I Average profits and losses on a roulette table

[BiGyElLoWhAt]

Recently, youtube suggested a video to me on roulette betting theory. The essence of the idea:
Start with a small bet. If you win, bet the same, if you lose, double up. If you lose a second time, double again (or bet the sum of your previous bets) until you win. Start over at the small bet. The important part is to bet on 5050 odds.
Example:
Bet $1 -lose (-$1)
Bet $2-win (+$1)
Bet $1- win(+$2)
Bet $1- lose (+$1)
Bet $2 -lose (-$1)
Bet $3 -lose (-$4)
Bet $6 - win (+$2)
Etc.
So with $192, you have to lose 8 times in a row to be bankrupt, which has a 1 in 256 chance.
Now if I played this for 2 people, one on black and red, we should essentially win everytime.
My question, as a function of the number of spins, how much money should we win on average? I'm not even sure how to go about figuring this out, my probability theory sucks. I would really like help in figuring this out. Answers are fine, but only with explanations.

To clarify, I have a partner. I bet red using this method, my partner bets black using this method.
Bet 1 black bet 1 red - roll red (+0)
Bet 2 black bet 1 red - roll black (+1)
Bet 1 black bet 2 red - roll black (+0)
Bet 1 black bet 3 red - roll red (+2)
Bet 2 black bet 1 red... etc.
Obviously assume a fair board, no wears, balanced and level. 50 50 odds for either roll. Bets are 1,2,3,6,12,24,48,96,etc. For the sake of argument, let's say you are bankrupt if you lose the 96 bet (thats the 8 bets with $192 bank).

 

[Orodruin]

You will win no money at all on average. It is a common fallacy to believe this. Your assumption is that you have an infinite amount of money to bet, but this is not the case.

In addition, the roulette wheel does not give you 50-50 odds by construction. The 0 is added to benefit the house, which effectively means that every time you make a bet you should expect a deficit.

 

[StoneTemplePython]

The original strategy here, I believe is a martingale strategy. Martingales mean something a bit more than that in modern probability, but this betting 'strategy' is the origin.

The 2 player proposal is quite nice because it simplifies things. You're basically running a Dutch book to guaranty that you and a friend lose (i.e. gain nothing on almost all spins and lose on the residuals linked to zero).

 

[BiGyElLoWhAt]

Ahh there's a 0? How many numbers are on the wheel?

 

[StoneTemplePython]

Ahh there's a 0? How many numbers are on the wheel?

Use Google. I think even Dostoevsky's The Gambler would give you this.

 

[BiGyElLoWhAt]

I don't see how we gain nothing on all spins. If it alternates red black, we are plus 1 on each alternation (after the first bet where it's net even). We of course lose on the zero.

 

[BiGyElLoWhAt]

In american roulett, it seems there is zero and 00. So there are 38 numbers, 2 of which are zeros. 1 in 19 spins on average is a loss for both, 9/19 red wins 9/19 black wins.

After looking at some cases (assuming no zeros) as long as we both stop after one of us "wins big", we always come out ahead. I am going to make a spreadsheet, but I venture to say that you can still win with zeros. I may very well be wrong. I will post my sheet later.

 

[jedishrfu]

I’ve heard the best strategy is to never double your bet but instead halve it. I remember reading about it in this book Fortunes Formula. In that way, you can control your losses with a slight chance of winning.

https://www.amazon.com/dp/0809045990/?tag=pfamazon01-20

 

[BiGyElLoWhAt]

Hmmm, that seems odd. You never recoup your losses except with an appropriately sized winning streak. If you double, you recoup all losses on the next win.

 

[BiGyElLoWhAt]

So I didn't make a spreadsheet, I made a python script. Other than the occasional error (out of bounds), it seems to be between 60 and 70 dollars profit each time. I had one that was 37. If it hits out of bounds, it's a bust. I thought I took care of it, but apparently not.

Spoiler: Code

"""
Created on Wed Jan 03 00:44:52 2018

@author: Tyler
"""
from random import randint
numbers = 38
pb = 0 #profit black
pr = 0 #profit red
pn = 0 #profit net
br = 0 #red bet index
bb = 0 #black bet index
games = 100
bets = [1,2,3,6,12,24,48,96]

for i in range(games):
    spin = randint(0,37)
    if spin < 2: # 0 or 00
        pb -= bets[bb]
        bb += 1
        pr -= bets[br]
        br +=1
    elif spin-2 <18: #black wins
        pb += bets[bb]
        bb = 0
        pr -= bets[br]
        br += 1
        leave = randint(0,19)
        if leave < 10 and bb >2 and br ==0:
            print i
            i = games
    else: #red wins
        pb -= bets[bb]
        bb += 1
        pr += bets[br]
        br = 0
        leave = randint(0,19)
        if leave < 10 and br >2 and bb ==0:
            print i
            i = games
    if br >= 7 or bb >= 7: #bust
        print i
        i = games
   
    pn = pr + pb
print pn

Basically I use rand ints between 0 and 37 (since there are 38 options on the table). 0 and 1 represent 0 and 00. 2-19 represent black and 20-37 are red. The distribution is a bit off, but with enough takes, this won't matter. Can anyone point out where I'm not catching my index out of bounds? I'm thinking that last if statement should catch it. Actually I should be able to be >= 8, since there is 0-7 in my array. My pc is garbo otherwise I would just debug (took me 10 mins just to save =/)

 

[jedishrfu]

Python rand numbers are pseudo random meaning they won’t reflect reality so be prepared to be surprised.

 

[BiGyElLoWhAt]

how can i generate "more random" numbers?

 

[jedishrfu]

There’s a site that uses atmospheric noise

Random.org

And here’s some detail on

https://en.m.wikipedia.org/wiki/Random_number_generation

 

[StoneTemplePython]

I’ve heard the best strategy is to never double your bet but instead halve it. I remember reading about it in this book Fortunes Formula. In that way, you can control your losses with a slight chance of winning.

https://www.amazon.com/dp/0809045990/?tag=pfamazon01-20

It's a good book with a nice background on Shannon and Ed Thorp.

The Edge Over Odds formula in that book is derived from entropy concerns, but its really just geometric mean optimization in a binary case. If you have no edge, then don't bet, says the formula. (Put differently, when optimizing a geometric mean, we can immediately see to not bet if the edge is negative.) OP's martingale proposal still has no edge.

People can read more about Thorp and his views on gambling directly on his website. Note: Thorp has at least 9 articles on roullete on this webpage

http://www.edwardothorp.com/articles/

From "SystemsForRoulette_l.pdf", found on that link.

The doubline-up system... to see how ridiculous the system would be... Is this system any good

and the answer is no and he proves other 'systems' with negative expectation are bad too. Read the articles for more discussion.

Python rand numbers are pseudo random meaning they won’t reflect reality so be prepared to be surprised.

It's been a while but last I heard, if users aren't concerned about cryptographic randomness, then Python's rand generator should be 'close enough' for any other purpose, as long as they aren't being seeded. Numpy's random number generators are generally preferred though.

 

[Orodruin]

I’ve heard the best strategy is to never double your bet but instead halve it. I remember reading about it in this book Fortunes Formula. In that way, you can control your losses with a slight chance of winning.

https://www.amazon.com/dp/0809045990/?tag=pfamazon01-20

The best strategy is to not play since the expectation value is negative.

 

[Erik Cox]

As well as the zero (or zeroes for US wheel), the Martingdale strategy won't work even if you did have an infinite amount of money, as the casinos will not allow such big bets. Typically, casinos limit these single bets to around £10,000 on a high stakes table...
If you're looking for 50/50 chance, choose baccarat - bet on the Banker for a 1.06% edge.

 

[Orodruin]

As well as the zero (or zeroes for US wheel), the Martingdale strategy won't work even if you did have an infinite amount of money, as the casinos will not allow such big bets. Typically, casinos limit these single bets to around £10,000 on a high stakes table...
If you're looking for 50/50 chance, choose baccarat - bet on the Banker for a 1.06% edge.

Indeed, the only casino games where you could theoretically carry a positive EV are games where you are betting against other players and not the house or games where the player edge is small enough and proper strategy rare enough that the actual edge based on the strategy used by the average player lies with the house. Otherwise the casinos would not offer those games.

A prime example where you can carry a positive EV is poker if you are (more) skilled at it (than your opponents). I made around $5000 on online poker as a diversion during the latter part of my PhD playing mostly at quite low stakes. The house still gains by taking a part of each pot (without even risking losses).

 

[CWatters]

Someone should change the thread title to average losses.

 

[willem2]

So I didn't make a spreadsheet, I made a python script. Other than the occasional error (out of bounds), it seems to be between 60 and 70 dollars profit each time. I had one that was 37. If it hits out of bounds, it's a bust. I thought I took care of it, but apparently not.
[.

Basically I use rand ints between 0 and 37 (since there are 38 options on the table). 0 and 1 represent 0 and 00. 2-19 represent black and 20-37 are red. The distribution is a bit off, but with enough takes, this won't matter. Can anyone point out where I'm not catching my index out of bounds? I'm thinking that last if statement should catch it. Actually I should be able to be >= 8, since there is 0-7 in my array. My pc is garbo otherwise I would just debug (took me 10 mins just to save =/)

The problem with your code is your horrible way of breaking out of the for loop bij setting i = games. This is NOT the way to do it, and often seems not to work at all. I don't know if python can also have demons fly out of your nose when doing something like that. . You can just use a break statement instead of i = games.
(Only the last break will actually work BTW, the if leave < 10 and bb >2 and br ==0: tests will never succeed.)

Unfortunately these are all cases where one of the players goes bust. This happens quite often, one of the players will be at about -90, and the other will be at about +30.
After repeating the entire game of maximum 100 rounds 10000 times, it appears the expected value for playing a maximum of hundred rounds and stopping when the red or the black player is bust, is about -10.

 

[BiGyElLoWhAt]

Ahh thanks for that catch. I don't really use python that much, but eclipse takes too long to load on this computer.

 

[BiGyElLoWhAt]

Here is my modified code:

Spoiler: Code

# -*- coding: utf-8 -*-
"""
Created on Wed Jan 03 00:44:52 2018

@author: Tyler
"""
from random import randint
numbers = 38
pb = 0 #profit black
pr = 0 #profit red
pn = 0 #profit net
br = 0 #red bet index
bb = 0 #black bet index
games = 100
cycles = 10000
bust = 0
bets = [1,2,3,6,12,24,48,96]

for i in range(cycles):
    pb = 0 #profit black
    pr = 0 #profit red
    pn = 0 #profit net
    br = 0 #red bet index
    bb = 0 #black bet index
    for i in range(games):
        spin = randint(0,37)
        if spin < 2: # 0 or 00
            pb -= bets[bb]
            bb += 1
            pr -= bets[br]
            br +=1
        elif spin-2 <18: #black wins
            pb += bets[bb]
            bb = 0
            pr -= bets[br]
            br += 1
            leave = randint(0,19)
            if leave < 10 and bb >2 and br <2:
                #print i
                break
        else: #red wins
            pb -= bets[bb]
            bb += 1
            pr += bets[br]
            br = 0
            leave = randint(0,19)
            if leave < 10 and br >2 and bb <2:
                #print i
                break
        if br >= 7 or bb >= 7: #bust
            bust+=1
            break
       
        pn += pr + pb
print "average profits: "
print pn/cycles
print "with"
print bust
print "busts"

if you change cycles to 1000 instead of 10,000, it still seems to lean slightly positive, like 1 or 2 dollars, however, on 10,000 it sits pretty close to zero.
Thanks for pointing out the flaws in my code.
I also "fixed" the lines with the "leave" statement, so there is a 50-50 chance that if one player wins big (to recoup losses) and the other player is on their first loss, that we both leave.

 

[BiGyElLoWhAt]

Whats really interesting, if you change games to 1000, you get almost 100% bust rate. (games is the number of allotted spins) At 100, it sits right around 66%, with 50 you get around 40%. The standard deviation on this seems quite low, as well, just from rerunning the program and eyeballing it.

Edit:
As expected, adding 1 additional bet term changes the bust rate by a factor of 2 (probably closer to 17/36, if I'm thinking about this correctly).

 

[BiGyElLoWhAt]

Is this part of the randomness issue? Slight variation to the code, I added in the next bet, we play at most 100 rounds, and 1000 cycles. This steadily climbs. The EV seems to be between 2 and 3. I just ran about 20 or 30 and totaled them by hand. Highest was +3, lowest was -1, and the total was around +67.

 

[BiGyElLoWhAt]

Hmmm... Someone might need to explain where I'm going wrong in my assumptions here, or where my coding error is.
I just reworked the code again. 1 game of roulette every night for a year (365 cycles). I assumed 2 minutes per spin, and a maximum of 4 hours per night, so 120 spins max. It's consistently coming out at around +4000 per year or so.

Spoiler: code

# -*- coding: utf-8 -*-
"""
Created on Wed Jan 03 00:44:52 2018

@author: Tyler
"""
from random import randint
from numpy import zeros
numbers = 38
pb = 0 #profit black
pr = 0 #profit red
pn = 0 #profit net
br = 0 #red bet index
bb = 0 #black bet index
games = 120
cycles = 365
rounds = 0
bust = 0
bets = [1,2,3,6,12,24,48,96,192,384]
roundsarr = zeros(cycles)

for i in range(cycles):
 
    pb = 0 #profit black
    pr = 0 #profit red
    pn = 0 #profit net
    br = 0 #red bet index
    bb = 0 #black bet index
    rounds = 0
    while(1):
        spin = randint(0,37)
        rounds +=1
        if spin < 2: # 0 or 00
            pb -= bets[bb]
            bb += 1
            pr -= bets[br]
            br +=1
        elif spin-2 <18: #black wins
            pb += bets[bb]
            bb = 0
            pr -= bets[br]
            br += 1
            leave = randint(0,19)
            if leave < 10 and bb >2 and br <2:
                #print i
                roundsarr[i] = rounds
                break
        else: #red wins
            pb -= bets[bb]
            bb += 1
            pr += bets[br]
            br = 0
            leave = randint(0,19)
            if leave < 10 and br >2 and bb <2:
                #print i
                roundsarr[i] = rounds
                break
        if br >= len(bets)-1 or bb >= len(bets)-1: #bust
            bust+=1
            roundsarr[i] = rounds
            break
     
        pn += pr + pb

roundsmax = 0
roundsavg = 0
for i in range(len(roundsarr)):
    roundsavg += roundsarr[i]
    if roundsarr[i] > roundsmax:
        roundsmax = roundsarr[i]
roundsavg = roundsavg/len(roundsarr)
print "average profits: "
print pn/cycles
print "total profits"
print pn
print "with"
print bust
print "busts"
print "average game length"
print roundsavg
print "longest game"
print roundsmax

Here in a bit I will try to make this more realistic. I'm still using 50-50 chance to see if we walk away on a big win, instead of forcing us to walk away on a big win, and the average length of the games are coming out to be 14-16. So I can lengthen the games, and force us to always walk away on a win, with high probability, by scaling the random chance of leaving to approach 100% as the night gets closer to an end. Every year I have run with this has been profitable, though. (one year turned a 47 dollar annual profit, which is the lowest I've seen.)

Edit:
I just made this change, and it didn't change hardly anything, which means that it's not the random factor, it's more likely the 33% bust rate that's causing the low length average games.

 

[Orodruin]

Why are you even trying to simulate this? The maths are trivial. You cannot make a +EV strategy composed of a series of bets with non-positive EV.

Any simulation that tells you otherwise has either too few repetitions or issues with the random numbers.

 

[mfb]

The "strategy" leads to a high probability to win a little bit and a small probability to lose all. If you don't run the simulation often enough you'll miss the second case and get an incorrect result.

Expectation values add. If you use 50/50 chances your expectation value is exactly zero in every game, therefore the overall expectation value is 0 as well. If you use realistic chances including 0 (or 0 and 00), your expectation value is negative in every game, therefore the overall expectation value is negative as well.

 

[BiGyElLoWhAt]

Yea it's really weird though. I'm actually wondering where my simulation went wrong. If I lose, the losses are always bigger than the profits, however, it seems I'm winning 2/3 games (coming out ahead). And averaging over a year gives a net profit. I'm using between 8 and 120 rolls per game, and 365 games per simulation. I know the sample isn't huge, but it's not negligably small. The years that are losses are very small, at most like 300 or so. The gain years (which are far more abundant with these numbers) are averaging what looks like 4500. I include 0 and 00. Here is my current code. I keep track of the number of nights that are wins and losses, the biggest win, the biggest loss, the average gain/loss (+/-) the average number of spins per game, the longest game, etc.

# -*- coding: utf-8 -*-
"""
Created on Wed Jan 03 00:44:52 2018

@author: Tyler
"""
from numpy.random import randint
from numpy import zeros
numbers = 38
pb = 0 #profit black
pr = 0 #profit red
pn = 0 #profit net
br = 0 #red bet index
bb = 0 #black bet index
games = 120
cycles = 365
rounds = 0
bust = 0
bets = [1,2,3,6,12,24,48,96,192,384]
roundsarr = zeros(cycles)
big =0
small = 0
win = 0
loss = 0

for i in range(cycles):
   
    pb = 0 #profit black
    pr = 0 #profit red
    pn = 0 #profit net
    br = 0 #red bet index
    bb = 0 #black bet index
    rounds = 0
    for j in range(games):
        spin = randint(0,37)
        rounds +=1
        if spin < 2: # 0 or 00
            pb -= bets[bb]
            bb += 1
            pr -= bets[br]
            br +=1
        elif spin-2 <18: #black wins
            pb += bets[bb]
            bb = 0
            pr -= bets[br]
            br += 1
            leave = randint(0,games -1)
            if leave < j and bb >2 and br <2:
                #print i
                roundsarr[i] = rounds
                break
        else: #red wins
            pb -= bets[bb]
            bb += 1
            pr += bets[br]
            br = 0
            leave = randint(0,games -1)
            if leave < j and br >2 and bb <2:
                #print i
                roundsarr[i] = rounds
                break
        if br >= len(bets)-1 or bb >= len(bets)-1: #bust
            bust+=1
            roundsarr[i] = rounds
            break
           
        pn += pr + pb
    if pr + pb > big:
        big = pr + pb
    elif pr +pb < small:
        small = pr + pb
    if pr + pb > 0:
        win += 1           
    elif pr + pb < 0:
        loss += 1

roundsmax = 0
roundsavg = 0
for i in range(len(roundsarr)):
    roundsavg += roundsarr[i]
    if roundsarr[i] > roundsmax:
        roundsmax = roundsarr[i]
roundsavg = roundsavg/len(roundsarr)
print "average profits: "
print pn/cycles
print "total profits"
print pn
print "with"
print bust
print "busts"
print "average game length"
print roundsavg
print "longest game"
print roundsmax
print "high/low"
print big
print small
print "win/loss"
print win
print loss

So far (30 or so rolls in) on some free online roulette, I'm up by about 12 wins worth. I am up 38 bucks, but I started off with 1,2,3 and switched to 10,20,30.

 

[mfb]

I know the sample isn't huge, but it's not negligably small.

It is too small. Run it for 100,000 days and see what happens. If you still get a profit then something is wrong with the code.

 

[Orodruin]

Also, including a partner is just a red herring. Without the zeros, all the partner does is to bet an amount that exactly negates your own bet. Your partner betting 1 on black and you 2 on red is exactly equivalent to you betting 1 on red for your common bankroll. Red would lead to +2-1=+1 and black to -2+1=-1.

The strategy is therefore exactly equivalent to a Martingale type strategy for your common roll.

 

[willem2]

.

pn += pr + pb

It is too small. Run it for 100,000 days and see what happens. If you still get a profit then something is wrong with the code.

There's another problem with the program, the line pn += pr + pb is indented too much, so it's in the inner loop and not in the outer loop, this means that the total winnings before one of the players goes bust get added to pn after each round. If you do it correctly, you'll see that most of the games, someone goes bust, and the total winnings are negative.

 

[BiGyElLoWhAt]

Yea that seemed to be my problem. I was counting everything multiple times. So if I was 1 dollar ahead on spin 1 and 2 dollars ahead on spin 2, my profits were 3 dollars instead of 2 dollars. Thanks for pointing that out.

 

[votingmachine]

Even if you are betting on a coin flip, the advantage lies with the side having more money. Say you start with $3 and are playing a house with $10,000. If you lose the first bet, and then bet $2 and lose again, you are done after 2 coin flips. If you win the second bet, you have $3 again, starting over. If you win the first bet, you are up to $4, and can follow your martingale for a $1 bet and a $2 double, and then are down to your last $1. If you win 4 flips in a row to start, you have $7 and could bet and lose a sequence of $1, $2, $4 and be wiped out. The house is only wiped out by a losing 10,000 bets more than it wins. The Martingale player is wiped out by a continuous losing streak.

There is a math treatment I could find for you, but the bottom line is that in an EVEN ODDS betting game, the side with the larger money reserve is less likely to lose a game to terminal bankruptcy. And of course, casino games are all NOT even odds, but favor the house.

Most games are not played to terminal bankruptcy. But a Martingale requires that the gambler treat his sequential losing bets with an exponential increase. And that is vulnerable to a losing streak, while the house is bullet-proof to the strategy. You only bet small amounts except when you lose. The house is ALWAYS playing with your money when you are doubling down. And eventually, you WILL have a losing streak that exceeds your pocket.

The average bettor loses money in casinoes. Some win a lot, some lose a lot, some win a little, some lose a little, some break even. But the odds are in the houses favor, and overall, the net is that the house wins a fraction more than it loses. There is no betting system to change that.

 

[mfb]

Even if you are betting on a coin flip, the advantage lies with the side having more money.

Not if you consider the expectation value.

Say you start with $3 and are playing a house with $10,000. If you lose the first bet, and then bet $2 and lose again, you are done after 2 coin flips. If you win the second bet, you have $3 again, starting over.

If you win the second one you have $4.

If you play as long as both sides can pay, you have a 3/10003 chance to win everything and a 10000/10003 chance to lose everything. That shouldn't be surprising, and the expectation value is $3. A 50/50 chance for both players to win everything would mean your initial money doesn't matter and you could make a lot of money.

 

[BWV]

While someone with infinite money would always win in a Martingale betting system, no one has infinite money. Also, the house always places limits on individual bets

 

[mfb]

While someone with infinite money would always win in a Martingale betting system

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

 

[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.