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Find the average score obtained

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Given a regular dice game in which each side has an equal probability of $1/6$, you roll the dice. If you get an even number, you roll two dice now and if the sum of the new values is again even, you roll three dice now and this continues until you get an odd sum. This odd sum is your score. What's your average score per turn?

This problem really confuses me as I've been learning probability. What if you continuously get even sums?

probability combinatorial-game-theory

share|cite|improve this question

asked Mar 12 at 21:22

mathwizard1mathwizard1

408

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  The probability that at some point, you get an odd score, is $1$, but the number of throws until this happens is unbounded. It could well be that the expected score is infinite , although my guess in this case is a finite expected score. It seems to be difficult to determine the expected score however.
  $endgroup$
  – Peter
  Mar 12 at 21:26
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  What might help : The probability to get an odd score is $frac12$ in every throw.
  $endgroup$
  – Peter
  Mar 12 at 21:29
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I had thought of that but couldn't proceed
  $endgroup$
  – mathwizard1
  Mar 13 at 3:59
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Is your main question about how to find the answer to the original problem, or about the issue of "what if you continuously get even sums?"? If the latter, just say so. If the former, then please provide more context of what you tried, where the problem comes from (a textbook? a course?), what recent facts or similar problems you've studied, etc.
  $endgroup$
  – Mark S.
  yesterday
  

  
  
  
  

add a comment | 

0

$begingroup$

Given a regular dice game in which each side has an equal probability of $1/6$, you roll the dice. If you get an even number, you roll two dice now and if the sum of the new values is again even, you roll three dice now and this continues until you get an odd sum. This odd sum is your score. What's your average score per turn?

This problem really confuses me as I've been learning probability. What if you continuously get even sums?

probability combinatorial-game-theory

share|cite|improve this question

asked Mar 12 at 21:22

mathwizard1mathwizard1

408

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  The probability that at some point, you get an odd score, is $1$, but the number of throws until this happens is unbounded. It could well be that the expected score is infinite , although my guess in this case is a finite expected score. It seems to be difficult to determine the expected score however.
  $endgroup$
  – Peter
  Mar 12 at 21:26
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  What might help : The probability to get an odd score is $frac12$ in every throw.
  $endgroup$
  – Peter
  Mar 12 at 21:29
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I had thought of that but couldn't proceed
  $endgroup$
  – mathwizard1
  Mar 13 at 3:59
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Is your main question about how to find the answer to the original problem, or about the issue of "what if you continuously get even sums?"? If the latter, just say so. If the former, then please provide more context of what you tried, where the problem comes from (a textbook? a course?), what recent facts or similar problems you've studied, etc.
  $endgroup$
  – Mark S.
  yesterday
  

  
  
  
  

add a comment | 

$begingroup$

Given a regular dice game in which each side has an equal probability of $1/6$, you roll the dice. If you get an even number, you roll two dice now and if the sum of the new values is again even, you roll three dice now and this continues until you get an odd sum. This odd sum is your score. What's your average score per turn?

This problem really confuses me as I've been learning probability. What if you continuously get even sums?

probability combinatorial-game-theory

share|cite|improve this question

asked Mar 12 at 21:22

mathwizard1mathwizard1

408

$endgroup$

share|cite|improve this question

asked Mar 12 at 21:22

mathwizard1mathwizard1

408

share|cite|improve this question

asked Mar 12 at 21:22

mathwizard1mathwizard1

408

asked Mar 12 at 21:22

mathwizard1mathwizard1

408

asked Mar 12 at 21:22

mathwizard1mathwizard1

408