Mixing
Magic the gathering welcome pack favorable color distribution
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I was wondering if there is a simple way to get this problem.
In magic the gathering there are five colors, { Red, White, Blue, Black, Geeen}. In a welcome deck you get to pick one of the five colors, the deck has both that color and a randomly chosen color that is different from it. If I get one pack from each color what is the probability that It evens out and I get two of each color as opposed to some other arrangement?
I have the total sample space as being 4^5 total outcomes
If it were total random For each color instead I would have 5^5 total outcomes and would be able to use the easy combination of 5! Total outcomes being the way you can arrange the 5 individual colors. Therefore having 5!/5^5 in the easier example
probability
asked Jan 16 at 0:49
Toni StackToni Stack
83
$endgroup$
|
show 1 more comment
$begingroup$
I was wondering if there is a simple way to get this problem.
In magic the gathering there are five colors, { Red, White, Blue, Black, Geeen}. In a welcome deck you get to pick one of the five colors, the deck has both that color and a randomly chosen color that is different from it. If I get one pack from each color what is the probability that It evens out and I get two of each color as opposed to some other arrangement?
I have the total sample space as being 4^5 total outcomes
If it were total random For each color instead I would have 5^5 total outcomes and would be able to use the easy combination of 5! Total outcomes being the way you can arrange the 5 individual colors. Therefore having 5!/5^5 in the easier example
probability
asked Jan 16 at 0:49
Toni StackToni Stack
83
$endgroup$
1
$begingroup$
I think this is a stars and bars problem. $x_1 + x_2 + x_3 + x_4 + x_5 = 5$ where the $x_i$'s can be zero. So the probability is one divided by $9$ choose $5$ or one in $126$. What do you think?
$endgroup$
– HJ_beginner
Jan 16 at 1:12
1
$begingroup$
Stars and bars is almost applicable, however it isn't because of the fact that there isn't $5$ choices for the second color of each pack. There are only $4$ choices for each, and the four choices are dependent upon which color deck you chose, as the second color is a "randomly chosen color that is different from it".
$endgroup$
– WaveX
Jan 16 at 1:21
$begingroup$
@WaveX ahhh that makes sense. Thanks for pointing that out for me!
$endgroup$
– HJ_beginner
Jan 16 at 1:57
$begingroup$
In practice, the distribution of colors in a case of welcome packs isn’t truly random. The cards are packed by cycling through some long fixed sequence.
$endgroup$
– amd
Jan 17 at 1:14
$begingroup$
What long fixed sequence is that?
$endgroup$
– Toni Stack
Jan 18 at 7:41
|
show 1 more comment
$begingroup$
I was wondering if there is a simple way to get this problem.
In magic the gathering there are five colors, { Red, White, Blue, Black, Geeen}. In a welcome deck you get to pick one of the five colors, the deck has both that color and a randomly chosen color that is different from it. If I get one pack from each color what is the probability that It evens out and I get two of each color as opposed to some other arrangement?
I have the total sample space as being 4^5 total outcomes
If it were total random For each color instead I would have 5^5 total outcomes and would be able to use the easy combination of 5! Total outcomes being the way you can arrange the 5 individual colors. Therefore having 5!/5^5 in the easier example
probability
asked Jan 16 at 0:49
Toni StackToni Stack
83
$endgroup$
I was wondering if there is a simple way to get this problem.
In magic the gathering there are five colors, { Red, White, Blue, Black, Geeen}. In a welcome deck you get to pick one of the five colors, the deck has both that color and a randomly chosen color that is different from it. If I get one pack from each color what is the probability that It evens out and I get two of each color as opposed to some other arrangement?
I have the total sample space as being 4^5 total outcomes
If it were total random For each color instead I would have 5^5 total outcomes and would be able to use the easy combination of 5! Total outcomes being the way you can arrange the 5 individual colors. Therefore having 5!/5^5 in the easier example
probability
probability
asked Jan 16 at 0:49
Toni StackToni Stack
83
asked Jan 16 at 0:49
Toni StackToni Stack
83
asked Jan 16 at 0:49
Toni StackToni Stack
83
asked Jan 16 at 0:49
Toni StackToni Stack
83
asked Jan 16 at 0:49
Toni StackToni Stack
83
83
1
$begingroup$
I think this is a stars and bars problem. $x_1 + x_2 + x_3 + x_4 + x_5 = 5$ where the $x_i$'s can be zero. So the probability is one divided by $9$ choose $5$ or one in $126$. What do you think?
$endgroup$
– HJ_beginner
Jan 16 at 1:12
1
$begingroup$
Stars and bars is almost applicable, however it isn't because of the fact that there isn't $5$ choices for the second color of each pack. There are only $4$ choices for each, and the four choices are dependent upon which color deck you chose, as the second color is a "randomly chosen color that is different from it".
$endgroup$
– WaveX
Jan 16 at 1:21
$begingroup$
@WaveX ahhh that makes sense. Thanks for pointing that out for me!
$endgroup$
– HJ_beginner
Jan 16 at 1:57
$begingroup$
In practice, the distribution of colors in a case of welcome packs isn’t truly random. The cards are packed by cycling through some long fixed sequence.
$endgroup$
– amd
Jan 17 at 1:14
$begingroup$
What long fixed sequence is that?
$endgroup$
– Toni Stack
Jan 18 at 7:41
|
show 1 more comment
1
$begingroup$
I think this is a stars and bars problem. $x_1 + x_2 + x_3 + x_4 + x_5 = 5$ where the $x_i$'s can be zero. So the probability is one divided by $9$ choose $5$ or one in $126$. What do you think?
$endgroup$
– HJ_beginner
Jan 16 at 1:12
1
$begingroup$
Stars and bars is almost applicable, however it isn't because of the fact that there isn't $5$ choices for the second color of each pack. There are only $4$ choices for each, and the four choices are dependent upon which color deck you chose, as the second color is a "randomly chosen color that is different from it".
$endgroup$
– WaveX
Jan 16 at 1:21
$begingroup$
@WaveX ahhh that makes sense. Thanks for pointing that out for me!
$endgroup$
– HJ_beginner
Jan 16 at 1:57
$begingroup$
In practice, the distribution of colors in a case of welcome packs isn’t truly random. The cards are packed by cycling through some long fixed sequence.
$endgroup$
– amd
Jan 17 at 1:14
$begingroup$
What long fixed sequence is that?
$endgroup$
– Toni Stack
Jan 18 at 7:41
1
1
$begingroup$
I think this is a stars and bars problem. $x_1 + x_2 + x_3 + x_4 + x_5 = 5$ where the $x_i$'s can be zero. So the probability is one divided by $9$ choose $5$ or one in $126$. What do you think?
$endgroup$
– HJ_beginner
Jan 16 at 1:12
$begingroup$
I think this is a stars and bars problem. $x_1 + x_2 + x_3 + x_4 + x_5 = 5$ where the $x_i$'s can be zero. So the probability is one divided by $9$ choose $5$ or one in $126$. What do you think?
$endgroup$
– HJ_beginner
Jan 16 at 1:12
1
1
$begingroup$
Stars and bars is almost applicable, however it isn't because of the fact that there isn't $5$ choices for the second color of each pack. There are only $4$ choices for each, and the four choices are dependent upon which color deck you chose, as the second color is a "randomly chosen color that is different from it".
$endgroup$
– WaveX
Jan 16 at 1:21
$begingroup$
Stars and bars is almost applicable, however it isn't because of the fact that there isn't $5$ choices for the second color of each pack. There are only $4$ choices for each, and the four choices are dependent upon which color deck you chose, as the second color is a "randomly chosen color that is different from it".
$endgroup$
– WaveX
Jan 16 at 1:21
$begingroup$
@WaveX ahhh that makes sense. Thanks for pointing that out for me!
$endgroup$
– HJ_beginner
Jan 16 at 1:57
$begingroup$
@WaveX ahhh that makes sense. Thanks for pointing that out for me!
$endgroup$
– HJ_beginner
Jan 16 at 1:57
$begingroup$
In practice, the distribution of colors in a case of welcome packs isn’t truly random. The cards are packed by cycling through some long fixed sequence.
$endgroup$
– amd
Jan 17 at 1:14
$begingroup$
In practice, the distribution of colors in a case of welcome packs isn’t truly random. The cards are packed by cycling through some long fixed sequence.
$endgroup$
– amd
Jan 17 at 1:14
$begingroup$
What long fixed sequence is that?
$endgroup$
– Toni Stack
Jan 18 at 7:41
$begingroup$
What long fixed sequence is that?
$endgroup$
– Toni Stack
Jan 18 at 7:41
|
show 1 more comment
1 Answer
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$begingroup$
The desired outcome is a derangement of $(r,w,u,b,g)$, thus there are $44$ ways to get one of each color from the randomly chosen colors.
$$frac{44}{1024}=4.296875%$$
answered Jan 16 at 1:24
Daniel MathiasDaniel Mathias
1,25318
$endgroup$
add a comment |
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$begingroup$
The desired outcome is a derangement of $(r,w,u,b,g)$, thus there are $44$ ways to get one of each color from the randomly chosen colors.
$$frac{44}{1024}=4.296875%$$
answered Jan 16 at 1:24
Daniel MathiasDaniel Mathias
1,25318
$endgroup$
add a comment |
$begingroup$
The desired outcome is a derangement of $(r,w,u,b,g)$, thus there are $44$ ways to get one of each color from the randomly chosen colors.
$$frac{44}{1024}=4.296875%$$
answered Jan 16 at 1:24
Daniel MathiasDaniel Mathias
1,25318
$endgroup$
add a comment |
$begingroup$
The desired outcome is a derangement of $(r,w,u,b,g)$, thus there are $44$ ways to get one of each color from the randomly chosen colors.
$$frac{44}{1024}=4.296875%$$
answered Jan 16 at 1:24
Daniel MathiasDaniel Mathias
1,25318
$endgroup$
The desired outcome is a derangement of $(r,w,u,b,g)$, thus there are $44$ ways to get one of each color from the randomly chosen colors.
$$frac{44}{1024}=4.296875%$$
answered Jan 16 at 1:24
Daniel MathiasDaniel Mathias
1,25318
answered Jan 16 at 1:24
Daniel MathiasDaniel Mathias
1,25318
answered Jan 16 at 1:24
Daniel MathiasDaniel Mathias
1,25318
answered Jan 16 at 1:24
Daniel MathiasDaniel Mathias
1,25318
1,25318
add a comment |
add a comment |
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1
$begingroup$
I think this is a stars and bars problem. $x_1 + x_2 + x_3 + x_4 + x_5 = 5$ where the $x_i$'s can be zero. So the probability is one divided by $9$ choose $5$ or one in $126$. What do you think?
$endgroup$
– HJ_beginner
Jan 16 at 1:12
1
$begingroup$
Stars and bars is almost applicable, however it isn't because of the fact that there isn't $5$ choices for the second color of each pack. There are only $4$ choices for each, and the four choices are dependent upon which color deck you chose, as the second color is a "randomly chosen color that is different from it".
$endgroup$
– WaveX
Jan 16 at 1:21
$begingroup$
@WaveX ahhh that makes sense. Thanks for pointing that out for me!
$endgroup$
– HJ_beginner
Jan 16 at 1:57
$begingroup$
In practice, the distribution of colors in a case of welcome packs isn’t truly random. The cards are packed by cycling through some long fixed sequence.
$endgroup$
– amd
Jan 17 at 1:14
$begingroup$
What long fixed sequence is that?
$endgroup$
– Toni Stack
Jan 18 at 7:41