Probability when cards are removed and when you know what they are

[SumofSigma]

Hi guys,

1.) So we just had a discussion on determining probability given a card is removed and not shown to the player. It was proved with the law of total probability that the probability would not change. So 1 card is removed and then 1 card is drawn, the probability of a 2 is 4/52 = 4/52.

2.) I believe this extends to more cards being removed. So if 10 are removed and then 1 is drawn, the chance of getting a is 2 still 4/52.


For case 1.) Now what if the card that is removed is known, say it is a 2, then the probability of getting another 2 is 3/51?

For case 2.) If 10 are removed, one is shown to be a 2, and then one is drawn, the chance of getting a 2 is 3/51?

Are these statements correct?

 

[Steven G]

What you wrote for case 1 is true.

What you wrote for case 2 is not very clear. Do you get to see exactly one of the cards which is a two or do you get to see all 10 cards where exactly one of them is a two? If it is the first case then your answer is correct.

There are 51 cards left over after you pick a two from the deck. The chances of any of the remaining 51 cards being a two is 3/51. The position of the card you pick is meaningless. Think about this-- Do you really think that there is a better chance that a two is in the 1st 10 cards or middle 10 cards or the last 10 cards? The cards are all identical, they can't arrange themselves in numerical order! Picking the 3rd card or the 33rd card in the deck still has a 3/51 chance of being a two. Is this clear?

 

[SumofSigma]

What you wrote for case 1 is true.

What you wrote for case 2 is not very clear. Do you get to see exactly one of the cards which is a two or do you get to see all 10 cards where exactly one of them is a two? If it is the first case then your answer is correct.

There are 51 cards left over after you pick a two from the deck. The chances of any of the remaining 51 cards being a two is 3/51. The position of the card you pick is meaningless. Think about this-- Do you really think that there is a better chance that a two is in the 1st 10 cards or middle 10 cards or the last 10 cards? The cards are all identical, they can't arrange themselves in numerical order! Picking the 3rd card or the 33rd card in the deck still has a 3/51 chance of being a two. Is this clear?

Yes out of the 10 cards, your'e simply shown that one is a two and the others are unknown.

It's both intuitive to me and also not at the same time to think this way. I understand that the deck is "random" so equal opportunity for each card throughout the deck. So it makes since to me that by removing cards, the probability is still the same. At the same time, I can't help thinking of running a ridiculous number of conditional probability problems based on all the possibilities of the unknown cards. I becomes quite impossible to account for that though...

Also interesting that when we are actually positive of a new piece of information, ie the 2 is in the 10 removed cards, we can reduce our odds.

Point being, from the classical poker examples, if I remove cards to deal to other players, then the probability of me getting a card out of the deck is still the same. If I then look at my cards and see I have (n) 2s in my hand, then I can say I have a 4-n/52-n of drawing a 2. If this is correct, then I think I have an understanding of the concept.

Thanks for your help!

 

[Steven G]

Yes out of the 10 cards, your'e simply shown that one is a two and the others are unknown.

It's both intuitive to me and also not at the same time to think this way. I understand that the deck is "random" so equal opportunity for each card throughout the deck. So it makes since to me that by removing cards, the probability is still the same. At the same time, I can't help thinking of running a ridiculous number of conditional probability problems based on all the possibilities of the unknown cards. I becomes quite impossible to account for that though...

Also interesting that when we are actually positive of a new piece of information, ie the 2 is in the 10 removed cards, we can reduce our odds.

Point being, from the classical poker examples, if I remove cards to deal to other players, then the probability of me getting a card out of the deck is still the same. If I then look at my cards and see I have (n) 2s in my hand, then I can say I have a 4-n/52-n of drawing a 2. If this is correct, then I think I have an understanding of the concept.

Thanks for your help!

If I then look at my cards and see I have (n) 2s in my hand, then I can say I have a 4-n/52-n of drawing a 2. I suspect that you meant (4-n)/(52-n). This is not correct as maybe you 5 cards containing n-2s. Then the probability of the next card you get is a 2 will be (4-n)/(52-5) assuming that you did not see any other cards. If you saw other cards then the probability will probably change.