is a five-element subset of a 52-element set. Thus,or about 1.5 in a million.
Example 1.22 Texas hold 'em. In Texas hold 'em poker, players are initially dealt two cards from a standard deck. What is the probability of being dealt at least one ace?Consider the complementary event of being dealt no aces. There are ways of being dealt two cards neither of which are aces. The desired probability is
Example 1.23 Twenty-five people will participate in a clinical trial, where 15 people receive the treatment and 10 people receive the placebo. In a group of six people who participated in the trial, what is the probability that four received the treatment and two received the placebo?There are ways to pick the four who received the treatment, and ways to pick the two who received the placebo. There are possible subgroups of six people. The desired probability is
Example 1.24 Powerball lottery. In the Powerball lottery, the player picks five numbers between 1 and 59 and then a single “powerball” number between 1 and 35. To win the jackpot, you need to match all six numbers. What is the probability of winning the jackpot?There are possible plays. Of these, one is the jackpot winner. The desired probability isor almost 1 out of 200 million.You can win $10,000 in the Powerball lottery if you match the powerball and exactly four of the other five numbers. The number of ways to make such a match is . This is accomplished by selecting the matching powerball , four of the five selected nonpowerball numbers , and one of the remaining nonselected nonpowerball numbers . Note that the coefficient for selecting the powerball winner is not shown in the computation. This is a common convention for coefficients that evaluate to 1 or , as shown below with . The probability of winning $10,000 in Powerball is thusabout the same as the probability of being dealt a royal straight flush in poker.
Example 1.25 Bridge. In the game of bridge, all 52 cards are dealt out to four players. Each player gets 13 cards. A perfect bridge hand is getting all cards of the same suit. (i) What is the probability of being dealt a perfect hand? (ii) What is the probability that all four players will be dealt perfect hands?
1 There are possible bridge hands. Of those, four contain all the same suit. Thus, the probability of being dealt a perfect hand is
2 There are ways for the first player to be dealt 13 cards. Then ways to deal the remaining 39 cards to the second player, and so on. There arepossible ways to deal out the deck. Perfect hands differ by the suit. And there are ways to permute the four suits among the four players. The desired probability is
Perfect Bridge Hands
According to the BBC News [2003], on January 27, 1998, in a whist club in Bucklesham, England, four card players at a table were dealt perfect bridge hands.
“Witnesses in the village hall where the game was being played,” reported the BBC, “have confirmed what happened. Eighty-seven-year-old Hilda Golding was the first to pick up her hand. She was dealt 13 clubs in the pack…. Hazel Ruffles had all the diamonds. Alison Chivers held the hearts….”
“Chivers insists that the cards were shuffled properly. ‘It was an ordinary pack of cards,’ she said.”
Another report of a perfect bridge hand appeared in The Herald of Sharon, Pa., on August 16, 2010.
An article in 2011 (https://aperiodical.com/2011/12/four-perfect-hands-an-event-never-seen-before-right/) reported several instances of perfect bridge hands being dealt starting in 1929.
Do you think these reports are plausible?
Example 1.26 In 20 coin tosses, what is the probability of getting exactly 10 heads?Here is a purely counting approach. There are possible coin tosses. Of those, there are sequences of H's and T's with exactly 10 H's. The desired probability isA slightly different approach first counts the number of possible outcomes and then computes the probability of each. There are possible outcomes. By independence, each outcome occurs with probability . This gives the same result.
Example 1.27 A DNA strand can be considered a sequence of As, Cs, Gs, and Ts. Positions on the DNA sequence are called sites. Assume that the letters at each site on a DNA strand are equally likely and independent of other sites (a simplifying assumption that is not true with actual DNA). Suppose we want to find (i) the probability that a DNA strand of length 20 is made up of 4 As and 16 Gs and (ii) the probability that a DNA strand of length 20 is made up of 4 As, 5 Gs, 3 Ts, and 8 Cs.
1 There are binary sequences of length 20 with 4 As and 16 Gs. By independence, each sequence occurs with probability . The desired probability is
2 Consider the positions of the letters. There are choices for the positions of the As. This leaves 16 positions for the Gs. There are choices for those. Of the remaining 11 positions, there are positions for the Ts. And the last eight positions are fixed for the Cs. This givesThis last expression is an example of a multinomial probability, discussed in Chapter 5.
Binomial theorem. The classic binomial theorem describes the algebraic expansion of powers of a polynomial with two terms. The algebraic proof uses induction and is somewhat technical. Here is a combinatorial proof.
Binomial Theorem
For nonnegative integer n, and real numbers x and y,
Proof: It will help the reader in following the proof to choose a small
Observe that all the terms of the expansion have the form
