**More Nontransitive Paradoxes**

This chapter was written independently of the last one on intransitive dice, and includes different examples of transitivity (if A = B, and B = C, then A = C) and intransitivity (if A is the father of B, and B is the father of C, A is NOT the father of C; or, if A loves B and B loves C, it’s not guaranteed that A loves C (just ask any father what they think about their daughter’s latest boyfriend). Gardner then comments that intransitive relationships are sometimes so counterintuitive in probability theory and decision theory that they’re referred to as intransitive paradoxes. One of the oldest ones is a voting paradox called the Arrow paradox after Kenneth J. Arrow’s use of it in his “impossibility theorem” (for which he shared a Nobel prize in economics in 1972).

Arrow gave 5 criteria that he deemed essential in a “democracy in which social decisions are based on individual preferences expressed by voting” in his work “Social Choice and Individual Values.” The problem is, the conditions are logically inconsistent, and in most instances it is impossible to devise a voting system that doesn’t violate at least one of the conditions. That is, “a perfect democratic voting system is in principle impossible.”

This is something very relevant in the current political environment in the U.S. these days…

In Martin’s words: [Our current system] “frequently puts in office a man who is cordially disliked by a majority of voters but who has an enthusiastic minority following.” Say that 40% of the voters really like candidate A. The opposition gets split between 30% for B and 30% for C. A is elected even though 60% of the people dislike him.

(All rights belong to their owners. Images used here for review purposes only. The voter’s paradox.)

One suggestion for resolving this problem is by letting voters rank the candidates in order of preference, but this works poorly, too. Using the chart above, The top row has 1/3 of the voters preferring candidates A, B and C in that order. The middle row shows another third ranks them BCA, and the bottom row gives the remaining third as liking the candidates in the order CAB. If you rank the candidates in pairs, 2/3rds prefer A to B, 2/3rds prefer B to C, and 2/3rds prefer C to A. If the candidates are paired up, A would defeat B, or B would defeat C, but C would defeat A. If instead of candidates, you vote on proposals, then it’s easy to rig the system to get the proposals you want to win by determining how to pair them up. This flaw has been recognized as far back as the 1700’s, but people keep rediscovering it, including Lewis Carroll. It turns out that the best solution when you get a paradox like this is to select a “dictator” to break the ties.

The same situation can occur when pitting teams against each other in a round-robin tournament, as shown in the above matrix. The paradox arises when picking two alternatives from a set of three or more. Say a woman ranks three men based on intelligence (A), physical attractiveness (B) and wealth (C). If, taken in pairs, she prefers A over B, B over C and C over A, she may have difficulty in selecting simultaneously from three suitors that fall into only one category each. Alternatively, we have Paul R. Halmos’s pie problem. If a restaurant can only serve two pies at any given meal (Apple, Blueberry and Cherry) and customers can make decisions over freshness, taste and size of slice, then it may be reasonable for someone to prefer apple over blueberry, blueberry over cherry, and cherry over apple.

Nontransitivity in gambling games is actually rather common and can lead to some major sucker bets. One kind of nontransitive game is the above top designed by Andrew Lenard. The lower disk is fixed while the upper one rotates. Two players pick different arrows (A, B or C) and spin the spinner. The player whose arrow points to the section with the highest number wins. A will beat B, B will beat C and C will beat A in 2/3rds of the cases.

Mathematician Walter Penney found one of the better betting systems. If you have a coin and flip it three times, the equally possible 8 outcomes are: HHH, HHT, HTH, HTT, THH, THT, TTH and TTT. One player selects one of these triplets, and the other player takes a different one. Now, start flipping the coin. The first person whose triplet appears first as a run wins. Example, A picks HHT and B takes THT. If the coin flips are THHHT, A wins. You would think that all 8 triplets are equally likely, but first let’s look at two flips: HH, HT, TH and TT. Say that you flip the coin and get H. From here, HH and HT are equally possible (HH = HT). But, starting from scratch, HH = TT, and with the same symmetry, TT = TH, and HT = TH. But, TH beats HH by 3 to 1. Looking at a longer run from the game, say we’re comparing HT and TT. TT is always preceded by HT except when TT shows up on the first 2 flips. I.e – HHHT versus TT on the first two flips. The odds are shown in the above matrix for triplets, and HT is much more likely to appear first in the run than TT is.

Followed by the odds for quadruplets.

John Conway developed a procedure for calculating the odds that one n-tuplet will precede another (above figure). To show how this works, let’s use the 7-tuplets (A) HHTHHHT and (B) THHTHHH. Can B beat A? First, write A above A, B above B, A above B, and B above A. Above each pair, you build up a binary number using the following rule: Do the first 7 letters match the seven letters of the tuplet below it? For A/A, the answer is “yes”, so write “1” above the first letter. Do the 6 letters starting with position 2 match the first 6 letters of the second tuplet? No, so write “0” above the second letter. Do the 5 letters starting with the third letter of the top line match the first 5 letters of the second line? No, so also write “0” here. Keep doing this for all 7 positions, then translate the binary value “1000100” to decimal to get 68. And repeat the entire process for the other three pairings. The odds of B beating A are given in the ratio AA – AB : BB – BA. In the above example, that’s 68-1 : 64-35 = 67:29.

If you look at the run of heads and tails, waiting for a particular string to show up, the number of tosses expected is called the waiting time. If you’re standing on a street corner waiting for a bus, the longer you wait, the shorter the time expected before the bus shows up. Coins, however, don’t have a memory, so the waiting time for a particular string of flips remains the same no matter how many times you’ve already flip the coin. The waiting time for H and T is 2. For doublets, it’s 4 for HT and TH, and 6 for HH and TT. For triplets, it’s 8 for HHT, HTT, THH and TTH; 10 for HTH and THT, and 14 for HHH and TTT. So far, so good. “But with a quadruplet, THTH has a waiting time of 20 and HTHH has a waiting time of 18, yet THTH is more likely to show up than HTHH with a probability of 9/14. That is, an event that is less likely to occur, is more likely to occur first. There’s no logical contradiction here, but it does look paradoxical.

**Problem for the week:** TTHH has a waiting time of 16, and HHH has a waiting time of 14. Which tuplet is most likely to appear first, and with what probability?