Section 4.4 Exercises
Determine the probabilities associated with the various 5-card hands. That is
- P(one pair)
- P(two pair)
- P(three of a kind)
- P(full house)
- P(four of a kind
- P(straight)
- P(flush)
- P(royal flush)
Determine the 36 possible outcomes related to the rolling a pair of fair dice. Justify why each of these outcomes is equally likely. Determine the probabilities associated with each possible sum.
Solution
Remember, when using equally likely outcomes |A|/|S| assumes that the items counted for A are also in the sample space S. In this case, for example, to determine the Probability of getting a sum of (say) 4 includes the rolls (1,3), (2,2), and (3,1). These three "successes" from the 36 possible ordered pairs gives P(4) = 3/36. Similarly, P(5) = |dice rolls with a sum of 5|/36 = |(1,4), (2,3), (3,2), (4,1)| / 36 = 4/36. Continue in this manner to determine the other possibilities and then compare to the experimental sage cell seen earlier for the sum of two dice .
Suppose you have one die which only has three possible sides labeled 1, 2, or 3. Suppose a second die has twelve equally likely sides with labels 1,2,3,4,4,5,5,6,6,7,8,9. Justify that the probabilities associate with each possible sum is the same as the probabilities when using two normal 6-sided dice.
Solution
Consider the outcome space
\begin{equation*}
S = {(1,1), (1,2), (1,3), (1,4), (1,4), (1,5), (1, 5), \\
(1,6), (1,6), (1,7), (1,8), (1,9), (2,1) ... (3,9)}
\end{equation*}
Then P(5) = |(1,4), (1,4), (2,3), (3,2) |/36 = 4/36. Compare this to the exercise with regular dice performed above. Similarly, compute the remaining probabilities.
Analyze the dice game known as "craps": Roll a pair of dice and consider the sum. If that sum is 7 or 11, the one who rolls wins and can roll again. If the sum is 2, 3, or 12 -- known as craps -- the one who rolls loses but keeps the dice. For any other outcome (called the "point"), the one who rolls continues hoping to roll the point value again before rolling a 7. If successful, then the roller wins and starts the game anew. If a 7 appears first, the roller loses and the next person gets to be the roller.
So, a win can be obtained in two ways: 7 or 11 on first roll or getting the point before the 7 thereafter. Therefore, determine the probability of a win and the probability of a loss.