Probability in games
The key and the main difficulty in calculating probabilities to do with games is understanding and representing the entire sample space, or the entire list of possible outcomes.
We may attempt to do this by constructing a list or perhaps with the help of a tree diagram. This can be unwieldy when there are several stages or when there are many possibilities for each stage. Counting the number of possible hands of some specified type in a game of cards is a typically difficult problem.
Worked examples
Example 1
Calculate the probability of a two-of-a-kind hand occurring when two cards are dealt from a standard deck of $52$52 cards.
Think: This means two cards of the same type are dealt. Something like this: 
So we can have any card for the first card, but the second card must be the same as the first. We know that there are $52$52 cards in a pack and there are $4$4 of each type of card.
Do: First calculate the total number of possible two-card hands without listing them all. The first card can be any one of the $52$52, and then there are $51$51 possibilities for the second card (because one card has already been used) so that in all there are $52\times51=2652$52×51=2652 possible hands.
Important! A hand of 
is the same as so we must divide by $2$2.
So, there are $\frac{2652}{2}=1326$26522=1326 different two-card hands.
Next, we count the number of two-of-a-kind hands. As before, there are $52$52 possibilities for the first card. But, for each of these there are only $3$3 cards that would complete the pair. Again, we must divide by $2$2.
Hence, the number of outcomes that would be counted as successes is $\frac{52\times3}{2}=78$52×32=78
Finally, the probability of the event 'two-of-a-kind' is:
| $P\left(\text{two-of-a-kind}\right)$P(two-of-a-kind) | $=$= | $\frac{\text{number of favourable outcomes}}{\text{total number of outcomes}}$number of favourable outcomestotal number of outcomes |
| $=$= | $\frac{78}{1326}$781326 | |
| $=$= | $\frac{1}{17}$117 |
Example 2
Calculate the probability that a player rolls a double on a die when playing a board game.
Think: This means rolling a die twice and getting the same number on both rolls. We could draw an array to to represent the sample space of rolling two dice and count the number of outcomes where a double appears. We could also draw a tree diagram to represent the situation - however, with $6$6 options at each stage this will be quite a large tree.
Let's instead picture the tree diagram in our mind and decide what calculations we would need to perform to determine the probability of rolling a double:
- To calculate the probability of rolling a particular double we would multiply across the branches.
For example, the probability of $3$3 on the first die and $3$3 on the second die is $\frac{1}{6}\times\frac{1}{6}=\frac{1}{36}$16×16=136.
- Then we would add up the probability of each outcome that was a double.
Do:
| $P\left(\text{rolling double}\right)$P(rolling double) | $=$= | $P\left(\text{two 1's}\right)+P\left(\text{two 2's}\right)+P\left(\text{two 3's}\right)+P\left(\text{two 4's}\right)+P\left(\text{two 5's}\right)+P\left(\text{two 6's}\right)$P(two 1's)+P(two 2's)+P(two 3's)+P(two 4's)+P(two 5's)+P(two 6's) |
| $=$= | $\frac{1}{36}+\frac{1}{36}+\frac{1}{36}+\frac{1}{36}+\frac{1}{36}+\frac{1}{36}$136+136+136+136+136+136 | |
| $=$= | $\frac{6}{36}$636 | |
| $=$= | $\frac{1}{6}$16 |
Reflect: The result of the first roll does not affect the result of the second roll so we have independent events. Whenever we have independent events and we encounter "and" it means multiply. For example, a three on the first die and a three on the second die. We multiply the two probabilities as this would be equivalent to multiplying across branches in a tree diagram.
Practice questions
QUESTION 1
A local raffle prize will be given to the person who has the winning number from one of the $200$200 tickets sold. What is the probability of winning the raffle if you purchase $30$30 tickets?
QUESTION 2
In a game of Monopoly, rolling a double means rolling the same number on both dice. When you roll a double this allows you to have another turn.
What is the probability that Sarah rolls:
a double $1$1?
a double $5$5?
any double?
two doubles in a row?
QUESTION 3
In a card game, Aaron is dealt a hand of two cards without replacement from a standard deck of $52$52 cards. How many different hands are possible?
QUESTION 4
In a game of Blackjack, a player is dealt a hand of two cards from the same standard deck. What is the probability that the hand dealt:
Is a Blackjack?
(A Blackjack is an Ace paired with 10, Jack, Queen or King.)
Has a value of 20?
(10, Jack, Queen and King are all worth 10. An Ace is worth 1 or 11.)
Traffic light problems
Probability can be applied to situations involving time in the same way as practical events. For example: If a light in a room is turned on for $45$45 seconds in every five minutes, we would say that the probability of a person who enters the room at a random moment encountering the light 'on' is $\frac{45}{5\times60}=0.15$455×60=0.15
Worked example
example 3
On a stretch of road a set of traffic lights has a sequence where in each $2$2-minute period, the traffic light is red for $30$30 seconds, yellow for $10$10 seconds and then green for $80$80 seconds.
(a) If Emma passes through the intersection on the way to and from work, what is the probability she gets green lights on both occasions?
Think: We can represent this situation with a tree diagram with two columns, one representing the light she encounters in the morning and one representing the light she encounters in the evening. Let's first simplify the probability of encountering the different coloured lights on arrival at the intersection.
| $P\left(\text{Green}\right)$P(Green) | $=$= | $\frac{80}{120}$80120 | $P\left(\text{Yellow}\right)$P(Yellow) | $=$= | $\frac{10}{120}$10120 | $P\left(\text{Red}\right)$P(Red) | $=$= | $\frac{30}{120}$30120 | ||
| $=$= | $\frac{2}{3}$23 | $=$= | $\frac{1}{12}$112 | $=$= | $\frac{1}{4}$14 |
Do:
To find the probability of $2$2 green lights we multiply across the branches in the uppermost path:
| $P\left(GG\right)$P(GG) | $=$= | $\frac{2}{3}\times\frac{2}{3}$23×23 |
| $=$= | $\frac{4}{9}$49 |
(b) What is the probability Emma encounters at least one red light?
Think: Using the tree diagram above we want to add the probability of the options $GR$GR, $YR$YR, $RG$RG, $RY$RY and $RR$RR.
Do:
| $P\left(GR,YR,RG,RY\text{ or }RR\right)$P(GR,YR,RG,RY or RR) | $=$= | $\frac{2}{3}\times\frac{1}{4}+\frac{1}{12}\times\frac{1}{4}+\frac{1}{4}\times\frac{2}{3}+\frac{1}{4}\times\frac{1}{12}+\frac{1}{4}\times\frac{1}{4}$23×14+112×14+14×23+14×112+14×14 |
| $=$= | $\frac{2}{12}+\frac{1}{48}+\frac{2}{12}+\frac{1}{48}+\frac{1}{16}$212+148+212+148+116 | |
| $=$= | $\frac{7}{16}$716 |
(c) If Emma was to pass through the the intersection three times, what is the probability she arrives at a red light all three times?
Think: Rather than drawing a three stage tree diagram, picture it and the calculation we would require to determine the probability of the outcome $RRR$RRR. This would be multiplying across the branches with the probability of encountering a red light, $P\left(\text{Red}\right)=\frac{1}{4}$P(Red)=14.
Do:
| $P\left(RRR\right)$P(RRR) | $=$= | $\frac{1}{4}\times\frac{1}{4}\times\frac{1}{4}$14×14×14 |
| $=$= | $\frac{1}{64}$164 |
Practice questions
QUESTION 5
Traffic lights showing green, red and yellow lights are set up on a race track. A race car approaching the lights is equally likely to see a green, red or yellow light.
This is because:
the three lights stay on for the same amount of time.
Athe race car driver is driving at a constant speed.
Bthe lights are the same size.
C
QUESTION 6
Every $60$60 seconds, a traffic light remains green for $34$34 seconds, yellow for $3$3 seconds and red for $23$23 seconds.
Which outcome is more likely?
Arriving at the traffic light when it is yellow.
AArriving at the traffic light when it is red.
BArriving at the traffic light when it is green.
COn her way to work, Nadia passes through $3$3 sets of such lights. What is the probability that none of the lights are green?
What is the probability that the first two sets of lights are green and red at the third set?
QUESTION 7
On the island of Timbuktoo the probability that a set of traffic lights shows red, yellow or green is equally likely. Christa is travelling down a road where there are two sets of traffic lights.
Construct a tree diagram to indicate the possible pairs of traffic lights.
What is the probability that both sets of traffic lights will be yellow?