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I'm not sure if this answer is right, but I'll do it anyways. For the grade 8 POTW: Alice must move the 3rd coin from the left to the most right. This means that there are only two coins that Beth can move, which are the first and second. However, the first coin is blocked by the second coin, so for the second turn Beth can only move the second coin.
Once we know this, We can figure out the answer.
Every time that Beth moves the second coin, Alice can follow the second coin to the same number of steps right behind it. For example, if Beth moves the second coin 3 units, then Alice can move the first coin 3 units to the right, following right behind. This will go until Beth moves the coin to the second right square, where Alice will follow, so Beth will not be able to move. -Alan
Grade 7 POTW: I think this was more complicated than the grade 8 POTW(Probably because I did the grade 8 one wrong) but here is how I did it: First of all, I need all the different possibilities. There are two different ones: One for the two numbers rolled and one for the digits. Note: I can have repeats because I am trying to find the probability. So, the first one, which is the digit sum, is:
And the second one is: 2,1 2,3 2,5 2,7 2,9 2,11 4,1 4,3 4,5 4,7 4,9 4,11 6,1 6,3 6,5 6,7 6,9 6,11 8,1 8,3 8,5 8,7 8,9 8,11 10,1 10,3 10,5 10,7 10,9 10,11 12,1 12,3 12,5 12,7 12,9 12,11
So, Now we have to see how many of these values fit the requirements. For multiples of 4(Anna): There are 6 4s and 4 8s, adding up to be a 10. The fraction would then be a 10/36
For multiples of each other(Elle) There would be: 2,1 4,1 6,1 6,3 8,1 10,1 10,5 12,1 12,3 This is a total of 9, which the fraction would be 9/36. Since one fraction is 10/36, and one fraction is 9/36, these two fractions are different so the game is not fair.
-Alan P.S. This question took me like 20 mins to do, so good luck to others!
All Alice has to do is move the penny on the very right to the last spot on the right. Beth can then only move the second penny to the 2nd last space. Alice moves the untouched penny to the 3rd last space, leaving Beth with no other move available. Then she wins. Obviously.
After repeating this process (Not writing it down, since it would be WAY too long), I found that Anna won 10/36 times, and Elle only won 9/36 times, so the game is not fair.
In order to figure this out, I created a tree diagram! On a piece of paper but it essentially goes like this, if the first die lands on 1, I create branches for the even numbers. And that would create a total of 6 combinations. And since there are six sides and there are six combinations for one tree diagram, the number of possibilities would be 36. From there, we just write that all the possibilities and determine whether Elle or Anna gets a point. In the end, the game is not fair because Anna has a more likelihood of winning than Elle.
Anne and Elle both have 9 possibilities in which they can win. This means that there is a total of 18 possibilities. Therefore, they both have a 9/18 or 1/2 chance in winning. This means that the game is fair.
Avi and Maxwell We made a mistake as we missed a possibility for Anna. So there should be 10 possibilities of winning for Anna. Therefore, Anna should have a higher chance of winning.
I'm not sure if this answer is right, but I'll do it anyways.
ReplyDeleteFor the grade 8 POTW: Alice must move the 3rd coin from the left to the most right. This means that there are only two coins that Beth can move, which are the first and second. However, the first coin is blocked by the second coin, so for the second turn Beth can only move the second coin.
Once we know this, We can figure out the answer.
Every time that Beth moves the second coin, Alice can follow the second coin to the same number of steps right behind it. For example, if Beth moves the second coin 3 units, then Alice can move the first coin 3 units to the right, following right behind. This will go until Beth moves the coin to the second right square, where Alice will follow, so Beth will not be able to move.
-Alan
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DeleteGrade 7 POTW:
ReplyDeleteI think this was more complicated than the grade 8 POTW(Probably because I did the grade 8 one wrong) but here is how I did it:
First of all, I need all the different possibilities.
There are two different ones: One for the two numbers rolled and one for the digits.
Note: I can have repeats because I am trying to find the probability.
So, the first one, which is the digit sum, is:
3 5 7 9 11 13
5 7 9 11 13 15
7 9 11 13 15 17
9 11 13 15 17 19
11 13 15 17 19 21
13 15 17 19 21 23
And the second one is:
2,1 2,3 2,5 2,7 2,9 2,11
4,1 4,3 4,5 4,7 4,9 4,11
6,1 6,3 6,5 6,7 6,9 6,11
8,1 8,3 8,5 8,7 8,9 8,11
10,1 10,3 10,5 10,7 10,9 10,11
12,1 12,3 12,5 12,7 12,9 12,11
So, Now we have to see how many of these values fit the requirements.
For multiples of 4(Anna): There are 6 4s and 4 8s, adding up to be a 10.
The fraction would then be a 10/36
For multiples of each other(Elle) There would be:
2,1 4,1 6,1 6,3 8,1 10,1 10,5 12,1 12,3
This is a total of 9, which the fraction would be 9/36.
Since one fraction is 10/36, and one fraction is 9/36, these two fractions are different so the game is not fair.
-Alan
P.S. This question took me like 20 mins to do, so good luck to others!
took me 5 minutes to read question
Delete2 minutes to do
All Alice has to do is move the penny on the very right to the last spot on the right. Beth can then only move the second penny to the 2nd last space. Alice moves the untouched penny to the 3rd last space, leaving Beth with no other move available. Then she wins. Obviously.
ReplyDeleteElle has a huge advantage over Anna because Anna has 4 chances to win, and Elle has 11 chances.
ReplyDeleteNVM my previous one, that is if the die were normal
ReplyDeleteIn order to figure out the probability of Anna/Elle getting a point, I decided to find out who would get a point with every possible S numbers
ReplyDelete1+2=E
1+4=E
1+6=E
1+8=E
1+10=E
1+12=E/A
2+1=E
2+3=N
2+5=N
2+7=N
2+9=N
2+11=A
4+1=E
4+3=N
4+5=N
4+7=N
4+9=A
4+11=N......
After repeating this process (Not writing it down, since it would be WAY too long), I found that Anna won 10/36 times, and Elle only won 9/36 times, so the game is not fair.
In order to figure this out, I created a tree diagram! On a piece of paper but it essentially goes like this, if the first die lands on 1, I create branches for the even numbers. And that would create a total of 6 combinations. And since there are six sides and there are six combinations for one tree diagram, the number of possibilities would be 36. From there, we just write that all the possibilities and determine whether Elle or Anna gets a point. In the end, the game is not fair because Anna has a more likelihood of winning than Elle.
ReplyDeletehow do you fit that in one paper?
DeleteMaxwell and Avi POTW:
ReplyDeleteAnne and Elle both have 9 possibilities in which they can win. This means that there is a total of 18 possibilities. Therefore, they both have a 9/18 or 1/2 chance in winning. This means that the game is fair.
Avi and Maxwell
ReplyDeleteWe made a mistake as we missed a possibility for Anna. So there should be 10 possibilities of winning for Anna. Therefore, Anna should have a higher chance of winning.
Anna has a higher chance of winning.