Blind Person Ball Game
$begingroup$
You are blind and your friend presents you with 6 equal balls of which 3 are red and 3 are green.
You friend challenges you to game to find a green ball.
You must select 3 of the balls and your friend will tell you if they are all green or not. You repeat this as many times as you need to until you can identify a ball as being green.
In a worst case scenario what is the maximum number of submissions of 3 balls needed to then be able to identify a green ball with certainty. Assuming you are playing an optimal game. How do you know there is no better strategy?
mathematics logical-deduction
asked Dec 3 '18 at 9:50
Ben FranksBen Franks
50214
$endgroup$
add a comment |
$begingroup$
You are blind and your friend presents you with 6 equal balls of which 3 are red and 3 are green.
You friend challenges you to game to find a green ball.
You must select 3 of the balls and your friend will tell you if they are all green or not. You repeat this as many times as you need to until you can identify a ball as being green.
In a worst case scenario what is the maximum number of submissions of 3 balls needed to then be able to identify a green ball with certainty. Assuming you are playing an optimal game. How do you know there is no better strategy?
mathematics logical-deduction
asked Dec 3 '18 at 9:50
Ben FranksBen Franks
50214
$endgroup$
add a comment |
$begingroup$
You are blind and your friend presents you with 6 equal balls of which 3 are red and 3 are green.
You friend challenges you to game to find a green ball.
You must select 3 of the balls and your friend will tell you if they are all green or not. You repeat this as many times as you need to until you can identify a ball as being green.
In a worst case scenario what is the maximum number of submissions of 3 balls needed to then be able to identify a green ball with certainty. Assuming you are playing an optimal game. How do you know there is no better strategy?
mathematics logical-deduction
asked Dec 3 '18 at 9:50
Ben FranksBen Franks
50214
$endgroup$
You are blind and your friend presents you with 6 equal balls of which 3 are red and 3 are green.
You friend challenges you to game to find a green ball.
You must select 3 of the balls and your friend will tell you if they are all green or not. You repeat this as many times as you need to until you can identify a ball as being green.
In a worst case scenario what is the maximum number of submissions of 3 balls needed to then be able to identify a green ball with certainty. Assuming you are playing an optimal game. How do you know there is no better strategy?
mathematics logical-deduction
mathematics logical-deduction
asked Dec 3 '18 at 9:50
Ben FranksBen Franks
50214
asked Dec 3 '18 at 9:50
Ben FranksBen Franks
50214
asked Dec 3 '18 at 9:50
Ben FranksBen Franks
50214
asked Dec 3 '18 at 9:50
Ben FranksBen Franks
50214
asked Dec 3 '18 at 9:50
Ben FranksBen Franks
50214
50214
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
elias's strategy for
10
is optimal.
To see this, note that
there are 20 possible triples, which can be divided into 10 complementary pairs, i.e. each triple is paired with the triple consisting of the other three balls. Now suppose you follow any strategy, and after nine attempts, none of the triples you've tried was the all-green triple. There must be one pair of complementary triples, neither of which has been tried (since there are ten pairs). Either of these two triples could be the all-green triple; since these two cases have no green balls in common, you can't definitively identify a green ball.
answered Dec 3 '18 at 10:29
Especially LimeEspecially Lime
1,536413
$endgroup$
add a comment |
$begingroup$
I think I have an answer of
$binom53 = 10$
Approach:
Choose any 5 balls, and ask for every subset of 3 of them.
If none of these triplets are 'all green', that means ball 6 has to be green.
answered Dec 3 '18 at 10:05
eliaselias
8,73332254
$endgroup$
$begingroup$
+1 Nice answer.
$endgroup$
– hexomino
Dec 3 '18 at 10:10
1
$begingroup$
Thanks, but far from being complete. No idea if this is minimal, and if yes, how to prove.
$endgroup$
– elias
Dec 3 '18 at 10:11
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
elias's strategy for
10
is optimal.
To see this, note that
there are 20 possible triples, which can be divided into 10 complementary pairs, i.e. each triple is paired with the triple consisting of the other three balls. Now suppose you follow any strategy, and after nine attempts, none of the triples you've tried was the all-green triple. There must be one pair of complementary triples, neither of which has been tried (since there are ten pairs). Either of these two triples could be the all-green triple; since these two cases have no green balls in common, you can't definitively identify a green ball.
answered Dec 3 '18 at 10:29
Especially LimeEspecially Lime
1,536413
$endgroup$
add a comment |
$begingroup$
elias's strategy for
10
is optimal.
To see this, note that
there are 20 possible triples, which can be divided into 10 complementary pairs, i.e. each triple is paired with the triple consisting of the other three balls. Now suppose you follow any strategy, and after nine attempts, none of the triples you've tried was the all-green triple. There must be one pair of complementary triples, neither of which has been tried (since there are ten pairs). Either of these two triples could be the all-green triple; since these two cases have no green balls in common, you can't definitively identify a green ball.
answered Dec 3 '18 at 10:29
Especially LimeEspecially Lime
1,536413
$endgroup$
add a comment |
$begingroup$
elias's strategy for
10
is optimal.
To see this, note that
there are 20 possible triples, which can be divided into 10 complementary pairs, i.e. each triple is paired with the triple consisting of the other three balls. Now suppose you follow any strategy, and after nine attempts, none of the triples you've tried was the all-green triple. There must be one pair of complementary triples, neither of which has been tried (since there are ten pairs). Either of these two triples could be the all-green triple; since these two cases have no green balls in common, you can't definitively identify a green ball.
answered Dec 3 '18 at 10:29
Especially LimeEspecially Lime
1,536413
$endgroup$
elias's strategy for
10
is optimal.
To see this, note that
there are 20 possible triples, which can be divided into 10 complementary pairs, i.e. each triple is paired with the triple consisting of the other three balls. Now suppose you follow any strategy, and after nine attempts, none of the triples you've tried was the all-green triple. There must be one pair of complementary triples, neither of which has been tried (since there are ten pairs). Either of these two triples could be the all-green triple; since these two cases have no green balls in common, you can't definitively identify a green ball.
answered Dec 3 '18 at 10:29
Especially LimeEspecially Lime
1,536413
answered Dec 3 '18 at 10:29
Especially LimeEspecially Lime
1,536413
answered Dec 3 '18 at 10:29
Especially LimeEspecially Lime
1,536413
answered Dec 3 '18 at 10:29
Especially LimeEspecially Lime
1,536413
1,536413
add a comment |
add a comment |
$begingroup$
I think I have an answer of
$binom53 = 10$
Approach:
Choose any 5 balls, and ask for every subset of 3 of them.
If none of these triplets are 'all green', that means ball 6 has to be green.
answered Dec 3 '18 at 10:05
eliaselias
8,73332254
$endgroup$
$begingroup$
+1 Nice answer.
$endgroup$
– hexomino
Dec 3 '18 at 10:10
1
$begingroup$
Thanks, but far from being complete. No idea if this is minimal, and if yes, how to prove.
$endgroup$
– elias
Dec 3 '18 at 10:11
add a comment |
$begingroup$
I think I have an answer of
$binom53 = 10$
Approach:
Choose any 5 balls, and ask for every subset of 3 of them.
If none of these triplets are 'all green', that means ball 6 has to be green.
answered Dec 3 '18 at 10:05
eliaselias
8,73332254
$endgroup$
$begingroup$
+1 Nice answer.
$endgroup$
– hexomino
Dec 3 '18 at 10:10
1
$begingroup$
Thanks, but far from being complete. No idea if this is minimal, and if yes, how to prove.
$endgroup$
– elias
Dec 3 '18 at 10:11
add a comment |
$begingroup$
I think I have an answer of
$binom53 = 10$
Approach:
Choose any 5 balls, and ask for every subset of 3 of them.
If none of these triplets are 'all green', that means ball 6 has to be green.
answered Dec 3 '18 at 10:05
eliaselias
8,73332254
$endgroup$
I think I have an answer of
$binom53 = 10$
Approach:
Choose any 5 balls, and ask for every subset of 3 of them.
If none of these triplets are 'all green', that means ball 6 has to be green.
answered Dec 3 '18 at 10:05
eliaselias
8,73332254
answered Dec 3 '18 at 10:05
eliaselias
8,73332254
answered Dec 3 '18 at 10:05
eliaselias
8,73332254
answered Dec 3 '18 at 10:05
eliaselias
8,73332254
8,73332254
$begingroup$
+1 Nice answer.
$endgroup$
– hexomino
Dec 3 '18 at 10:10
1
$begingroup$
Thanks, but far from being complete. No idea if this is minimal, and if yes, how to prove.
$endgroup$
– elias
Dec 3 '18 at 10:11
add a comment |
$begingroup$
+1 Nice answer.
$endgroup$
– hexomino
Dec 3 '18 at 10:10
1
$begingroup$
Thanks, but far from being complete. No idea if this is minimal, and if yes, how to prove.
$endgroup$
– elias
Dec 3 '18 at 10:11
$begingroup$
+1 Nice answer.
$endgroup$
– hexomino
Dec 3 '18 at 10:10
$begingroup$
+1 Nice answer.
$endgroup$
– hexomino
Dec 3 '18 at 10:10
1
1
$begingroup$
Thanks, but far from being complete. No idea if this is minimal, and if yes, how to prove.
$endgroup$
– elias
Dec 3 '18 at 10:11
$begingroup$
Thanks, but far from being complete. No idea if this is minimal, and if yes, how to prove.
$endgroup$
– elias
Dec 3 '18 at 10:11
add a comment |
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