HOW I TRIED TO BEAT VEGA$
An Exercise in the Right and Wrong Ways to
Use Probability
Dr. Shane Redmond
September 23, 2013
Disclaimer:
No EKU funds were used at casinos
to research this talk.
This talk is also not an endorsement of
gambling.
The game: Roulette
Invention has been attributed to mathematicians
Blaise Pascal, Don Pasquale, and many others.
The wheel: 38 numbered slots
18 red
18 black
0 and 00 are green in American roulette
(European roulette has only one 0, 37 slots)
The bets: Bet on one individual number
Payout is 35 to 1
Meaning if you bet $1 and win, you will be given
$36, for a net gain of $35.
P(lose) = 37/38 =.9737
The bets: Corner bets
Payout is 8 to 1
P(lose) = 34/38 = .8947
Outside bets: Columns
Payout is 2 to 1
P(lose) = 26/38 = .6842
Outside bets: Color, parity, high/low
Payout is 1 to 1
P(lose) = 20/38 = .5263
So our odds are best with the outside bets
Let’s consider playing against a streak.
For example, if we bet red, let’s bet red again
because the probability of not seeing red twice
in a row is (20/38)*(20/38) = .2770.
In fact, the probabilities of not seeing red
STREAK
PROBABILITY
1
.5263
2
.2770
3
.1458
4
.0767
5
.0404
6
.0213
7
.0112
8
.0059
9
.0031
10
.0016
11
.0009
20
38
𝑛
Gambler’s Fallacy
Just because the wheel came up black last time,
it does NOT make red MORE likely to come up
the next time. The probability of red is the SAME
on each spin, no matter the previous outcomes.
This is called the “Gambler’s Fallacy” or “Monte
Carlo Fallacy” after one crazy night in 1913
where black came up on the roulette wheel 26
times in a row.
However, we’re not falling for the
Monte Carlo Fallacy
Flip a coin. P(heads) = ½
Flip a coin twice, P(heads twice) = ¼
So how do we take advantage of these
favorable probabilities?
If we bet $1, lose, and bet $1 again, we
do not come out ahead.
But, if we double the second bet to $2,
we do come out ahead:
-$ 1 - $ 2 + 2($ 2) = $ 1
Loss on
Bet #1
Amount of
Bet #2
Payout
Bet #2
Overall
profit
Similarly, if we lose twice, double the third
bet again to cover previous losses and
ensure a profit.
-$ 1 - $ 2 - $ 4 + 2($ 4) = $ 1
Loss on
Bet #1
Loss on
Bet #2
Amount of
Bet #3
Payout
Bet #3
Overall
profit
We can continue this process of doubling
our bets to cover previous losses for streaks
of any length.
This method is an example of a
Martingale System.
First gained popularity in 18th century France.
First studied mathematically by
Paul Levy (1934).
Named “martingale” by Jean Ville (1939).
Table of bets to make betting against a streak
STREAK
PROBABILITY
BET
1
.5263
$1
2
.2770
$2
3
.1458
$4
4
.0767
$8
5
.0404
$ 16
6
.0213
$ 32
7
.0112
$ 64
8
.0059
$ 128
9
.0031
$ 256
10
.0016
$ 512
11
.0009
$ 1024
𝑛−1
2
What problems do you see with our
Martingale System?
Practical problems: You need a
large initial bankroll.
STREAK
PROBABILITY
BET
TOTAL BET
PAID
1
.5263
$1
$1
2
.2770
$2
$3
3
.1458
$4
$7
4
.0767
$8
$ 15
5
.0404
$ 16
$ 31
6
.0213
$ 32
$ 63
7
.0112
$ 64
$ 127
8
.0059
$ 128
$ 255
9
.0031
$ 256
$ 511
10
.0016
$ 512
$ 1023
11
.0009
$ 1024
$ 2047
Sum of
previous
bets, or
𝑛
2 -1
Practical problems: The payout is
small.
STREAK
PROBABILITY
BET
TOTAL BET
PAID
1
.5263
$1
$1
2
.2770
$2
$3
3
.1458
$4
$7
4
.0767
$8
$ 15
5
.0404
$ 16
$ 31
6
.0213
$ 32
$ 63
7
.0112
$ 64
$ 127
8
.0059
$ 128
$ 255
9
.0031
$ 256
$ 511
10
.0016
$ 512
$ 1023
11
.0009
$ 1024
$ 2047
In all cases,
the profit
is only $ 1.
The Martingale System is like a
“reverse lottery”
Lottery
• We know the odds of losing
are high.
• Profit when we win:
great (millions of $$)
• Consequences of losing:
small ($1-$2)
Martingale system
• We think the odds of losing
become low.
• Profit when we win:
small (usually the initial bet)
• Consequences of losing:
great (may hundreds times
the initial bet)
Some ways around the small profit
*Start with bets larger than $ 1
(Of course, this means a much bigger initial
bankroll is necessary.)
*Run several bets at once
(For example, bet red and odd and high numbers,
continuing the doubling only for the bets that lose)
*Use other patterns other than doubling
(For example, $ 1 -> $ 2 -> $ 5 -> $ 10 -> $ 20 ->
$ 50 -> $ 100 -> $ 200 -> $400 …)
Practical problems: Complementary
Drinks
The Hangover, 2009, Warner Bros.
Practical problems: Table Limits
Classic sign
Modern digital sign
Practical problems: Pit Boss
1995 – Universal Pictures
Practical problems: Pit Boss
$ 400 bet on red
represented a risk of
$ 1 + $ 2 + $ 5 + $ 10
+ $ 20 + $ 50 + $ 100
+ $ 200 + $400 = $788
for a profit of $ 12
Mathematical problems: The
probability is never zero …
because we do not have an infinite amount of
time to play the game or an infinite budget.
Mathematical problems
*The probability of a streak occurring is not the
best way to look at the outcomes.
*Long streaks are not as uncommon as the
probabilities suggest.
Expected Value
The expected value is the value one would
“expect” to find if one could repeat a random
process over an very large number of trials
and take the average of the values obtained.
EV = (profit from win) × (probability of win)
+ (amount of loss) × (probability of loss)
Expected Value: one play
EV = (profit from win) × (probability of win)
+ (amount of loss) × (probability of loss)
$1
18
+ −$ 1
38
20
−2
−1
=
=
38
38
19
= −$ .05
Expected Value: streak of n plays
Probability of win Probability of
Amount of loss
before a streak
streak of n losses
of n losses
$1
20
1−
38
20
=1−
38
40
=1−
38
𝑛
20
+
38
𝑛
𝑛
−$ 2𝑛 − 1
𝑛
20
20
−2
+
38
38
𝑛
𝒏
𝟐𝟎
=𝟏−
𝟏𝟗
𝑛
𝑛
Expected Value: 1:1 outside bets
STREAK
PROBABILITY BET
TOTAL BET
PAID
EXPECTED
VALUE
1
.5263
$1
$1
- $ .05
2
.2770
$2
$3
- $ .11
3
.1458
$4
$7
- $ .17
4
.0767
$8
$ 15
- $ .23
5
.0404
$ 16
$ 31
- $ .29
6
.0213
$ 32
$ 63
- $ .36
7
.0112
$ 64
$ 127
- $ .43
8
.0059
$ 128
$ 255
- $ .51
9
.0031
$ 256
$ 511
- $ .59
10
.0016
$ 512
$ 1023
- $ .67
11
.0009
$ 1024
$ 2047
- $ .76
Outside bets: Columns
Payout is 2 to 1
P(lose) = 26/38 = .6842
STREAK
PROBABILITY
BET
TOTAL BET PROFIT
PAID
EXPECTED
VALUE
1
.6842
$1
$1
$2
- $ .05
2
.4681
$1
$2
$1
- $ .09
3
.3203
$2
$4
$2
- $ .14
4
.2192
$3
$7
$2
- $ .19
5
.1500
$4
$ 11
$1
- $ .23
6
.1026
$6
$ 17
$1
- $ .28
7
.0702
$9
$ 26
$1
- $ .33
8
.0480
$ 14
$ 40
$2
- $ .38
9
.0329
$ 21
$ 61
$2
- $ .44
10
.0225
$ 31
$ 92
$1
- $ .49
11
.0154
$ 47
$ 139
$2
- $ .54
12
.0105
$ 70
$ 209
$1
- $ .60
13
.0072
$ 105
$ 314
$1
- $ .66
14
.0049
$ 158
$ 472
$2
- $ .72
15
.0034
$ 237
$ 709
$2
- $ .78
2:1
outside
bets
Expected Value: a recursive formula
𝑒𝑛 =
Expected value
for spin n
𝑒𝑛−1 =
Expected value
for spin n - 1
𝑝 = probability
of a loss on this
spin
𝑒𝑛 = 𝑒𝑛−1 + 𝑝𝑛−1 (𝑤𝑛 1 − 𝑝 + 𝑏𝑛−1 − 𝑏𝑛 𝑝)
𝑤𝑛 =
profit for win
on spin n
𝑏𝑛 =
Sum of bets
for spin n
𝑏𝑛−1 =
Sum of bets
for spin n -1
8:1 corner bets
STREAK
PROBABILITY
BET
TOTAL BET PROFIT
PAID
EXPECTED
VALUE
1
.8947
$1
$1
$8
- $ .05
2
.8006
$1
$2
$7
- $ .10
3
.7163
$1
$3
$6
- $ .14
4
.6409
$1
$4
$5
- $ .18
5
.5734
$1
$5
$4
- $ .21
6
.5131
$1
$6
$3
- $ .24
7
.4591
$1
$7
$2
- $ .27
8
.4107
$1
$8
$1
- $ .30
9
.3675
$2
$ 10
$8
- $ .34
…
…
…
…
…
…
39
.0131
$ 57
$ 510
$3
- $ 1.54
STREAK
PROBABILITY
BET
TOTAL BET PROFIT
PAID
EXPECTED
VALUE
1
.9737
$1
$1
$ 35
- $ .05
2
.9481
$1
$2
$ 34
- $ .10
3
.9231
$1
$3
$ 33
- $ .15
…
…
…
…
…
…
35
.3932
$1
$ 35
$1
- $ 1.21
36
.3829
$2
$ 37
$ 35
- $ 1.26
…
…
…
…
…
…
53
.2433
$2
$ 71
$1
- $ 1.81
54
.2369
$3
$ 74
$ 34
- $ 1.85
…
…
…
…
…
…
…
…
…
…
…
…
116
.0453
$ 15
$ 508
$ 32
- $ 3.97
35:1
single
number
bets
Are streaks that rare?
We calculated earlier with a $ 511 bankroll, you could
bet on red/black for a streak of up to 9 spins with a
probability of not winning in those 9 spins of
20 9
=.0031
38
But how likely is it to hit a streak of 9 or more and
losing our bankroll before we earn back our $ 511?
Likelihood of hitting a bad streak
So, our chances of not bankrupting (not hitting a streak
of 9 or more against us) on the first 9 spins is
1 - .0031 = .9971.
The probability of going bankrupt on each
subsequent spin is
18
× .0031 = .001468
38
So, the probability of not going bankrupt after n > 9
spins is
.9971 1 − .001468 𝑛−9
Likelihood of staying solvent after n
spins
Number of spins
9
Probability Not
Bankrupt
.9971
100
200
300
.8723
.7531
.6502
400
500
750
.5614
.4847
.3357
1000
.2325
Lessons learned
Thanks!!!
and …
This Saturday 9/28
Bluegrass Undergraduate Mathematics Symposium
(B.U.M.S.)
Centre College in Danville
FREE to undergraduates if register online by
Sept. 24 (tomorrow!)
Grad students – contact me for FREE registration
FREE LUNCH TO ALL WHO PRE-REGISTER!
http://web.centre.edu/bums/Home.html