Average profits and losses on a roulette table

[StoneTemplePython]

Infinite money plus a dollar is still infinite money - the same amount.

As usual, when $\infty$ shows up, some extra care is needed.

There is an interesting case, perhaps best interpretted where we fix the betting size increment, where you have a finite bankroll and the house has an infinite one. This is a stopping trial problem. If your probability of winning a round, $p \gt \frac{1}{2}$ then you 'win' overall with probability $\gt 0$.

(What does it mean to 'win' here? It means to not 'lose'. What does it mean to 'lose'? For the house to take all of your money i.e. you go bankrupt. Thus winning means playing the game forever and escaping the clutches of the house. If you want, you can make it so you bet a dollar each time and your starting bankroll is only a dollar, thus your probability of winning overall is the probability of never being down -- which of course queues up ideas related to ballot problems... From this vantage point we can get to the house being always 'down' vs you with positive probability despite having an $\infty$ bankroll... this is probably as close to a 'lose' by the house as we can make it given an infinite bankroll. )

However, if $p \leq \frac{1}{2}$ you lose with probability one. The $p = \frac{1}{2}$ case is is questionable though, as the expected time until bankruptcy $= \infty$.

FiveThirtyEight had a variant of this as classic challenge (micro-organisms multiply) last year.

https://fivethirtyeight.com/features/can-you-rule-riddler-nation/
https://fivethirtyeight.com/features/can-you-rule-riddler-nation/

 

[mfb]

You can do a lot of mess even with a finite but large number of bets. Let's do it as story:

The devil appears to you and tells you that you only have one year left to live. But you are allowed to gamble for your time. Each time you do so, you have a 60% chance to reduce your remaining time by 30% and a 40% chance to increase it by 50% (the next round uses the result of the previous round). The only downside: You have to determine how often you want to gamble in advance.
You calculate the expectation value: 0.6*0.7 + 0.4*1.5 = 1.02. Your expected remaining time increases by 2% each round. Happily you decide to take this offer 1000 times, for an expected time of 400 million years. The devil accepts, uses a perfect random number generator and tells you that you have 96 femtoseconds left to live. Maybe you were unlucky? The devil gives you the option to try again. You accept. This time the devil calculates that you have 47 attoseconds left.

What went wrong? All the calculations are correct. The expectation value is indeed huge, but it comes from a tiny probability to live longer than the heat death of the universe, while most of the time (>99%) you get a lifetime shorter than one second, and the probability to live shorter than a year is larger than 99.99%.
The most likely value is 0.7⁶⁰⁰*1.5⁴⁰⁰ years = 10^-15 seconds.

This is similar to the systems discussed before: If you keep playing you have a very large chance to lose all and a small chance to win a lot. That is typically not what people want.

A related game is the St. Petersburg paradox, where a game has an infinite expectation value for the player - but if you ask people how much they would pay to play it, you get very small numbers.

Instead of the expectation value, it is often better to look at the expected logarithm of the outcome. This leads to the Kelly criterion. The summary: If you don't have a positive expectation value, bet nothing. If you have a positive expectation value, only bet a small fraction of what you have. If you could do this in the game above, you would only bet 2.5% of your time every time. That way you are very likely to get a large result after a large number of games.

 

[cave man]

There are a number of strategies, but I find simply watching the wheel for some time can help one to win. Given the 50-50, or even the odds in favor of the house that uses 0 and 00, you can always observe a wheel favoring one half of something on the table. For instance, I have seen a wheel hit eleven times on red. When should I play black? Never. But take a wheel that seems to vacillate between red and black, only at times favoring one color over the other. One can only guess when a wheel will start to run on one color. But generally it is in the vacillations in preference where one can make a little money. The betting systems will tell you that it is in the runs where you will makea lot of money. True. But I have not seen a run too often where it stays on one half for 11 times.

Black is B. Red is R. The wheel does something like this...R B R B B B R B R B B R R/ B R R B R R R B B B B. The time to bet red is where the slash is because of the wheel balancing its dispersion. But if you watch and you see something like R B R BB R R R B R R R 0 R B 0 R R R R R 00 B R R B B /R B B B R R The time to bet black is again at the slash as the wheel is beginning to balance itself out. When this balancing happens you bet one chip (whatever the minimum is) and leave it there. If it comes out red and you lose, bet black again with one chip. If you win leave it there but only for a double B or, if you dare, a triple B. Do not risk more of you will lose all the doubling you had after you be the B. When you win $2 or $4 stop. Even if the wheel comes up with a five B. If it does then wait for the wheel to re balance and begin the small bet again. It will take time to build any kind of large winnings, but playing the balance of the wheel is better than trying to outsmart it.

Another strategy is to switch to thirds, or play them at the same time you watch the opposites come up. The opposites are high-low, odd-even, red black. Watch all three to assume the balancing bet. The thirds has a similar strategy. Watch for one of the thirds to hit for some time. At a point when one of the thirds starts to change to the other two bet one chip(again the minimum bet) on both of the other two thirds. Once you win wait for other imbalances before playing the same again. When you put the two chips down and win, you will be one chip, or five or ten dollars, ahead.

There is no GUARANTEE that this will work. But remember that if the wheel is balanced, and the casino takes pains to see that it is, you are playing the strategy of numbers balancing back to the 50-50 ratio.

Another strategy is again watch the wheel to see what number does NOT come up for 18 to 38 spins. Put the minimum bet down on that number and replace it as it loses. Remember, each time you lose will deduct the bet from the total you will win when it hits. This is more risky than playing the balance, but each time after the number has not come up after multiple spins increases the odds that it will. When it does you win 36 times that number, or $180 if the minimum bet is $5. And, of course you have to deduct all the money you spent getting there. The longer you wait and the number does not come up, the odds increase that it will. But, of course, it is possible that it will take a much longer time to hit as you lose all your money. If you have the time, see how often one number takes to come up. Sometimes one will come up more frequently. But remember what the wheel will do-balance itself.

There is basically no system possible to insure that you will will using any strategy. That is why it is advisable to set a limit on how much you are willing to lose when you gamble. Also set a limit on how much you might win. Walk away if either reaches your planned amount and chalk it up, win or lose, to FUN. Do not play if you get too nervous playing. Then gambling is not fun.

 

[CWatters]

Personally I think you are seeing patterns where none exist. I'd be very surprised if wheels are prone to "runs" of a particular colour any more than would occur by chance.

What physical reason could there be?

 

[mfb]

The time to bet black is again at the slash as the wheel is beginning to balance itself out.

There is no such process. This is the classical gambler's fallacy, and you are running directly into it.

 

[votingmachine]

You can do a lot of mess even with a finite but large number of bets. Let's do it as story:

The devil appears to you and tells you that you only have one year left to live. But you are allowed to gamble for your time. Each time you do so, you have a 60% chance to reduce your remaining time by 30% and a 40% chance to increase it by 50% (the next round uses the result of the previous round). The only downside: You have to determine how often you want to gamble in advance.
You calculate the expectation value: 0.6*0.7 + 0.4*1.5 = 1.02. Your expected remaining time increases by 2% each round. Happily you decide to take this offer 1000 times, for an expected time of 400 million years. The devil accepts, uses a perfect random number generator and tells you that you have 96 femtoseconds left to live. Maybe you were unlucky? The devil gives you the option to try again. You accept. This time the devil calculates that you have 47 attoseconds left.

What went wrong? All the calculations are correct. The expectation value is indeed huge, but it comes from a tiny probability to live longer than the heat death of the universe, while most of the time (>99%) you get a lifetime shorter than one second, and the probability to live shorter than a year is larger than 99.99%.
The most likely value is 0.7⁶⁰⁰*1.5⁴⁰⁰ years = 10^-15 seconds.

This is similar to the systems discussed before: If you keep playing you have a very large chance to lose all and a small chance to win a lot. That is typically not what people want.

A related game is the St. Petersburg paradox, where a game has an infinite expectation value for the player - but if you ask people how much they would pay to play it, you get very small numbers.

Instead of the expectation value, it is often better to look at the expected logarithm of the outcome. This leads to the Kelly criterion. The summary: If you don't have a positive expectation value, bet nothing. If you have a positive expectation value, only bet a small fraction of what you have. If you could do this in the game above, you would only bet 2.5% of your time every time. That way you are very likely to get a large result after a large number of games.

Are you sure your calculation is correct?

When I put it in a spreadsheet and iteratively split the outcomes each time, the EVEN number outcomes are favorable:

1st outcome:
0.6 of the time is 0.7
0.4 of the time is 1.5

2nd outcome:
0.36 of the time is 0.49 ... (0.6*0.6, 0.7*0.7)
0.24 of the time is 1.05... (0.6*0.4, 0.7*1.35)
0.24 of the time is 1.05... (0.4*0.6, 1.5*0.7)
0.16 of the time is 2.25... (0.4*0.4, 1.5*1.5)

3rd outcome:
0.216 of the time is 0.343 ... (0.6*0.6*0.6, 0.7*0.7*0.7)
0.144 of the time is 0.735 ... (0.6*0.6*0.4, 0.7*0.7*1.5)
0.144 of the time is 0.735 ... (0.6*0.4*0.6, 0.7*1.5*0.7)
0.096 of the time is 1.575 ... (0.6*0.4*0.4, 0.7*1.5*1.5)
0.144 of the time is 0.735 ... (0.4*0.6*0.6, 1.5*0.7*0.7)
0.096 of the time is 1.575 ... (0.4*0.6*0.4, 1.5*0.7*1.5)
0.096 of the time is 1.575 ... (0.4*0.4*0.6, 1.5*1.5*0.7)
0.064 of the time is 3.375 ... (0.6*0.4*0.4, 0.7*1.5*1.5)

That nets to:
0.216 of the time is 0.343
0.432 of the time is 0.735
0.288 of the time is 1.575
0.064 of the time is 3.375
The 4th outcome will have a positive outcome for the last two as 0.7*1.575=1.1025. Positive also results 40% of the time for the input of 0.735. And 0.4*0.432=0.1728.

0.064+0.288+0.1728=0.5248. So over half the time one wagered on 4 rounds it is a positive result. I think the 5th round was again a worse one, but the 6th round improves again.

My intuition is that you've made a mistake. The spreadsheet approach of number crunching says the gamble is not great, but not as bad as your calculation indicates.

 

[mfb]

Don't run it with 4 rounds. You'll need many rounds to get a pronounced effect.

After 6 rounds your probability to gain is 46%, after 10 rounds it is 37%, after 20 rounds it is 24%. After 100 rounds it is 9% and after 500 it is 0.12% (and just 15% to live longer than a second).
After 1000 rounds your chance to gain time is 0.0008% and your chance to live longer than a second is 0.17%.

The key here is a positive expectation value for the lifetime but a negative expectation value for the logarithm of it. On a logarithmic scale you get a wide distribution with the mean somewhere at 10^-12 seconds and tails to very short and very long lifetimes, producing a big expectation value but a tiny probability for long lifetimes.

 

[StoneTemplePython]

My intuition is that you've made a mistake. The spreadsheet approach of number crunching says the gamble is not great, but not as bad as your calculation indicates.

If you want an easy intuitive take, consider that

$\text{Arithmetic Mean} = 0.6(0.7) + 0.4(1.5) = 1.02$
$\text{Geometric Mean} = (0.7)^{0.6} (1.5)^{0.4} \approx 0.95$

since $GM \leq AM$ this should not surprise, you, but the key relation is

$GM \lt 1 \lt AM$

hence the simple gains (AM) grow over time, as they exceed one, but the compounded return (GM) shrinks because it is less than one.

So if you bet your entire bankroll every time, then you basically get the compound rate of return -- i.e. $GM^n$ or in log space, n times the expected logarithm. Since the compound rate of return is less than 1, after a large amount of turns you can be nearly certain that you've driven your bank roll to zero.

The exception is the very rare case where you get a mixture of wins $\gt \approx 0.47$ (i.e. significantly higher than the mixture of 0.4 that almost surely occurs after a very large amount of bets). And is in this very collection of very rare cases where you come out ahead (and significantly so).

There's lot of nitpicks from economists on this approach, but most successful practitioners, including some very math sophisticated ones like Thorpe and Simmons, do have an eye on Kelly criterion and variants.

 

[votingmachine]

Sorry to be slow to respond, but I see the result. It surprised me because a gambler not obliged to let it ride could profit from the odds presented. And letting it ride a few times seemed profitable. Obviously a random result of 10 bets of $1 at the odds given would win $0.7 6-times and $1.5 4-times. Which is profitable. I did not carry the spreadsheet far enough, and see that beyond more than a few rounds of letting it ride, the profitable situation disappears.

 

[BWV]

If you move this to continuous, the difference between AM and GM is half the variance (Ito's Lemma)

or exponating

So, for example, if there was a (bad) investment strategy with an expected return of 3% and an annualized standard deviation of returns of 20%, you would expect to lose 1% per year

 

[PhanthomJay]

There are just 2 ways to win at Roulette.

The first way I discovered when I was in Vegas 15 years ago with my son , when he got married. We found a 'biased' wheel...one that is slightly out of balance so that a certain number comes up much more frequently than probability otherwise dictate for an unbiased wheel. You watch the wheel for 20 spins or so, and if a certain number comes out say once every 5 spins or so, you found the biased wheel! We then played the number repeatedly and sure enough, it came out often and we won pretty big until the banker got nervous and upset and started giving us a hard time. But we had to leave for his wedding anyway.
The other way to win is described in the movie 'War Games'. The computer plays 'tic tac do' against itself, and a tie always results, from which it then concludes "The only way to win ...is not to play"!

 

[StoneTemplePython]

So, for example, if there was a (bad) investment strategy with an expected return of 3% and an annualized standard deviation of returns of 20%, you would expect to lose 1% per year

A lot more care is needed when saying things like this. Expectation is well defined linear operator, hence you must have things like $\alpha E\big[X \big] = E\big[ \alpha X\big]$ and $E\big[X \big] + E\big[Y \big] = E\big[X + Y\big]$. Suppose I find some alternative investment $Y$ that has same expected return as $X$ and same variance, but these investments are not i.i.d. and in fact there is a correlation coefficient of $-1$.

In this case consider a convex combination where $0 \leq \alpha \leq 1$

$3\% = E\big[\alpha X+(1-\alpha)Y\big] =\alpha E\big[X\big] + (1 - \alpha) E\big[Y\big] \neq -1\%$

In particular consider where $\alpha := 0.5$. In this case the geometric return exactly equals the arithmetic mean return.

Linearity is a big deal... and especially so with the expectation operator as it is the bridge between linearity and convexity.

So what I think you intended to say is that while working in log space the expected value of an investment after some positive time period is negative which is what @mfb has been doing. The underlying asset price distribution here is implicitly log-normal, so again emphasizing expectations in logspace is quite natural.

 

[BWV]

but the formal expectation of a lognormal variable is μ + σ²/2 - the arithmetic expectation includes the σ²/2 term so in your portfolio example it is already embedded there if you are blending arithmetic expecations.

 

[CWatters]

You watch the wheel for 20 spins or so, and if a certain number comes out say once every 5 spins or so, you found the biased wheel!

I think it would have to be rigged somehow not accidentally biased to get anything like that result.

 

[PhanthomJay]

I think it would have to be rigged somehow not accidentally biased to get anything like that result.

It sure was a frequent occurrence, whether rigged or biased. I don't know if the 'illegals' were still involved with the casinos at the time (most likely, yes). I can't recall if we played just one number (10 was the number, if you're feeling lucky) , or some nearby numbers as well. We didn't bet too heavy, as our cash was limited. I do know that when we started winning, and softly shouted "we found the biased wheel!", it was still within earshot of the banker or whatever you call that guy on the other side of the table , and he was upset and started yelling at us like "keep your hands off the table when the wheel is spinning!"', and then he called over the bar girl who gave us a couple of martinis or something, and then it was wedding time. I dunno, but I haven't been back since, realizing that indeed it was a strange game, and that the best way to win was not to play it again.

 

[cave man]

I waited for one of the thirds to lose six times in a row at Trump Plaza. On the nex roll someone bet on the two thirds that were coming up while I put $100 on the one that lost six times. I won $200. And yes, a third can lose for longer than six times. But the next time you are at a casino, find out for yourself how many wheels have this happen in 5 or 6 hours. Not many.

 

[cave man]

I waited for one of the thirds to lose six times in a row. On the next throw, a fellow put his chips on both the thirds that were winning for the last six throws. I put $100 on the third that lost six times in a row, right in the middle between his two stacks of chips. I won $200.
See for yourself how many times in a two to five hour period any wheel has one third lose for six times in a row. If this bothers you to make a large bet after six loses, wait until a wheel has one third lose eight times in a row (if you have the patience to wait that long) before betting on the losing third.
I know...the odds of any third or 0 and 00 are the same with each spin. But it seems that runs of length on a losing third do not happen with any kind of frequency. Try it some time. After a couple of hours, track the most number of times a third will lose in a row...

 

[Orodruin]

I waited for one of the thirds to lose six times in a row at Trump Plaza. On the nex roll someone bet on the two thirds that were coming up while I put $100 on the one that lost six times. I won $200. And yes, a third can lose for longer than six times. But the next time you are at a casino, find out for yourself how many wheels have this happen in 5 or 6 hours. Not many.

Again, gambler's fallacy.

You are not interested in the first six times - you bet no money on it. You are interested in the odds at the present time, i.e., the odds of getting something seven times in a row given that it has already occurred six times in a row.

 

[mfb]

I know...the odds of any third or 0 and 00 are the same with each spin.

So what exactly is your point?