If the win probability is 55% and the odds are +900.00 kelly says to bet $500, but if you enter odds of +2000.00 with the same probability of 55%, kelly says to only bet a extra $28 ($528). Why does kelly only increase the bet by $28 when the odds have doubled?

I would assume it takes into account some game day variables, but this seems weird if we are entering the correct true probabilities into the kelly calculator.

A kelly calculator should tell you what fraction of your bankroll to bet, not a dollar amount

I don't do sports betting, but does "+900" mean "If I bet 100 I profit 900"? i.e. the payout is 9:1?

f = (bp - q) / b
p is the probability of winning, here 0.55
q is the probability of losing, here 0.45
b is the payout in b:1 odds, here, if I understand properly, it's 9 in one case and 20 in the other

f1 = (9*.55 - .45) / 9 = 0.5
f2 = (20*.55 - .45) / 20 = .528

So yeah, it goes from suggesting you bet half your bankroll to 52.8% of your bankroll. I guess the deal here is that the payout is so astounding that the results get somewhat skewed. The most it will ever recommend you bet is 55% of your bankroll in this case, when b tends to infinity.

It should be kind of clear why - as b gets very large, you basically get
f = .55b / b
because q is very small compared to .55b

This calculator allows you to enter bankroll amount hence the dollar amount.

https://www.sportsbookreview.com/pic...ly-calculator/


Yes


OK, because that's weird as I read some articles where they said one could lose nearly their whole bankroll on a single bet, hmm.

Actually if I enter +100 or $2.0 decimal odds into all the online kelly calculators with a 99% probability of the bet winning, kelly says to bet 98% of your bankroll. Are these calculators over betting? Also why I posted the thread.

Sorry, to be clear, I mean that the highest fraction you can get is going to be your odds of winning, and that happens when the payout is extremely large. It's not 55% in general, it's 55% in your initial case because that were your odds of winning.

With 9:1 payout, you were already really close to 55%, which sort of "explains" why getting a bigger payout doesn't move the fraction much. It approach 55% asymptotically so the closer you are already to the max, the less an increase in payout moves the kelly fraction.

The calculators are doing no more, surely, than the formula I have given above, and as such are probably not very useful.

Lots of people consider using the full kelly fraction to be a bit of a roller coaster ride and will use a fractional kelly bet - like, whatever value of "f" you get, they might use f/2 or 2f/3 or something. This also helps a bit to compensate for when your estimate of your win chances is wrong - overbetting is worse than underbetting, so intentionally underbetting can keep you from accidentally overbetting.

thanks for explaining.

OK, I think I am giving incorrect theory over in the Poker Theory sub-forum. Can I get an expert in Kelly to answer this particular scenario:

You are playing no limit holdem.
Your hand is QQ.
You are all-in versus AKo.
Your stack is 5 percent of your roll.
Your odds of winning are known to be 57.2 percent.

My understanding is that 5 percent of your roll is comfortably below Kelly.

My question is this:

If you are already below Kelly, then why would you "run it twice" and split the bet into two separate bets that are even farther below Kelly?

Part of a non-rigorous proof of the simple Kelly is to show that delta from Kelly leads to smaller growth as N goes to infinity.

Should not that same proof hold when N goes from 1 to 2?

Thanks.

-Rob

If you could simply be handed your expectation of 57.2% of the pot with no variance, this would be the best possible outcome, right? In fact you'd be safe with 100% of your roll in the pot since it's literally printing money risk-free. The more times you run it, the closer the payout distribution is to simply being handed your equity, and for an equal expectation, lowering variance is always better.

How could I not risk demise by putting all of my roll (assuming no way to replenish it) on a 57.2 chance to win?

I thought Kelly was about managing variance in the most optimal manner?

TomC is correct.

Using the Kelly criterion is equivalent to wanting to maximize your expected log stack. So the concept of "risk-aversion" is built into Kelly.

When choosing between running it once, twice, four times, or ten times, you are not changing the total amount of your "exposure" (or bet). So the principles underlying Kelly would suggest that you would choose running it as many times as you can.

As TomC points out, under the Kelly framework a sure thing is worth more than any gamble with equal expected value but positive variance.

(The confusion may stem from Kelly being used to choose how much to wager on a positive-EV bet. This is not really what is going on in run-it-twice decisions.)

Edit: I suppose it could be pointed out that when you choose to run-it-twice, this is essentially making two simultaneous half-size bets. The simultaneous-bet Kelly criterion is slightly different from the single-bet Kelly criterion. But I am quite sure that this subtlety is irrelevant to this discussion. There is another thread in this forum on simultaneous kelly you may want to look at.

I found the simultaneous bet thread, and would have put this there, but I don't want to tard up multiple threads at once...

I am missing an underlying concept probably because I am overly simplifying Kelly.

Bare with me, as I refirmulate the question in more and more simple ways until it "clicks".

So we can not change the amount of our exposure once we are all-in.
BUT, the separate deals from the dealer are quasi independent (winning one makes you less likely to win the other).

So if we only organize our future betting to include all the time this happens, is it not the same as two bets with half the exposure each? Why would Kelly not apply independently to each bet?

Maximizing the fraction of Kelly that you're betting (up to 1x) is only guaranteed to be a meaningful goal when the only variable is your choice of wager size. For example, somebody could add $100 to the pot for free and you'd be betting a smaller fraction of Kelly than you were before the $100, but it's still obviously good for you.

Ok, in the simultaneous bet thread, you proved that making two bets simultaneously is only slightly different than making them one at a time, such that a gambler who is dealing in estimates of edges should not really take notice at all.

But in those cases, the bettor is allowed to choose the size of each bet.

Here, our exposure is locked and we are presented with the option to move down from Kelly in two separate bets. The bets are only quasi independent such that the total overall EV is 57.2 percent.

Poker bets are (apparently) usually way below full Kelly as a consequence of heuristics of poker bank roll management.

The curve as Kelly maximizes growth is gradual, and humans experience less variance and better enjoyment by fudging below Kelly and allowing for human error in edge calculation.

But here, the edge is rigged. Full Kelly should be implemented by an optimal Econ bettor.

Since the poker player is already at about 1/3 Kelly, that gradual curve is now actually quite noticable.

Moving farther down away from Kelly has to be a mistake right?

OR...

Is Kelly no longer applicable because we are past the "bet sizing" decision?

I would say that the precepts underlying Kelly make you want to run it as many times as you can. So it is more of your second option than your first.

Consider the following stylized example:

You have $10,000 in front of you. Villain shoves for $1,000 and prematurely exposes her hand. You have QQ and can see that you have 60% equity if you call.

Clearly you should call. Assume no other money in the pot so you are getting exactly 1:1 payoff odds.

The Kelly criterion formula is:

f = (bp-q)/b

= (1*.60-.40)/1

= .20

So Kelly says you should be willing to call up to 20% of your stack (ignore any other bankroll for the purposes of this stylized example) or $2,000. But you can only wager $1,000 since that is all that Villain had in her bankroll.

Now Villain asks how many times do you want to run it? Once or twice or more?

Remember that Kelly is tantamount to maximizing the expected log of your ending chip stack. So let's calculate these in your various options.

Option 1: Run it Once

Then U = .6*ln(11000) + .4*ln(9000) = 9.225.

Option 2: Run it Twice

Assume "probabilistic independence" to make the math easier (it does not qualitatively change the result).

Then U = .36*ln(11000) + .48*ln(10000) + .16*ln(9000) = 9.228

Option 3: Run it Infinity times (realize your exact equity with no variance)

Then U = ln(.6*11000 + .4*9000) = ln(1020) = 9.230

Clearly, with a "concave" utility function, among options with the same EV you'll always prefer the option with the smallest variance.

I don't know if the above is helpful or not. The way I think about is that the decision on how many times to run out a board is not affected by Kelly (other than saying that you'll always want to minimize variance).

You proved, by just a fraction of a percent, that growth is improved by running it twice.

Very very helpful. Thanks I will sleep better!

I think I went wacko when I saw a graph of that exact probability (60 percent EV) and that 20 percent of your net worth should be on the table. Only a Skynet Econobot 3000 would bring 20 percent of its net worth to a deep stack poker table.

Alas, we are stuck over in the extremely sub-optimal nitty world of poker.

Thank you whosnext

Just re-read this. I do not agree with the conclusion and would like to set the record straight (in the unlikely event that anyone ever reads these threads in the future).

I spent several hours deriving the analytical formula for the optimal amounts to wager on two different independent and simultaneous binary bets.

It turns out that the optimal amounts to wager in a few numerical examples were found to be slightly more conservative than the amounts recommended by single kelly (simultaneous kelly will always be less than single kelly).

However, I did not say nor would I ever say that simultaneous kelly should be ignored or that single kelly should be used in all cases.

There is a great deal of confusion surrounding the kelly criterion and I do not wish to add to the fog.