Roulette

[henrikrh]

Ages ago I read a little bit on this forum about roulette, and In seem to recall a strategy where one bets on all-black or all-red and then raises tehir bet everytime they lose, eventually winning the 50-50 chance, and making up for their lower bets, then going back down and startign again.

I dismissed this since roulette is regarded as a -ev game, no matter the strategy it can't be profitable I figured.

Recently I've been thinking about this alot, and I can't seem to figure out why that strategy doesn't work. It can't possibly work or everyone would use it. Can anyone enlighten me?

Sorry if this has been discussed, didn't see anything bout it on the first page.

[Komodo]

It doesnt work because you dont have unlimited money to double the bet forever when you lose.

[punter11235]

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It doesnt work because you dont have unlimited money

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Its not true. It doesnt work even if you had unlimited money. While its pretty obvious for anybody who have some contact with mathematic its not for average person. You can try to search a thread "less obvious martingale fallacy" with more discussion about the topic.

[Komodo]

</font><blockquote><font class="small">Svar till:</font><hr />
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It doesnt work because you dont have unlimited money

[/ QUOTE ]

Its not true. It doesnt work even if you had unlimited money. While its pretty obvious for anybody who have some contact with mathematic its not for average person. You can try to search a thread "less obvious martingale fallacy" with more discussion about the topic.

[/ QUOTE ]

No, its not obvious. Could you explain?
I dont see how you can lose with this system if you have unlimited money as long as the winning chance is bigger than 0

[bpb]

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It doesnt work because you dont have unlimited money

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Its not true. It doesnt work even if you had unlimited money. While its pretty obvious for anybody who have some contact with mathematic its not for average person. You can try to search a thread "less obvious martingale fallacy" with more discussion about the topic.

[/ QUOTE ]

No, its not obvious. Could you explain?
I dont see how you can lose with this system if you have unlimited money as long as the winning chance is bigger than 0

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Because if you can assume infinite bankroll, the casino can assume an infinite string of losses.

[Komodo]

</font><blockquote><font class="small">Svar till:</font><hr />
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It doesnt work because you dont have unlimited money

[/ QUOTE ]

Its not true. It doesnt work even if you had unlimited money. While its pretty obvious for anybody who have some contact with mathematic its not for average person. You can try to search a thread "less obvious martingale fallacy" with more discussion about the topic.

[/ QUOTE ]

No, its not obvious. Could you explain?
I dont see how you can lose with this system if you have unlimited money as long as the winning chance is bigger than 0

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Because if you can assume infinite bankroll, the casino can assume an infinite string of losses.

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I make 4 assumtions here:
1, We have unlimited money.
2, We have unlimited time to complete each session.
3, The casino also also got unlimited time and money.
4, Winning chance is bigger than zero, so a string of losses can never be infinite like you suggest.

Every session will land on +EV after completion under these conditions. These conditions are 100% impossible to apply of course.

[Mr. Curious]

Betting on red or black is not a 50-50 proposition because there are one or two green spots on the wheel. Your true odds are something like 48% (red) - 48% (black) - 4% (green), so the house would need to pay you 52-1 on red or black to make the bet even odds.

This is the same as someone who bets the same number every time. There are 38 spots, but you only get 35 to 1 on your money, so if you hit 1 time out of 38, you will only have 36 to show for it.

[henrikrh]

Yes but you don't need an infinite bankroll, a 48% chance will come up pretty darn quick.

Bet $10 on black, lose $10
Bet $30 on black, lose $40
Bet $90 on black, win total of about $80 (not sure what the payout is on a colour bet, never actually played, see the question at the end)

Other than the fact that variance might eventually lose you 5 or 6 in a row it seems bullet-proof even though coming an expect value view point it just can't be.

I want the to see the flaw in the system clearly.

Whats the pay put on black or red? 1-1? Or worse?

[TM1212]

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Yes but you don't need an infinite bankroll, a 48% chance will come up pretty darn quick.

Bet $10 on black, lose $10
Bet $30 on black, lose $40
Bet $90 on black, win total of about $80 (not sure what the payout is on a colour bet, never actually played, see the question at the end)

Other than the fact that variance might eventually lose you 5 or 6 in a row it seems bullet-proof even though coming an expect value view point it just can't be.

I want the to see the flaw in the system clearly.

Whats the pay put on black or red? 1-1? Or worse?

[/ QUOTE ]

eventually you will see 10 striaght blacks or reds, and your bankroll is gone, simple ev

[The Goober]

This has been dicussed many, many times on here and lots of other places. Do a search for "martingale" either on here or on google.

The short answer is that even though intuitively it seems really "unlikely" that you will hit a long enough losing streak to wipe you out, eventually you will. With a simple bet on black, you have a medium chance of winning a bit, and a slightly larger chance of losing a bit. Employing a martingale strategy gives you a large chance of winning a small amount, and a small chance of losing a very large amount. In the end, you still get the exact same EV.

The fallacy of the martingale is essentially that we humans have a hard time understanding very small numbers - in our mind, a 0.0001% chance of losing seems "impossible", but its not and it will happen.