Probability of picking a card randomly vs always picking the first card?
I shuffle a deck of cards. Each card appears there only once. Now, I need to pick a card randomly. I can do this in two ways:
- literally, pick a card somewhere in the beginning/middle/end of the deck randomly.
- always pick the first card
Context: for many games, you start by picking a random card to determine various things: how to start, who starts etc.
My friend claims that these two ways yield the same result, essentially they are the same i.e. each card has an equal probability of being at the very first place in the deck since we are shuffling randomly.
I claim that:
1) it COULD only be the case IF we could ensure "universally equal" shuffling, meaning that based on the person's shuffling, some cards might never make it to the top,
2) since we do not have such "universally equal shuffling", the probabilities of the cards being picked are 1 for the first card, and 0 for all others; while if you randomly pick a card (way #1) the probabilities are indeed all equal to $1/52$ for each card (assuming there are 52 cards in the deck).
Where is the truth?
probability card-games
asked Nov 24 at 13:09
Oleksandra
1235
|
show 5 more comments
I shuffle a deck of cards. Each card appears there only once. Now, I need to pick a card randomly. I can do this in two ways:
- literally, pick a card somewhere in the beginning/middle/end of the deck randomly.
- always pick the first card
Context: for many games, you start by picking a random card to determine various things: how to start, who starts etc.
My friend claims that these two ways yield the same result, essentially they are the same i.e. each card has an equal probability of being at the very first place in the deck since we are shuffling randomly.
I claim that:
1) it COULD only be the case IF we could ensure "universally equal" shuffling, meaning that based on the person's shuffling, some cards might never make it to the top,
2) since we do not have such "universally equal shuffling", the probabilities of the cards being picked are 1 for the first card, and 0 for all others; while if you randomly pick a card (way #1) the probabilities are indeed all equal to $1/52$ for each card (assuming there are 52 cards in the deck).
Where is the truth?
probability card-games
asked Nov 24 at 13:09
Oleksandra
1235
2
If the shuffling is biased then, to get a randomized selection, you do need to randomize the card you select.
– lulu
Nov 24 at 13:11
3
If you are going to pick a card from a random position in the deck, then there's no need to shuffle. By the way, what method do you have in mind to randomize what card you pick?
– bof
Nov 24 at 13:16
What probsbility is ‘1 for the first card, and 0 for the others’...?
– Bram28
Nov 24 at 13:17
2
At the end of the selection procedure, whatever it is, the probability of being picked is always $1$ for the card you pick. The question is what is the probability of each card in the original deck to be the one picked. Regarding the bias of the shuffle, why not shuffle as usual, then pick a random point in the deck, cut the deck there, put the bottom part on top of the top part, and call that the last step of the shuffle? This gives you a shuffle whose top card is equivalent to one selected at random from a shuffled deck.
– David K
Nov 24 at 13:55
1
Do you play cards? In real card games, after one player shuffles, another player cuts the deck, creating a new top card. If it were possible to cut the deck randomly, this would amount to your method #1, no?
– bof
Nov 25 at 11:07
|
show 5 more comments
I shuffle a deck of cards. Each card appears there only once. Now, I need to pick a card randomly. I can do this in two ways:
- literally, pick a card somewhere in the beginning/middle/end of the deck randomly.
- always pick the first card
Context: for many games, you start by picking a random card to determine various things: how to start, who starts etc.
My friend claims that these two ways yield the same result, essentially they are the same i.e. each card has an equal probability of being at the very first place in the deck since we are shuffling randomly.
I claim that:
1) it COULD only be the case IF we could ensure "universally equal" shuffling, meaning that based on the person's shuffling, some cards might never make it to the top,
2) since we do not have such "universally equal shuffling", the probabilities of the cards being picked are 1 for the first card, and 0 for all others; while if you randomly pick a card (way #1) the probabilities are indeed all equal to $1/52$ for each card (assuming there are 52 cards in the deck).
Where is the truth?
probability card-games
asked Nov 24 at 13:09
Oleksandra
1235
I shuffle a deck of cards. Each card appears there only once. Now, I need to pick a card randomly. I can do this in two ways:
- literally, pick a card somewhere in the beginning/middle/end of the deck randomly.
- always pick the first card
Context: for many games, you start by picking a random card to determine various things: how to start, who starts etc.
My friend claims that these two ways yield the same result, essentially they are the same i.e. each card has an equal probability of being at the very first place in the deck since we are shuffling randomly.
I claim that:
1) it COULD only be the case IF we could ensure "universally equal" shuffling, meaning that based on the person's shuffling, some cards might never make it to the top,
2) since we do not have such "universally equal shuffling", the probabilities of the cards being picked are 1 for the first card, and 0 for all others; while if you randomly pick a card (way #1) the probabilities are indeed all equal to $1/52$ for each card (assuming there are 52 cards in the deck).
Where is the truth?
probability card-games
probability card-games
asked Nov 24 at 13:09
Oleksandra
1235
asked Nov 24 at 13:09
Oleksandra
1235
asked Nov 24 at 13:09
Oleksandra
1235
asked Nov 24 at 13:09
Oleksandra
1235
asked Nov 24 at 13:09
Oleksandra
1235
1235
2
If the shuffling is biased then, to get a randomized selection, you do need to randomize the card you select.
– lulu
Nov 24 at 13:11
3
If you are going to pick a card from a random position in the deck, then there's no need to shuffle. By the way, what method do you have in mind to randomize what card you pick?
– bof
Nov 24 at 13:16
What probsbility is ‘1 for the first card, and 0 for the others’...?
– Bram28
Nov 24 at 13:17
2
At the end of the selection procedure, whatever it is, the probability of being picked is always $1$ for the card you pick. The question is what is the probability of each card in the original deck to be the one picked. Regarding the bias of the shuffle, why not shuffle as usual, then pick a random point in the deck, cut the deck there, put the bottom part on top of the top part, and call that the last step of the shuffle? This gives you a shuffle whose top card is equivalent to one selected at random from a shuffled deck.
– David K
Nov 24 at 13:55
1
Do you play cards? In real card games, after one player shuffles, another player cuts the deck, creating a new top card. If it were possible to cut the deck randomly, this would amount to your method #1, no?
– bof
Nov 25 at 11:07
|
show 5 more comments
2
If the shuffling is biased then, to get a randomized selection, you do need to randomize the card you select.
– lulu
Nov 24 at 13:11
3
If you are going to pick a card from a random position in the deck, then there's no need to shuffle. By the way, what method do you have in mind to randomize what card you pick?
– bof
Nov 24 at 13:16
What probsbility is ‘1 for the first card, and 0 for the others’...?
– Bram28
Nov 24 at 13:17
2
At the end of the selection procedure, whatever it is, the probability of being picked is always $1$ for the card you pick. The question is what is the probability of each card in the original deck to be the one picked. Regarding the bias of the shuffle, why not shuffle as usual, then pick a random point in the deck, cut the deck there, put the bottom part on top of the top part, and call that the last step of the shuffle? This gives you a shuffle whose top card is equivalent to one selected at random from a shuffled deck.
– David K
Nov 24 at 13:55
1
Do you play cards? In real card games, after one player shuffles, another player cuts the deck, creating a new top card. If it were possible to cut the deck randomly, this would amount to your method #1, no?
– bof
Nov 25 at 11:07
2
2
If the shuffling is biased then, to get a randomized selection, you do need to randomize the card you select.
– lulu
Nov 24 at 13:11
If the shuffling is biased then, to get a randomized selection, you do need to randomize the card you select.
– lulu
Nov 24 at 13:11
3
3
If you are going to pick a card from a random position in the deck, then there's no need to shuffle. By the way, what method do you have in mind to randomize what card you pick?
– bof
Nov 24 at 13:16
If you are going to pick a card from a random position in the deck, then there's no need to shuffle. By the way, what method do you have in mind to randomize what card you pick?
– bof
Nov 24 at 13:16
What probsbility is ‘1 for the first card, and 0 for the others’...?
– Bram28
Nov 24 at 13:17
What probsbility is ‘1 for the first card, and 0 for the others’...?
– Bram28
Nov 24 at 13:17
2
2
At the end of the selection procedure, whatever it is, the probability of being picked is always $1$ for the card you pick. The question is what is the probability of each card in the original deck to be the one picked. Regarding the bias of the shuffle, why not shuffle as usual, then pick a random point in the deck, cut the deck there, put the bottom part on top of the top part, and call that the last step of the shuffle? This gives you a shuffle whose top card is equivalent to one selected at random from a shuffled deck.
– David K
Nov 24 at 13:55
At the end of the selection procedure, whatever it is, the probability of being picked is always $1$ for the card you pick. The question is what is the probability of each card in the original deck to be the one picked. Regarding the bias of the shuffle, why not shuffle as usual, then pick a random point in the deck, cut the deck there, put the bottom part on top of the top part, and call that the last step of the shuffle? This gives you a shuffle whose top card is equivalent to one selected at random from a shuffled deck.
– David K
Nov 24 at 13:55
1
1
Do you play cards? In real card games, after one player shuffles, another player cuts the deck, creating a new top card. If it were possible to cut the deck randomly, this would amount to your method #1, no?
– bof
Nov 25 at 11:07
Do you play cards? In real card games, after one player shuffles, another player cuts the deck, creating a new top card. If it were possible to cut the deck randomly, this would amount to your method #1, no?
– bof
Nov 25 at 11:07
|
show 5 more comments
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2
If the shuffling is biased then, to get a randomized selection, you do need to randomize the card you select.
– lulu
Nov 24 at 13:11
3
If you are going to pick a card from a random position in the deck, then there's no need to shuffle. By the way, what method do you have in mind to randomize what card you pick?
– bof
Nov 24 at 13:16
What probsbility is ‘1 for the first card, and 0 for the others’...?
– Bram28
Nov 24 at 13:17
2
At the end of the selection procedure, whatever it is, the probability of being picked is always $1$ for the card you pick. The question is what is the probability of each card in the original deck to be the one picked. Regarding the bias of the shuffle, why not shuffle as usual, then pick a random point in the deck, cut the deck there, put the bottom part on top of the top part, and call that the last step of the shuffle? This gives you a shuffle whose top card is equivalent to one selected at random from a shuffled deck.
– David K
Nov 24 at 13:55
1
Do you play cards? In real card games, after one player shuffles, another player cuts the deck, creating a new top card. If it were possible to cut the deck randomly, this would amount to your method #1, no?
– bof
Nov 25 at 11:07