Two main types of coin tossing games, one based on either a specific sequence of throws of a single coin, the other on throwing several coins at once and counting the results. One side of the coin is called “heads” (on traditional coins, the side with the face of a person), and the other side is called “tails”. Coin tosses are presumed to be mutually independent.
The probability of getting heads in one toss of a coin.
The number of results of any one coin toss is 2 (heads or tails), so the probability of any single toss resulting in heads is
The probability of getting two heads in one toss of two coins, (without regard to the order in which the coins are tossed).
(Of course the answer is 1/4, but let’s try to put in it the form of events).
The event whose probability we want is
As the coin tosses are independent, we just multiply the probabilities of successive tosses:
Generally, the probability of getting all heads in tossing n coins is 1/2n.
The probability of getting a head and a tail in one toss of two coins.
careful!
Why doesn’t this thinking work? The probability of a head and tails. So multiply the two probabilities to get 1/4.
We have enough tools that more than one method answers the question. A counting approach with the definition of probability, or counting plus probability rules.
One right way of thinking is to divide the number of ways to get two heads, counting order, by the number of different tosses of two coins, counting order
Counting order, there are two ways to get a head and a tail. Those are: head and then a tail, or a tail and then a head. There are four possible throws of two coins, counting order.
Another way of looking at it:
Since the two cannot both happen in only one throw, the joint probability is just the sum of the probabilities:
There is more than one kind of die: the usual kind is a cube with 6 sides, but there are ones with 4 or 12 sides. Here, let’s consider only the usual 6-sided die.
The probability of any particular value coming up with one throw of one die.
There are 6 sides on the die, and only one side comes up in a throw, so the probability of any one side coming up is
The number of distinct throws of 2 dice, not counting order.
The probability of throwing a 3 and then a 5 with two consecutive throws of one die.
The rolls of the two dice are independent of one another. So the probability of any given roll, considering order, is just the product of the probabilities of any one value on each die:
The probability of throwing a 3 and a 5 with one throw of two dice.
The probability of any given roll, considering order, as in the preceding exercise, is
Counting order, there are two ways to throw a 3 and a 5. that is, (3 then 5) or (5 then 3). So the answer is
The probability of throwing “snake eyes” (both dice showing 1) with one throw of two dice.
This is no different from the probability of any particular roll, not counting order. So, as in the previous exercise, the probability is
The probability of throwing “doubles” (both dice the same value) with one throw of two dice.
The six possible doubles are mutually exclusive. So the probability of drawing
is
The “Anglo-American” deck has 54 cards,
The 1 card is called an ace and labeled with an A. In many games, either one or both jokers are not used.
Many card games involve successive draws and showings of cards, where the cards are not returned to the deck. In these cases, the probability of a card of a certain kind being drawn, depends on whether a card of that card has been drawn already.
The jokers are not used, so there are 52 cards in the deck. The ace card is ranked higher than the king. Cards are dealt from the deck and not replaced. The players at some point receive a “hand” of 5 cards.
The order of cards in a hand does not matter.
The number of poker hands
The probability of a “flush”, that is, five cards of the same suite.
First, let’s calculate the probability of drawing 5 cards from a particular suite, say hearts. Now, each time a heart is drawn, there is one fewer heart remaining in the deck. We need to calculate the joint probability of
The probability of each draw is the number of ♥ remaining divided by the number of cards remaining:
So the probability of a flush in hearts comes to
The probability of getting any flush is the probability of a flush in hearts or a flush in spades or a flush in diamonds or a flush in clubs. In a single draw of 5 cards from a deck, of course you can’t get two flushes (that would take 10 cards), that is, any one flush is mutually exclusive of another. So to find the probability of any one flush, just add the probabilities of the 4.
This is a bit better than one in a thousand.
The probability of a “royal flush”, that is, A, K, Q, J, 10, all from the same suite.
There are four royal flushes, one for each suite.
Any one of them is just a specific poker hand, of which there are C(52, 5). So the probability of a specific royal flush is
Since royal flushes are mutually exclusive, the probability of any of the four happening is just the sum of their individual probabilities:
This is a bit better than 1.5 in a million.
The probability of a “straight”, that is, five cards of consecutive rank (for instance, 8, 9, 10, J, Q), but with ace taking either the value of 1 or 14, without regard of suite.
First, let’s count the runs of five consecutive numbers from 1 to 14. The first is (1, 2, 3, 4, 5) and the last is (10, 11 , 12, 13, 14). So there are 10 of them.
Now, any of the five cards in the hand could be of any of those 4 suites. So for each consecutive list of numbers, there are
different arrangements of the suites among the 5 cards.
So there are a total of 1024 × 10 = 10240 straights in the deck.
The probability of a straight is this number divided by the total number of draws of 5 cards from a deck of 52:
about 0.00394.
What is the probability in a room of 30 people that at least two people will share a birthday?
For simplicity, let’s say all birthdays are equally likely. This isn’t really true, but for the purposes of this problem, the assumption will suffice. Also, let’s not worry about the complication of leap years, and pretend there are 365 days in the year.
A trick will make it easier: calculate the probability that no birthdays are shared among the people in the room. This probability can be treated as that of pulling 30 distinct birthdays pulled out of the bag of 365 birthdays, regardless of order.
The number of such combinations of birthdays is just
The number of all combinations of birthdays (with replacement-- any one birthday can happen any number of times) is just