Segal's law

Supposedly, the saying was coined by the San Diego Union on September 20, 1930: "Confusion.—Retail jewelers assert that every man should carry two watches. But a man with one watch knows what time it is, and a man with two watches could never be sure." Later this was — mistakenly — attributed to Lee Segall of KIXL, then to be misquoted again in Arthur Bloch's book as "Segal's Law". [2]

In reality a man possessing one watch has no idea whether it is the correct time unless he is able to compare it to a known Time standard.[3] This situation is not made any worse by having two watches, because the probability of all combinations of states of these watches needs to be taken into account in order to know the right time. Two watches may, depending on the size of the errors of each watch, let you determine which is closer to the correct time based on estimated solar time (or other non-standard but trustworthy time sources). Let there be two states: W (working—showing the correct time), and B (broken—showing the incorrect time). The set of possible states of the two watches are then:

${\displaystyle S={\text{(WW, WB, BW, BB)}}}$

If the probability of a watch being in the W state is p and in the B state is q, and assuming both watches have the same probability of working, then the total probability of all possible states is

${\displaystyle p^{2}+pq+qp+q^{2}=p^{2}+2pq+q^{2}=1}$

since it is certain the watches are in one of these states. The first term, p² represents both watches in the working state so this state will unconditionally yield the correct time. The second term 2pq represents one watch working and the other not. Since it is impossible to know which one is correct one can only guess. Half the time the guess will be right and half wrong so the effective probability of having the right time from this state is only pq. The last term represents both watches not working which will never yield the correct time. The total probability, P, of having the correct time is thus

${\displaystyle P=p^{2}+pq}$

and since q = 1 − p

${\displaystyle P=p^{2}+p(1-p)=p}$

that is, the same probability as one watch. An improved probability of obtaining the correct time is only possible with at least three watches since majority voting logic can then be applied.[4] The case of three watches has a total probability of

${\displaystyle p^{3}+3p^{2}q+3pq^{2}+q^{3}=1}$

The second term will always yield the correct time by majority voting. The third term represents two malfunctioning watches. It is possible to tell that there is a problem but not which watch is correct. Thus again, the best solution is a simple guess that will only be right one third of the time. Thus, the total probability of having the correct time is

${\displaystyle {\begin{aligned}P&=p^{3}+3p^{2}q+pq^{2}\\&=p+p^{2}(1-p)\end{aligned}}}$

which is clearly greater than p. Likewise, the probability function of n watches can be found from the binomial expansion of (p + q)ⁿ.

This reasoning is not valid if there are systematic errors present in the watches. For instance, if all the watches start to gain at high temperature in the same way this is an error that cannot be either corrected or even detected by majority voting.[3]

• List of chronometers on HMS Beagle
• The four-faced liar
• Triple modular redundancy in chronometers
• wikt:a stopped clock is right twice a day

• Farmer, Dan; Wietse Venema. "2". Forensic Discovery. Addison-Wesley. ISBN 978-0-201-63497-6.