Lost first game – can I still make my mother proud?












10















I am one of the greatest finger wrestlers of my generation. Only 0.1% of all players are better.



When I play against someone who's worse than me, I win 99% of the time. Against someone who's better than me, I always lose.



It's the first round of a tournament. We're playing best of three games (i.e. two wins needed). I hate to say it, but I have just lost the first game! What are my chances of still making it to the second round?



Assuming that the players who are better than me always beat the players who are worse than me, what are my chances of making it to the third round?











share|improve this question


















  • 8




    This appears more like a math problem to me.
    – Ahmed Abdelhameed
    Dec 19 at 9:19






  • 1




    Or not... since it was already answered...
    – jafe
    Dec 19 at 11:11






  • 1




    @AhmedAbdelhameed It is a math problem, but it's one that has a surprising paradox at its heart. I think that makes it an interesting puzzle. A puzzle that can be solved using straightforward math isn't automatically not a puzzle.
    – Kevin
    Dec 19 at 14:28






  • 1




    Are all players entered into this tournament? If not then the "only 0.1% of all players are better" is a useless statistic.
    – AndyT
    Dec 19 at 16:12






  • 1




    @jafe - I don't understand your comment. For a fuller explanation of my comment, please see my longer comment under Kruga's answer.
    – AndyT
    Dec 19 at 16:41
















10















I am one of the greatest finger wrestlers of my generation. Only 0.1% of all players are better.



When I play against someone who's worse than me, I win 99% of the time. Against someone who's better than me, I always lose.



It's the first round of a tournament. We're playing best of three games (i.e. two wins needed). I hate to say it, but I have just lost the first game! What are my chances of still making it to the second round?



Assuming that the players who are better than me always beat the players who are worse than me, what are my chances of making it to the third round?











share|improve this question


















  • 8




    This appears more like a math problem to me.
    – Ahmed Abdelhameed
    Dec 19 at 9:19






  • 1




    Or not... since it was already answered...
    – jafe
    Dec 19 at 11:11






  • 1




    @AhmedAbdelhameed It is a math problem, but it's one that has a surprising paradox at its heart. I think that makes it an interesting puzzle. A puzzle that can be solved using straightforward math isn't automatically not a puzzle.
    – Kevin
    Dec 19 at 14:28






  • 1




    Are all players entered into this tournament? If not then the "only 0.1% of all players are better" is a useless statistic.
    – AndyT
    Dec 19 at 16:12






  • 1




    @jafe - I don't understand your comment. For a fuller explanation of my comment, please see my longer comment under Kruga's answer.
    – AndyT
    Dec 19 at 16:41














10












10








10


2






I am one of the greatest finger wrestlers of my generation. Only 0.1% of all players are better.



When I play against someone who's worse than me, I win 99% of the time. Against someone who's better than me, I always lose.



It's the first round of a tournament. We're playing best of three games (i.e. two wins needed). I hate to say it, but I have just lost the first game! What are my chances of still making it to the second round?



Assuming that the players who are better than me always beat the players who are worse than me, what are my chances of making it to the third round?











share|improve this question














I am one of the greatest finger wrestlers of my generation. Only 0.1% of all players are better.



When I play against someone who's worse than me, I win 99% of the time. Against someone who's better than me, I always lose.



It's the first round of a tournament. We're playing best of three games (i.e. two wins needed). I hate to say it, but I have just lost the first game! What are my chances of still making it to the second round?



Assuming that the players who are better than me always beat the players who are worse than me, what are my chances of making it to the third round?








mathematics probability






share|improve this question













share|improve this question











share|improve this question




share|improve this question










asked Dec 19 at 8:51









jafe

16.3k142160




16.3k142160








  • 8




    This appears more like a math problem to me.
    – Ahmed Abdelhameed
    Dec 19 at 9:19






  • 1




    Or not... since it was already answered...
    – jafe
    Dec 19 at 11:11






  • 1




    @AhmedAbdelhameed It is a math problem, but it's one that has a surprising paradox at its heart. I think that makes it an interesting puzzle. A puzzle that can be solved using straightforward math isn't automatically not a puzzle.
    – Kevin
    Dec 19 at 14:28






  • 1




    Are all players entered into this tournament? If not then the "only 0.1% of all players are better" is a useless statistic.
    – AndyT
    Dec 19 at 16:12






  • 1




    @jafe - I don't understand your comment. For a fuller explanation of my comment, please see my longer comment under Kruga's answer.
    – AndyT
    Dec 19 at 16:41














  • 8




    This appears more like a math problem to me.
    – Ahmed Abdelhameed
    Dec 19 at 9:19






  • 1




    Or not... since it was already answered...
    – jafe
    Dec 19 at 11:11






  • 1




    @AhmedAbdelhameed It is a math problem, but it's one that has a surprising paradox at its heart. I think that makes it an interesting puzzle. A puzzle that can be solved using straightforward math isn't automatically not a puzzle.
    – Kevin
    Dec 19 at 14:28






  • 1




    Are all players entered into this tournament? If not then the "only 0.1% of all players are better" is a useless statistic.
    – AndyT
    Dec 19 at 16:12






  • 1




    @jafe - I don't understand your comment. For a fuller explanation of my comment, please see my longer comment under Kruga's answer.
    – AndyT
    Dec 19 at 16:41








8




8




This appears more like a math problem to me.
– Ahmed Abdelhameed
Dec 19 at 9:19




This appears more like a math problem to me.
– Ahmed Abdelhameed
Dec 19 at 9:19




1




1




Or not... since it was already answered...
– jafe
Dec 19 at 11:11




Or not... since it was already answered...
– jafe
Dec 19 at 11:11




1




1




@AhmedAbdelhameed It is a math problem, but it's one that has a surprising paradox at its heart. I think that makes it an interesting puzzle. A puzzle that can be solved using straightforward math isn't automatically not a puzzle.
– Kevin
Dec 19 at 14:28




@AhmedAbdelhameed It is a math problem, but it's one that has a surprising paradox at its heart. I think that makes it an interesting puzzle. A puzzle that can be solved using straightforward math isn't automatically not a puzzle.
– Kevin
Dec 19 at 14:28




1




1




Are all players entered into this tournament? If not then the "only 0.1% of all players are better" is a useless statistic.
– AndyT
Dec 19 at 16:12




Are all players entered into this tournament? If not then the "only 0.1% of all players are better" is a useless statistic.
– AndyT
Dec 19 at 16:12




1




1




@jafe - I don't understand your comment. For a fuller explanation of my comment, please see my longer comment under Kruga's answer.
– AndyT
Dec 19 at 16:41




@jafe - I don't understand your comment. For a fuller explanation of my comment, please see my longer comment under Kruga's answer.
– AndyT
Dec 19 at 16:41










1 Answer
1






active

oldest

votes


















15















If you play 100.000 times versus random players, how many of them would be better/worse than you, and how many would you win/lose? We can put this in a grid.


     Worse   Better
Win 98901 0
Lose 999 100

Since you lost you first match there is a $999/1099$ chance that the your opponent is worse than you, and $100/1099$ chance that your opponent is better.


If your opponent is better, your chance of winning the series is 0. If your opponent is worse, your chance of winning is $0.99*0.99=0.9801$


You total chance of winning is then $0.9801*999/1099 + 0*100/1099 = 9791199/10990000 = 89.1%$





share|improve this answer

















  • 1




    Yep, this is correct! Nice work.
    – jafe
    Dec 19 at 11:19






  • 5




    This is known as the false positive paradox
    – BlueRaja - Danny Pflughoeft
    Dec 19 at 12:45






  • 3




    But what's the likelihood that the players entered into the tournament are evenly distributed across all skill ranges? Gary Kasparov doesn't play in tournaments at my local chess club, and a beginner at my local chess club doesn't play in the Chess World Cup. I suspect that any tournament the OP enters is likely to have a higher than average skill level among the entrants, hence his chances of winning are lower than you calculate.
    – AndyT
    Dec 19 at 16:11






  • 3




    @AndyT - in the cutthroat world of finger wrestling, you don't choose to play. You're conscripted. :-)
    – Kevin
    Dec 19 at 19:29










  • @AndyT You're right that that bit is an assumption, but in the absence of the OP explicitly providing that information, assuming it's random and uniformly distributed is the only sane assumption if you're trying to produce a number. But yea, OP probably should have stated we should assume that.
    – Shufflepants
    Dec 19 at 20:55











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1 Answer
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active

oldest

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1 Answer
1






active

oldest

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active

oldest

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active

oldest

votes









15















If you play 100.000 times versus random players, how many of them would be better/worse than you, and how many would you win/lose? We can put this in a grid.


     Worse   Better
Win 98901 0
Lose 999 100

Since you lost you first match there is a $999/1099$ chance that the your opponent is worse than you, and $100/1099$ chance that your opponent is better.


If your opponent is better, your chance of winning the series is 0. If your opponent is worse, your chance of winning is $0.99*0.99=0.9801$


You total chance of winning is then $0.9801*999/1099 + 0*100/1099 = 9791199/10990000 = 89.1%$





share|improve this answer

















  • 1




    Yep, this is correct! Nice work.
    – jafe
    Dec 19 at 11:19






  • 5




    This is known as the false positive paradox
    – BlueRaja - Danny Pflughoeft
    Dec 19 at 12:45






  • 3




    But what's the likelihood that the players entered into the tournament are evenly distributed across all skill ranges? Gary Kasparov doesn't play in tournaments at my local chess club, and a beginner at my local chess club doesn't play in the Chess World Cup. I suspect that any tournament the OP enters is likely to have a higher than average skill level among the entrants, hence his chances of winning are lower than you calculate.
    – AndyT
    Dec 19 at 16:11






  • 3




    @AndyT - in the cutthroat world of finger wrestling, you don't choose to play. You're conscripted. :-)
    – Kevin
    Dec 19 at 19:29










  • @AndyT You're right that that bit is an assumption, but in the absence of the OP explicitly providing that information, assuming it's random and uniformly distributed is the only sane assumption if you're trying to produce a number. But yea, OP probably should have stated we should assume that.
    – Shufflepants
    Dec 19 at 20:55
















15















If you play 100.000 times versus random players, how many of them would be better/worse than you, and how many would you win/lose? We can put this in a grid.


     Worse   Better
Win 98901 0
Lose 999 100

Since you lost you first match there is a $999/1099$ chance that the your opponent is worse than you, and $100/1099$ chance that your opponent is better.


If your opponent is better, your chance of winning the series is 0. If your opponent is worse, your chance of winning is $0.99*0.99=0.9801$


You total chance of winning is then $0.9801*999/1099 + 0*100/1099 = 9791199/10990000 = 89.1%$





share|improve this answer

















  • 1




    Yep, this is correct! Nice work.
    – jafe
    Dec 19 at 11:19






  • 5




    This is known as the false positive paradox
    – BlueRaja - Danny Pflughoeft
    Dec 19 at 12:45






  • 3




    But what's the likelihood that the players entered into the tournament are evenly distributed across all skill ranges? Gary Kasparov doesn't play in tournaments at my local chess club, and a beginner at my local chess club doesn't play in the Chess World Cup. I suspect that any tournament the OP enters is likely to have a higher than average skill level among the entrants, hence his chances of winning are lower than you calculate.
    – AndyT
    Dec 19 at 16:11






  • 3




    @AndyT - in the cutthroat world of finger wrestling, you don't choose to play. You're conscripted. :-)
    – Kevin
    Dec 19 at 19:29










  • @AndyT You're right that that bit is an assumption, but in the absence of the OP explicitly providing that information, assuming it's random and uniformly distributed is the only sane assumption if you're trying to produce a number. But yea, OP probably should have stated we should assume that.
    – Shufflepants
    Dec 19 at 20:55














15












15








15







If you play 100.000 times versus random players, how many of them would be better/worse than you, and how many would you win/lose? We can put this in a grid.


     Worse   Better
Win 98901 0
Lose 999 100

Since you lost you first match there is a $999/1099$ chance that the your opponent is worse than you, and $100/1099$ chance that your opponent is better.


If your opponent is better, your chance of winning the series is 0. If your opponent is worse, your chance of winning is $0.99*0.99=0.9801$


You total chance of winning is then $0.9801*999/1099 + 0*100/1099 = 9791199/10990000 = 89.1%$





share|improve this answer













If you play 100.000 times versus random players, how many of them would be better/worse than you, and how many would you win/lose? We can put this in a grid.


     Worse   Better
Win 98901 0
Lose 999 100

Since you lost you first match there is a $999/1099$ chance that the your opponent is worse than you, and $100/1099$ chance that your opponent is better.


If your opponent is better, your chance of winning the series is 0. If your opponent is worse, your chance of winning is $0.99*0.99=0.9801$


You total chance of winning is then $0.9801*999/1099 + 0*100/1099 = 9791199/10990000 = 89.1%$






share|improve this answer












share|improve this answer



share|improve this answer










answered Dec 19 at 9:30









Kruga

2,672926




2,672926








  • 1




    Yep, this is correct! Nice work.
    – jafe
    Dec 19 at 11:19






  • 5




    This is known as the false positive paradox
    – BlueRaja - Danny Pflughoeft
    Dec 19 at 12:45






  • 3




    But what's the likelihood that the players entered into the tournament are evenly distributed across all skill ranges? Gary Kasparov doesn't play in tournaments at my local chess club, and a beginner at my local chess club doesn't play in the Chess World Cup. I suspect that any tournament the OP enters is likely to have a higher than average skill level among the entrants, hence his chances of winning are lower than you calculate.
    – AndyT
    Dec 19 at 16:11






  • 3




    @AndyT - in the cutthroat world of finger wrestling, you don't choose to play. You're conscripted. :-)
    – Kevin
    Dec 19 at 19:29










  • @AndyT You're right that that bit is an assumption, but in the absence of the OP explicitly providing that information, assuming it's random and uniformly distributed is the only sane assumption if you're trying to produce a number. But yea, OP probably should have stated we should assume that.
    – Shufflepants
    Dec 19 at 20:55














  • 1




    Yep, this is correct! Nice work.
    – jafe
    Dec 19 at 11:19






  • 5




    This is known as the false positive paradox
    – BlueRaja - Danny Pflughoeft
    Dec 19 at 12:45






  • 3




    But what's the likelihood that the players entered into the tournament are evenly distributed across all skill ranges? Gary Kasparov doesn't play in tournaments at my local chess club, and a beginner at my local chess club doesn't play in the Chess World Cup. I suspect that any tournament the OP enters is likely to have a higher than average skill level among the entrants, hence his chances of winning are lower than you calculate.
    – AndyT
    Dec 19 at 16:11






  • 3




    @AndyT - in the cutthroat world of finger wrestling, you don't choose to play. You're conscripted. :-)
    – Kevin
    Dec 19 at 19:29










  • @AndyT You're right that that bit is an assumption, but in the absence of the OP explicitly providing that information, assuming it's random and uniformly distributed is the only sane assumption if you're trying to produce a number. But yea, OP probably should have stated we should assume that.
    – Shufflepants
    Dec 19 at 20:55








1




1




Yep, this is correct! Nice work.
– jafe
Dec 19 at 11:19




Yep, this is correct! Nice work.
– jafe
Dec 19 at 11:19




5




5




This is known as the false positive paradox
– BlueRaja - Danny Pflughoeft
Dec 19 at 12:45




This is known as the false positive paradox
– BlueRaja - Danny Pflughoeft
Dec 19 at 12:45




3




3




But what's the likelihood that the players entered into the tournament are evenly distributed across all skill ranges? Gary Kasparov doesn't play in tournaments at my local chess club, and a beginner at my local chess club doesn't play in the Chess World Cup. I suspect that any tournament the OP enters is likely to have a higher than average skill level among the entrants, hence his chances of winning are lower than you calculate.
– AndyT
Dec 19 at 16:11




But what's the likelihood that the players entered into the tournament are evenly distributed across all skill ranges? Gary Kasparov doesn't play in tournaments at my local chess club, and a beginner at my local chess club doesn't play in the Chess World Cup. I suspect that any tournament the OP enters is likely to have a higher than average skill level among the entrants, hence his chances of winning are lower than you calculate.
– AndyT
Dec 19 at 16:11




3




3




@AndyT - in the cutthroat world of finger wrestling, you don't choose to play. You're conscripted. :-)
– Kevin
Dec 19 at 19:29




@AndyT - in the cutthroat world of finger wrestling, you don't choose to play. You're conscripted. :-)
– Kevin
Dec 19 at 19:29












@AndyT You're right that that bit is an assumption, but in the absence of the OP explicitly providing that information, assuming it's random and uniformly distributed is the only sane assumption if you're trying to produce a number. But yea, OP probably should have stated we should assume that.
– Shufflepants
Dec 19 at 20:55




@AndyT You're right that that bit is an assumption, but in the absence of the OP explicitly providing that information, assuming it's random and uniformly distributed is the only sane assumption if you're trying to produce a number. But yea, OP probably should have stated we should assume that.
– Shufflepants
Dec 19 at 20:55


















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