# A Proposition Bet Walks Into a Bar ...

There's time for another quick paradox before the season commences.

Consider the following proposition bet. Each week we'll look at the total points scored in the first game of the round and look at whether the total is even or odd. I win if, across consecutive rounds, the sequence (even,odd,odd) occurs before the sequence (even,odd,even) and you win if the converse occurs. So, for example, if the total scores in the first game of Rounds 1, 2 and 3 were (146, 171, 155) then I'd win. If, instead, they were (132, 175, 162) then you'd win. If no result had been achieved after Round 3 then we'd keep going, starting with the aggregate score for game 1 of Round 4, until one or other of the winning sequences occurred.

Now were I to bet you that my sequence would occur before yours it probably wouldn't surprise you to learn that this is a fair bet at even money odds (recognising that points aggregates for games are as likely to be odd as to be even). But what if, instead, I said that we would play this game repeatedly over the next 3 seasons, with each game ending and another commencing only once both sequences had occurred and that the overall winner of the bet would be the person whose sequence had, on average across all of the completed games, taken the fewest number of games to occur?

So, for example, we might over the course of the three seasons complete 6 games, with my sequence taking 8,6,9,7,11 and 9 games to occur and yours taking 10,11,8,6,10 and 15 games to occur. The average time for my sequence is 50/6 = 8.33 and for your sequence is 60/6 = 10 so, in this case, I'd win.

Given that it's an even money bet whose sequence occurs first would you also be willing to accept even money odds that the average number of games it takes for my sequence to occur will be less than the average number of games it takes for your sequence to occur?

Well if you would, you shouldn't. On average my sequence will take 8 games to occur and yours will take 10 games.

The reason for this apparently paradoxical result is subtle and hinges on how much longer, on average, it takes for the losing sequence to occur after the winning sequence has just been completed. If your sequence - (even,odd,even) - has just occurred then I'm already one-third of the way to completing my sequence of (even,odd,odd), but if my sequence has just occurred then the last result was a game with an odd number of points, so you're still at least three games away from completing your sequence. When you do the maths it turns out that this makes the average number of games required to generate your sequence equal to 10 games while it's only 8 games for my sequence. This despite the fact that it's an even money bet whose sequence turns up first.

You might need to run some simulations with a coin to convince yourself of this, but it is true. A discussion of the result is included in this TED talk from Peter Donnelly. (There are some other fantastic talks on the TED site. While you're visiting you might also want to take a look at the talks by Elizabeth Gilbert, Ken Robinson, Malcolm Gladwell, and a stack of others.)

This result is not, I'll acknowledge, a cracking way to win bar bets - unless, I suppose, you're contemplating a long session, but still expecting to remain sufficiently clear-headed to track a hundred or so coin tosses and to, frankly, give a proverbial about the outcome - but it does have a geeky charm to it.

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