# Probability : Dependent Events

1. Apr 8, 2007

### Beryligos

Hello all, this is my first post here.

Got into a debate over when a deck of cards drawn without replacement is dependent. My opinion is that when there is no replacement, it is always dependent. That even before you draw the first card, you know already what the odds are.

My "esteemed colleague" (I think he's nuts) believes it is only when you know what the first event is. The following is his argument.

<quoting>
---
My argument is that until there is information on what any of the initial cards are, all of the cards in any of the positions are independent. As soon as any information is available on the initial cards, they become dependent.
---
</quoting>

I prefer to draw a tree diagram to map out what the chance of an event is.

The question that sparked this was "What is the probability of the second card drawn without replacement from a 52-card deck being a spade?"

My friend thinks that without any information the first card has no influence on the odds of getting a spade, and is independent. That if you were to pick out the second from the top (The one that would be drawn in the second event) the probability is 1/4 that it is a spade, because you don't have information on the first trial.

My apologies if I was a bit odd with my post. Its very late. :zzz: and I think all the arguing has gotten both me and my friend confused.

2. Apr 8, 2007

### Data

The answer of 1/4 is correct, and you will get that answer if you treat the problem using a tree as you suggest:

Either the first card is a spade (prob. 1/4) or not (prob. 3/4).

In the first case the either the second card is a spade (prob. 12/51) or not (prob. 39/51). In the second case the second card is a space (prob. 13/51) or not (prob. 38/51).

Thus the probability that the second card is a spade is

$$\left(\frac{1}{4}\right)\left(\frac{12}{51}\right) + \left(\frac{3}{4}\right)\left(\frac{13}{51}\right) = \frac{1}{4}.$$

Now, fairly obviously, the second card's probability of being a spade is dependent on the first card drawn. If what the question means is "after drawing one card from a 52-card deck, what is the probability, assuming no replacement, that the next card is a spade?," then I would take that to mean that an answer in each eventuality for the first card's identity is what they want (so you'd reply, "the probability is 12/51 if the first card was a spade, or 13/51 if the first card was not a spade. If we don't know whether the first card was a spade or not, then the probability is 1/4.").

Last edited: Apr 8, 2007
3. Apr 8, 2007

### christianjb

I'm confused.

Do you mean if you remove one card without looking at it then it changes the probability for what the second card can be?

4. Apr 8, 2007

### arildno

No. Data said that if you take away the first card, without looking at it, then this does not change the probabilities for any other card drawn, for whatever purpose.

5. Apr 9, 2007

### Piriformis

I thank you all for your responses. Despite my colleague Beryligos's concerns regarding my sanity I must at this time reflect on his delusional interpretation of events. For not only did he originally get the final answer fundamentally wrong he is using these arguments to claim a draw in our debate and avoid the consequences of our agreement.

My point to him was that the probability of any given card in the deck being of a particular suit was not influenced by, dependent on or contingent to any other cards unless information was available identifying the other cards. I empasized that without such information the probability was the same whether the card was the first, second, middle or last in the deck. Unless the information on the other card is available the probability of any individual card being of a particular suit can be calculated independently of the other cards or treated as a first event.

I trust he will see the error of his ways and admit defeat despite the consequences of his loss. Have no fear of his financial state as no money was wagered. The positive side of his loss is that a 2 year abstinence from gaming will allow him to renew his focus on math and physics.

Thank you once again.

6. Apr 9, 2007

### Data

Yes, that interpretation is correct in every respect! The answer to the particular problem depends on what exactly is being asked for. However, if pressed to give a single numerical answer, then the only one that makes any sense is 1/4.

Last edited: Apr 9, 2007
7. Apr 3, 2010

### Rukh

I know this is an old thread, but it was a similar problem to something I'm currently trying to solve! Can someone help me with my logic? Thanks!

I'm trying to show (similar to post #2) dependency... (I've also used nCr notation - hope it's clear!)

The question is (and it's derived from a texas holden poker game), there are 46 cards, of which 9 are hearts. Two players have 2 cards each, but we don't know what they are. What is the probability that the next draw is a heart?

Theory 1: There are 46 cards. The chance of any of those cards (be it a card in someone else's hand, the bottom card on the pack, or the next card drawn) is 9/46 (19.6%).

Theory 2: How can we break it down?
The other players have 4 hearts = 4c4*4c4*42c0/46c4 = 0.001%
The other players have 3 hearts = 4c3*3c3*42c1/46c4 = 1%
The other players have 2 hearts = 4c2*2c2*42c2/46c4 = 3%
The other players have 1 hearts = 4c1*1c1*42c3/46c4 = 28%
The other players have 0 hearts = 4c0*0c0*42c4/46c4 = 69%

So that all adds to 100%, so on the right track.

Now I then wanted to multiply each of those by the chance of the next card being a heart (e.g. 5/42, 6/42, 7/42, 8/42, 9/42) but when I sum it, I get 20.6%.

Shouldn't it be the same as Theory 1? What am I missing?

Cheers,
A slightly confused poker player - Warren

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8. Aug 6, 2010

### schwanky7

Rukh,

There is a problem with your probability calculations in Theory 2. Here is how I would approach it (I'll use your nCr for combinations and nPr for permutations):

The other players have 4 hearts = 4c4*9p4*37p0/46p4 = 0.0772%
The other players have 3 hearts = 4c3*9p3*37p1/46p4 = 1.90%
The other players have 2 hearts = 4c2*9p2*37p2/46p4 = 14.7%
The other players have 1 heart = 4c1*9p1*37p3/46p4 = 42.9%
The other players have 0 hearts = 4c0*9P0*37p4/46p4 = 40.5%

These are, of course, rounded percents, but their exact values add up to 100%.

Now, when you multiply each of these probabilities by the respective probability of drawing a heart with the next card (5/42, 6/42, 7/42, 8/42, and 9/42), you will get exactly 9/46, just as you surmised you should!

Probability can be a tricky thing. It often helps me to begin to write out the individual probabilities of each event occurring before I try to summarize with permutations, combinations, and/or factorials.