Hypothetical Gamble

twerg_85

jameshayes wrote: »

Some people dont have that long.

My quote was in the context of the 1/n chance of your bankroll falling to 1/n.

As regards the Kelly concept, it's for serial bettors. Clearly if you're only ever going to make 1 bet in your entire life, then you should bet your entire bankroll on the best bet you can find.
If you intend to continue betting however, then you need to take into acocunt the possibility of going bust, the cost of recovery from losing bets (e.g. as per the +50%, -50% example given before) etc.

Robin1982

me wrote:

...Kelly won't start...the realities of the underlying mathematics...

Mellor wrote: »

...As I said, you said stake, not roll in the original, this makes a huge difference, if the stake was unchanged, the bet would be neutral EV after 500k bets or so...

Not sure why a fixed stake was assumed since Kelly was being discussed (fixed % of roll, best choice = greatest geometric mean). The full bankroll was used for clarity rather than including another variable for % of roll staked.

Mellor wrote: »

..Are you trying to suggest that your -ev coinflip example is in anyway related to Kellys formula?..fail to see how the example is relevant.

Generalising the formula outlined:

(asymptotic) return rate = ((return on win)^prob.win) * ((return on loss)^prob.loss)

which is same as (assuming x = optimal stake, y = return rate when win, z = return rate when lose)

RR = [(1-x) + ((x*y)^prob.win)] * [(1-x) + ((x*z)^prob.loss)]

Finding x that maximises RR involves the use of differential calculus. In the normal sports betting scenario (where you lose entire stake per loss), the derivation simplifies to:

x = (y*prob.win) - 1 / (y - 1)

which is Kelly.

Mellor

Robin1982 wrote: »

Not sure why a fixed stake was assumed since Kelly was being discussed (fixed % of roll, best choice = greatest geometric mean).

Well thats why I got mixed up, we were discussing Kelly, so why did your example suggest 50% when it was an even money bet?

Generalising the formula outlined:

(asymptotic) return rate = ((return on win)^prob.win) * ((return on loss)^prob.loss)

which is same as (assuming x = optimal stake, y = return rate when win, z = return rate when lose)

RR = [(1-x) + ((x*y)^prob.win)] * [(1-x) + ((x*z)^prob.loss)]

Finding x that maximises RR involves the use of differential calculus. In the normal sports betting scenario (where you lose entire stake per loss), the derivation simplifies to:

x = (y*prob.win) - 1 / (y - 1)

which is Kelly.

right, and if you plug in the values, what does Kelly suggest to bet (as a % of BR)
for the coinflip given even odds.

Kelly says bet 0%
In the example you bet 50%,
If 0% (not betting) is the optimum. Them it makes perfect sense that betting 50% would be -EV

Just like if 100/1 on the coinflip. And betting 100%. On a single bet, it is +EV,
but in series it is -EV. Here Kelly suggests an amount approaching 50%, which has the greatest +EV in series.




I understand your point, but all it did was highlight a -EV bet that was arrive at by not using Kelly.

Mellor

twerg_85 wrote: »

Or you could win some, then go on a losing streak. Or you could lose 3 for every 1 you win over a long period etc. Doesn't have to be all losses at the start.

If you run some simulations you'll get a number pretty close to 10% chance of your roll falling to 10% or less (I got 9.8% chance). Same for 50% loss (I got 48% chance on my simulations).

Sorry, yeah I mis-read the OP i think. I thought it meant being ground down to X% from start.

That is a downfall, its not a flawless betting system, nor the best way to exploit an edge. It is mearly the % that yields the highest return. This doesn't not meant it has the lowest ROR. In extreme examples (10/1 on a 50/50) It would possibly make more sense betting a sub optimal amount with a lower ROR.
Max EV doesn't not always mean the best bet. (or a good bet as in St Petersburg )