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Best of Donald Catlin  # Ships, Captain and Crew Part One

30 November 2013

Ships, Captain & Crew – Part 1

I recently received the following email from one of my readers, Pat Ballew:

“An associate just introduced me to a game that is apparently common to sailors around the world, but poses an interesting probability problem. The game is called Ships, Captain & Crew and is played with five dice. The object is to roll a 6, a 5, and a 4 in three rolls of the set of five dice under the following rules. The six must be rolled on or before the roll on which the five comes up, and similarly for the five and the four. In other words if you roll the six, five and four and the first roll of the five dice you win. If you roll the six but not the five then you can set out the six and make your next two rolls with only four of the remaining dice. If you roll a six and a five then you can set them both out and roll with only the three remaining dice.

It is possible to win on any of the three rolls but not equally likely. The questions then are: What is the probability that a person would complete the task in three or less rolls? What is the probability of completing the task on each of the three rolls?

I found the problem posed an interesting challenge. Enjoy.

Hakke Yoi!
Pat Ballew”

Thanks Pat. I am breaking the analysis into two parts. In part one (this article) I’ll develop the relevant probabilities and in part two I’ll put the information together to answer the first of the above questions. I’ll leave the second question to you the reader however you should have enough information to easily answer it.

I started the analysis by replacing the dice symbols as follows. The 1, 2 , and 3 are all replaced by the numeral 0, six is replaced by 1, five is replaced by 2, and four is replaced by 3. My reason for doing this was to (hopefully) find representations that would be convenient to program. Unfortunately I saw no way to program anything and my analysis was, as is often the case, reduced to brute force and awkwardness. The replaced symbols were, nevertheless, still helpful.

The outcomes for the first roll are five tuples (a1, a2, a3, a4, a5) where ai is the numeral 0, 1, 2, or 3 that is upmost on the ith die. Since there are six possible numerals (you can think of the 0s as being three different colors if that is helpful) there are 65 or 7776 possible outcomes for the first roll. For example one outcome could be (2, 3, 0, 1, 0). The probability of this outcome is 1 x 1 x 3 x 1 x 3 divided by 1776 or 9/7776. How many outcomes consist of two 0s, and 1, 2, and 3? The answer is all possible arrangements of the 5 tuple (0, 0, 1, 2, 3). It turns out that there are 60 of them so the probability of getting two 0s and a single 1, 2, and 3 is 9 x 60/7776 or 540/7776.

How many different combinations are there? It turns out that an easy way to catalog all the distinct combinations is to list the numerals in each 5 tuple in ascending order. For example (0, 0, 2, 2, 3), (0, 1, 1, 1, 3), and (1, 2, 2, 3, 3) are listed as such. This can be done systematically by starting with (0, 0, 0, 0, 0) and cycling through all of the possibilities like an odometer (with this ascending order restriction) and ending with (3, 3, 3, 3, 3). There are 56 distinct 5 tuples in such a list. For example (0, 2, 2, 3, 3) represents the set of outcomes having one 0, two 2s and two 3s and there are 30 of these. Hence the probability of getting such an outcome is the probability of getting any one of them 3/1776 times 30 or 90/1776.

The next step is to partition these outcomes into sets that are relevant to the game. Probabilities are frequencies divided by 7776. Here is a table that lists the facts.

Type Outcome 1, 2, and 3 rolled / Frequency 1230
Type Outcome 1 and 2, no 3 / Frequency 1320
Type Outcome 1 and no 2 / Frequency 2101
Type Outcome No 1 / Frequency 3125

Total - 7776

Table 1 – Five Dice Rolled

If 1 is rolled on the first Roll and no 2 is rolled then four dice are rolled on the second roll. The outcomes are 4 tuples so there are 64 or 1296 outcomes. Using an analysis analogous to that for the 5 tuples we can obtain the following table.

Type Outcome 2 and 3 rolled / Frequency 299
Type Outcome 2 and no 3 / Frequency 372
Type Outcome No 2 / Frequency 625

Totals - 1296

Table 2 – Four Dice Rolled

If both 1 and 2 have been rolled then on the second or perhaps third roll three dice are rolled. The outcomes are 3 tuples and there are 216 of them. The relevant table is

Outcome Frequency
Outcome 3 rolled / Frequency 91
Outcome No 3 / Frequency 125
Totals - 216

Table 3 – Three Dice Rolled

I think this is enough to keep you busy for now. I’ll organize the information in these tables next month and complete the solution to the Ships, Captain & Crew puzzle. See you then.

Don Catlin can be reached at 711cat@comcast.net
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Best of Donald Catlin
Donald Catlin Don Catlin is a retired professor of mathematics and statistics from the University of Massachusetts. His original research area was in Stochastic Estimation applied to submarine navigation problems but has spent the last several years doing gaming analysis for gaming developers and writing about gaming. He is the author of The Lottery Book, The Truth Behind the Numbers published by Bonus books.

#### Books by Donald Catlin:

Lottery Book: The Truth Behind the Numbers
Donald Catlin
Don Catlin is a retired professor of mathematics and statistics from the University of Massachusetts. His original research area was in Stochastic Estimation applied to submarine navigation problems but has spent the last several years doing gaming analysis for gaming developers and writing about gaming. He is the author of The Lottery Book, The Truth Behind the Numbers published by Bonus books.

#### Books by Donald Catlin:

Lottery Book: The Truth Behind the Numbers