Mixing
Odds of drawing each card from partially known subset of cards
$begingroup$
Say you have $6$ decks of $52$ suitless cards (so $24$ indistinguishable copies of each rank). From that, you're given a randomly chosen, shuffled subset $S$ of size $52$, of which half of the cards are made known to you (though how these known cards are selected is random, and you don't know how they're ordered or distributed within $S$). Call this group of cards $K$ and call the other cards in $S$ $U$. Then, each card of $S$ is revealed and discarded one at a time. My question is, after having seen $n$ cards revealed and discarded, can you compute the probability of what the $(n+1)th$ revealed card will be without enumerating all the different ways that the revealed cards could have been taken from $K$ and $U$?
Here's a link to Python script I wrote that solves the problem using such an enumeration.
probability computational-complexity
asked Jan 6 at 2:40
Nathan SchmidtNathan Schmidt
13
$endgroup$
add a comment |
$begingroup$
Say you have $6$ decks of $52$ suitless cards (so $24$ indistinguishable copies of each rank). From that, you're given a randomly chosen, shuffled subset $S$ of size $52$, of which half of the cards are made known to you (though how these known cards are selected is random, and you don't know how they're ordered or distributed within $S$). Call this group of cards $K$ and call the other cards in $S$ $U$. Then, each card of $S$ is revealed and discarded one at a time. My question is, after having seen $n$ cards revealed and discarded, can you compute the probability of what the $(n+1)th$ revealed card will be without enumerating all the different ways that the revealed cards could have been taken from $K$ and $U$?
Here's a link to Python script I wrote that solves the problem using such an enumeration.
probability computational-complexity
asked Jan 6 at 2:40
Nathan SchmidtNathan Schmidt
13
$endgroup$
$begingroup$
The problem is underdefined, based on how you're told the 26 cards out of 52. If you're told "the 26 lowest ranks in the 52 cards" (and you know this), the answer will be dramatically different from you're told "the 26 highest ranks in the 52 cards".
$endgroup$
– obscurans
Jan 6 at 3:49
$begingroup$
The 26 cards out of the 52 are a randomly selected
$endgroup$
– Nathan Schmidt
Jan 6 at 3:57
add a comment |
$begingroup$
Say you have $6$ decks of $52$ suitless cards (so $24$ indistinguishable copies of each rank). From that, you're given a randomly chosen, shuffled subset $S$ of size $52$, of which half of the cards are made known to you (though how these known cards are selected is random, and you don't know how they're ordered or distributed within $S$). Call this group of cards $K$ and call the other cards in $S$ $U$. Then, each card of $S$ is revealed and discarded one at a time. My question is, after having seen $n$ cards revealed and discarded, can you compute the probability of what the $(n+1)th$ revealed card will be without enumerating all the different ways that the revealed cards could have been taken from $K$ and $U$?
Here's a link to Python script I wrote that solves the problem using such an enumeration.
probability computational-complexity
asked Jan 6 at 2:40
Nathan SchmidtNathan Schmidt
13
$endgroup$
Say you have $6$ decks of $52$ suitless cards (so $24$ indistinguishable copies of each rank). From that, you're given a randomly chosen, shuffled subset $S$ of size $52$, of which half of the cards are made known to you (though how these known cards are selected is random, and you don't know how they're ordered or distributed within $S$). Call this group of cards $K$ and call the other cards in $S$ $U$. Then, each card of $S$ is revealed and discarded one at a time. My question is, after having seen $n$ cards revealed and discarded, can you compute the probability of what the $(n+1)th$ revealed card will be without enumerating all the different ways that the revealed cards could have been taken from $K$ and $U$?
Here's a link to Python script I wrote that solves the problem using such an enumeration.
probability computational-complexity
probability computational-complexity
asked Jan 6 at 2:40
Nathan SchmidtNathan Schmidt
13
asked Jan 6 at 2:40
Nathan SchmidtNathan Schmidt
13
edited Jan 8 at 5:45
Nathan Schmidt
asked Jan 6 at 2:40
Nathan SchmidtNathan Schmidt
13
asked Jan 6 at 2:40
Nathan SchmidtNathan Schmidt
13
asked Jan 6 at 2:40
Nathan SchmidtNathan Schmidt
13
13
$begingroup$
The problem is underdefined, based on how you're told the 26 cards out of 52. If you're told "the 26 lowest ranks in the 52 cards" (and you know this), the answer will be dramatically different from you're told "the 26 highest ranks in the 52 cards".
$endgroup$
– obscurans
Jan 6 at 3:49
$begingroup$
The 26 cards out of the 52 are a randomly selected
$endgroup$
– Nathan Schmidt
Jan 6 at 3:57
add a comment |
$begingroup$
The problem is underdefined, based on how you're told the 26 cards out of 52. If you're told "the 26 lowest ranks in the 52 cards" (and you know this), the answer will be dramatically different from you're told "the 26 highest ranks in the 52 cards".
$endgroup$
– obscurans
Jan 6 at 3:49
$begingroup$
The 26 cards out of the 52 are a randomly selected
$endgroup$
– Nathan Schmidt
Jan 6 at 3:57
$begingroup$
The problem is underdefined, based on how you're told the 26 cards out of 52. If you're told "the 26 lowest ranks in the 52 cards" (and you know this), the answer will be dramatically different from you're told "the 26 highest ranks in the 52 cards".
$endgroup$
– obscurans
Jan 6 at 3:49
$begingroup$
The problem is underdefined, based on how you're told the 26 cards out of 52. If you're told "the 26 lowest ranks in the 52 cards" (and you know this), the answer will be dramatically different from you're told "the 26 highest ranks in the 52 cards".
$endgroup$
– obscurans
Jan 6 at 3:49
$begingroup$
The 26 cards out of the 52 are a randomly selected
$endgroup$
– Nathan Schmidt
Jan 6 at 3:57
$begingroup$
The 26 cards out of the 52 are a randomly selected
$endgroup$
– Nathan Schmidt
Jan 6 at 3:57
add a comment |
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$begingroup$
The problem is underdefined, based on how you're told the 26 cards out of 52. If you're told "the 26 lowest ranks in the 52 cards" (and you know this), the answer will be dramatically different from you're told "the 26 highest ranks in the 52 cards".
$endgroup$
– obscurans
Jan 6 at 3:49
$begingroup$
The 26 cards out of the 52 are a randomly selected
$endgroup$
– Nathan Schmidt
Jan 6 at 3:57