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)
Checkpoint 4.4.2. Dice.
Determine the 36 possible outcomes related to the rolling a pair of fair dice. Justify why each of these outcomes is equally likely 4.3.11. 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.Checkpoint 4.4.3. Skew 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 4.3.11 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.
Checkpoint 4.4.4. Craps.
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.
mathworld.wolfram.com/Craps.html
