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View Full Version : Kelly Criterion with Parlay Rotations

Matty
08-19-2007, 12:13 PM
Math time! I could x-post this in Sports but I assume most of the sports gamblers are here...

Say you are betting on 3 games A,B,C by using a set of 2-game parlays AB,AC,BC (this is a "2/3 rotation"). Suppose you have a \$1000 bankroll and that each parlay has a 10% edge and 25-1 odds. Then each of the 3 bets would have a Kelly wager of \$4 if treated independently.

So the question is - what is the correct Kelly wager for the 2/3 rotation? We know that the 3 bets are positively correlated, so the variance of the rotation > the variance of 3 independent bets...that leads me to believe that the answer is something less than \$4 per ticket.

Help me out and I'll buy you a beer next Vegas trip...make it 2 beers if you can give me a generalization for x/y rotations for any x and y :)

Why do I care about this? One of the many great things about living in Canada is that our government offers a "sports lottery" that is inefficient enough to be +EV if played properly!

Thanks!

Brutus Buckeye
08-19-2007, 02:15 PM
matty - depends on how correlated that positive correlation is.

I think if the kelly bet is \$4 for 3 independent bets, you can't go wrong with \$2 bets....the answer should be in the middle, probably higher than \$3.

Just a guess at this point.

jayhawk
08-20-2007, 11:44 AM
Say you are betting on 3 games A,B,C by using a set of 2-game parlays AB,AC,BC (this is a "2/3 rotation").

If they are 3 separate games, there shouldn't be any coorelation so using a parlay instead of a strait bet is -EV.

Matty
08-20-2007, 12:56 PM
If they are 3 separate games, there shouldn't be any coorelation so using a parlay instead of a strait bet is -EV.

I don't have a choice - it's parlays or nothing.

jayhawk
08-20-2007, 05:16 PM
I don't have a choice - it's parlays or nothing.

When I was more into sports betting, I would figure out how much to bet by figuring out how much my target winning amount was...
So if you roll is \$1k, and you think you have a 10% edge, then you should bet enough to win \$100 on your first bet. Normally, people overestimate their edge, so I would cut that in half... So you think you have a 5% edge, you make all your bets to win \$50 (that amount of course changes after each wager). If the odds are 5-1, bet \$10, if 1-2, bet \$100.
When I was betting, I assumed my edge was closer to 3%, but to simplify things, bet to win 1% of my bankroll - 4% of my bankroll, depending on how confident I was in the bet, and how far off the bookie odds were from my calculations.

This doesn't work perfectly, but since no one really knows what their true edge is, is a good enough approximation.

Matty
08-20-2007, 06:30 PM
This doesn't work perfectly, but since no one really knows what their true edge is, is a good enough approximation.

I do :)

I can use the real-time odds offered by Pinnacle to back into the true probabilities (assuming efficient odds) and run those probabilities against the odds that are being offered to me.

Still doesn't solve the Kelly problem though...your "bet to win a certain amount" system would cause me to overbet my 2/3 rotation.

Sotally Tober
08-20-2007, 07:22 PM
I do :)

I can use the real-time odds offered by Pinnacle to back into the true probabilities (assuming efficient odds) and run those probabilities against the odds that are being offered to me.

Still doesn't solve the Kelly problem though...your "bet to win a certain amount" system would cause me to overbet my 2/3 rotation.
Please stop all this geeky, math talk and go drink a beer or two, find a nice, young sloot, and have some real fun. And at all costs, don't ask her what the optimal bet amount is in your Kelly problem. And if you do, and she knows, she's the wrong kind of girl.

Matty
08-20-2007, 08:44 PM