how i tried to beat vega$ an exercise in the right and wrong ways to use probability

HOW I TRIED TO BEAT VEGA$An Exercise in the Right and Wrong Ways

to Use Probability

Dr. Shane RedmondSeptember 23, 2013

Disclaimer:

No EKU funds were used at casinosto research this talk.

This talk is also not an endorsement ofgambling.

The game: Roulette

Invention has been attributed to mathematiciansBlaise Pascal, Don Pasquale, and many others.

The wheel: 38 numbered slots

18 red18 black0 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 1Meaning 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 againbecause the probability of not seeing red twicein a row is (20/38)*(20/38) = .2770.

In fact, the probabilities of not seeing redSTREAK PROBABILITY

1 .5263

2 .2770

3 .1458

4 .0767

5 .0404

6 .0213

7 .0112

8 .0059

9 .0031

10 .0016

11 .0009

( 2038 )𝑛

Gambler’s FallacyJust because the wheel came up black last time, it does NOT make red MORE likely to come upthe next time. The probability of red is the SAMEon each spin, no matter the previous outcomes.

This is called the “Gambler’s Fallacy” or “MonteCarlo Fallacy” after one crazy night in 1913where black came up on the roulette wheel 26times 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) = $ 1Loss onBet #1

Amount ofBet #2

PayoutBet #2

Overallprofit

Similarly, if we lose twice, double the thirdbet again to cover previous losses andensure a profit.

Loss onBet #1

Loss onBet #2

PayoutBet #3

Overallprofit

-$ 1 - $ 2 - $ 4 + 2($ 4) = $ 1

Amount ofBet #3

We can continue this process of doubling our bets to cover previous losses for streaksof any length.

This method is an example of a Martingale System.

First gained popularity in 18th century France.

First studied mathematically byPaul Levy (1934).

Named “martingale” by Jean Ville (1939).

Table of bets to make betting against a streakSTREAK 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

2𝑛− 1

What problems do you see with ourMartingale System?

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

-1

Sum ofprevious bets, or

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

In all cases,the profitis only $ 1.

Practical problems: The payout is small.

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 thebest way to look at the outcomes.

*Long streaks are not as uncommon as theprobabilities suggest.

Expected Value

EV = (profit from win) × (probability of win) + (amount of loss) × (probability of loss)

The expected value is the value one would“expect” to find if one could repeat a randomprocess over an very large number of trials and take the average of the values obtained.

Expected Value: one play

EV = (profit from win) × (probability of win) + (amount of loss) × (probability of loss)

Expected Value: streak of n plays

($1 )(1−( 2038 )𝑛)+( 2038 )

𝑛(−$ (2𝑛−1 ) )

Probability of win before a streak of n losses

Probability of streak of n losses

Amount of loss

¿1−( 2038 )𝑛−2𝑛( 2038 )

𝑛+( 2038 )

𝑛=1−( 4038 )

𝑛=𝟏−(𝟐𝟎𝟏𝟗 )

𝒏

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

Expected Value: 1:1 outside bets

Outside bets: Columns

Payout is 2 to 1

P(lose) = 26/38 = .6842

STREAK PROBABILITY BET TOTAL BET PAID

PROFIT 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

2:1outsidebets

Expected Value: a recursive formula

𝑒𝑛=𝑒𝑛−1+𝑝𝑛−1(𝑤𝑛 (1−𝑝 )+𝑏𝑛−1−𝑏𝑛𝑝 )

= Expected value for spin n

= Expected value for spin n - 1

= probability of a loss on this spin

= profit for win on spin n

= Sum of betsfor spin n

= Sum of betsfor spin n -1

STREAK PROBABILITY BET TOTAL BET PAID

PROFIT 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

8:1 corner bets

STREAK PROBABILITY BET TOTAL BET PAID

PROFIT 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

35:1singlenumberbets

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

=.0031

But how likely is it to hit a streak of 9 or more andlosing our bankroll before we earn back our $ 511?

Likelihood of hitting a bad streakSo, our chances of not bankrupting (not hitting a streakof 9 or more against us) on the first 9 spins is 1 - .0031 = .9971.The probability of going bankrupt on each subsequent spin is

So, the probability of not going bankrupt after n > 9spins is

.9971

Likelihood of staying solvent after n spins

Number of spins Probability Not Bankrupt

9 .9971100 .8723200 .7531300 .6502400 .5614500 .4847750 .33571000 .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 bySept. 24 (tomorrow!)

Grad students – contact me for FREE registrationFREE LUNCH TO ALL WHO PRE-REGISTER!

http://web.centre.edu/bums/Home.html