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I was wondering what rules the user true blue anil used to get the answer they gave: Card Game Bridge Probability

I am thinking we assume all the cards are identical and so the probability of any one of them is $\frac{52}{52} = 1$ by the discrete probability law. Since the first Ace is one of these identical cards, the probability of this Ace is $1$. Now we let the first thirteen cards to be distinct from the rest of the $39$ identical ones. Each of the remaining $39$ has probability of $\frac {39}{51}$. Since the second Ace is in this bunch of $39$, the probability of the second Ace is $\frac {39}{51}$. We continue in this way till we exhaust all Aces. Granted all of this makes sense, why do we multiply the probabilities? Because the outcomes are independent?

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Imagine $52$ slots, delimited by three barriers to create $4$ sections of $13$ slots

We consider $A_k$ denoting the event that we put an Ace on the $(4-k)$ remaining sections .

We want to know $$P(\cap_{k=1}^{4}A_k)$$

This can be rewritten as $$P(\cap_{k=1}^{4}A_k)=P(A_4|\cap_{k=1}^{3}A_k)P(A_3|\cap_{k=1}^{2}A_k)P(A_2|A_1)P(A_1)$$.

The probability $P(A_1)$ to put an ace in the $4$ remaining section is $1$. We have now $51$ cards, and $3$ sections( 39 slots) because $A_1$ occured. The probability $P(A_2|A_1)$to put an Ace in the $3$ remaining section is obviously $\frac{39}{51}$, and so on...

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  • $\begingroup$ @ Canardini, your answer is exactly what I was looking for. Thanks. $\endgroup$
    – JJJ Doe
    Dec 20 '16 at 17:52
  • $\begingroup$ The event's definition must be clarified. $A_2$ is the event that the second ace is put in a section different from the first ace. $A_3$ is the event that the third ace is in a section different from the two first aces, etc. There is no $A_1$ and this reasoning works. However it hides a part of the problem: why is there a multiplication in the conditional's probability formula, that is to say why does concrete experimentation obey this law ? $\endgroup$ Dec 20 '16 at 18:07
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In these types of questions, the order of draws is not important. However, it always helps to think about the draws occurring in an ordered matter. In this example, we can consider the drawing of aces to occur in a sequential order (i.e. the first ace is drawn, the second ace is drawn, etc.) before we draw any other cards.

In order for all 4 players to receive an ace, we need the following events to occur:

1) The first ace can go to any of the 4 players (so any one of the 52 slots out of 52 available slots), i.e. $ P_1 = \frac{52}{52} = 1 $

2) The second ace must go to one of the 3 players without an ace (so any one of the 39 slots out of 51 available slots), i.e. $ P_2 = \frac{39}{51} $

3) The third ace must go to one of the 2 players without an ace (so any one of the 26 slots out of 50 available slots), i.e. $ P_3 = \frac{26}{50} $

4) The third ace must go to the 1 player without an ace (so any one of the 13 slots out of 49 available slots), i.e. $ P_4 = \frac{13}{49} $

Since these events are independent, we can multiply their probabilities to get:

$$ P = P_1 \times P_2 \times P_3 \times P_4 = 1 \times \frac{39}{51} \times \frac{26}{50} \times \frac{13}{49} $$

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  • $\begingroup$ @ Michael R, I didn't see your answer before I chose one. But one thing i don't get is why these events are independent. Suppose the first Ace we choose is that of hearts. Shouldn't that affect the probability of other outcomes? $\endgroup$
    – JJJ Doe
    Dec 20 '16 at 17:59
  • $\begingroup$ @JJJDoe The ace's suit is not important. It's best to think about this situation as follows: you have four aces (in a deck of 52 cards), and you want to distribute all 52 cards among the four players. What is the probability that each player receives an ace? If the first ace (hearts) is given to player 1, then what is the probability that the second ace (spade, clubs, diamonds - it doesn't matter) is given to either player 2, 3, or 4? $\endgroup$
    – Michael R
    Dec 20 '16 at 21:12

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