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.

[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).

[jason1990]

I just started covering martingales in a class I'm teaching and this betting strategy was an example. Also, coincidentally, I was just discussing this with my wife at dinner last night.

I had a friend who was also convinced of this. She has a masters degree in mathematics. I was also unable to convince her it was a bad idea. Strange how stubborn some folks can be.

Here's how I tried to explain it. Suppose for simplicity that you start with $127 and your initial bet is $1 on black. If you lose 7 times in a row, you will go broke. If you do not, then you will win $1.

First, let's pretend there are no green numbers, so that the game is fair and your chance of winning is 1/2. Then the probability of losing 7 in a row is 1/2^7=1/128. So your EV is

1*(127/128) + (-127)*(1/128) = 0.

This is what you expect from a fair game. But notice the nature of this expectation: you have a high probability of winning very little and a low probability of losing a lot. It's like a "reverse lottery ticket".

To make an extreme example, I asked my friend if she would be willing to play the following game: we pick some lottery numbers. If they don't win, I give her $1. If they do win, she gives me everything she owns -- her house, her car, all her money, everything. She's almost certain to get a dollar out of this bargain (and in fact the game for her is +EV), but why put everything on the line for a dollar?

Alright, now put the green squares back. Now the probability of losing a single bet is 20/38=10/19. So the probability of losing 7 in a row is (10/19)^7. This makes your EV

1*(1-(10/19)^7) + (-127)*(10/19)^7
= 1 - 128*(10/19)^7
= 1 - (2^7)*(10/19)^7
= 1 - (20/19)^7.

So not only are you risking a lot to win a little, you also have a negative EV of -43 cents. You should be able to work out that if you start with enough of a bankroll to lose n in a row, then your EV is

1 - (20/19)^n.

In other words, as your bankroll increases, your EV decreases. To some, this might seem counterintuitive, since you would think that this strategy works better as you increase your bankroll. But it doesn't. In fact the limit of your EV as n goes to infinity is minus infinity. In other words, from a certain perspective, with an infinite bankroll you have an EV of minus infinity.

This also suggests that the optimum bankroll for this strategy is n=0, i.e. don't play at all!

[Jman28]

Thanks for the replies.

This is like the example I came up with last night, so I'll see if it works today.

-Jman28

[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

[blank frank]

[ 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.

[/ QUOTE ]

If the math isn't convincing him, he may not be seeing the limit. That is, he thinks each bet is -EV, but eventually he wins and has +EV. The thing is, the only way to be sure you win is to play forever. In reality you are limited, either by your bankroll or the table limits. So there is always a chance you won't win even once. Now, if you can work it so you have a lot of bets before you hit the limit, that chance is really small. However, the ammount you have to risk is really big, which balances things out and keeps you at -EV.

[CMonkey]

Yeah, I know; people get hung up on the fact that L^n approaches 0 as n increases and so they think they can't lose in the long run. What they ignore is that L^n never actually equals zero so there is always some chance of losing after n trials. Also, they ignore the fact that 2^n grows to infinity as n increases.

Since EV is determined by the (Outcome)*(Probability of that outcome occurring), the two terms, (2^n) and (L^n) are competing against each other. (2^n), the amount you are risking, is racing off to infinity and (L^n), your overall chance of losing, is racing towards zero. In roulette, the casino stacks the odds so that L^n decreases more slowly than 2^n increases (by making L > 0.50) to such a degree that you can never reach a situation where L^n is small enough to make up for the ever-increasing 2^n, not to mention all the money you lost on previous trials. So you wind up with an -EV situation if you use the Martingale system. Of course, this feature of the Martingale system in roulette is not unique; all betting systems in roulette are -EV.