Ole Peters' coin flip problem

Bloomberg has a recent article about Ole Peters, who challenges traditional economic theory. One of his ideas is a coin flip game where you win +50% of your bankroll for heads, and -40% of your bankroll for tails. Peters posits that this is a bad game, even though it has a positive expected value, because (perhaps surprisingly), the most common result is losing over 99% of the starting bankroll.

Discuss.

Perhaps intuition tells people the problem is the violation of the Kelly Criterion, the article notes most people decline the bet. Putting up your entire bankroll to start betting is asking for it, and each bet afterwards has to be the entire bankroll [according to the graph that illustrates the below]

Quote:

Suppose in the same game, heads came up half the time. Instead of getting fatter, your $100 bankroll would actually be down to $59 after 10 coin flips. It doesn’t matter whether you land on heads the first five times, the last five times or any other combination in between

This is the most interesting part to me. Winning exactly half the time doesn't cut it, evidently. Not if you keep putting it all on the line. The article does go on to say some people would win big if allowed to keep winning, which is another deviation from reality that every gambler should know: no one is allowed to just keep winning!

Quote: MichaelBluejay

Bloomberg has a recent article about Ole Peters, who challenges traditional economic theory. One of his ideas is a coin flip game where you win +50% of your bankroll for heads, and -40% of your bankroll for tails. Peters posits that this is a bad game, even though it has a positive expected value, because (perhaps surprisingly), the most common result is losing over 99% of the starting bankroll.

In a Kelly world, the proportion of bankroll risked would be much smaller and the most likely case closer to breakeven. People should intuitively understand this.

What happens in the real world is that wealthy people invest in diversified government and corporate bonds with low risk and +EV. Low wealth people invest in bankrupt companies on Robinhood and lottery tickets with negative EV and a chance for a big score. This is an anti-Kelly world. 'People Are Idiots and I Can Prove It!'

The author suggests a 50/50 bet that pays 6 to 5.

If find the optimal Kelly bet is 1/12 of bankroll.

Should you take the bet if forced to bet half your bankroll, thus increasing wealth by 60% or losing wealth by 50%? The Kelly Criterion says definitely not as there is a drop in expected utility. I simulated doing this 1,000 times many times and every time a starting bankroll of $1,000,000 was less than a penny by the end, which illustrates why.

I majored in math and economics. The economics half of that never did me much good. However, every teacher I had taught that intelligent people act in a way to maximize expected utility, not expected value. This explains why people buy insurance, despite being a bad bet.

The topic goes to show the danger of over-betting your bankroll.

Since you are increasing your bets with each win and decreasing them with each loss, this is a Reverse D'Alembert, which normally results in losing money whenever the wins and losses balance, but that assumes the bets are even money.

In this case, after N wins and N losses, in any order, a bankroll of 1 becomes 1.5^N x 0.6^N = 0.9^N.

ev= +5%, Var =0.2025. If BR = bet size, Kelly ratio = 0.2025/0.05 = 4.05, ROR = e^(-2/4.05) = 0.61

So your risk of ruin = 61%.

I am not sure, please correct me if you don't agree

Quote: ssho88

ev= +5%, Var =0.2025. If BR = bet size, Kelly ratio = 0.2025/0.05 = 4.05, ROR = e^(-2/4.05) = 0.61

So your risk of ruin = 61%.

I am not sure, please correct me if you don't agree

Technically, I thought the ROR is zero. Perhaps your ROR is a reasonable estimate or is truncated at some minimum bankroll. For example, if you can't bet less than a penny, you are are ruined.

Is the variance of the outcome of the infinite bet stream infinite when expressed as starting bankroll? I would think you would need to do MC sims to get the variance of a finite number of flips based on min and max bankroll stopping conditions. I am not a math guy, so I don't know if you can do it analytically for N flips.

Your variance agrees with mine, assuming you mean single flip variance stated in units of BR^2 . The problem does not state how much each wager is. One could use different units.

Quote: Mental

Technically, I thought the ROR is zero.

I was tempted to post that, but wasn't really sure what the post in question was saying.

Quote: ssho88

ev= +5%, Var =0.2025. If BR = bet size, Kelly ratio = 0.2025/0.05 = 4.05, ROR = e^(-2/4.05) = 0.61

So your risk of ruin = 61%.

I am not sure, please correct me if you don't agree

Monte Carlo sims don't converge very well because the distribution of results for N flips is horribly skewed toward zero bankroll with a long tail to very high bankrolls. After 100M trials, the results for N=100 flips is nowhere near converging. Treat these as estimates. This is the average BR after N flips based on a $100 initial BR. VAR is the variance of the final BR after N flips in units of initial BR².

N	Ave BR	VAR
1	$104.99	0.2025
3	$115.76	2941.066
10	$162.87	11672.05
30	$432.50	992336
100	$11613.87	258954827488

Anything seem out of place here?

EDIT: I used the wrong normalization for the variance. Ave BR is not changed.

N	Ave BR	VAR
1	$105.00	0.2025
3	$115.76	0.8823
10	$162.89	11.66
20	$265.33	199.25
30	$431.50	2,841
40	$704.43	33,101
50	$1143.30	271,264
75	$3849.67	29,796,153
100	$13737.59	5,186,109,723

Quote: Wizard

I was tempted to post that, but wasn't really sure what the post in question was saying.

I have no problem with what ssho88 is saying about ROR.

I would consider that turning $100 into 0.1 pennies in a few minutes to be be ruin, especially if that was my entire BR. At that point, I will just steal the coin that Ole is flipping and walk out without being technically ruined.

Quote: Mental

Monte Carlo sims don't converge very well because the distribution of results for N flips is horribly skewed toward zero bankroll with a long tail to very high bankrolls. After 100M trials, the results for N=100 flips is nowhere near converging. Treat these as estimates. This is the average BR after N flips based on a $100 initial BR. VAR is the variance of the final BR after N flips in units of initial BR².

N	Ave BR	VAR
1	$104.99	0.2025
3	$115.76	2941.066
10	$162.87	11672.05
30	$432.50	992336
100	$11613.87	258954827488

Anything seem out of place here?

Could I trouble you to run this same sim but with each bet made being the correct Kelly fraction of bankroll?

If your bankroll goes up 50% on heads and down 40% on tails, that means you’re wagering 40% of your bankroll with a winning payoff of 5 to 4. Your expected bankroll after n wagers would be (1.5 * 0.6)^(n/2)= 0.9^(n/2). So the more you play the lower it goes.

In this case you should wager (5/4 - 1) / (2 * 5/4) = 10% of your bankroll. Your expected balance after n wagers is (1.125 * 0.9)^(n/2)= 1.0125^(n/2) times your beginning balance

Where the bankroll percentage wagered is b and the winning payoff on a “to 1” basis is t, the growth rate is (1 + bt)(1 - b). Take the derivative of that with respect to b and set to zero (highest growth rate) and you get the Kelly formula for a 50/50 wager : (t - 1) / 2t.

With all due respect, I suspect that everyone is massively overcomplicating the problem.

1.) This IS a good bet...at least once.

First of all, there are two assumptions being made (that are impractical) that even make this a negative EV proposition in the first place:

A.) The assumption of a non-replenishable bankroll.

-Simply put, if you could keep going back and wager $100 over and over again, then you would want to do this forever as long as you never had to wager more than $100 if you didn't want to. The only reason it doesn't work in this example is because of the assumption that the player's entire bankroll is $100 and can never be replenished.

B.) The other side has an infinite bankroll (or, effectively does).

-The other assumption being made is that the other side has an infinite bankroll. Obviously, if both sides were limited to a $100 bankroll, then you'd want the 50% side of this all day long. (Well, my opponent would have to bring a bankroll of $125, actually)

***But, let's get to this being a good bet at least once. 50/50 proposition, if you win, you gain $50, if you lose, you lose $40. This yields a positive expectation of $5.

Thus, if you only had to make the bet once...you would want to do it.

2.) It's actually a good bet until you lose.

-The way the math works on this one is that we see +50% and -40%, so it seems like a no-brainer than you would always want to take this, but it's a bit more complicated than that.

Where the math works against the player is actual loss which is to say that the player ALWAYS loses more (if he loses) in actual dollars than the player stands to win on the next outcome.

Imagine if the player won the first toss and now has a bankroll of $150. On the next toss, the player loses and gets hit for $60 leaving a total of $90. Now, even if the player wins, the player will only have a bankroll of $135.

If the player loses the next toss, the $135 bankroll becomes $81, which means that a win after that will only give the player $121.50.

What you end up with here is that the player eventually needs to win well over 50% of the time to be at a profitable point. (Relative to the $100 if the first result was a loss or relative to his most recent win) When you combine this with the two assumptions from above, you see that the player is eventually going to get chipped down to almost nothing most of the time.

Even with that, it's a good bet until the player loses. At which point, the player should quit.

The first toss is a good bet for the player in and of itself, that's obvious. Not only does the player have an advantage, but the player also can win more than he stands to lose.

My conclusion is that the player should play until the first loss. Since the coin doesn't care about the results of previous flips, we could do this forever...but let's just go to a total of ten flips (last one is always tails) to demonstrate:

(T) -$40 * .5 = -$20

(HT) -$10 * .25 = -$2.50

(HHT) $35 * .125 = $4.375

(HHHT) $102.50 * .0625 = $6.40625

(HHHHT) $203.75 * .03125 = $6.3671875

(HHHHHT) $355.625 * .015625 = $5.556640625

(HHHHHHT) $583.4375 * 0.0078125 = $4.55810546875

At this point, our expected value of winning series' of results exceeds all possible losing series' of results, assuming we are going to stop as soon as we lose.

The key here is that 40%, which was selected for a reason. At every point, the 40% loss of total bankroll depletes us of more than we stand to gain if the very next fliup after that goes our way. Specifically:

100 * 1.5 * .6 = $90

150 * 1.5 * .6 = $135

225 * 1.5 * .6 = $202.5

The point is that even if we ignore multiple losses in a row, at all times, a win followed by a loss will leave you with only 90% of the bankroll that you had before the win happened.

Because of that, any win that follows a loss increases your bankroll only by 35% compared to where it was before the win. Let's look at the scenario of $150

150 * 1.5 * .6 * 1.5 = $202.50

202.5/150 = 1.35 or 135%

Now, lets look at Win-Loss-Loss-Win from $150

150 * 1.5 * .6 * .6 * 1.5 = $121.50

So, even though we had two losses and two wins, we are left with only 81% of the $150.

These percentages will be constant.

It is for that reason that so many players end up with almost nothing. All losses have to be countered with multiple wins, in pretty short fashion, to get back to the original starting point of prior to the previous win.

So...there might be a better strategy than mine, but my quick strategy would simply be to play until you lose, then stop. Despite the positive expected value, the way the percentage of bankroll works out is such that you NOW need to have a win rate exceeding 50% to get back to where you were prior to the most recent win.

3.) The principle is not even really true.

Again, the 40% was chosen for a pretty clear reason, because it results in losses that will not be overcome with an immediate win. If instead you made the results:

WIN: +50%

LOSS: -33.33333333333333%

Then it would be much different because results W-L-W would bring you to the exact point that you were before the first win of that series, but not below it.

QUICK DISCLAIMER: In the post above, I'm not suggesting that my strategy is optimal compared to Kelly betting, or anything of that nature. I'm suggesting that operating within the original parameters that the entire bankroll or nothing must be bet (even though only 40% is actually risked) that it can be played positively. Specifically, the player has an overall expected value that will exceed the $5 of just doing it once (unless the player loses, then he WILL only do it once...and it actually would have exceeded on the next series beyond the ones I did) AND a profitability rate of 25%.

50%: Lose $40
25%: Lose $10
25%: Some amount of profit.

Quote: Mission146

My conclusion is that the player should play until the first loss.

The expected value, starting with a bankroll of 1, is the sum of:
1/2 x 3/5
(1/2 x 3/2) x 1/2 x 3/5
(1/2 x 3/2) x (1/2 x 3/2) x 1/2 x 3/5
(1/2 x 3/2) x (1/2 x 3/2) x (1/2 x 3/2) x 1/2 x 3/5
...
= 3/10 x (1 + 3/4 + (3/4)^2 + (3/4)^3 + ...)
= 3/10 x (1 / (1 - 3/4))
= 6/5
In other words, for each individual play, the player advantage is 20% of the initial bankroll.

"So why aren't you expected to win if you keep playing?" Because as you lose, your bankroll decreases. The plays are not independent, since each play's starting bankroll depends on the previous play's result. If the bankroll is replenished to its starting point after each loss, then it's a different result.

Quote: ThatDonGuy

The expected value, starting with a bankroll of 1, is the sum of:
1/2 x 3/5
(1/2 x 3/2) x 1/2 x 3/5
(1/2 x 3/2) x (1/2 x 3/2) x 1/2 x 3/5
(1/2 x 3/2) x (1/2 x 3/2) x (1/2 x 3/2) x 1/2 x 3/5
...
= 3/10 x (1 + 3/4 + (3/4)^2 + (3/4)^3 + ...)
= 3/10 x (1 / (1 - 3/4))
= 6/5
In other words, for each individual play, the player advantage is 20% of the initial bankroll.

"So why aren't you expected to win if you keep playing?" Because as you lose, your bankroll decreases. The plays are not independent, since each play's starting bankroll depends on the previous play's result. If the bankroll is replenished to its starting point after each loss, then it's a different result.

I agree with all of that.

Deleted the rest. Misunderstood previous post.

Also, thanks! Much easier than the way I'd have done it (which would have been just to continue what I had been doing until it got ridiculous.)

when I first started investing in the stock market I wasn't really thinking analytically
I had a bad year and lost 20% of my portfolio
the next year was a good year and my portfolio earned 23%

my first thought - this is good - I made it all back and then some - but when I looked at the dollar value it didn't look right

bad analysis

have $100 and lose 20% - you now have $80

then you gain 23% on your $80 - you now have $98.40

you're still down overall


reverse it - same result

gain 23% on your $100 you now have $123
then lose 20% of your $123 -.....................................................you've got $98.40


*

Quote: lilredrooster

when I first started investing in the stock market I wasn't really thinking analytically
I had a bad year and lost 20% of my portfolio
the next year was a good year and my portfolio earned 23%

my first thought - this is good - I made it all back and then some - but when I looked at the dollar value it didn't look right

bad analysis

have $100 and lose 20% - you now have $80

then you gain 23% on your $80 - you now have $98.40

you're still down overall


reverse it - same result

gain 23% on your $100 you now have $123
then lose 20% of your $123 -.....................................................you've got $98.40


*

Another way to look at it: Your portfolio can go up 20% or 20000%, but no matter how much it’s gone up, a 100% drop brings it to zero.

Which is why one of the golden rules of investing is : don’t lose capital. If you lose too much, it can be difficult or impossible to recover

Quote: Mission146

With all due respect, I suspect that everyone is massively overcomplicating the problem.

1.) This IS a good bet...at least once.

It's not complicated. It is a huge +EV bet and you want to get money down. I Ole lets me quit after a loss and start again with $100, that will be great. I get more money down at good odds.

But Ole won't let me quit and start a new series until I lose everything, which will never happen.

Quote: unJon

Could I trouble you to run this same sim but with each bet made being the correct Kelly fraction of bankroll?

Which fractional bet do you want me to use?

Ole Peters' original proposition: "Starting with $100, your bankroll increases 50% every time you flip heads. But if the coin lands on tails, you lose 40% of your total."

Change to: "Starting with $100, your bankroll increases 10% every time you flip heads. But if the coin lands on tails, you lose 8% of your total."

For one flip, you get one-fifth of the edge and 1/25th of the variance.

N	Ave BR	VAR
1	$101.00	0.0081
30	$134.79	0.4864
100	$270.50	8.8291

After 100 flips, you see your earning rate is drastically reduced, that is Ave BR is only $270 vs $11K.

Quote: Mental

Which fractional bet do you want me to use?

Ole Peters' original proposition: "Starting with $100, your bankroll increases 50% every time you flip heads. But if the coin lands on tails, you lose 40% of your total."

Change to: "Starting with $100, your bankroll increases 10% every time you flip heads. But if the coin lands on tails, you lose 8% of your total."

For one flip, you get one-fifth of the edge and 1/25th of the variance.

N	Ave BR	VAR
1	$101.00	0.0081
30	$134.79	0.4864
100	$270.50	8.8291

After 100 flips, you see your earning rate is drastically reduced, that is Ave BR is only $270 vs $11K.

Thanks! My point was more about the convergence. At N=100 the difference in variance relative to EV is stark between the two (as would be expected).

By flipping a coin, we say that the probability of getting “heads” or “tails” is 50 to 50. This means that out of 100 attempts, the coin will land 50 times “heads” up and the same amount - “tails”. However, talking about the probability of 50:50 is not entirely correct, since the chance or probability of a given event is the number of events that have occurred divided by the total number of results obtained. Thus, both “heads” and “tails” can come up 50 times out of 100. The degree of probability can be expressed as 50%, 0.5.1 out of 2, or 1/2.

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Results from my Monte Carlo simulations, 10,000 sessions, 100 flips each session, starting bankroll $200:

Bet entire bankroll each flip:
Best: $93,746,624.90
Worst: <$0.000001
Median: $1.03
Average: $18,804.53

Flat bet $5:
Best: $319.50
Worst: $144
Median: $225
Average: $224.81

The discussion has been interesting and illuminating. There are many cases in which focusing on the house edge exclusively while ignoring all other variables results in an erroneous conclusion. This is one of them.

Quote: MichaelBluejay

Results from my Monte Carlo simulations, 10,000 sessions, 100 flips each session, starting bankroll $200:

Bet entire bankroll each flip:
Best: $93,746,624.90
Worst: <$0.000001
Median: $1.03
Average: $18,804.53

Flat bet $5:
Best: $319.50
Worst: $144
Median: $225
Average: $224.81

The discussion has been interesting and illuminating. There are many cases in which focusing on the house edge exclusively while ignoring all other variables results in an erroneous conclusion. This is one of them.

In order to achieve "long term" betting, my opinion is that you should flip at least(3 standard deviation) 3^2 * Var/ev^2 = 9 * 0.2025/0.05/0.05 = 729 times per session.

If bet entire bankroll each flip, what is the probability you lose ALL( BR < .00000001) ?

I am more interested in case of Betting Entire Bankroll(each flip).

the following is the comment about the article from Don Schlesinger, BJ HOF member and author of the highly regarded book "Blackjack Attack"........... :


"Really dumb example. The whole world knows that if you overbet your bankroll with respect to your advantage, you will eventually go broke. Kelly tells us that overbetting by more than twice your edge leads to eventual ruin, despite having an advantage. In this case, for every two flips, you gain 10%, so your edge is 5% per flip. But, he has us betting our entire bankroll on every flip!! OF COURSE you're going to go bust.

Let me play this game every minute of every day for the rest of my life, wagering 5% of my current bankroll, instead of all of it, and getting paid 50% of my bet when I win and losing 40% when I lose, and I'll own Peters, the horse he rode in on, and the entire rest of the world!"



*

Quote: MichaelBluejay

Results from my Monte Carlo simulations, 10,000 sessions, 100 flips each session, starting bankroll $200:

Bet entire bankroll each flip:
Best: $93,746,624.90
Worst: <$0.000001
Median: $1.03
Average: $18,804.53

Flat bet $5:
Best: $319.50
Worst: $144
Median: $225
Average: $224.81

The discussion has been interesting and illuminating. There are many cases in which focusing on the house edge exclusively while ignoring all other variables results in an erroneous conclusion. This is one of them.

10 million sessions simulation, Bankroll = $200, no of flips = 3^2 * Var/ev^2 = 9 * 0.2025/0.05/0.05 = 729 flips per session OR Bankroll < $0.000001.

1) Bankroll >= $200, No of sessions = 9257
2) $0.000001<= Bankroll < $200, No of sessions = 358,932
3) Bankroll < $0.000001(broke) , No of sessions = 9,631,811

Sim ROR = 96.32%.

Quote: erinsflo88

By flipping a coin, we say that the probability of getting “heads” or “tails” is 50 to 50. This means that out of 100 attempts, the coin will land 50 times “heads” up and the same amount - “tails”. However, talking about the probability of 50:50 is not entirely correct, since the chance or probability of a given event is the number of events that have occurred divided by the total number of results obtained. Thus, both “heads” and “tails” can come up 50 times out of 100. The degree of probability can be expressed as 50%, 0.5.1 out of 2, or 1/2.

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wow, that is a masterpiece of circular argument

does it take a bot to create that? I'm not sure a human could do it