Random port tenure

From MegaTravellerWiki

Some travellers have no fixed address and prefer to live aboard starships, buying and selling goods directly from their mobile home and engaging in their skilled trade under the protection of the ship's captain and crew. Such clients include stock brokers, computer programmers, architects, jewellers, and various other skilled persons who can conduct a fair proportion of their business by correspondence.

These clients need to periodically visit ports in order to acquire personal effects, set up temporary shops, or otherwise conduct their business.

It can be useful to know how long a charter client aboard a ship will want to spend in port. The ship's crew and captain will either have to refuse this tenure (and risk offending their client) or remain in port for the duration of their tenure.

The following system provides a system for computing a requested tenure period.

• Roll 1d6. This is the number of "clusters" of time spent.
• For each cluster, roll 1d6 and compare against the following chart:

Die   Tenure
---  ------------
 1    1d6/3 days
 2    1d6/2 days
 3    1d6 days
 4    2d6-2 days
 5    0 days
 6    0 days

• Retain all fractions and total the rolled values to come up with a resulting figure. Fractional days will always work out to a number in hours.

The highest possible time spent in port is 60 days, but this is phenomenally rare (less than one in 600 trillion!)*. The average time a client will ask to spend in port is 6-7 days. On occasion you will see clients requesting up to 15 days. The highest you are likely ever to see is 25 days.

A TableSmith generator for port tenure is available on the TableSmith page.

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(*) If you're interested in the math behind that assertion:

P1(rolling a 6 on 1d6 for number of clusters) = 1/6
P2(rolling a 4 on 1d6 for largest cluster range) = 1/6
P3(rolling largest cluster range six times in a row) = P2 * P2 * P2 * P2 * P2 * P2
                                                     = 1/6 * 1/6 * 1/6 * 1/6 * 1/6 * 1/6
P4(10 on 2d6-2 for highest possible day span) = 1/36
P5(rolling highest possible day span six times in a row) = P4 * P4 * P4 * P4 * P4 * P4
                                                         = 1/36 * 1/36 * 1/36 * 1/36 * 1/36 * 1/36

P(rolling a 6 for clusters, rolling 4 six times in a row, and rolling 10 six times in a row)
    = P1 * P3 * P5
    = 1/6 * (1/6 * 1/6 * 1/6 * 1/6 * 1/6 * 1/6) * (1/36 * 1/36 * 1/36 * 1/36 * 1/36 * 1/36)
    = 1 / 609 359 740 010 496