[bonanova]

So I was reading in Wikipedia about expectation values, and they analyzed roulette.

There are 38 outcomes and guessing it pays $35.

So on average you lose about a nickel on a dollar bet.

So it's dumb to play for a long time: the law of large numbers sends you home broke.

Which reminds me of the guy in desperate financial trouble who

went to Vegas to gamble. On the way, he asked for supernatural

intervention: "Lord, help me to at least break even. I need the money."

But I digress.

I could win on the first spin of the wheel. Or the second. Or the

third, and so on. By the 38th spin, I should expect to have won

at least once. But don't the odds favor the case where my one

win will occur before the 35th spin? If so then there exists a

moment in time when I will be ahead. And I win, just by quitting.

So doesn't that mean I can win, in the short term at roulette? http://brainden.com/forum/uploads/emoticons/default_wink.png' alt=';)'>

And if that works, couldn't I do it again, and win in the long term?

An answer should include an idea of the number of spins I

should expect to wait before winning once.

[Guest]

... you will always have a 1/38 chance of winning regardless of how long you play and "the win" may come on any roll. The HOUSE however is always "playing" in the long run and will therefore always (statistically speaking) win. And they love it when somenone wins big and often, it attracts more players... who will not be so lucky!

What you can do is play until you get a win (hoping it takes less than 34 turns) and quit as soon as you are ahead...

[Guest]

Could you increase your odds of winning by watching the board and lets say, waiting until you see black come up 3 times in a row and betting red? Or odd, etc. By not betting 38 times consecutively.

[Guest]

I haven't thought of it as the problem demanded, but there may be another way to ensure that you win at Roulette. For the first 34 plays, you bet $1 each time, and if you still haven't won, then from 35th to the 51st turn, you bet $2 each time, and if you still haven't won, then from 53rd to the 63rd bet, you bet $3 each time. This will ensure that whenever you win on a given turn, you will always make a profit; the smallest profit being $1.

[bonanova]

I haven't thought of it as the problem demanded, but there may be another way to ensure that you win at Roulette. For the first 34 plays, you bet $1 each time, and if you still haven't won, then from 35th to the 51st turn, you bet $2 each time, and if you still haven't won, then from 53rd to the 63rd bet, you bet $3 each time. This will ensure that whenever you win on a given turn, you will always make a profit; the smallest profit being $1.

Something like Martingale.

[Guest]

I haven't thought of it as the problem demanded, but there may be another way to ensure that you win at Roulette. For the first 34 plays, you bet $1 each time, and if you still haven't won, then from 35th to the 51st turn, you bet $2 each time, and if you still haven't won, then from 53rd to the 63rd bet, you bet $3 each time. This will ensure that whenever you win on a given turn, you will always make a profit; the smallest profit being $1.

This reminds me of another idea i thought up a few years ago:

If you were betting on a coin toss for example with 50:50 odds you simply start by betting by $1 then if you lose bet $2 then $4 etc. At some point you must win and will have made $1 profit.

This would be fine you you had an infinite amount of money. However, assuming you don't, there may be a time if you are unlucky when you have run out of money to gamble and have just made a huge loss.

Mathematically - if you start with $x

P(win 1 $) = 1 - 0.5^log₂(x) = 1 - 1/x

P(loose $x) = 0.5 ^log₂(x) = 1/x

If we calculate the profit as: P(winning) * amount won

Expected winnings = [ (1- 1/x)*1 ] + [(1/x)* (-x) ]

= -$(1/x)

So you would not make a profit if you kept doing this.

[bonanova]

This reminds me of another idea i thought up a few years ago:

If you were betting on a coin toss for example with 50:50 odds you simply start by betting by $1 then if you lose bet $2 then $4 etc. At some point you must win and will have made $1 profit.

This would be fine you you had an infinite amount of money. However, assuming you don't, there may be a time if you are unlucky when you have run out of money to gamble and have just made a huge loss.

Mathematically - if you start with $x

P(win 1 $) = 1 - 0.5^log₂(x) = 1 - 1/x

P(loose $x) = 0.5 ^log₂(x) = 1/x

If we calculate the profit as: P(winning) * amount won

Expected winnings = [ (1- 1/x)*1 ] + [(1/x)* (-x) ]

= -$(1/x)

So you would not make a profit if you kept doing this.

Precisely Martingale.

[Guest]

This reminds me of another idea i thought up a few years ago:

If you were betting on a coin toss for example with 50:50 odds you simply start by betting by $1 then if you lose bet $2 then $4 etc. At some point you must win and will have made $1 profit.

This would be fine you you had an infinite amount of money. However, assuming you don't, there may be a time if you are unlucky when you have run out of money to gamble and have just made a huge loss.

Mathematically - if you start with $x

P(win 1 $) = 1 - 0.5^log₂(x) = 1 - 1/x

P(loose $x) = 0.5 ^log₂(x) = 1/x

If we calculate the profit as: P(winning) * amount won

Expected winnings = [ (1- 1/x)*1 ] + [(1/x)* (-x) ]

= -$(1/x)

So you would not make a profit if you kept doing this.

That is where this particular case is different becuase you are not increasing the bet amount on each turn, but after 34 turns or on 52nd turn or on the 63rd turn and so on. Since, the probability of getting the bet right is 1/38, statistically speaking, with so many turns at the same bet amount, you are not likely to leave the casino broke.

[Guest]

I haven't thought of it as the problem demanded, but there may be another way to ensure that you win at Roulette. For the first 34 plays, you bet $1 each time, and if you still haven't won, then from 35th to the 51st turn, you bet $2 each time, and if you still haven't won, then from 53rd to the 63rd bet, you bet $3 each time. This will ensure that whenever you win on a given turn, you will always make a profit; the smallest profit being $1.

I tried this in Excel.

I started with $50. Each roll, I bet one dollar. For every 35 consecutive times I lost, I added another dollar to the bet. If I won then the number of losses started back at 0.

I decided to make exactly 250 bets. And record my cash at the end of the run.

I did this 25 times.

My average return for each "trip" to the casino was 259.12%

2 times, I went home no money.

2 other times, I went home with less than 50 dollars.

Subtracting my orginal $50 seed money, I netted $3189.

[bushindo]

I tried this in Excel.

I started with $50. Each roll, I bet one dollar. For every 35 consecutive times I lost, I added another dollar to the bet. If I won then the number of losses started back at 0.

I decided to make exactly 250 bets. And record my cash at the end of the run.

I did this 25 times.

My average return for each "trip" to the casino was 259.12%

2 times, I went home no money.

2 other times, I went home with less than 50 dollars.

Subtracting my orginal $50 seed money, I netted $3189.

That is excellent result. 259% gain on your bankroll in roulette is great. Let's go to Vegas together this weekend.

please consider checking your excel code

Edited April 1, 2010 by bushindo

[Guest]

I tried this in Excel.

I started with $50. Each roll, I bet one dollar. For every 35 consecutive times I lost, I added another dollar to the bet. If I won then the number of losses started back at 0.

I decided to make exactly 250 bets. And record my cash at the end of the run.

I did this 25 times.

My average return for each "trip" to the casino was 259.12%

2 times, I went home no money.

2 other times, I went home with less than 50 dollars.

Subtracting my orginal $50 seed money, I netted $3189.

Can you attach the excel doc?

[preflop]

I tried this in Excel.

I started with $50. Each roll, I bet one dollar. For every 35 consecutive times I lost, I added another dollar to the bet. If I won then the number of losses started back at 0.

I decided to make exactly 250 bets. And record my cash at the end of the run.

I did this 25 times.

My average return for each "trip" to the casino was 259.12%

2 times, I went home no money.

2 other times, I went home with less than 50 dollars.

Subtracting my orginal $50 seed money, I netted $3189.

I would double check your spreadsheet. I wrote a quick simulation program. I was down $548 after 25 "visits" to the casino.

I then modified the simulation to match the OP. Using the quit while i was ahead methodology

first, in the quit while i'm ahead methodology here are the rules i used.

1. I "visited" the casino with $50.

2. I played 1 number on the wheel for a $1 bet (paying 35 to one if i hit)

3. I bet until I had either 0(zero) money, or $51 or more, or i reached a max bets value (i used 250 - although this never hit). and then left the casino

4. My profit was my ending amount - my starting amount(50)

the results after 100 "visits" I was down $20. So while i'm not getting rich, I was playing very cheaply and I figure I made that up in free drinks from the house

I would imagine you could go on runs where you were up a little, but I doubt over time this style would result in any appreciable gains.

The other problem with this approach is it requires walking away whenever you get ahead, which few gamblers ever do.

I'll stick to poker, where I know I can make money.

Here are the run results if you are interested:

-50

12

9

-50

19

-50

1

9

-50

3

7

13

29

-50

13

8

32

-50

-50

24

16

16

26

33

29

-50

12

22

-50

-50

18

-50

32

32

-50

9

11

24

31

18

31

7

31

20

28

35

-50

34

31

11

22

35

30

20

20

-50

21

10

18

-50

14

-50

28

1

23

5

-50

-50

7

10

24

5

-50

15

14

14

8

20

-50

24

12

35

-50

28

-50

10

-50

11

11

33

-50

2

14

8

-50

28

7

-50

7

-50

The code in visual basic if you are interested :

Public Function iSpinWheel() As Integer

Dim fRandNum As Single

Dim fResult As Single

Randomize

fRandNum = Rnd()


fResult = Round((fRandNum * 38), 0)

If fResult = 0 Then

    fResult = 38

End If


iSpinWheel = Int(fResult)


End Function




Public Function cSimulateVisit(cStartAmount As Currency, iLuckyNumber As Integer, iMaxBets As Integer) As Currency



Dim i As Integer

Dim cCurrentBet As Currency

Dim iResult As Integer

Dim cAmountOnHand As Currency


cCurrentBet = 1

cAmountOnHand = cStartAmount

For i = 1 To iMaxBets

    iResult = iSpinWheel()


    If iResult = iLuckyNumber Then

        'we have a winner

        cAmountOnHand = cAmountOnHand + (35 * cCurrentBet)

    Else

        'we did not win

        cAmountOnHand = cAmountOnHand - cCurrentBet

    End If



    If cAmountOnHand <= 0 Or cAmountOnHand > cStartAmount Then

        Exit For

    End If


Next

cSimulateVisit = cAmountOnHand - cStartAmount


End Function

[bonanova]

There are 38 outcomes and guessing [the outcome correctly] pays $35.

So on average you lose about a nickel on a dollar bet.

So it's dumb to play for a long time: the law of large numbers sends you home broke.

I could win on the first spin of the wheel. Or the second. Or the

third, and so on. By the 38th spin, I should expect to have won

at least once. But don't the odds favor the case where my one

win will occur before the 35th spin? If so then there exists a

moment in time when I will be ahead. And I win, just by quitting.

So doesn't that mean I can win, in the short term at roulette? http://brainden.com/forum/uploads/emoticons/default_wink.png' alt=';)'>

And if that works, couldn't I do it again, and win in the long term?

An answer should include an idea of the number of spins I

should expect to wait before winning once.

In a previous post, I described a town where child-bearing couples stopped having children immediately after the birth of their first son, and I asked for the resulting gender distribution of the town's children. Some people thought the odds in this case favored male children.

That each set of parents stops having children at a predetermined point has no effect on the gender distribution of their children. The initial supposition was that each birth had equal likelihood of being a boy or a girl. That probability, and nothing else, determines the overall distribution. After the first birth, the children are all-male or all-female. But you can't make money betting they will be all-male then stop betting after the first birth.

In roulette the expected number of spins to win is 38 - one for each outcome. After your first win you are ahead or you are behind. But you can't make money if you stop betting at that point. The odds favor your being behind. There are more cases of winning after the 35th spin than before.

No matter what stopping algorithm you use, eventually it will cost you $38 to win $35.

[Guest]

So I was reading in Wikipedia about expectation values, and they analyzed roulette.

There are 38 outcomes and guessing it pays $35.

So on average you lose about a nickel on a dollar bet.

So it's dumb to play for a long time: the law of large numbers sends you home broke.

Which reminds me of the guy in desperate financial trouble who

went to Vegas to gamble. On the way, he asked for supernatural

intervention: "Lord, help me to at least break even. I need the money."

But I digress.

I could win on the first spin of the wheel. Or the second. Or the

third, and so on. By the 38th spin, I should expect to have won

at least once. But don't the odds favor the case where my one

win will occur before the 35th spin? If so then there exists a

moment in time when I will be ahead. And I win, just by quitting.

So doesn't that mean I can win, in the short term at roulette? http://brainden.com/forum/uploads/emoticons/default_wink.png' alt=';)'>

And if that works, couldn't I do it again, and win in the long term?

An answer should include an idea of the number of spins I

should expect to wait before winning once.

The odds do favor the case where your one win will occur before the 35th spin. But whether this strategy pays in the long run depends on how much you expect to win. If you win on the first spin you leave $34 ahead. If you win on the 35th spin you leave breaking even. If you lose on the 35th spin you leave $35 behind.

It turns out that you would win by the 35th spin about 61% of the time, and on average you would win about $19.68. You would lose all $35 about 39% of the time. Your expected net earnings per trip would then be about -$1.65. So this strategy would lose money in the long run. (In other words, the number of spins you should expect to wait before winning once is 38.)

P(L) = (37/38)^35 = .39

P(W) = 1 - P(L) = .61

E(net|W) = Sum s = 1 to 35 { (1/38)(37/38)^(s-1) / P(W) * (35-s) } = $19.68

E(net|L) = -$35

E(net) = E(net|W) P(W) + E(net|L) P(L) = -$1.65