Mixing
Two players multiply a number until it reaches 1000
$begingroup$
Two players start with $1$ and take turns multiplying it by $2$ or $3$ or $4$ ... or $9$. The first player to make the number $ge 1000$ wins. Who has a winning strategy?
My attempt: obviously $112$ is a losing position since $112*9=1008$. Then
it's a winning position for all numbers between $56$ to $111$. Then the first player can choose to multiply by $5$ on the first turn, then on his second turn he can always multiply a number so it's between $56$ and $111$. Therefore the first player has a winning strategy.
Is this solution correct. Is there another way of doing it? Also, in general, who has a winning strategy if the first player to make the number $ge n$ wins?
combinatorics proof-verification game-theory
edited Jan 30 at 10:48
Asaf Karagila♦
307k33441774
asked Jan 30 at 5:10
abc...abc...
3,237739
$endgroup$
add a comment |
$begingroup$
Two players start with $1$ and take turns multiplying it by $2$ or $3$ or $4$ ... or $9$. The first player to make the number $ge 1000$ wins. Who has a winning strategy?
My attempt: obviously $112$ is a losing position since $112*9=1008$. Then
it's a winning position for all numbers between $56$ to $111$. Then the first player can choose to multiply by $5$ on the first turn, then on his second turn he can always multiply a number so it's between $56$ and $111$. Therefore the first player has a winning strategy.
Is this solution correct. Is there another way of doing it? Also, in general, who has a winning strategy if the first player to make the number $ge n$ wins?
combinatorics proof-verification game-theory
edited Jan 30 at 10:48
Asaf Karagila♦
307k33441774
asked Jan 30 at 5:10
abc...abc...
3,237739
$endgroup$
$begingroup$
Do you think "generalization" is a good tag?
$endgroup$
– YuiTo Cheng
Jan 30 at 7:33
$begingroup$
I think the usual terminology is to call $112$ a winning position (i.e. it is winning for the player to move).
$endgroup$
– TonyK
Jan 30 at 10:52
add a comment |
$begingroup$
Two players start with $1$ and take turns multiplying it by $2$ or $3$ or $4$ ... or $9$. The first player to make the number $ge 1000$ wins. Who has a winning strategy?
My attempt: obviously $112$ is a losing position since $112*9=1008$. Then
it's a winning position for all numbers between $56$ to $111$. Then the first player can choose to multiply by $5$ on the first turn, then on his second turn he can always multiply a number so it's between $56$ and $111$. Therefore the first player has a winning strategy.
Is this solution correct. Is there another way of doing it? Also, in general, who has a winning strategy if the first player to make the number $ge n$ wins?
combinatorics proof-verification game-theory
edited Jan 30 at 10:48
Asaf Karagila♦
307k33441774
asked Jan 30 at 5:10
abc...abc...
3,237739
$endgroup$
Two players start with $1$ and take turns multiplying it by $2$ or $3$ or $4$ ... or $9$. The first player to make the number $ge 1000$ wins. Who has a winning strategy?
My attempt: obviously $112$ is a losing position since $112*9=1008$. Then
it's a winning position for all numbers between $56$ to $111$. Then the first player can choose to multiply by $5$ on the first turn, then on his second turn he can always multiply a number so it's between $56$ and $111$. Therefore the first player has a winning strategy.
Is this solution correct. Is there another way of doing it? Also, in general, who has a winning strategy if the first player to make the number $ge n$ wins?
combinatorics proof-verification game-theory
combinatorics proof-verification game-theory
edited Jan 30 at 10:48
Asaf Karagila♦
307k33441774
asked Jan 30 at 5:10
abc...abc...
3,237739
edited Jan 30 at 10:48
Asaf Karagila♦
307k33441774
asked Jan 30 at 5:10
abc...abc...
3,237739
edited Jan 30 at 10:48
Asaf Karagila♦
307k33441774
edited Jan 30 at 10:48
Asaf Karagila♦
307k33441774
edited Jan 30 at 10:48
Asaf Karagila♦
307k33441774
307k33441774
asked Jan 30 at 5:10
abc...abc...
3,237739
asked Jan 30 at 5:10
abc...abc...
3,237739
asked Jan 30 at 5:10
abc...abc...
3,237739
3,237739
$begingroup$
Do you think "generalization" is a good tag?
$endgroup$
– YuiTo Cheng
Jan 30 at 7:33
$begingroup$
I think the usual terminology is to call $112$ a winning position (i.e. it is winning for the player to move).
$endgroup$
– TonyK
Jan 30 at 10:52
add a comment |
$begingroup$
Do you think "generalization" is a good tag?
$endgroup$
– YuiTo Cheng
Jan 30 at 7:33
$begingroup$
I think the usual terminology is to call $112$ a winning position (i.e. it is winning for the player to move).
$endgroup$
– TonyK
Jan 30 at 10:52
$begingroup$
Do you think "generalization" is a good tag?
$endgroup$
– YuiTo Cheng
Jan 30 at 7:33
$begingroup$
Do you think "generalization" is a good tag?
$endgroup$
– YuiTo Cheng
Jan 30 at 7:33
$begingroup$
I think the usual terminology is to call $112$ a winning position (i.e. it is winning for the player to move).
$endgroup$
– TonyK
Jan 30 at 10:52
$begingroup$
I think the usual terminology is to call $112$ a winning position (i.e. it is winning for the player to move).
$endgroup$
– TonyK
Jan 30 at 10:52
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Yes, this is a good strategy and demonstration that it works. $4$ and $6$ also work.
You might compare with the game where you add instead of multiply and are allowed to add $1,2,3,4$ at each turn and the first to reach some total, say $27$ wins. A player can always guarantee that two successive turns add to $5$, so the first player wins by taking $2$, then hitting $7,12,17,22,27$.
In your game a player can guarantee that two turns multiply by at least $18$ but one turn cannot multiply by more than $9$. Your $56$ comes from $lceil frac {1000}{18}rceil$ and $111$ from $lfloor frac {1000}9 rfloor$. This suggests that for target $T$ and $n$ pairs of turns before the end you should leave between $lceil frac T{18^n} rceil$ and $lfloor frac T{18^{n-1}cdot 9}rfloor$ The worry is that you can't guarantee that two turns will multiply by exactly $18$. If your opponent multiplies by $7$ you have to choose between $2$ and $3$, making the pair multiply by $14$ or $21$. I believe you are OK because the range is a factor $2$. In this example if multiplying by $3$ is too much, $2$ will be enough to stay in range.
answered Jan 30 at 5:32
Ross MillikanRoss Millikan
301k24200375
$endgroup$
add a comment |
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1 Answer
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$begingroup$
Yes, this is a good strategy and demonstration that it works. $4$ and $6$ also work.
You might compare with the game where you add instead of multiply and are allowed to add $1,2,3,4$ at each turn and the first to reach some total, say $27$ wins. A player can always guarantee that two successive turns add to $5$, so the first player wins by taking $2$, then hitting $7,12,17,22,27$.
In your game a player can guarantee that two turns multiply by at least $18$ but one turn cannot multiply by more than $9$. Your $56$ comes from $lceil frac {1000}{18}rceil$ and $111$ from $lfloor frac {1000}9 rfloor$. This suggests that for target $T$ and $n$ pairs of turns before the end you should leave between $lceil frac T{18^n} rceil$ and $lfloor frac T{18^{n-1}cdot 9}rfloor$ The worry is that you can't guarantee that two turns will multiply by exactly $18$. If your opponent multiplies by $7$ you have to choose between $2$ and $3$, making the pair multiply by $14$ or $21$. I believe you are OK because the range is a factor $2$. In this example if multiplying by $3$ is too much, $2$ will be enough to stay in range.
answered Jan 30 at 5:32
Ross MillikanRoss Millikan
301k24200375
$endgroup$
add a comment |
$begingroup$
Yes, this is a good strategy and demonstration that it works. $4$ and $6$ also work.
You might compare with the game where you add instead of multiply and are allowed to add $1,2,3,4$ at each turn and the first to reach some total, say $27$ wins. A player can always guarantee that two successive turns add to $5$, so the first player wins by taking $2$, then hitting $7,12,17,22,27$.
In your game a player can guarantee that two turns multiply by at least $18$ but one turn cannot multiply by more than $9$. Your $56$ comes from $lceil frac {1000}{18}rceil$ and $111$ from $lfloor frac {1000}9 rfloor$. This suggests that for target $T$ and $n$ pairs of turns before the end you should leave between $lceil frac T{18^n} rceil$ and $lfloor frac T{18^{n-1}cdot 9}rfloor$ The worry is that you can't guarantee that two turns will multiply by exactly $18$. If your opponent multiplies by $7$ you have to choose between $2$ and $3$, making the pair multiply by $14$ or $21$. I believe you are OK because the range is a factor $2$. In this example if multiplying by $3$ is too much, $2$ will be enough to stay in range.
answered Jan 30 at 5:32
Ross MillikanRoss Millikan
301k24200375
$endgroup$
add a comment |
$begingroup$
Yes, this is a good strategy and demonstration that it works. $4$ and $6$ also work.
You might compare with the game where you add instead of multiply and are allowed to add $1,2,3,4$ at each turn and the first to reach some total, say $27$ wins. A player can always guarantee that two successive turns add to $5$, so the first player wins by taking $2$, then hitting $7,12,17,22,27$.
In your game a player can guarantee that two turns multiply by at least $18$ but one turn cannot multiply by more than $9$. Your $56$ comes from $lceil frac {1000}{18}rceil$ and $111$ from $lfloor frac {1000}9 rfloor$. This suggests that for target $T$ and $n$ pairs of turns before the end you should leave between $lceil frac T{18^n} rceil$ and $lfloor frac T{18^{n-1}cdot 9}rfloor$ The worry is that you can't guarantee that two turns will multiply by exactly $18$. If your opponent multiplies by $7$ you have to choose between $2$ and $3$, making the pair multiply by $14$ or $21$. I believe you are OK because the range is a factor $2$. In this example if multiplying by $3$ is too much, $2$ will be enough to stay in range.
answered Jan 30 at 5:32
Ross MillikanRoss Millikan
301k24200375
$endgroup$
Yes, this is a good strategy and demonstration that it works. $4$ and $6$ also work.
You might compare with the game where you add instead of multiply and are allowed to add $1,2,3,4$ at each turn and the first to reach some total, say $27$ wins. A player can always guarantee that two successive turns add to $5$, so the first player wins by taking $2$, then hitting $7,12,17,22,27$.
In your game a player can guarantee that two turns multiply by at least $18$ but one turn cannot multiply by more than $9$. Your $56$ comes from $lceil frac {1000}{18}rceil$ and $111$ from $lfloor frac {1000}9 rfloor$. This suggests that for target $T$ and $n$ pairs of turns before the end you should leave between $lceil frac T{18^n} rceil$ and $lfloor frac T{18^{n-1}cdot 9}rfloor$ The worry is that you can't guarantee that two turns will multiply by exactly $18$. If your opponent multiplies by $7$ you have to choose between $2$ and $3$, making the pair multiply by $14$ or $21$. I believe you are OK because the range is a factor $2$. In this example if multiplying by $3$ is too much, $2$ will be enough to stay in range.
answered Jan 30 at 5:32
Ross MillikanRoss Millikan
301k24200375
answered Jan 30 at 5:32
Ross MillikanRoss Millikan
301k24200375
answered Jan 30 at 5:32
Ross MillikanRoss Millikan
301k24200375
answered Jan 30 at 5:32
Ross MillikanRoss Millikan
301k24200375
301k24200375
add a comment |
add a comment |
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$begingroup$
Do you think "generalization" is a good tag?
$endgroup$
– YuiTo Cheng
Jan 30 at 7:33
$begingroup$
I think the usual terminology is to call $112$ a winning position (i.e. it is winning for the player to move).
$endgroup$
– TonyK
Jan 30 at 10:52