More from the postbag and an experiment

Guess who?

Not a maths fight, just a genuine puzzlement, as (as i said) your recommendation of the double/triple/whatever flies in the face of accepted gambling wisdom. Obviously that doesn't mean they must be correct, but did mean I was intrigued anyway, the point I was trying to make was not about value - the double clearly offers more (that did blow my mind, but I accept it :)) but that, it seemed, over time the singles staker would grow his bankroll by more (which has to be good, yes?), and also that the double is actually a hidden bit of poor bankroll management, as the amount you'll be effectively whacking on the second outcome is (usually) too much in relation to your bankroll. I don't know if those two considerations are important or not.

I probably owe Splittter an apology, in which case sorry. On reflection, I think he is right - that even though the value offered by the double is better, it's a worse bet for your bankroll than the two singles.

How can that be so? Isn't a better-value bet necessarily better than a worse one? It would seem not - it's also necessary to factor in the probability. Which leads, of course, to the question of what constitutes the best bet in a a situation - would you rather have 1% value at 1.5 or 10% at 15? As Splittter's original maths showed (you wouldn't know it, because I didn't show it), the key is not value per se, but expectation.

For a given market, the fractional Kelly stake k is (p - (1-p)/(o-1)). If you win - which you'll do p% of the time - you pick up (p(o-1) - (1-p) = (po - 1). If you lose, as you do (1-p)% of the time, you drop your stake. Your expectation is the total of (the probabilities times the outcomes): p(po-1) - (1-p)(p - (1-p)/(o-1)). If you want to do the algebra, that comes out to be E = k(po-1).

The larger this is, the better the bet. Of the two examples above:
1) p = 0.6733, o = 1.50 => v = po-1 = 0.01*. k = 0.02, E = 0.0002.
2) p = 0.0733, o = 15.00 => v = po-1 = 0.10. k = 0.0071, E = 0.0007.

So the outsider - in this case - is a better bet. Returning to the bets in the last example (two singles with p = 0.37 and p=0.35, both with odds of 3.00, versus a double of p = 0.1295 and o = 9.00)

3) p = 0.37, o = 3.00 -> v = 0.11. k = 0.055, E = 0.0061 (much better than 2) above)
4) p = 0.35, o = 3.00 -> v = 0.05. k = 0.025, E = 0.0013
5) p = 0.1295, o = 9.00 -> v = 0.166. k = 0.021, E = 0.0034.

What does all of this imply? In general, if you have two different bets offering the same value, the one with the shorter odds (i.e., the more probable of the two) gives the higher expectation. The key equation is E=k(po-1): the higher the value of E, the larger your expected return.

Next time I'll hopefully generate some non-postbag-related material, unless Splittter provides more thought-provoking analysis and argument. I may get moving on probability estimation techniques. But that's hard...


* po-1 is, of course, our definition of value. Greater than 0, we're looking at a value bet.