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Probability of full deck of cards being counted through without drawing card being named… [duplicate]
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Derangements with repetitive numbers
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Probability of winning the game 1-2-3-4-5-6-7-8-9-10-J-Q-K [duplicate]
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Suppose I have a deck of cards and start counting Ace, (draw card to see if it matches, if not discard) then 2, then 3... then king, and repeat sequence until the entire deck has been gone through. What are the chances that I will get through the entire deck WITHOUT drawing the card being named?
probability
asked Jul 31 at 0:19
Vance Wilson
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marked as duplicate by amWhy, Xander Henderson, JMoravitz, joriki probability
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Jul 31 at 2:50
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Derangements with repetitive numbers
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Probability of winning the game 1-2-3-4-5-6-7-8-9-10-J-Q-K [duplicate]
4 answers
Suppose I have a deck of cards and start counting Ace, (draw card to see if it matches, if not discard) then 2, then 3... then king, and repeat sequence until the entire deck has been gone through. What are the chances that I will get through the entire deck WITHOUT drawing the card being named?
probability
asked Jul 31 at 0:19
Vance Wilson
61
marked as duplicate by amWhy, Xander Henderson, JMoravitz, joriki probability
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Jul 31 at 2:50
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
1
This is exactly the same as drawing the card on the first draw. There are $52!$ possible orderings of all of the cards in the deck. Name a card and fix a position in the deck. There are $51!$ possible orderings of the deck with that card in that position (first, last, or otherwise). Thus the probability of your card being in the chosen position is $$frac51!52! = frac152. $$ That is, the chance of drawing your chosen card on the first draw is the same as it being in the last position, i.e. $1/52$.
– Xander Henderson
Jul 31 at 0:22
To be clear., you are asking for the probability that the $1$st, $14$th, $27$th, and $40$th cards are all not aces, the $2$nd, $15$th, $28$th, $41$st cards all not being twos, etc..., on up to the $13$th, $26$th, $39$th, and $52$nd cards all not being kings?
– JMoravitz
Jul 31 at 0:25
Or, are you asking that a specifically named card (e.g. Ace of spades) does not appear within the first $51$ cards? (This is the interpretation that @XanderHenderson used in his comment, but I have trouble believing that this is the intended interpretation)
– JMoravitz
Jul 31 at 0:28
1
For interpretation in the first comment, although not exact, a decent approximation can be obtained by looking at the related problem where we were dealing cards from an infinite shoe (i.e. drawing with replacement), where we get an approximation of $left(frac1213right)^52approx .0156$. To calculate an exact value for this, some very tedious inclusion-exclusion coupled with casework would likely need to be done, more work than anyone in their right minds should probably do.
– JMoravitz
Jul 31 at 0:34
1
@joriki oooh, it has been solved? Looks like I have some light reading to do tonight. Yay! The naive solution I was thinking of just isn't at all feasible so I am interested to read the paper.
– JMoravitz
Jul 31 at 21:56
 |Â
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up vote
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This question already has an answer here:
Derangements with repetitive numbers
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Probability of winning the game 1-2-3-4-5-6-7-8-9-10-J-Q-K [duplicate]
4 answers
Suppose I have a deck of cards and start counting Ace, (draw card to see if it matches, if not discard) then 2, then 3... then king, and repeat sequence until the entire deck has been gone through. What are the chances that I will get through the entire deck WITHOUT drawing the card being named?
probability
asked Jul 31 at 0:19
Vance Wilson
61
This question already has an answer here:
Derangements with repetitive numbers
1 answer
Probability of winning the game 1-2-3-4-5-6-7-8-9-10-J-Q-K [duplicate]
4 answers
Suppose I have a deck of cards and start counting Ace, (draw card to see if it matches, if not discard) then 2, then 3... then king, and repeat sequence until the entire deck has been gone through. What are the chances that I will get through the entire deck WITHOUT drawing the card being named?
This question already has an answer here:
Derangements with repetitive numbers
1 answer
Probability of winning the game 1-2-3-4-5-6-7-8-9-10-J-Q-K [duplicate]
4 answers
probability
asked Jul 31 at 0:19
Vance Wilson
61
asked Jul 31 at 0:19
Vance Wilson
61
asked Jul 31 at 0:19
Vance Wilson
61
asked Jul 31 at 0:19
Vance Wilson
61
61
marked as duplicate by amWhy, Xander Henderson, JMoravitz, joriki probability
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Jul 31 at 2:50
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
1
This is exactly the same as drawing the card on the first draw. There are $52!$ possible orderings of all of the cards in the deck. Name a card and fix a position in the deck. There are $51!$ possible orderings of the deck with that card in that position (first, last, or otherwise). Thus the probability of your card being in the chosen position is $$frac51!52! = frac152. $$ That is, the chance of drawing your chosen card on the first draw is the same as it being in the last position, i.e. $1/52$.
– Xander Henderson
Jul 31 at 0:22
To be clear., you are asking for the probability that the $1$st, $14$th, $27$th, and $40$th cards are all not aces, the $2$nd, $15$th, $28$th, $41$st cards all not being twos, etc..., on up to the $13$th, $26$th, $39$th, and $52$nd cards all not being kings?
– JMoravitz
Jul 31 at 0:25
Or, are you asking that a specifically named card (e.g. Ace of spades) does not appear within the first $51$ cards? (This is the interpretation that @XanderHenderson used in his comment, but I have trouble believing that this is the intended interpretation)
– JMoravitz
Jul 31 at 0:28
1
For interpretation in the first comment, although not exact, a decent approximation can be obtained by looking at the related problem where we were dealing cards from an infinite shoe (i.e. drawing with replacement), where we get an approximation of $left(frac1213right)^52approx .0156$. To calculate an exact value for this, some very tedious inclusion-exclusion coupled with casework would likely need to be done, more work than anyone in their right minds should probably do.
– JMoravitz
Jul 31 at 0:34
1
@joriki oooh, it has been solved? Looks like I have some light reading to do tonight. Yay! The naive solution I was thinking of just isn't at all feasible so I am interested to read the paper.
– JMoravitz
Jul 31 at 21:56
 |Â
show 14 more comments
1
This is exactly the same as drawing the card on the first draw. There are $52!$ possible orderings of all of the cards in the deck. Name a card and fix a position in the deck. There are $51!$ possible orderings of the deck with that card in that position (first, last, or otherwise). Thus the probability of your card being in the chosen position is $$frac51!52! = frac152. $$ That is, the chance of drawing your chosen card on the first draw is the same as it being in the last position, i.e. $1/52$.
– Xander Henderson
Jul 31 at 0:22
To be clear., you are asking for the probability that the $1$st, $14$th, $27$th, and $40$th cards are all not aces, the $2$nd, $15$th, $28$th, $41$st cards all not being twos, etc..., on up to the $13$th, $26$th, $39$th, and $52$nd cards all not being kings?
– JMoravitz
Jul 31 at 0:25
Or, are you asking that a specifically named card (e.g. Ace of spades) does not appear within the first $51$ cards? (This is the interpretation that @XanderHenderson used in his comment, but I have trouble believing that this is the intended interpretation)
– JMoravitz
Jul 31 at 0:28
1
For interpretation in the first comment, although not exact, a decent approximation can be obtained by looking at the related problem where we were dealing cards from an infinite shoe (i.e. drawing with replacement), where we get an approximation of $left(frac1213right)^52approx .0156$. To calculate an exact value for this, some very tedious inclusion-exclusion coupled with casework would likely need to be done, more work than anyone in their right minds should probably do.
– JMoravitz
Jul 31 at 0:34
1
@joriki oooh, it has been solved? Looks like I have some light reading to do tonight. Yay! The naive solution I was thinking of just isn't at all feasible so I am interested to read the paper.
– JMoravitz
Jul 31 at 21:56
1
1
This is exactly the same as drawing the card on the first draw. There are $52!$ possible orderings of all of the cards in the deck. Name a card and fix a position in the deck. There are $51!$ possible orderings of the deck with that card in that position (first, last, or otherwise). Thus the probability of your card being in the chosen position is $$frac51!52! = frac152. $$ That is, the chance of drawing your chosen card on the first draw is the same as it being in the last position, i.e. $1/52$.
– Xander Henderson
Jul 31 at 0:22
This is exactly the same as drawing the card on the first draw. There are $52!$ possible orderings of all of the cards in the deck. Name a card and fix a position in the deck. There are $51!$ possible orderings of the deck with that card in that position (first, last, or otherwise). Thus the probability of your card being in the chosen position is $$frac51!52! = frac152. $$ That is, the chance of drawing your chosen card on the first draw is the same as it being in the last position, i.e. $1/52$.
– Xander Henderson
Jul 31 at 0:22
To be clear., you are asking for the probability that the $1$st, $14$th, $27$th, and $40$th cards are all not aces, the $2$nd, $15$th, $28$th, $41$st cards all not being twos, etc..., on up to the $13$th, $26$th, $39$th, and $52$nd cards all not being kings?
– JMoravitz
Jul 31 at 0:25
To be clear., you are asking for the probability that the $1$st, $14$th, $27$th, and $40$th cards are all not aces, the $2$nd, $15$th, $28$th, $41$st cards all not being twos, etc..., on up to the $13$th, $26$th, $39$th, and $52$nd cards all not being kings?
– JMoravitz
Jul 31 at 0:25
Or, are you asking that a specifically named card (e.g. Ace of spades) does not appear within the first $51$ cards? (This is the interpretation that @XanderHenderson used in his comment, but I have trouble believing that this is the intended interpretation)
– JMoravitz
Jul 31 at 0:28
Or, are you asking that a specifically named card (e.g. Ace of spades) does not appear within the first $51$ cards? (This is the interpretation that @XanderHenderson used in his comment, but I have trouble believing that this is the intended interpretation)
– JMoravitz
Jul 31 at 0:28
1
1
For interpretation in the first comment, although not exact, a decent approximation can be obtained by looking at the related problem where we were dealing cards from an infinite shoe (i.e. drawing with replacement), where we get an approximation of $left(frac1213right)^52approx .0156$. To calculate an exact value for this, some very tedious inclusion-exclusion coupled with casework would likely need to be done, more work than anyone in their right minds should probably do.
– JMoravitz
Jul 31 at 0:34
For interpretation in the first comment, although not exact, a decent approximation can be obtained by looking at the related problem where we were dealing cards from an infinite shoe (i.e. drawing with replacement), where we get an approximation of $left(frac1213right)^52approx .0156$. To calculate an exact value for this, some very tedious inclusion-exclusion coupled with casework would likely need to be done, more work than anyone in their right minds should probably do.
– JMoravitz
Jul 31 at 0:34
1
1
@joriki oooh, it has been solved? Looks like I have some light reading to do tonight. Yay! The naive solution I was thinking of just isn't at all feasible so I am interested to read the paper.
– JMoravitz
Jul 31 at 21:56
@joriki oooh, it has been solved? Looks like I have some light reading to do tonight. Yay! The naive solution I was thinking of just isn't at all feasible so I am interested to read the paper.
– JMoravitz
Jul 31 at 21:56
 |Â
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1
This is exactly the same as drawing the card on the first draw. There are $52!$ possible orderings of all of the cards in the deck. Name a card and fix a position in the deck. There are $51!$ possible orderings of the deck with that card in that position (first, last, or otherwise). Thus the probability of your card being in the chosen position is $$frac51!52! = frac152. $$ That is, the chance of drawing your chosen card on the first draw is the same as it being in the last position, i.e. $1/52$.
– Xander Henderson
Jul 31 at 0:22
To be clear., you are asking for the probability that the $1$st, $14$th, $27$th, and $40$th cards are all not aces, the $2$nd, $15$th, $28$th, $41$st cards all not being twos, etc..., on up to the $13$th, $26$th, $39$th, and $52$nd cards all not being kings?
– JMoravitz
Jul 31 at 0:25
Or, are you asking that a specifically named card (e.g. Ace of spades) does not appear within the first $51$ cards? (This is the interpretation that @XanderHenderson used in his comment, but I have trouble believing that this is the intended interpretation)
– JMoravitz
Jul 31 at 0:28
1
For interpretation in the first comment, although not exact, a decent approximation can be obtained by looking at the related problem where we were dealing cards from an infinite shoe (i.e. drawing with replacement), where we get an approximation of $left(frac1213right)^52approx .0156$. To calculate an exact value for this, some very tedious inclusion-exclusion coupled with casework would likely need to be done, more work than anyone in their right minds should probably do.
– JMoravitz
Jul 31 at 0:34
1
@joriki oooh, it has been solved? Looks like I have some light reading to do tonight. Yay! The naive solution I was thinking of just isn't at all feasible so I am interested to read the paper.
– JMoravitz
Jul 31 at 21:56