrow{1}$ (4)

... where $\displaystyle \overrightarrow{1}$ is the 9 x 1 'all 1' column vector. Of course the main task is the matrix inversion in (2), normally very time expensive to do by hand... Kind regards $\chi$ $\sigma$
 
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  • #5
chisigma said:
The presence of the final state 10 makes the Markov chain an Adsorbing Markov chain and, because the state 10 is the only adsorbing state we can compute the fundamental matrix N as...

$\displaystyle N = \sum_{k=0}^{\infty} Q^{k} = (I_{9}-Q)^{-1}$ (2)

Thankx for ur great help.
But one question, I can't find any transient state in the matrix. Is it possible to implement the above equation if there is no transient state?

Could you please suggest me a book to read and understand this scenario of marcov chain in detail?

Regards
Concept
 
  • #6
concept said:
Thankx for ur great help.
But one question, I can't find any transient state in the matrix. Is it possible to implement the above equation if there is no transient state?

Could you please suggest me a book to read and understand this scenario of marcov chain in detail?

Regards
Concept

The states from 1 to 9 are all 'transient states' and the matrix P represents the probability to pass from the state i to the state j for i=1,2,...,10 and j=1,2,...,10...

Regarding a 'good text' about Markov chain I can suggest You the following tutorial article ...

http://www.dartmouth.edu/~chance/teaching_aids/books_articles/probability_book/Chapter11.pdf

... that includes a large number of exercizes and references...

Kind regards

$\chi$ $\sigma$
 
  • #7
chisigma said:
The states from 1 to 9 are all 'transient states'
Thankx you.
From the book of Wayne winston, I find that,a state i is transient if there is a way to leave state i that never returns to state i.
I can move from state 1 to state 0 by giving 4 consecutive wrong answers and again can come back to state 1 by giving right answer. So, is it a transient state?

Could you please explain?

Another approach: I need three chances to answer a correct question. So, it will take 5*3=15 questions to get 5 points. Is it right?
 
  • #8
concept said:
Thankx you.
From the book of Wayne winston, I find that,a state i is transient if there is a way to leave state i that never returns to state i.
I can move from state 1 to state 0 by giving 4 consecutive wrong answers and again can come back to state 1 by giving right answer. So, is it a transient state?

Could you please explain?

Another approach: I need three chances to answer a correct question. So, it will take 5*3=15 questions to get 5 points. Is it right?

In your case for the states from 1 to 9 there is at least one emerging path that conducts to the adsorbing state 10, so that they are all transient states...

Regarding 'other approaches' what I can say is that the problem can't [probably] be solved with 'conventional' statistic approaches...

Kind regards

$\chi$ $\sigma$
 

FAQ: Expected number of questions to win a game

What is the expected number of questions to win a game?

The expected number of questions to win a game refers to the average number of questions that a player would need to answer correctly in order to win the game. This value is calculated based on the probability of correctly answering a question and the total number of questions in the game.

How is the expected number of questions to win a game calculated?

The expected number of questions to win a game is calculated by multiplying the probability of correctly answering a question by the total number of questions in the game. For example, if there are 10 questions in a game and the probability of answering a question correctly is 0.5, the expected number of questions to win would be 5 (0.5 x 10).

Does the expected number of questions to win a game change depending on the difficulty of the questions?

Yes, the expected number of questions to win a game can change depending on the difficulty of the questions. If the questions are easier, the probability of correctly answering them increases, resulting in a lower expected number of questions to win. On the other hand, if the questions are more difficult, the probability of correctly answering them decreases, resulting in a higher expected number of questions to win.

Can the expected number of questions to win a game be used to predict the outcome of a game?

No, the expected number of questions to win a game is a theoretical calculation based on probability and does not guarantee a specific outcome. Other factors such as luck and skill of the players can also influence the outcome of a game.

What are the limitations of using the expected number of questions to win a game?

The expected number of questions to win a game is based on assumptions and can only provide an estimate. It does not take into account factors such as player fatigue or changing probabilities as the game progresses. Additionally, it may not accurately reflect the actual outcome of a game due to the unpredictable nature of chance events.

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