Betting 3 even chance at $30each instead of 1 at $90
How much would this lower the variance?
How would I calculate it?
Answer: yes, you lower variance.
I don't have the roulette layout in memory, so here is an approximation for (pseudo-)independent division of the board in Odd, Black and Manque.
I suppose American roulette (2 zeroes).
P = 18/38 of winning one of the bets.
Winning all three : P = (18/38)³ W = +$90
Winning two : P = 3 (18/38)²(20/38) W = +$30
Winning one : P = 3 (18/38)(20/38)² W = -$30
Winning none : P = (20/38)³ W = -$90
Variance = SUM {PjWj²} - {SUM PjWj}² = 2692.52 (squared dollars)
Standard deviation = $51.89
All $90 on one bet would have yielded a SD of $89.87, that is sqrt(3)=1.73 times more.
Not that it is unsound. It depends on personal preferences. But are you sure that's what you want?
I do not necesarily want to lower my variance/ more so just curious as to the maths behind it.
So I suppose a good way to increase variance would be to play one even chance and always go for double or triple parlay?
Quote: kubikulannWhat is "parlay"?
Parley = "Let it Ride" Pressing your winnings all the way in hopes to win again.
Betting a single number is already increasing your variance, so your positive progressive (if I got what parlay is) could be used there for more impact. Enormous variance is available in State lotteries, so you might be wary of Expected gain too.Quote: Walkinshaw30tSo I suppose a good way to increase variance would be to play one even chance and always go for double or triple parlay?
Also remember that variance is symmetric: a negative progressive (a martingale) increases your variance the same way. Only, it is the negative result that is huge.
See? Increasing or decreasing variance (second moment) by itself is not relevant. Look at expectation (first moment) and direction (third moment) as well.
Quote: kubikulannWhat is "parlay"?
Pirate code, it means he wants a meeting. I can understand pirate code, I just can't speak it.
First, define the variable. Number of tosses is n (here, 80). Say X is the number of successes, and G is the associated gain (bX - cn, according to your chosen bet; for example, betting $10 on a simple chance would mean G = 20X - 10n).Quote: Walkinshaw30tCould you tell me the maths to calculate the standard deviation of wins/ losses for say 80 spins?
Variable X has a binomial distribution with parameters n=80 and p=(probability of winning in your game; for simple roulette p = 18/38). Never mind the formula of probabilities in a binomial: it is known that the Expectation is np and the Variance is np(1-p). So in the example, EVX = 80(18/38) = 37.8947 successes and Standard dev = square root of variance = sqrt[80(18/38)(20/38)] = 4.4659 successes.
Variable G has EVG = 20 EVX - 10*80 = $ -42.1053, and SDG = 20 SDX = $ 89.3186.
Your EV is .4714 Sd below zero, so a table of Normal-Gauss values tells you you have an approximate 18.08% probability of ending up positive. (And consequently a 81.92% chance of overall loss.)
EDIT: those two last values are incorrect. See below.
Yes, that was the only meaning I knew (Pirates of the Caribbean). Now I saw the meaning in sports bet, and then this use here. Confusing.Quote: Face"Parlay": Pirate code, it means he wants a meeting. I can understand pirate code, I just can't speak it.
Huh?Quote: kubikulannVariable G has EVG = 20 EVX - 10*80 = $ -42.1053, and SDG = 20 SDX = $ 89.3186.
Youe EV is .4714 Sd below zero, so a table of Normal-Gauss values tells you you have an approximate 18.08% probability of ending up positive. (And consequently a 81.92% chance of overall loss.)
please show that table you used and whom and what produced it.
I assume you read the table correctly
I get the same ev and sd
(now using the binomial probability formula -
I cheated and used a function and a quick simulation)
for 18/38 and 80 equal bets made
probability of a net loss = 0.6408883
even or up = 0.3591117
Excel shows, without correction,
ev/sd = -0.471404521 = 0.318675944 probability of even or up
why do you think your approximate values
of a net win/loss (18.08%/81.92%)
are so far off from a direct calculation in two different programs and a quick simulation
that I used?
BTW
for 18/37 and 80 equal bets made
probability of a net loss = 0.5520498
even or up = 0.4479502
This may be my 1,324,854th error
You assume incorrectly. I was too fast in my reading. It is not 18.08% it is (50-18.08)=31.82% Normal approximation.Quote: 7crapsHuh?
please show that table you used and whom and what produced it.
I assume you read the table correctly
Your prob of 0.318676 is the exact one (0.47140531936... instead of just 0.47).
The discrepancy between this 0.32 and the true 0.36 shows how even n=80 tosses is not fulfilling the conditions for the Central Limit Theorem.
I figured that is what you did.Quote: kubikulannYou assume incorrectly. I was too fast in my reading. It is not 18.08% it is (50-18.08)=31.82% Normal approximation.
Your prob of 0.318676 is the exact one (0.47140531936... instead of just 0.47).
something I have done many times before
many think that when n=80, that should be "enough"Quote: kubikulannThe discrepancy between this 0.32 and the true 0.36 shows how even n=80 tosses
is not fulfilling the conditions for the Central Limit Theorem.
the rule of thumb should work
and it does
Just matters how much error one will accept.
even for a fair coin flip when n=80
that shows about a 50% chance of even or up
when calculated is 0.5444639
But was the OP looking really for the wins/losses for betting 3 even money bets for 80 spins
A bit more fun on the math. Was not that specific.
BOH (black, odd, high) has a slightly different probability distribution than does BEH
(ROL instead of REL)
Experienced Roulette players know this, so I have been told
A bit more fun on the math. Was not that specific.
BOH (black, odd, high) has a slightly different probability distribution than does BEH
(ROL instead of REL)
Experienced Roulette players know this, so I have been told
Why is this so??
Is it do do with the layout of the numbers on the wheel?
Quote: Walkinshaw30tBut was the OP looking really for the wins/losses for betting 3 even money bets for 80 spins
A bit more fun on the math. Was not that specific.
BOH (black, odd, high) has a slightly different probability distribution than does BEH
(ROL instead of REL)
Experienced Roulette players know this, so I have been told
Why is this so??
Is it do do with the layout of the numbers on the wheel?
There are 5 even numbers and 4 odd numbers that are black and on the high end (18-36). That would change the odds depending on whether you're combining the other 2 with odd or even.
exactlyQuote: Walkinshaw30tWhy is this so??
Is it do do with the layout of the numbers on the wheel?
Black
2,4,6,8,10,11,13,15,17,20,22,24,26,28,29,31,33,35
how many black #s are even?
how many are odd?
they look to be not equal
the probability of a net win is still equal between BOH (black, odd, high) and BEH
just a bit different on the # of wins (1st column, 2nd column = ways)
BOH
0 6 0.157894737
1 14 0.368421053
2 14 0.368421053
3 4 0.105263158
38
BEH
0 7 0.184210526
1 13 0.342105263
2 13 0.342105263
3 5 0.131578947
38
The ev remains the same but the sd is different
Quote: 7crapsexactly
Black
2,4,6,8,10,11,13,15,17,20,22,24,26,28,29,31,33,35
how many black #s are even?
how many are odd?
they look to be not equal
the probability of a net win is still equal between BOH (black, odd, high) and BEH
just a bit different on the # of wins (1st column, 2nd column = ways)BOH
0 6 0.157894737
1 14 0.368421053
2 14 0.368421053
3 4 0.105263158
38
BEH
0 7 0.184210526
1 13 0.342105263
2 13 0.342105263
3 5 0.131578947
38
The ev remains the same but the sd is different
Ok interesting I had not noticed this before.
If it is not to hard or tedious to calculate, would someone be able to tell me the standard deviation for wins/losses at 80 spins betting black/even/high?
Quote: Walkinshaw30t
Ok interesting I had not noticed this before.
If it is not to hard or tedious to calculate, would someone be able to tell me the standard deviation for wins/losses at 80 spins betting black/even/high?
Is it single-zero roulette & no La Partage rule?
Paytable: odds
+3 units: 5/37
+1 unit: 13/37
-1 unit: 13/37
-3 units: 6/37
Standard deviation per round: 1.84 units (relative to 3-unit bets)
Standard deviation per 80 rounds: 1.84*SQRT(80) = 16.46 units (relative to 3-unit bets)
Total wagered in 80 rounds: 3*80 = 240 units
Total house edge in 80 rounds: 0.0270*240 = 6.48 units
So after 80 rounds your expectation is to be down 6.48 units.
68% of time your result will be within one standard devation of expectation, that is between -22.94 ... +9.98 units.
compare the values to just one 18/37 bet of 3 unitsQuote: Jufo81I Standard deviation per round: 1.84 units (relative to 3-unit bets)
Standard deviation per 80 rounds: 1.84*SQRT(80) = 16.46 units (relative to 3-unit bets)
Total wagered in 80 rounds: 3*80 = 240 units
Total house edge in 80 rounds: 0.0270*240 = 6.48 units
So after 80 rounds your expectation is to be down 6.48 units.
68% of time your result will be within one standard deviation of expectation, that is between -22.94 ... +9.98 units.
we can even look at a graph of 80 bets (for both)
(probs less than .000001% are not plotted)
notice how smooth the curve looks at it's peak
for the 3 even money bets compared to the one even money bet.
from that the actual values of intervals and using the central limit theorem will be quite close
meaning the error will be small.
only 3 even money bets for 80 rounds
and both
the probability of one even money bet showing a loss after 80 rounds of play - flat betting = 55.2%
for the 3 even money bets = 63.1%
cumulative view
Quote: 7crapscompare the values to just one 18/37 bet of 3 units
we can even look at a graph of 80 bets
(probs less than .000001% are not plotted)
notice how smooth the curve looks at it's peak
for the 3 even money bets compared to the one even money bet.
from that the actual values of intervals and using the central limit theorem will be quite close
meaning the error will be small.
To clarify: Is the red line for the 3-way bet (black,even,passe) and green line for single even money bets (both 80 rounds)? It cannot be seen very clearly from the graph but you should have much higher chance of ending up ahead with the single bets because you get more variance to escape the house edge.
If you want to minimize variance, ie. get a lot wagering action with low risk, you should bet on either red+even or black+odd. This will result in a push 21 out of 37 times.
yesQuote: Jufo81To clarify: Is the red line for the 3-way bet (black,even,passe) and green line for single even money bets (both 80 rounds)? It cannot be seen very clearly from the graph
but you should have much higher chance of ending up ahead with the single bets because you get more variance to escape the house edge.
I added the cumulative graph that shows that better
still
ev/sd
which one is closest to 0 has the greater chance of being even or ahead over X trials played
assumes enough money to make every bet
Quote: 7crapsyes
I added the cumulative graph that shows that better
still
ev/sd
which one is closest to 0 has the greater chance of being even or ahead over X trials played
assumes enough money to make every bet
I think the equation is ev/variance and not ev/sd? (variance = sd^2)
That standard deviation is much higher than I though it would be
Quote: kubikulann(Place this in Math or Gamble or Roulette, not in Forum info)
Answer: yes, you lower variance.
Standard deviation = $51.89
All $90 on one bet would have yielded a SD of $89.87, that is sqrt(3)=1.73 times more.
Would that mean that betting 3 even chances (eg black, even, high) would be 1.73 times less likely to fall 21wins behind the other 3 even chances and 0 (eg red, odd, low),
then say black to fall 7 wins behind red and 0?
Quote: Jufo81To clarify: Is the red line for the 3-way bet (black,even,passe) and green line for single even money bets (both 80 rounds)? It cannot be seen very clearly from the graph but you should have much higher chance of ending up ahead with the single bets because you get more variance to escape the house edge.
If you want to minimize variance, ie. get a lot wagering action with low risk, you should bet on either red+even or black+odd. This will result in a push 21 out of 37 times.
Is this strategy a better way of minimizing variance than BEH?
I have been interested in ways to reduce variance also for various reasons
