Probabilities of Games

The key and the main difficulty in calculating probabilities to do with games is the enumeration of the outcome space.

We may attempt to do this by constructing a list, perhaps with the help of a tree diagram. This can be unwieldy, however, when there are several stages in the outcomes and when there are many possibilities for each of them. Counting the number of possible hands of some specified type in a game of cards is a typically difficult problem.

Examples

Example 1

We may be interested in the probability of a two-of-a-kind hand occurring when two cards are dealt from a standard deck of 52 cards. In this case, we can count the number of possible two-card hands without listing them all. We would argue that the first card can be any one of the 52, and for each of these there are 51 possibilities for the second card so that in all there are $52\times51=2652$52×51=2652 hands formed in this way. Now, for each of these there will be another that has the same cards but in the opposite order. The reversed hand is considered to be the same hand as the order does not matter. So, there are $\frac{2652}{2}=1326$26522​=1326 distinct two-card hands.

Next, we must count the number of two-of-a-kind hands. As before, there are 52 possibilities for the first card. But, for each of these there are only 3 cards that would complete the pair. Again, the order in which the cards were dealt does not matter. So, only half the pairs counted in this way are distinct. 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: $\frac{\text{number of successful outcomes}}{\text{number of outcomes in the sample space}}$number of successful outcomesnumber of outcomes in the sample space​. This is, $\frac{78}{1326}=\frac{1}{17}$781326​=117​.

Example 2

Continuing from the first example, we might wonder about the situation in which two players are each dealt a two-card hand. What then would be the probability that both players receive two-of-a-kind?

This is a more difficult problem than in the previous case where just one hand was dealt because the cards that are dealt to one of the players affect what can be dealt to the other player. The events are not independent.

We now have a sample space of groups of four cards arranged as pairs of two-card hands. As before 78 of the two-card hands are pairs. We note that each card in the deck occurs in three different two-of-a-kind pairs. One player might be dealt any one of the 78 pairs but then there are only 75 pairs remaining for the other player. The order of the players does not matter, so there will be a division by two in the following calculation. The number of ways in which two-of-a-kind hands can be dealt to two players must be $\frac{78\times75}{2}=2925$78×752​=2925. 

But, how big is the sample space? We need to count the number of ways four cards can be selected from 52 bearing in mind that the order of cards in each two-card hand does not matter and neither does the order of the two players. The number we seek is $\frac{52\times51\times50\times49}{2\times2\times2}=812175$52×51×50×492×2×2​=812175.

Finally, the probability that both players are dealt a pair is:

$\frac{\text{number of successful outcomes}}{\text{number of outcomes in the sample space}}=\frac{2925}{812175}=\frac{3}{833}$number of successful outcomesnumber of outcomes in the sample space​=2925812175​=3833​.

As a check, we might compare this probability with the probability of getting pairs twice in succession if the experiment in Example 1 were to be performed twice. This would be $\left(\frac{1}{17}\right)^2=\frac{1}{289}$(117​)2=1289​ or $\frac{3}{867}$3867​ which is not very different from the value calculated for the non-independent case.

Worked Examples

QUESTION 1

QUESTION 2

QUESTION 3