Probability of Cards: One at a time vs simultaneous

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rpalmtree

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I don't think this is a duplicate post, and I haven't been able to find an answer anywhere.

If there are 10 cards (all different) and 10 people drawing them, does the probability of drawing a particular card (say there is only one ace in the deck) change if the 10 people all draw the cards at the same time vs if one person draws then the deck is shuffled and then another person draws, shuffle, repeat?

I'm not necessarily in need of the equation at play here, but that may help. I just need to know if the probability of drawing the Ace changes when drawing all at once vs drawing one at a time.
 
Are you really unsure about this? You really should think about this carefully.
Case 1: If you replace the chosen cards, then the probability of choosing the Ace is 1/10
Case 2: You do not replace the card. If the 1st person picks the Ace, then what is the probability that any of the 9 other people get an Ace. Suppose the 1st nine people do not pick the Ace. What is the probability that the 10th person picks the Ace?
 
This should be fairly simple fraction multiplication and division. What is different about this problem?
 
I don't think in the serial case they are replacing cards. It's best to just skip the shuffle. It doesn't affect anything and just confuses matters.

Assuming all the drawers are independent all the probabilities are identical whether they draw 10 cards at once or 10 separated in time.

I suppose if it's a long enough time between draws there might be substantial probability that the next card will magically morph into a different card
just via a massive quantum fluctuation but the universe will probably be dead before that happens. :D

Note that this assumes perfect shuffling which doesn't exist in real life but is always assumed unless otherwise stated.
 
Romsek said:
yeah, it should be pretty obvious that replacement vs. non-replacement yield different distributions
Click to expand...

But as to finding the probability, the factorial of replacement affects the fraction that was found already.
 
rpalmtree said:
If there are 10 cards (all different) and 10 people drawing them, does the probability of drawing a particular card (say there is only one ace in the deck) change if the 10 people all draw the cards at the same time vs if one person draws then the deck is shuffled and then another person draws, shuffle, repeat?

I'm not necessarily in need of the equation at play here, but that may help. I just need to know if the probability of drawing the Ace changes when drawing all at once vs drawing one at a time.
Click to expand...
The first thing we need here is a clear statement of what is happening, and what you are asking for.

One scenario is drawing one at a time. I will assume there is no replacement, so A draws a card, then B draws a card (shuffling doesn't matter), and so on until J draws the last card.

The other scenario is drawing all at the same time. What does that mean, exactly? I'll suppose they are standing in a row, and each puts his hand out to take a card from a fanned deck. A gets the first card from the left, B gets the second, and so on until J gets the card on the right.

But what probability are you asking for -- that someone gets the ace (100%), or that A gets the ace (10%), or that J gets the ace (10%), or what???

In any case, the answer is that it doesn't matter which way you do it (each person has the same probability of getting the ace); but the important thing is the question you said didn't matter -- why?

What thought have you put into the question?
 
In any case, the answer is that it doesn't matter which way you do it (each person has the same probability of getting the ace); but the important thing is the question you said didn't matter -- why?

What thought have you put into the question?
[/QUOTE]

There are 10 cards and 10 people. One card is an Ace. In situation A all 10 people simultaneously pick a card. Imagine all 10 people stick their hands in at the same time and take a card. In situation B, the 10 people stand in a line and one by one come up to the deck to pick a card. The cards are not replaced.

In the one at a time scenario, I guess what I'm thinking is that person A has 10 cards to choose from so has a 1 in 10 chance of getting the ace, so if you want the ace, picking first maybe isn't the best? Then person two has a 1 in 9 chance of getting the ace (assuming the Ace wasn't selected, and since odds were against it they probably didn't).

All I'm asking is if the probability in selecting the Ace changes based on whether all 10 pick the cards simultaneously or one by one.

I'm not writing a problem for a math text book here, folks. No need to get snarky
 
rpalmtree said:
In any case, the answer is that it doesn't matter which way you do it (each person has the same probability of getting the ace); but the important thing is the question you said didn't matter -- why?

What thought have you put into the question?
Click to expand...

There are 10 cards and 10 people. One card is an Ace. In situation A all 10 people simultaneously pick a card. Imagine all 10 people stick their hands in at the same time and take a card. In situation B, the 10 people stand in a line and one by one come up to the deck to pick a card. The cards are not replaced.

In the one at a time scenario, I guess what I'm thinking is that person A has 10 cards to choose from so has a 1 in 10 chance of getting the ace, so if you want the ace, picking first maybe isn't the best? Then person two has a 1 in 9 chance of getting the ace (assuming the Ace wasn't selected, and since odds were against it they probably didn't).

All I'm asking is if the probability in selecting the Ace changes based on whether all 10 pick the cards simultaneously or one by one.

I'm not writing a problem for a math text book here, folks. No need to get snarky
[/QUOTE]
It depends on what exactly you mean.
When the 10 people take there cards at the same time, well let's stop right there.
If no one looks at their cards then each person has a 1 in 10 chance of having the Ace
If they look at their cards all at once, each person has a 1 in 10 chance of having the Ace.

If they look at it one at a time then things start to change.
If person 1 got the Ace, at that point the chance of any of the remaining having the Ace is 0.
If person 1 did not get the Ace, then at that point the chance of any of the remaining having the Ace is 1 in 9.
If person 2 got the Ace, at that point the chance of any of the remaining having the Ace is 0.
If person 1 and 2 did not get the Ace, then at that point the chance of any of the remaining having the Ace is 1 in 8.
....

Then you started to talk about which position is best to be in at the beginning. There is no position that is better than any other position. The reason is because the cards do not know about position. That is there is a 1 in 10 chance that the card is in any of the 10 position.

Learning information AFTER picking your card will not chance the 1 in 10 chance that you had at the beginning.
Sure if the 1st 9 people did not have the Ace, then with certainty the 10th person has the Ace. So you want to be number 10, because you will know with certainty that you have the Ace? The problem with that is you will know with certainty 1 in 10 times (on average) that you have the Ace.

Seeing the cards of the people will not change the fact that from the start that you have a 1 in 10 chance of winning.
 
rpalmtree said:
There are 10 cards and 10 people. One card is an Ace. In situation A all 10 people simultaneously pick a card. Imagine all 10 people stick their hands in at the same time and take a card. In situation B, the 10 people stand in a line and one by one come up to the deck to pick a card. The cards are not replaced.

In the one at a time scenario, I guess what I'm thinking is that person A has 10 cards to choose from so has a 1 in 10 chance of getting the ace, so if you want the ace, picking first maybe isn't the best? Then person two has a 1 in 9 chance of getting the ace (assuming the Ace wasn't selected, and since odds were against it they probably didn't).

All I'm asking is if the probability in selecting the Ace changes based on whether all 10 pick the cards simultaneously or one by one.

I'm not writing a problem for a math text book here, folks. No need to get snarky
Click to expand...

I hope I'm not being snarky. I'm pointing out that probability problems have to be stated clearly, and I did answer your question. But I wanted to see what you are thinking, in order to clearly point out why your apparent guess that there is a difference is wrong. You haven't really answered my questions; "the probability in selecting the ace" is no clearer than before. So let's move on:

Presumably you are asking about the probability for each of the ten people. Is each of them equally likely to get the ace, in both scenarios? Is that what you want to know?

Let's take the one-by-one scenario, since you started to show your work for that. Person A has probability 1/10 to get the ace. For person B, the probability is 1/9, given that A did not get it. The overall probability is there fore 9/10 * 1/9 = 1/10 again: he gets the ace in 1/9 of the 9/10 of all cases where A did not get it. Similarly, C's probability is 9/10*8/9*1/8 = 1/10. Keep going, and you find that each of them has a 1/10 chance of getting the ace.

Now, in the all-at-once scenario, each person clearly has an equal chance at the ace, since on of the ten hands will be pointing at the ace. So, again, each person has a 1/10 chance.

There's your answer!
 
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