How much would you bet to win one dollar?

If you bet $100, you have a 1-in-100 chance of losing $100 and a 99-in-100 chance to win $1. If you put up $1,000, you have a 1-in-1,000 chance of losing $1,000 and a 999-in-1,000 chance to win $1. And so on. You can bet any amount of money you want.

Do you bet less money, lowering your potential loss but also lowering your chance to win? Or do you bet a lot of money, making it less likely you'll lose but also IF you lose, then you'll lose a lot? Do you even bet at all? How much would you personally bet? If you had $100k bankroll, how much should you bet?

EDIT: The poll answers are jokes. Don't go trynna debate that stuff.

Okay, I'll file this under jokes, but if anyone responds with arguments about bringing back banned members, I will probably issue a suspension for trolling.

Bridge jumper bet?

I remember taking my companion thru the mechanics of a bet. I told her you are betting twenty dollars in order to win ten dollars She was instantly opposed to such an absurd notion but the very next roll was a four and she won, so therefore she became less perturbed.

A lot would depend upon timing but if I had 100 grand as a bankroll i'd go no more than 40 grand.

Quote: RS

If you bet $100, you have a 1-in-100 chance of losing $100 and a 99-in-100 chance to win $1. If you put up $1,000, you have a 1-in-1,000 chance of losing $1,000 and a 999-in-1,000 chance to win $1. And so on. You can bet any amount of money you want.

Do you bet less money, lowering your potential loss but also lowering your chance to win? Or do you bet a lot of money, making it less likely you'll lose but also IF you lose, then you'll lose a lot? Do you even bet at all? How much would you personally bet? If you had $100k bankroll, how much should you bet?

EDIT: The poll answers are jokes. Don't go trynna debate that stuff.

The win is still $1 either way.

I would wager $10 for a 9 out of 10 chance to win that same dollar.

Lot less risk for same reward

Quote: RS

If you bet $100, you have a 1-in-100 chance of losing $100 and a 99-in-100 chance to win $1. If you put up $1,000, you have a 1-in-1,000 chance of losing $1,000 and a 999-in-1,000 chance to win $1. And so on. You can bet any amount of money you want.

Do you bet less money, lowering your potential loss but also lowering your chance to win? Or do you bet a lot of money, making it less likely you'll lose but also IF you lose, then you'll lose a lot? Do you even bet at all? How much would you personally bet? If you had $100k bankroll, how much should you bet?

EDIT: The poll answers are jokes. Don't go trynna debate that stuff.

On the flip side if you bet 1 USD is it 50/50? That would seem the best option.

No support for MickeyCrimm

Quote: darkoz

Quote: RS

If you bet $100, you have a 1-in-100 chance of losing $100 and a 99-in-100 chance to win $1. If you put up $1,000, you have a 1-in-1,000 chance of losing $1,000 and a 999-in-1,000 chance to win $1. And so on. You can bet any amount of money you want.

Do you bet less money, lowering your potential loss but also lowering your chance to win? Or do you bet a lot of money, making it less likely you'll lose but also IF you lose, then you'll lose a lot? Do you even bet at all? How much would you personally bet? If you had $100k bankroll, how much should you bet?

EDIT: The poll answers are jokes. Don't go trynna debate that stuff.

The win is still $1 either way.

I would wager $10 for a 9 out of 10 chance to win that same dollar.

Lot less risk for same reward

I guess I didn't word it how I wanted to word it. What is the absolute MOST that you would wager on this? You say $10, but would you do $100? What about $10,000 or $100,000?

Quote: Gandler

Quote: RS

If you bet $100, you have a 1-in-100 chance of losing $100 and a 99-in-100 chance to win $1. If you put up $1,000, you have a 1-in-1,000 chance of losing $1,000 and a 999-in-1,000 chance to win $1. And so on. You can bet any amount of money you want.

Do you bet less money, lowering your potential loss but also lowering your chance to win? Or do you bet a lot of money, making it less likely you'll lose but also IF you lose, then you'll lose a lot? Do you even bet at all? How much would you personally bet? If you had $100k bankroll, how much should you bet?

EDIT: The poll answers are jokes. Don't go trynna debate that stuff.

On the flip side if you bet 1 USD is it 50/50? That would seem the best option.

No. Betting $1 would be a guaranteed loss. I mean, if you are that opposed to variance and want to play a game without any variance at all, that's not a bad option.

How many times can you bet? Could I risk $1B for 10k rounds? That's less risk and more reward than bonds.

Quote: Ayecarumba

How many times can you bet? Could I risk $1B for 10k rounds? That's less risk and more reward than bonds.

Yeah sure, if you have a billion dollars! And no one’s gonna loan that to you.

I voted to outcast four members 😆, how did that work? Multiple votes cast simultaneously were accepted?

Can everybody please announce who they actually voted for so we can see how many sock puppets were involved in voting? Didn't Nathan log on with all her f****** sock puppets on one particular day?

Quote: AxelWolf

Can everybody please announce who they actually voted for so we can see how many sock puppets were involved in voting? Didn't Nathan log on with all her f****** sock puppets on one particular day?

Yes, she did, yesterday, with at least 5 accounts that i saw. And banned accounts can still vote in a poll. So chances are "they" all voted in this poll, but there's no way to tell who voted for what in any poll, at least at my access level, and I doubt at any level.

So, I would suggest anyone evaluating these results discount that choice by at least 5 votes.

Quote: beachbumbabs

. And banned accounts can still vote in a poll.

I have one question. WHY!!!!!!????
I have to admit, that's some AP thinking right there. I guess I never been suspended long enough to find out.

Come to think of it, she probably didn't even realize she was banned until after she cast a vote and then went to post. Then that light bulb went off in her head. I guess some of the best ideas come from mistakes.

Quote: AxelWolf

I have one question. WHY!!!!!!????
I have to admit, that's some AP thinking right there. I guess I never been suspended long enough to find out.

Come to think of it, she probably didn't even realize she was banned until after she cast a vote and then went to post. Then that light bulb went off in her head. I guess some of the best ideas come from mistakes.

So true (bold mine):
1) penicillin
2) vulcanized rubber
3) post it notes
4) the pacemaker
5) super glue
6) the microwave
7) X ray machine
8) teflon
9) Viagra
10) Dynamite
11) Ivory soap

And many, many others.

To Axel's point. I didn't know outcast good be a verb. I was gonna say you should say "exile," but just before making an ass of myself I looked it up.

deleted

Looks like no one is interested in this actual topic. Although, I think part of that might be because I worded it (partially) wrong.

If you win, then you would win $2, not $1. This would make the game +EV.

EG: If you bet $100, then you have a 99% chance to win $2 and a 1% chance to lose $100. Over 100 equally distributed games, you'd have won $198 and lost $100, for a profit of $98 on $10,000 in action, or a 0.98% edge. With a $1,000,000 wager, you'd win $1,999,998 and lose $1,000,000 for a net profit of $999,998 over $10,000,000,000,000 in action, for an edge of 0.0000000999998% edge. I might have messed up on the zeroes though (back off me chief). IOW: It's pretty much worth $1 per bet.

I'm sure everyone here would play the game for $100. But what if it was only offered for $1k, $10k, or $100k? Or better yet, would the most you'd be willing to bet be 1% of your current bankroll? 5%, 10%, 25%, 50%, or 100%?

Quote: RS

Show Spoiler

Lol at Rigindux not knowing English

Are you shaming someone's lack of proficiency in understanding or writing in English?
#notin2019

Quote: aceofspades

deleted

Damn, swift edit. And no, that had nothing to do with my post or intention. I originally was gonna post up %'s, like 1, 5, 10, 25, 50, or 100% of your bankroll (or cash amounts like $100, $1k, $10k, etc.), but I figured that's lame and I'll just write something else because #YOLO.

Glad to see you're back AOS. Anything interesting happenin' lately?

Quote: RS

Damn, swift edit. And no, that had nothing to do with my post or intention. I originally was gonna post up %'s, like 1, 5, 10, 25, 50, or 100% of your bankroll (or cash amounts like $100, $1k, $10k, etc.), but I figured that's lame and I'll just write something else because #YOLO.

Glad to see you're back AOS. Anything interesting happenin' lately?

Just a soap opera that finally ended:

http://diversitytomorrow.com/thread/2875/0/

Quote: RS

Looks like no one is interested in this actual topic. Although, I think part of that might be because I worded it (partially) wrong.

If you win, then you would win $2, not $1. This would make the game +EV.

EG: If you bet $100, then you have a 99% chance to win $2 and a 1% chance to lose $100. Over 100 equally distributed games, you'd have won $198 and lost $100, for a profit of $98 on $10,000 in action, or a 0.98% edge. With a $1,000,000 wager, you'd win $1,999,998 and lose $1,000,000 for a net profit of $999,998 over $10,000,000,000,000 in action, for an edge of 0.0000000999998% edge. I might have messed up on the zeroes though (back off me chief). IOW: It's pretty much worth $1 per bet.

I'm sure everyone here would play the game for $100. But what if it was only offered for $1k, $10k, or $100k? Or better yet, would the most you'd be willing to bet be 1% of your current bankroll? 5%, 10%, 25%, 50%, or 100%?

Show Spoiler

Lol at Rigindux not knowing English

Isnt this basically the same thing as voting no safety except with no data? So it is bet 100 to win 1 or 10k to win 1 and so on. At current financial I could only risk 100, but I am not risking 100 for $1 even with 99% chance of win.

Quote: GWAE

Quote: RS

Looks like no one is interested in this actual topic. Although, I think part of that might be because I worded it (partially) wrong.

If you win, then you would win $2, not $1. This would make the game +EV.

EG: If you bet $100, then you have a 99% chance to win $2 and a 1% chance to lose $100. Over 100 equally distributed games, you'd have won $198 and lost $100, for a profit of $98 on $10,000 in action, or a 0.98% edge. With a $1,000,000 wager, you'd win $1,999,998 and lose $1,000,000 for a net profit of $999,998 over $10,000,000,000,000 in action, for an edge of 0.0000000999998% edge. I might have messed up on the zeroes though (back off me chief). IOW: It's pretty much worth $1 per bet.

I'm sure everyone here would play the game for $100. But what if it was only offered for $1k, $10k, or $100k? Or better yet, would the most you'd be willing to bet be 1% of your current bankroll? 5%, 10%, 25%, 50%, or 100%?

Show Spoiler

Lol at Rigindux not knowing English

Isnt this basically the same thing as voting no safety except with no data? So it is bet 100 to win 1 or 10k to win 1 and so on. At current financial I could only risk 100, but I am not risking 100 for $1 even with 99% chance of win.

Yes, it's a bridge jumper as I think FleaStiff said earlier in the thread.

And it'd be to win $2, not $1.

Quote: RS

If you win, then you would win $2, not $1. This would make the game +EV

So no matter what I bet, the value is +$1? I would just reach into my pocket and bet whatever amount I pulled out. And if I could repeat that bet, I would keep doing so until I was rich.

With a minimum bet of $1,000, I would probably just keep betting $1,000 until all those $2 wins added up to a real nice net worth -- assuming it's like a slot machine with hundreds of decisions per hour. My rough estimates show up to 50% of bankroll should be ok (lets say 40% for being conservative), I would still prefer less. At a certain point it becomes almost exactly like a reverse powerball. Or more accurately, it is just booking powerball bets (only the jackpot never grows, but the winners don't have to split their prizes), which we know is a great way to make money.

Quote: TomG

So no matter what I bet, the value is +$1? I would just reach into my pocket and bet whatever amount I pulled out. And if I could repeat that bet, I would keep doing so until I was rich.

With a minimum bet of $1,000, I would probably just keep betting $1,000 until all those $2 wins added up to a real nice net worth -- assuming it's like a slot machine with hundreds of decisions per hour. My rough estimates show up to 50% of bankroll should be ok (lets say 40% for being conservative), I would still prefer less. At a certain point it becomes almost exactly like a reverse powerball. Or more accurately, it is just booking powerball bets (only the jackpot never grows, but the winners don't have to split their prizes), which we know is a great way to make money.

I came to the same conclusion, that the maximum (full kelly) bet would 50% of BR. Although it is kinda weird, since you normally want to bet the most you can, in this case, that isn’t really applicable because betting more doesn’t get you (that much) more EV.

There’s a point where a large enough wager makes your chance of losing, “the same as 0%”. How large would the wager need to be?

I ask because it was established some time ago in another thread that 0.99999.... = 1.

Quote: Ayecarumba

There’s a point where a large enough wager makes your chance of losing, “the same as 0%”. How large would the wager need to be?

I ask because it was established some time ago in another thread that 0.99999.... = 1.

0.999 has to repeat forever, which would mean the wager would have to be infinite.

Quote: RS

I came to the same conclusion, that the maximum (full kelly) bet would 50% of BR. Although it is kinda weird, since you normally want to bet the most you can, in this case, that isn’t really applicable because betting more doesn’t get you (that much) more EV.

Kelly criterion is for long term growth. If horizon is finite, the formula is Max E[G^(1/n)], with G=resulting capital and n=number of plays. Hence for one try, it says Bet All when return is proportional to wager (w).

Édit: and positive! The original game had a negative EV.
The new one has (1)/w * (-w) + (w-1)/w * 2 = (w-2)/w = 1 - 2/w which is not proportional to w.
In Kelly parlance, EG = 1/w * (B-w) + (w-1)/w * (B+2) = (B-w+wB-B+2w-2)/w = (wB+w-2)/w = B+1-2/w.
Maximum for largest w as possible.

This result shows the weakness of Kelly play. It does not take into account the variance. Or equivalently, it ignores risk-aversion and loss-aversion. What human would risk all their capital to win 2$? They do the reverse: buy for 2$ a lottery ticket ; pay 2$ premium to insure against the loss of a capital.

RS’s question is precisely : « what is your risk-aversion? »
His game is akin to Bernoulli’s St-Petersburg bet.

- -

Furthermore, the Kelly criterion is based on a traditional bet, i.e. one with fixed probabilities. Here, probabilities are dependent on w. This voids all the mathematical development of the Kelly formula. (Because the derivative of the expectation is NOT anymore the expectation of the derivative.)

Development:
Max E[G^(1/n)] = (1/w){ (B-w)^(1/n) + (w-1)*(B+2)^(1/n) }

Derivative w.r.t. w = (-1/ww) {…} + (1/w) { -(1/n)(B-w)^(1/n-1) + (w-1)*(1/n)(B+2)^(1/n-1) + (B+2)^(1/n) }
= (1/w) * { (B-w)^(1/n) * [-1/w - (B-w)^(-1)/n] + (B+2)^(1/n) * [1/w + (w-1) (B+2)^(-1)/n] }
= (1/ww) { (B+2)^(1/n) - (B-w)^(1/n) } + (1/nw) { (w-1) (B+2)^(1/n)/(B+2) - (B-w)^(1/n)/(B-w) }
= 0

Limit for n-> inf
(1/ww) {epsilon} + (1/nw) { (w-1)/(B+2) - 1/(B-w) } = 0
<=> (w-1)/(B+2) = 1/(B-w)
<=> (w-1)(B-w) = (B+2) Quadratic in w: ww - (B+1) w - 2 = 0
<=> w = (B+1)/2 +/- sqroot[((B+1)/2)^2 + 2]
This outside the bounds 0<w<B so to find a maximum we look at the sign of (w-1)/(B+2) - 1/(B-w)
= sign of (w-1)(B-w) - (B+2) if w<B
= positives between the roots
So the maximum is reached by increasing w.

A long term risk-neutral Kelly bettor invests his whole bankroll on this game!
Ain’t the Kelly criterion silly?😋
we all know this leads to assured ruin.

Quote: kubikulann

Kelly criterion is for long term growth. If horizon is finite, the formula is Max E[G^(1/n)], with G=resulting capital and n=number of plays. Hence for one try, it says Bet All when return is proportional to wager (w).

Édit: and positive! The original game had a negative EV.
The new one has (1)/w * (-w) + (w-1)/w * 2 = (w-2)/w = 1 - 2/w which is not proportional to w.
In Kelly parlance, EG = 1/w * (B-w) + (w-1)/w * (B+2) = (B-w+wB-B+2w-2)/w = (wB+w-2)/w = B+1-2/w.
Maximum for largest w as possible.

This result shows the weakness of Kelly play. It does not take into account the variance. Or equivalently, it ignores risk-aversion and loss-aversion. What human would risk all their capital to win 2$? They do the reverse: buy for 2$ a lottery ticket ; pay 2$ premium to insure against the loss of a capital.

RS’s question is precisely : « what is your risk-aversion? »
His game is akin to Bernoulli’s St-Petersburg bet.

- -

Furthermore, the Kelly criterion is based on a traditional bet, i.e. one with fixed probabilities. Here, probabilities are dependent on w. This voids all the mathematical development of the Kelly formula. (Because the derivative of the expectation is NOT anymore the expectation of the derivative.)
Development:
Max E[G^(1/n)] = (1/w){ (B-w)^(1/n) + (w-1)*(B+2)^(1/n) }
Derivative w.r.t. w = (-1/ww) {…} + (1/w) { -(1/n)(B-w)^(1/n-1) + (w-1)*(1/n)(B+2)^(1/n-1) + (B+2)^(1/n) }
= (1/w) * { (B-w)^(1/n) * [-1/w - (B-w)^(-1)/n] + (B+2)^(1/n) * [1/w + (w-1) (B+2)^(-1)/n] }
= (1/ww) { (B+2)^(1/n) - (B-w)^(1/n) } + (1/nw) { (w-1) (B+2)^(1/n)/(B+2) - (B-w)^(1/n)/(B-w) }
=

I don't understand all of that math. But wouldn't it be fair to say if someone asked you if you'd make a certain bet, you'd apply the kelly criterion, regardless of any other potential bets? Why does it matter if the probability is based on the size of the wager? Wouldn't it be the same as if you wanted to determine if taking a coin flip at +105 odds (1.05-to-1 payout) for $10k is within your risk tolerance? But in this case, you want to see if betting $100k to win $2 in a reverse lottery is within your risk tolerance.

Quote: RS

I don't understand all of that math. But wouldn't it be fair to say if someone asked you if you'd make a certain bet, you'd apply the kelly criterion, regardless of any other potential bets?

No. As I said, the Kelly criterion is designed for optimizing the long term growth of a capital/bankroll in a repeated positive EV situation. And a situation where the outcomes are proportional to the wager.
If someone asks me whether I’d make a bet, I don’t think in long term (it’s a one shot) and, more importantly, I am not risk-neutral. And here, the outcome (2$) is not proportional.
In this here game, Kelly betting leads to ruin in the long run.

Quote: RS

Why does it matter if the probability is based on the size of the wager? .

Because the demonstration of the formula requires to pass from the derivative of E(x) to E(derivative of x).

In other words, it assumes repetition of the same bet: same probability - yet here the probability depends on w, which depends on B, which varies along the way.

Quote: kubikulann

No. As I said, the Kelly criterion is designed for optimizing the long term growth of a capital/bankroll in a repeated positive EV situation. And a situation where the outcomes are proportional to the wager.
If someone asks me whether I’d make a bet, I don’t think in long term (it’s a one shot) and, more importantly, I am not risk-neutral. And here, the outcome (2$) is not proportional. Because the demonstration of the formula requires to pass from the derivative of E(x) to E(derivative of x).
In other words, it assumes repetition of the same bet: same probability - yet here the probability depends on w, which depends on B, which varies along the way.

Okay, so if you can't apply the kelly criterion, then what's the solution? How do you figure out what is the maximum amount to wager, if offered at that amount? IOW, what size of wager brings too high of a risk?

There is no ‘‘solution’’. Because there is no exact definition of optimal.

What is optimal depends on the individual preferences. In gambling or investing, that includes risk aversion level.
IOW, what do you mean by ‘‘too high’’? There is no objective answer. Each his own. « Risk neutrality » means just that: no risk is deemed too high.

Kelly answered one specific goal: optimizing safe long term growth. In a specific environment (repetition, indépendance, proportionality, risk neutrality,…).
This game, or the St Petersburg one, do not apply to Kelly reasoning, because wager sizes affect the form of the bet.

Long term reasoning and targeting only Expectation leads to infinity paradoxes. A Kelly bet is never the whole bankroll. In these two examples, using it leads to wagering the whole lot. Paradox.

Quote: kubikulann

There is no ‘‘solution’’. Because there is no exact definition of optimal.

What is optimal depends on the individual preferences. In gambling or investing, that includes risk aversion level.
IOW, what do you mean by ‘‘too high’’? There is no objective answer. Each his own. « Risk neutrality » means just that: no risk is deemed too high.

Kelly answered one specific goal: optimizing safe long term growth. In a specific environment (repetition, indépendance, proportionality, risk neutrality,…).
This game, or the St Petersburg one, do not apply to Kelly reasoning, because wager sizes affect the form of the bet.

Long term reasoning and targeting only Expectation leads to infinity paradoxes. A Kelly bet is never the whole bankroll. In these two examples, using it leads to wagering the whole lot. Paradox.

Can’t we still intelligibly ask the question about this game of what bet sizing strategy maximizes long term growth while holding the risk of ruin to zero?

INTRODUCING VARIANCE
It was shown that
EV = 1/w * (-w) + (w-1)/w * 2 = (w-2)/w = 1 - 2/w .
(Please note this is negative for w<2)

What is the variance?
E(V²) = 1/w * w² + (w-1)/w * 2² = (w²+4(w-1))/w
VarV = E(V²) - (EV)² = (w²+4w-4)/w - (w-2)²/w²
= w+4- 4/w - 1 +4/w -4/w² = w + 3 - (2/w)² .

The variance ratio is
VarV / (EV)² = [w + 3- (2/w)²] / [ 1 - 4/w + (2/w)² ]
= [w³ + 3w² - 4] / [ w² - 4w + 4]
= (w-2) + 9 + 24/(w-2) + 20/(w-2)²
It increases with w (w>2).

This is the usual case: to increase return you have to increase variance. Except that here, your return rises marginally to 1 while your variance rises proportionally with w. Going to the max is definitely not a good idea.

Maximizing some sort of « utility function » F= aEV - bSD would lead to a much more reasonable
a/b is the individual’s preference.

Quote: unJon

Can’t we still intelligibly ask the question about this game of what bet sizing strategy maximizes long term growth while holding the risk of ruin to zero?

One has to define precisely « long term growth » and « risk of ruin ».
With Kelly’s definition of long term growth (i.e. average relative rate of growth of the capital), there is incompatibility with the second condition.
Why is that? Well, grossly, you can win 2$ max. Amassing those 2+2+2+…, your bankroll increases. Yet the gain does not, so it represents an ever lower share of your current bankroll. Your relative rate of growth is diminishing. So the Kelly criterion decides to favor the case with the maximal rate, i.e. the soonest.

On the other hand, technically whenever you keep 0.01$ of your capital untouched, you are not ruined. So I guess there should be some more explicit condition on how much you want to safeguard... Suppose you possess a capital K and you want to protect part p from loss. You then simply establish that your gambling (or investment) bankroll is B=(1-p)K. This bankroll then can meet ruin.

So, in the end, the so-called Kelly answer to RS’s question seems to be:
Establish the amount that you are comfortable with losing, and bet it entirely.

Quote: kubikulann

One has to define precisely « long term growth » and « risk of ruin ».
With Kelly’s definition of long term growth (i.e. average relative rate of growth of the capital), there is incompatibility with the second condition.
Why is that? Well, grossly, you can win 2$ max. Amassing those 2+2+2+…, your bankroll increases. Yet the gain does not, so it represents an ever lower share of your current bankroll. Your relative rate of growth is diminishing. So the Kelly criterion decides to favor the case with the maximal rate, i.e. the soonest.

On the other hand, technically whenever you keep 0.01$ of your capital untouched, you are not ruined. So I guess there should be some more explicit condition on how much you want to safeguard... Suppose you possess a capital K and you want to protect part p from loss. You then simply establish that your gambling (or investment) bankroll is B=(1-p)K. This bankroll then can meet ruin.

Ah. I see. I guess what I would do is call my utility function equal to ln(bankroll).

Quote: unJon

I guess what I would do is call my utility function equal to ln(bankroll).

Many economists do that. But it has been shown to be a less than accurate estimate of real people utility functions — admitting there are such things, which other research has proven false.
Actually, there has to be a utility for negative values (debts and the such) and U should be concave on some domain to account for lottery and casino gambling behaviour.

Quote: kubikulann

With Kelly’s definition of long term growth (i.e. average relative rate of growth of the capital)

CORRECTION
geometric average of relative growth factor.

1+R = [(1+r₁)(1+r₂)…(1+r_n)]^(1/n)

and not

R = [(r₁)+(r₂)+…+(r_n)]/n
.

Quote: kubikulann

There is no ‘‘solution’’. Because there is no exact definition of optimal.

What is optimal depends on the individual preferences. In gambling or investing, that includes risk aversion level.
IOW, what do you mean by ‘‘too high’’? There is no objective answer. Each his own. « Risk neutrality » means just that: no risk is deemed too high.

Kelly answered one specific goal: optimizing safe long term growth. In a specific environment (repetition, indépendance, proportionality, risk neutrality,…).
This game, or the St Petersburg one, do not apply to Kelly reasoning, because wager sizes affect the form of the bet.

Long term reasoning and targeting only Expectation leads to infinity paradoxes. A Kelly bet is never the whole bankroll. In these two examples, using it leads to wagering the whole lot. Paradox.

Isn't the kelly criterion optimal betting for maximized bankroll growth? Certainly there is an optimal bet or at least there exists a bet which is too high. If you bet minimum ($2), then you make no EV. If you bet 100% of BR then you have 100% ROR.

Quote: RS

Isn't the kelly criterion optimal betting for maximized bankroll growth? Certainly there is an optimal bet or at least there exists a bet which is too high. If you bet minimum ($2), then you make no EV. If you bet 100% of BR then you have 100% ROR.

<sigh>
Do you read what I write?

Kelly optimizes one specific objective, under a series of sepcific conditions. Those conditions are not met by your game.

« Optimal » and « too high » are subjective concepts. What is better in the eye of one player is worse for another.

EV and Risk of ruin are two different measures. One is an amount, the other is a probability.
« If I spend nothing, I get no butter. If I spend all, I have no money anymore. There *must* be an optimal level of spending. »
Yes... depends on your personal preference between butter and money.