As England, have, remarkably enough, won five games in a row under new skipper Kevin Pieterson, there has been some debate on the traditional question of whether KP is a good captain or a lucky one...I’m not going to consider this here, but it did remind me of the time in 2000-2001 when then England captain Nasser Hussein somehow contrived to lose 14 tosses in a row. Surely you can’t get more unlucky than that? Maybe there’s even something spooky going on? (Some non-mathematically minded commentators even suggested he should stop calling “heads” each time to give him more chance of finally winning a toss.)
But. The chances of losing 14 tosses in a row is 1 in 2^14, or 1 in 16,384. This is indeed fairly astronomical. However. Hussein captained England in 45 Tests and 56 ODIs, meaning he contested 101 international tosses in total. There are, unless I’ve made an egregious blunder, 88 possible starting points for a sub-sequence of 14 in an overall sequence of 101, so the chance, sometime in his captaincy, of Hussein losing 14 tosses in a row is actually something of the order of 1 in 186 or thereabouts, which isn’t nearly so unlikely...
Mind you. If not exactly a case for Mulder and Scully, was this still, nevertheless, slightly odd and unexpected? How often might we expect a sequence of x consecutive heads or tails in a particular sequence of tosses? The chance of, say, x tails from a given start point is (½)^x. If there are N possible starting points for the sequence, the average number of times a sequence of x tails appear will be N x ((½)^x). So, with 200 possible start points for a sequence of 5 tails, we would expect, on average, about 200 x ((½)^5), or about 6, such sequences to turn up.
So. Let’s say we decide (as seems quite reasonable), that we wouldn’t be surprised at a particular sequence occurring, if, on average, we would expect at least one such sequence to occur; in other words when N x (½)^x is as least 1.
The expression will equal 1 when x = Log (base 2) N.
E.g. the base 2 log of 64 is 6, and 64 x ((½)^6) = 1, and the base 2 log of 512 is 9, and 512 x ((½)^9) = 1.
So, with 88 possible starting points for a sequence, the expression will equal one when x = 6.459, i.e. 88 x ((½)^6.459) = 1.000. In other words, a total lack of surprise at Hussein’s tossing performance (as it were), would only be preserved if he had lost about 7 tosses in a row, rather than 14.
(From the above, we can use the elegant, if admittedly approximate and somewhat arbitrary “rule”, that if we toss a coin N times, we should expect the longest sequence of heads or tails to be round about the base 2 logarithm of N. The base 2 log of 101 is about six and two-thirds, give or take.)
There we are then. Although not colossally surprising (especially given that Hussein is not the only international captain who has taken part in a great many coin-tosses, so before being amazed we need to think beyond the chances of just him having such a run of bad luck), it was indeed perhaps a slightly odd run. Not nearly as odd as England being 4-0 up in a one-day series against South Africa, though. Funny old game, isn’t it?
Due Credit Department: [For more of this sort of thing, see Beating the Odds by Rob Eastaway and John Haigh (Robson Books, 2007)]