Help me convince friend that Martingale Betting Strategy Doesn't work

[Jman28]

Hey guys.

So, I just had a conversation with a friend of mine about his 'roulette betting strategy' which is the martingale strategy. (bet 1, double if you lose till you run outta cash. Start over if you win)

He's a bright guy, but has no gambling experience and I just couldn't convince him that it is not a winning strategy. I need some numbers help.

Can anyone help me get a mathematical proof that will convince him that he will not win, even with a $1 million bankroll? I tried explaining it with words, but I need some numbers.

It's weird cause he definitely understands probability and mathematics fairly well, but is convinced that martingale can beat the house.

Thanks for any help.

-Jman28

[reubenf]

I'll give you some numbers. In the 60 years there have been casinos in Vegas, 0 people have busted the casino on roulette.

Is it okay if I pulled the numbers out of my ass? [img]/images/graemlins/laugh.gif[/img]

[reubenf]

Ask him what happens when you change all but two of the slots to greens.

If he can see that it doesn't work in that situation, ask him to try to figure out how many green slots need to be removed for it to work.

If he can't see why it doesn't work in that situation, BUY A ROULTETTE WHEEL.

[Jman28]

I could work out the numbers but I'm drunk. We had this argument in a bar.

[eastbay]

[ QUOTE ]
I could work out the numbers but I'm drunk. We had this argument in a bar.

[/ QUOTE ]

Plan for tomorrow:

1) Sober up.
2) Forget about it, as he no doubt will

eastbay

[wdbaker]

The first time I tried to understand this i just took an excel spreadsheet and placed formulas to generate random 0's and 1's in a column of 100, 1000, 10,000 cells long.

At the top or bottom of the columns put formula's to count the # of 0's and 1's

Each time you refresh the page you will get similiar results.

Then, and this was the hardest part for me to figure out and there may be easier ways to do it, was to count the groupings(clumping) of 0's and 1's for each column.

This will do two things, it will make you realize that small groupings are far more common than large groupings, and that large groupings do actually happen on a regular basis.

These large groupings would easily take you over the table limit. Refresh and run them over and over, they won't go away, sooner or later your going to hit a very long string of the same.

If you are just going to go for the night though and play for fun and it must be roullette, you might develop a strategy of betting or martingale that would start only after X number of 0's or 1's have clumped together as it would appear that the odds become greater that the clump will break up more quickly the larger the clump.

I would think that if you have to play roullette and must gamble that this would be your best chance of coming out a winner for a night, no guarantee's though [img]/images/graemlins/grin.gif[/img]

This game has no memory and each spin is independent of all others, so the clumping theory is really a fiction of our minds and not at all reality, the clump your betting will break doesn't have to, but could, last forever.

Thanks for listening to the babblings of a person who didn't find it important to finish grade school [img]/images/graemlins/blush.gif[/img]

One Street at a Time
wdbaker Denver, Co

[slavic]

I would think that if you have to play roullette and must gamble that this would be your best chance of coming out a winner for a night, no guarantee's though

You are correct in saying that this could give you a small win. However it can also give you a huge disasterous loss. This is not much different than the guy who always plays to get even in poker. I bunch of small wins followed by a bone crushing loss.

I've done the math on a matringale system before and your EV is actually less than if you just played the game in a non weighted fashion.

[Loci]

Wave a shiny object in front of him, throw it and say fetch. You'd be surprised at how often they'll chase, and how often they'll not recall the conversation in the thirty seconds that follow.
If you dont' have a shiny object, point at a pretty girl. If you don't see any pretty girls, find another bar.
Ez

[CMonkey]

I'm assuming you're referring to a Martingale strategy where you place a bet that pays out at even money but has odds somewhat worse than that due to the presence of 0 and 00.

Let:

B = the size of initial bet
L = the probability of losing at any one trial ( L = 20/38 for a roulette wheel with 0 and 00 )
W = the probability of winning at any one trial ( W = 1 – L = 18/38 for a roulette wheel with 0 and 00 )
EV = the expected value

After one spin,

EV = BW – BL .

However, under the Martingale system, you place a double-sized bet if you lose, so

EV = [BW – BL] + [2BWL – 2BLL] .

If you lose after the both the first and second spins, you place a quadruple-sized bet, so

EV = [BW – BL] + [2BWL – 2BLL] + [4BW(L^2) – 4BL(L^2)]

More generally,

EV = SUM(n=0,infinity) [(2^n)BW(L^n) – (2^n)BL(L^n)]
EV = SUM(n=0, infinity) [B(W-L) * (2^n)(L^n)]
EV = B(W-L) * SUM(n=0, infinity) [(2L)^n]

L > W, so B(W-L) is negative. (2L)^n is positive for all values of n. Therefore, SUM(n=0,infinity) [(2L)^n] is positive as well. Hence, EV must be negative.

Which is really just a long-winded way of saying that no combination ( B * SUM(n=0, infinity) [(2L)^n] ) of negative-expectation bets ( W-L ) can result in a positive expectation (EV).

[Jman28]

Thanks. Great response CMonkey.

[ QUOTE ]
Which is really just a long-winded way of saying that no combination ( B * SUM(n=0, infinity) [(2L)^n] ) of negative-expectation bets ( W-L ) can result in a positive expectation (EV).

[/ QUOTE ]

I tried explaining this to him, since he knows that each individual bet is -EV. It didn't work, so I'm worried that your mathematical explanation won't do it for him either.

I worked out an example with real numbers which I hope will convince him.

-Jman28