Confirming my Understanding of Kelly -diverted-

[cschaps]

Hello,

I am new to the forum. I have been reading at length the threads on the Kelly Criterion. I would appreciate it if someone would confirm my understanding. Here is how I think it is supposed to work.

Example:

Bankroll = $10,000

7/9 - Cardinals (+142) vs. Phillies (-152)

Cardinals:
Probability - 100/242 = 41.3%
Decimal Odds - 1+(142/100) = 2.42
Edge - .413 * 2.42 - 1 = -.05%

Phillies:
Probability - 152/252 = 60.3%
Decimal Odds - 1+ (100/152) = 1.66
Edge - .603 * 1.66 - 1 = .098%

Combined Probability - 41.3% + 60.3% = 101.60

New Info:

Cardinals:
New Probability - 41.3/101.60 = 40.65%
New Edge = .4065 * 2.42 - 1 = -.016

Phillies:
New Probability - 60.3/101.60 = 59.35%
New Edge - .5935 * 1.66 - 1 = -.014

Questions:

Do I base edge on
1) the calculated probability that includes vig (Cardinals -.05, Phillies .098)?
2) the calculated probability that does not include vig (Cardinals -.016 Phillies -.014)?
3) on a probability I would get from another source (i.e. espn.com)?

I am thinking it has to be the 3rd option. So assuming that is the case I would only wager if the following criteria are met:

1. If the probability of winning for either team found on espn.com, for example, is greater than the implied probability calculated.
2. If the probability of winning for each team taken from espn.com and their respective lines create a positive edge. Of course if they both have positive edges, I would choose the team with the higher edge.

Am I understanding this correctly?

So assuming the edge on the Phillies tonight is 3% and the line is -152 (this means I am relying on a probability of (1+
.03)/1.66 = 62.05% chance that the Phillies will win, the amount I would wager would be:
.03 (Edge)/1.66 (Odds) - 1 = 4.5% of $10,000, which is $450.

Again I would appreciate an confirmation that I am understanding all of this.

Thanks in advance!

cschaps

[Ganchrow]

Example:

Bankroll = $10,000

7/9 - Cardinals (+142) vs. Phillies (-152)

Cardinals:
Probability - 100/242 = 41.3%
Decimal Odds - 1+(142/100) = 2.42
Edge - .413 * 2.42 - 1 = -.05%

This is just rounding error. "Edge" calculated in this manner will be definition always be exactly 0.

"Edge" = 100/242 * 2.42 - 1 = 0

Phillies:
Probability - 152/252 = 60.3%
Decimal Odds - 1+ (100/152) = 1.66
Edge - .603 * 1.66 - 1 = .098%

"Edge" = 152/252 * 252/152 = 0

The implied probability of a money line is by definition the probability that implies zero edge.

See http://forum.sbrforum.com/handicappe...rcentages.html and http://forum.sbrforum.com/handicappe...ical-hold.html.

Do I base edge on
1) the calculated probability that includes vig (Cardinals -.05, Phillies .098)?

No. This would always imply 0 edge and hence no bet (unless hedging).

2) the calculated probability that does not include vig (Cardinals -.016 Phillies -.014)?

No. This would always imply negative edge and hence no bet (unless hedging).

3) on a probability I would get from another source (i.e. espn.com)?

Yes. Although ESPN.com might not be the most accurate source of such estimates.

1. If the probability of winning for either team found on espn.com, for example, is greater than the implied probability calculated.
2. If the probability of winning for each team taken from espn.com and their respective lines create a positive edge. Of course if they both have positive edges, I would choose the team with the higher edge.

Am I understanding this correctly?

You place a bet if the Kelly stake is > 0. This will be the case if and only if (once again, ignoring hedging) edge > 0.

Unless a scalping opportunity existed there'd be no way for there to be a positive edge
on both a team and its opponent.

So assuming the edge on the Phillies tonight is 3% and the line is -152 (this means I am relying on a probability of (1+
.03)/1.66 = 62.05% chance that the Phillies will win, the amount I would wager would be:
.03 (Edge)/1.66 (Odds) - 1 = 4.5% of $10,000, which is $450.

If you were a full Kelly bettor then yes, exactly.

[cschaps]

Thanks

Thanks for the quick response, Ganchrow.

So the edge should always be calculated using a probability determined through an outside handicapping strategy?

If so, what handicapping strategy do you recommend?

And that outside probability would need to be higher than the implied probability to create an edge greater than 0, right?


Thanks again for taking the time to help me understand!

cschaps

[Ganchrow]

So the edge should always be calculated using a probability determined through an outside handicapping strategy?

Exactly.

If so, what handicapping strategy do you recommend

Naturally, the one that works best.

That's the million-dollar question.

And that outside probability would need to be higher than the implied probability to create an edge greater than 0, right?

You got it.