6. Probability examples

"Sometimes I have believed as many as six impossible things before breakfast'

Lewis Carrol, 'Alice through the Looking Glass'

1. The mechanics of calculating probabilities are fortunately fairly straight-forward and can be exercised effectively by suitable examples without raising the abstract theoretical issues involved in interpretation too obtrusively provided care is taken not to misrepresent the latter in the process. The main practical issues are careful attention to the definition of 'events' and a systematic approach to enumerating possible outcomes - even in relatively simple cases it is easy to miss some possibilities or count them twice without a clear plan. The following seven examples cover most typical simple situations and include the specified NC 'vocabulary' while lending themselves to a 'hands-on' approach where this might be beneficial.

1. Dice 4. Card match (ii)
2. Collisions 6. Children
2. Card sequence 7. Commentary
3. Card match (i)   

1. Two men are gambling with dice. The bet is 36 ducats that Player A can roll a 1 in four or less throws. The game is interrupted after Player A has had 2 unsuccessful attempts. How should the stakes be divided fairly between the players ?

The answer is debatable because the precise meaning of 'fair' in this context is not defined. Possible alternatives are

(a) The match is abandoned so all bets are off.

(b) A has lost on half his attempts, so he should pay 18 ducats

(c) The appropriate player pays what he would lose on average in the two-try game left at the interruption ie A pays 14 ducats.

2. Two men are walking towards one another along a corridor in the dark. The corridor is wide enough for them to pass without touching if they move to opposite sides. Each man, on hearing the other approach, moves to one side, choosing left or right with equal probability.

(a) What is the probability that they will collide ?

If they do collide there is an even chance that either will drop the papers they are carrying.

(b) What is the probability that both will drop their papers ?

(c) What is the probability only one will drop his papers ?

A A
LeftRight DropHold
BLeftmisshit BDropbothone
RighthitmissHoldonenone

(a) They collide in 2 cases out of 4 ie Probability = 2/4 = 1/2

(b) Both drop their papers in a collision in 1 case in 4
Probability = P(collision) x P(both drop) = 1/2 x 1/4 = 1/8

(c) Each drops his papers with probability 1/2 in a collision
Probability that only one drops his papers
   = P(collision) x [P(A drops, B holds) + P(A holds,B drops)]
   = 1/2 x[1/4 + 1/4] = 1/2 x 1/2 = 1/4

'no collision' corresponds to 'neither drop their papers' and combines with 'neither drop despite colliding', giving P=5/8

3. Given four cards - ace, two, three, four - which have been thoroughly shuffled and dealt out, what is the probability that

(a) All the cards are in the correct (face value) order ?

(b) Only one card is in the correct place ?

(c) At least one card is in the correct place ?

(d) Exactly 3 cards are in the correct place ?

(e) No card is dealt in the right order ?

A1234
B 23413 41241 23
C34242334 1413241412 231312
D43423243 4131424121 323121

(a) All correct - (1,2,3,4) - 1 case only

(b) One card only - (1,3,4,2) (1,4,2,3) (2,3,1,4) (2,4,3,1)

(3,1,2,4) (3,2,4,1) (4,1,3,2) (4,2,1,3) - 8 cases

(c) More than one - (1,2,3,4) (1,2,4,3) (1,3,2,4) (1,4,3,2)

(2,1,3,4) (3,2,1,4) (4,2,3,1) - 7 cases

One or more = 7 + 8 = 15 cases

(d) None - (2,1,4,3) (2,3,4,1) (2,4,1,3) (3,1,4,2) (3,4,1,2)

(3,4,2,1) (4,1,2,3) (4,3,1,2) (4,3,2,1) - 9 cases

There are 24 cases in all so the probabilities are, in order

1/24, 8/24 = 1/3 ,15/24 = 5/8, 0, 9/24 = 3/8

4(i) Four cards, two red and two black are shuffled and dealt out. What is the probability that the first two are the same colour ?

let red = odd, black = even

(a) first two odd(red) (1,3,2,4) (1,3,4,2) (3,1,2,4) (3,1,4,2)

(b) first two even(black) (2,4,1,3) (2,4,3,1) (4,2,1,3) (4,2,3,1)

there are 24 cases in total so P = 8/24 = 1/3

4(ii) What is the probability the last two are the same colour ?

Whether the first pair are the same or different the last pair must be correspondingly the same or different ie P = 1/3 again

5(i) Three cards are drawn at random from a well-shuffled full pack of playing cards. What is the probability that

(a) All three are the same colour? (25/51)x(24/50)=12/51

(b) At least two are the same colour? 1

5(ii) The three cards are drawn one from each of three separate packs. What are the probabilities in this case ? 1/4 & 1

6. A family has two children. Given that boys and girls are equally likely what is the probability that

(a) both are boys
(b) both are boys given that at least one is a boy
(c) both are boys given that the younger is a boy

A A Older
BoyGirl BoyGirl BoyGirl
B Boy *  B Boy *  YoungerBoy * 
Girl    Girl  — Girl — —
(a) 1/4  (b) 1/3  (c) 1/2

3. Commentary on solutions

1. The problem is a tribute to the disreputable but brilliant Italian mathematician Giralamo Cardano, who was the first to write about probability theory in around 1545, dealing with problems of this kind.

Assuming the die is unbiased the odds are initially slightly in A's favour - his chance of not throwing a 1 on any throw is 5/6 so his chance of losing after 4 throws ie not throwing a 1 in any of the four attempts is (5/6)x(5/6)x(5/6)x(5/6) = 625/1296. Hence his chance of winning at the start is 671/1296 ie slightly better than even. When A has lost both the first two throws his chances of winning overall have been reduced to [1 - (5/6)x(5/6)] = 11/36. ie less than one in three.

Player B claims that 'all bets off' is unfair to him because he has risked losing twice while A has actually lost the first two throws without penalty. Player A claims that paying half the stake is unfair since he would still have had a chance of recovering the losses and winning the bet if the game had continued.

If the 'average' outcome of a series of repetitions of the two-throw game remaining after the interruption is agreed to be the fairest way of dividing the stake (as proposed by Cardano), A would win in 11/36 of them and B would win in 25/36 of them in the long run ie A would win 36x(11/36)=11 ducats and lose 36x(25/36)=25 ducats on average so he should pay B 14 ducats, and both are right in their objections. The same answer is obtained (not by coincidence) if A is regarded as losing 25 ducats by losing the two throws but has a chance of winning 36x(11/36) ducats on average in the abandoned two throws.

2. The solution illustrates the tabular form of enumeration usually used for independent events. The second part illustrates the rules for combining probability and the difference between dependent and independent events. The tabular form is less suitable for dependent events (eg the separate symbols needed for AND and EXCLUSIVE OR) and the tree form used in 3 is better for more complicated dependent situations. There is a minor quirk which may catch some pupils - the 'no collision' case corresponds to both men moving in the same direction (L or R) from their point of view

3. The solution illustrates the systematic enumeration of outcomes for dependent events and the need for careful consideration of the nature of the specified events. The tree form suggested is the most suitable for compound events which are sequential: the elementary events (ie dealing the next card) in this case are not independent since their outcome affects the probabilities of subsequent events (ie the card dealt is no longer in the pack to be dealt again). The rule used to construct the tree is that at each node the remaining available items are arranged in numerical order from left to right across the various levels.

If (b) were written as 'One card is in the right place' it would be uncertain whether or not 'more than one' is included - the 'Only' or 'Exactly' and 'At least' remove any ambiguity and are widely used in such problems. The ace is intended to be taken as low but the probabilities are correct if it is taken as high - in ordering problems the actual values can be replaced by a 'rank'.

For (d) the solution is easily found by examining the tree but 'exactly three' in the right order is impossible - if three are in the right place the fourth must be also because there is nowhere else for it to go. Some pupils who examine the tree might wonder why there are no such cases, and the moral is that sometimes a lot of work can be avoided by a modicum of thinking first. A further consequence is that the number of cases in which exactly two cards are in the right place is the number 'more than one' in (c) less the 1 case where all are in the right order ie 6.

As a check it can be pointed out that (c) + (e) must equal the total number of cases and the sum of the corresponding probabilities is 1.

4. This is a well-known example where the 'equally likely' fallacy or 'equal ignorance' syndrome frequently leads to error. One common version is that there are three 'equally likely' outcomes viz two red, two black, one red and one black, so the probability of a pair is 2/3 (sic). Another more sophisticated version is that there are four 'equally likely' outcomes - red red; red black; black red; black black - two of which produce a pair, so the probability is 1/2 (sic). In fact the 'equally likely' events, if that's what they are, are that each of the cards still in the pack has an equal chance of being dealt next. Thus whichever colour is dealt first the pack has one card which matches it and two that do not, so the probability of a match is the probability that the matching one appears next ie 1/3. This example illustrates again that in some cases a little fore-thought can save labour.

The word 'shuffled' here is a euphemism concealing the 'likely' spectre - to be unequivocal the question should be phrased '..shuffled so that all permutations of the cards are equally probable'. If it is not the answer should be, pedantically, 'IF shuffling makes every permutation of the cards equally likely the probability the first two cards dealt have the same colour is...'. It is the omission of the 'IF..' and its implied replacement by 'Since..' which renders the 'equally likely' formulation of probability theory untenable. When asked to 'pick a card' by a conjuror or 'Find-the-Lady' operator, one would have reasonable doubt as to whether all possible outcomes are equally likely even though the means by which the outcome is influenced are not apparent. Whether or not the 'IF..' can be replaced legitimately by 'Since..' depends entirely on the nature of the random processes involved and is beyond the bounds of probability theory. Verification depends on empirical observation (ie statistics) viz whether or not the proportions of the various outcomes tend in the limit of a large number of trials to the deduced probabilities resulting from the assumption.

5. This is similar to Problem 4 but the number of alternative outcomes for the first three cards (132600) is too large to write out explicitly and the problem has to be solved by analysis of the successive situations viz

For the first choice in (i), and all in (ii), there are 26 cards of one colour and 26 of the other in the pack ie the probability of selecting a red card is 1/2 and of picking a black 1/2, but since the requirement is only that colours match the outcome of the first choice is irrelevant.

For the second choice in (i) there are 25 cards with the same colour as the first choice and 26 of the opposite colour left in the pack ie the probability of matching the first card is 25/51.

For the third choice in (i) there are 24 cards of the same colour as the first two when these match and still 26 of the opposite colour. If the first pair did not match the outcome cannot satisfy the required condition (all three match) ie the probability of picking a third card matching the first two when these do match, is 24/50 and the probability of both occurring - the only way the condition can be satisfied - is (25/51) x (24/50) = 12/51

Note that this is also the probability of picking a second card of the same suit as the first. The probability of three of the same suit, automatically matching colour, is 66/1275 and hence the probability all three are from the same suit given that they match in colour is (66/1275) ÷ (12/51) ie 11/50. This can be derived directly as for matching colours by noting there are 11 cards out of the remaining 50 that are the same suit if the first two are.

For (b) there are three cards and only two colours so at least two always match ie the probability is 1.

Part (ii) converts the problem to one involving independent processes since what happens to one pack does not affect either of the others ie the triple-match probability is now (1/2) x (1/2) - the pupil still has to realise that the first choice is irrelevant or, if the 'red' and 'black' cases are analysed separately, remember to add their probabilities.

The same comments apply to the phrases 'at random' and 'well-shuffled' as in 4 : 'at random' in particular needs to be specified in more detail to be meaningful.

6. This problem appears so simple that many people will guess an answer immediately but wrongly. The main point is that the case which satisfies the condition is clear and it is the number of possible outcomes satisfying the constraint that has to be determined - once again the precise phrasing of the conditions has to be considered carefully. It is in fact a painless example of conditional probabilities without the usual formalities.

The em-dash in the diagrams indicates the boundary imposed by the constraints ie the number of possible outcomes disallowed, and the asterisk the case which satisfies the stated condition. All permitted combinations are equally probable by definition.

This is a case where the equally likely assumption is not justified in real life - the proportion of boys in the age range 5-14 exceeds that of girls by 5.2%, so the probabilty of two boys in (a) should be in reality 1/3.8 (and of two girls 1/4.2), with similar adjustments for (b) and (c) viz 1/2.9 and 1/1.95. Statistics for the distribution of boys and girls in sibling pairs are less readily available - in the present social climate the question which children are full siblings and which broken siblings can cause embarrassment or uncertainty - but there is probably a significant correlation ie a specific pair of parents are more likely to produce two boys or two girls than a 'pigeon pair'. In this case the probability of two boys would be higher in all cases.