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Homework - Part 211 June 2010
As promised this is the second article addressing the questions sent to me by an anonymous student. Here are the last two questions:
Let us look at question (a) first. One way to address this would be to look at all possible four-hand configurations and count those that contain all four Aces. Unfortunately there are 53,644,737,765,488,792,839,237,440,000 such configurations, so I don't think that is a very practical way to proceed. Rather let's do the following.
The chance that an Ace will be in one hand is just as likely that it will be in another and we really don't care what the rest of the cards are in any hand. So let's just look at the possible patterns of Aces in the four hands. For example (2, 2, 0, 0) is one possibility. There are five other Ace patterns that are like this, namely two Aces in each of two hands and no Aces in the other two hands. The possibilities are:
(2, 2, 0, 0)
Using the above idea we can construct the following table:
Since each of these patterns is equally likely, the probability the four Aces are all in one hand is 4/35 or approximately 11.4286%. Notice that there is only one chance that the four Aces will be in all four hands, so it is four times more likely that all four Aces will be in one hand than in four different hands.
Question (b) is considerably more difficult than question (a), but is handled in a similar fashion. Here we care only about suits. If the four-tuple (c, h, d, s) represents the number of clubs, heart, diamonds, spades respectively in A's hand, then a typical example would be (4, 4, 3, 2). Notice that c + h + d + s has to equal 13. There are 39 such patterns but, unlike the situation in question (a), not all of the patterns are equally likely. One must use a probability distribution known as the hypergeometric distribution. This is beyond the scope of this article. If you are interested in pursuing this, you can find information about this in Richard Epstein's book The Theory of Gambling and Statistical Logic, Revised Edition, Academic Press, 1977.
For the example at hand, the probability of this suit distribution is 0.01796. This suit pattern can be permuted 12 different ways, e.g. (3, 4, 2, 4), and one of these is as likely as the next. Hence all possible patterns of this type have a probability of 12 x 0.01796 or 0.21552. This type of calculation must be carried out for each of the 39 patterns. Fortunately on page 256 of Epstein's book this has been done for us. All one has to do is add up all of the probabilities for each pattern type that has at least one zero. The answer is approximately 0.051.
Tough professor! See you next month.
Don Catlin can be reached at email@example.com
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