Probability and Chance

When flipping a coin, it seems obvious that the chance (or probability) of head is ½ and the chance of tail is also ½. However, what is the probability of two times head in succession? One might think that the probability of this happening is ½ + ½ = 1. However, if the probability is 1 (i.e. 100%), we mean it is a certainty. We all know it is quite possible to have tails three times in succession, so the probability cannot be 1 and should be lower than that.

So, what is the probability of heads twice in succession?

$P=\frac{1}{2}\times\frac{1}{2}=\frac{1}{4}$

Probabilities are multiplied rather than added

The example of the coin above is rather unique, in that the probability of heads is equal to the probability of tails ( ½ ). Consequently, the probability of heads twice is equal to the probability of tails twice and indeed the probability of heads once and tails once in succession.

So what if we look at a slightly more complicated example; the die? When rolling a die, the probability of throwing “1” = 1/6. This is equal to the probability of throwing “2”, “3”, “4”, “5” and “6”.

As explained above, the probability of twice “1” in succession is:

$P=\frac{1}{6}\times\frac{1}{6}=\frac{1}{36}$

Similarly, the probability of throwing six times “6” is:

$P=\frac{1}{6}\times\frac{1}{6}\times\frac{1}{6}\times\frac{1}{6}\times\frac{1}{6}\times\frac{1}{6}=\frac{1}{46656}\approx 0.0000214$

What is the probability of once “1”, when throwing the die twice? The probability of this happening is the probability of:

First time “1”, second time not “1” : $P=\frac{1}{6}\times\frac{5}{6}$

First time not “1”, second time “1” : $P=\frac{5}{6}\times\frac{1}{6}$

So the probability is:

$P=\frac{1}{6}\times\frac{5}{6}+\frac{5}{6}\times\frac{1}{6}=\frac{5}{36}+\frac{5}{36}=\frac{10}{36}\approx 0.278\approx 28%$

So why are these probabilities suddenly added together (rather than multiplied)? The probability of the first time “1” depends on the second time not being “1”. These probabilities are therefore dependent and should be multiplied.

Probabilities that are not dependent on each other should be added

Similarly, we can calculate the probability of the die being at least once “1” in two throws. This probability is:

First time “1”, second time not “1” : $P=\frac{1}{6}\times\frac{5}{6}$

First time not “1”, second time “1” : $P=\frac{5}{6}\times\frac{1}{6}$

First time ”1” and second time “1” : $P=\frac{1}{6}\times\frac{1}{6}$

So the probability is:

$P=\frac{1}{6}\times\frac{5}{6}+\frac{5}{6}\times\frac{1}{6}+\frac{1}{6}\times\frac{1}{6}=\frac{5}{36}+\frac{5}{36}+\frac{1}{36}=\frac{11}{36}\approx 0.3056\approx 31%$

Let us look at a slightly more complex example. What is the probability on twice “1” on throwing a die six times? One might think this probability would be:

$P=\frac{1}{6}\times\frac{1}{6}\times\frac{5}{6}\times\frac{5}{6}\times\frac{5}{6}\times\frac{5}{6}$

However, this is the probability of throwing “1” in the first and second throw followed by four throws when the die is not “1”. But if we throw a “1” the first and third time we also have thrown “1” twice out of six:

$P=\frac{1}{6}\times\frac{5}{6}\times\frac{1}{6}\times\frac{5}{6}\times\frac{5}{6}\times\frac{5}{6}$

And if we throw a “1” the first and fourth throw:

$P=\frac{1}{6}\times\frac{5}{6}\times\frac{5}{6}\times\frac{1}{6}\times\frac{5}{6}\times\frac{5}{6}$

Obviously, there are many combinations possible. If “1” depicts the die being “1” and “x” the die not being “1”, the possible combinations are:

11xxxx x11xxx xx11xx xxx11x xxxx11

1x1xxx x1x1xx xx1x1x xxx1x1

1xx1xx x1xx1x xx1xx1

1xxx1x x1xxx1

1xxxx1

So there are in total 15 possible combinations. And the probability of throwing “1” twice out of six throws is:

$P=\frac{1}{6}\times\frac{1}{6}\times\frac{5}{6}\times\frac{5}{6}\times\frac{5}{6}\times\frac{5}{6}+$ $\frac{1}{6}\times\frac{5}{6}\times\frac{1}{6}\times\frac{5}{6}\times\frac{5}{6}\times\frac{5}{6}+$

$\frac{1}{6}\times\frac{5}{6}\times\frac{5}{6}\times\frac{1}{6}\times\frac{5}{6}\times\frac{5}{6}+....$ $+\frac{5}{6}\times\frac{5}{6}\times\frac{5}{6}\times\frac{5}{6}\times\frac{1}{6}\times\frac{1}{6}$

In other words the number of combinations times the probability of one of these combinations:

$P=15\times\frac{1}{6}\times\frac{1}{6}\times\frac{5}{6}\times\frac{5}{6}\times\frac{5}{6}\times\frac{5}{6}\approx 0.201$

So how do we calculate the number of possible combinations? The number of possible combinations of two out of six is:

${6\choose2}$ ‘six over two’

$\frac{6!}{4!\times2!}=\frac{6\times5\times4\times3\times2\times1}{(4\times3\times2\times1)\times(2\times1)}=\frac{6\times5}{2\times1}=\frac{30}{2}=15$

In general, the total number of combinations N from n (number of occurrences) out of t (total number of occurrences):

$N={t\choose n}=\frac{t!}{(t-n)!\times(n)!}$

If c is the probability on a single occurrence; the probability of this not happening is 1 – c:

$P={t\choose n}\times(c)^n\times(1-c)^{t-n}$

So in our example of the die, what is the probability of throwing twice “1” out of six throws?

t = 6, n = 2 and c = 1/6:

$P={6\choose 2}\times(\frac{1}{6}})^2\times(\frac{5}{6})^{6-2}=\frac{6!}{(6-2)!\times(2)!)}\times(\frac{1}{6}})^2\times(\frac{5}{6})^{6-2}$

$P=\frac{6\times5\times4\times3\times2\times1}{(4\times3\times2\times1)\times(2\times1))}\times(\frac{1}{6}})^2\times(\frac{5}{6})^{4}$

$P=\frac{6\times5}{2\times1}\times(\frac{1}{6}})^2\times(\frac{5}{6})^{4}=15\times(\frac{625}{46656})=\frac{9375}{46656}\approx 0.201$

What is the probability of throwing at least four “6” out of six throws? This is the probability of throwing 4 times “6” plus the probability of throwing 5 times “6” plus the probability of throwing six times “6”:

t = 6, n1 = 4, n2 = 5, n3 = 6, c = 1/6

$P={6\choose 4}\times(\frac{1}{6})^4\times(\frac{5}{6})^{6-4}+{6\choose 5}\times(\frac{1}{6})^5\times(\frac{5}{6})^{6-5}+{6\choose 6}\times(\frac{1}{6})^6\times(\frac{5}{6})^{6-6}$

$P=\frac{6!}{2!\times4!}\times(\frac{1}{6})^4\times(\frac{5}{6})^{2}+\frac{6!}{1!\times5!}\times(\frac{1}{6})^5\times(\frac{5}{6})^{1}+\frac{6!}{0!\times6!}\times(\frac{1}{6})^6\times(\frac{5}{6})^{0}$

$P=\frac{6\times5}{2}\times(\frac{1}{6}})^4\times(\frac{5}{6})^{2}+\frac{6}{1}\times(\frac{1}{6}})^5\times(\frac{5}{6})^{1}+1\times(\frac{1}{6}})^6\times(\frac{5}{6})^{0}$

$P=15\times(\frac{1^4\times5^2}{6^6})}+6\times(\frac{1^5\times5^1}{6^6}})+1\times(\frac{1^6}{6^6}})$

$P=\frac{15\times5^2+6\times5+1}{6^6}=\frac{406}{46656}\approx 0.0087\approx 1%$

Please note 0! = 1, ${6 \choose4}={6 \choose 2}=15$ and ${6 \choose6}={6 \choose 0}=1$

In general: ${t \choose n}={t \choose t-n}$

So far, we have only been concerned with probabilities that do not change. Our examples of the coin and die are similar to putting ten balls (numbered 1 to 10) in a hat; taking one out and putting it back afterwards. In that case, the probability of getting ball number “1” once is 1/10 and getting it twice in succession is 1/10 * 1/10 = 1/100. The probability of getting ball number “1” twice out of four draws is:

$P={t\choose n}\times(c)^n\times{(1-c)^{t-n})$

c = 1/10, t = 4, n=2:

$P={4\choose 2}\times(\frac{1}{10})^2\times{(\frac{9}{10})^{2}=\frac{4!}{2!\times2!}\times(\frac{1}{10})^2\times{(\frac{9}{10})^{2}$

$P=\frac{4\times3}{2\times1}\times(\frac{1}{10})^2\times{(\frac{9}{10})^{2}=6\times\frac{1\times9^2}{10^4}=\frac{486}{10000}\approx5%$

However, often we have to deal with situations were the probability depends on the previous history. In other words, taking the balls out of the hat without putting them back! If we start with all ten balls in the hat, the probability of drawing number “1” is 1/10. The probability on a successive ball (“2”, “3”, “4”, “5”, “6”, “7”, “8”, “9” 0r “10”) is now 1/9. The probability that we draw number “1” followed by number “5” is 1/10 × 1/9. This is equal to the probability we draw number “5” followed by number “1”. So in drawing two balls, the probability we draw ball number “1” and ball number “5” in any order is:

$P=2\times\frac{1}{10}\times\frac{1}{9}\approx0.022\approx2%$

So how do we calculate the probability in more complicated scenarios? It has to be remembered that the probability on X happening equals the number of possible combinations with X divided by the total number of possible combinations:

$Probability(X)=\frac{Number Of Combinations With X}{TotalNumberOf Combinations}$

So how do we calculate the possible number of combinations? We have already seen that the number of combinations N equals:

$N={t\choose n}=\frac{t!}{(t-n)!\times(n)!}$

The number of combinations of getting one ball out of the hat is obviously:

${1\choose 1}=1$

So the probability of getting the ball with number “1” out of the hat with ten balls is:

$\frac{1\choose 1}{10\choose 1}=\frac{1}{\frac{10!}{9!\times1!}}=\frac{9!}{10!}=\frac{1}{10}$

Which is what we calculated previously. Similarly, getting number “1” and number “5” in any order is:

$\frac{{1\choose 1 }\times{1\choose 1}}{10\choose 2}=\frac{1}{\frac{10!}{8!\times2!}}=\frac{8!\times2!}{10!}=\frac{2}{10\times9}\approx0.022\approx2%$

The number of combinations in the numerator should be multiplied, as they depend on each other. In more general terms:

$P(X)=\frac{{z\choose a }\times{x\choose b}\times{y \choose c}}{t\choose n}$

t = total number of occurrences, n = number of occurrences, t = z + x + y, n = a + b + c

So let us look at playing cards. What is the probability on getting one king when we draw five cards? There are four kings in the set, so there are ${4 \choose 1}$ possibilities. The remaining four cards should not be kings. There are 51 cards left, of which three are kings and 48 non kings. So the number of possibilities on selecting four cards that are not kings is ${48 \choose 4}$ . The total number of possibilities on selecting five cards out of 52 is obviously ${52 \choose 5}$ . So the probability of getting one King is:

$P=\frac{{4\choose 1 }\times{48\choose 4}}{52\choose 5}=\frac{\frac{4!}{3!\times1!}\times\frac{48!}{44!\times4!}}{\frac{52!}{47!\times5!}}$

$P=\frac{4!\times48!\times47!\times5!}{3!\times1!\times44!\times4!\times52!}=\frac{48!\times47!\times5!}{3!\times44!\times52!}$

$P=\frac{47\times46\times45\times5\times4}{52\times51\times50\times49}=\frac{1945800}{6497400}\approx0.299\appro30%$

It should be noted that 4 + 48 = 52 and 1 + 4 = 5.

Let us calculate the probability of drawing any one king, the queen of hearts and a further queen out of a total five playing cards. We start with the number of possibilities of drawing one king: ${4 \choose 1}$ . Now the Queen of hearts: ${1 \choose 1}$ . There are three queens remaining, so the number of possibilities on drawing one further queen is ${3 \choose 1}$ . Followed by two cards that are neither a king nor a remaining queen (there are in total 44 non kings and queens in the set): ${44 \choose 2}$ . It should be noted that 4 + 1 + 3 + 44 = 52 and 1 + 1 + 1 + 2 = 5. The total number of possible combinations is as previously ${52 \choose 5}$ . So the probability of drawing any one king, the queen of hearts and a further queen is:

$P=\frac{{4\choose 1 }\times{1\choose 1}\times{3 \choose 1}\times{44 \choose2}}{52\choose 5}$

$P=\frac{\frac{4!}{3!\times1!}\times1\times\frac{3!}{2!\times1!}\times\frac{44!}{42!\times2!}}{\frac{52!}{47!\times5!}}=\frac{4!\times3!\times44!\times47!\times5!}{3!\times2!\times42!\times2!\times52!}$

$P=\frac{44\times43\times5\times4\times3\times4\times3}{52\times51\times50\times49\times48}=\frac{1362240}{311875200}\approx0.0044\approx0.4%$