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Hello, I am in an Applied Statistics class at college and I am having some trouble with my homework. I am finding it difficult to understand the concept of probability. So therefore, if you could walk me through how to solve this problem, I would greatly appreciate it.

My question is:

Assume you are being dealt a heart with the first card and there are 4 more to come. What is the chance of being dealt a flush?

I am not certain how to start this question.

Hello Brittany,

There are a few principles required to understand how to solve this question:

A probability is a number between 0 and 1 assigned to some event. An event is an outcome or a set of outcomes of a random process. A random process is something with multiple outcomes of which one occurs each time the process occurs. Examples include flipping coins, rolling dice, or drawing cards from a deck.

Real life events that are not technically random are often treated as random, like the probability of having a particular disease -- either a person has it or does not have it, but if we do not know, we can base treatment on a probability-based analysis.

When you compute a simple probability, you count the number of outcomes you want to include in the event and divide by the total number of outcomes. For example, if you wanted to roll a five on one die, the probability is 1/6 (one outcome is five, six outcomes total). If you wanted to roll a five on

A deck of cards is 52 cards. There are four of each rank (1 through 10, jack, queen, king), and for each rank, there is a suit assigned (spades, hearts, clubs, diamonds). So there is (for example) exactly one jack of clubs, exactly one three of diamonds, but altogether there are 13 hearts and are four kings.

When a card is drawn from the deck at random, if you want a specific card to be drawn, the probability of drawing that card is 1/52.

If you want to draw a particular suit, that is 13/52 = 1/4.

If you want to draw a particular number, that probability is 4/52 = 1/13.

If one heart has already been drawn and you want to draw another, the probability changes. There is one fewer heart (only 12 left) and one fewer card in the whole deck (51) for a probability of 12/51 = 17/4.

When you draw one heart from the deck and then draw another, these two events are dependent. You know the probability of drawing a heart is 1/4, but once you have drawn a heart, that chances the probability of drawing a second heart on your next draw.

As mentioned above, the second heart has probability 12/51 = 17/4.

So you assume you already have one heart. The second heart is 12/51. The next heart is 11/50, then 10/49, then 9/48.

The total probability of getting the four hearts (beyond the one you already have) is:

(12/51) * (11/50) * (10/49) * (9/48) = 11880 / 5997600 = 33 / 16660 ≈ 0.00198079

That means there is approximately 0.198% probability of getting all four of these hearts that you want. (This is why a flush is so uncommon -- the probability of receiving a flush is very low, less than 1%.)

Another way to solve this problem is to consider all four cards at once.

Instead of counting each "event" as a sequence of four cards, we can think of the "event" as one bundle of four cards.

Using a type of mathematics called "combinatorics," we can count the number of combinations explicitly. There is a way of expressing "the number of rearrangements" of a thing called a factorial. The number of ways to shuffle the 51 cards we have left is expressed as:

51! = 51*50*49*48*47*....*4*3*2*1

There are 51 choices for the first card, 50 for the next, 49 for the next, etc.

This is the total number of ways to shuffle. Then you draw four cards, the top four, and their order does not matter, so you could shuffle them.

That means there are 4! ways to shuffle the top four cards, which has no effect on the hand you receive. You could also shuffle the bottom 47 cards and it would not affect the hand you receive.

Thus the total number of four-card hands is 51! / ( 4! 47! )

This is often written 51c4 or "51 choose 4."

Among these hands, you want the ones that have all hearts. Since one heart is already taken, there are only 12 left. There are 12c4 = 12! / ( 4! 8! ) possible ways of drawing a hand with all hearts in it.

If you want a quick answer, then, the answer is 12c4 / 51c4, which is equal to what we determined to be the probability in part

12c4 / 51c4 = ( 12! / (8! 4!) ) / ( 51! / (47! 4!) ) = ( 12! 47! 4! ) / ( 51! 8! 4! )

= ( 12! 47! ) / ( 51! 8! ) = (12!/8!) * (47!/51!) = (12*11*10*9)/(51*50*49*48)

= (12/51)*(11/50)*(10/49)*(9/48) = same as above.

This verifies that we have the same answer part

I can answer all questions up to, and including, graduate level mathematics. I do not have expertise in statistics (I can answer questions about the mathematical foundations of statistics). I am very much proficient in probability. I am not inclined to answer questions that appear to be homework, nor questions that are not meaningful or advanced in any way.

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