# Gambler's Ruin Quiestioned

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I've either just found a way or somebody else have told me how to hack the Roulette. If you bid on the same Even \ Odd only and constantly double the rates, in end you'll win the starting sum

1 - you've either lost 1 - or you win 2 - 1 = 1
2 - you've either lost 3 - or you win 4 - 3 = 1
4 - you've either lost 7 - or you win = 8 - 7 = 1
8 - you've either lost 15 - or you win 16 - 15 = 1 and so on

2 - you've either lost 2 - or you win 4 - 2 = 2
4 - you've either lost 6 - or you win 8 - 6 = 2
8 - you've either lost 14 - or you win 16 - 14 = 2
16 - you've either lost 30 - or you win 32 - 30 = 2 and so on

3 - you've either lost 3 - or you win 6 - 3 = 3
6 - you've either lost 9 - or you win 12 - 9 = 3
12 - you've either lost 21 - or you win 24 - 21 = 3
24 - you've either lost 45 - or you win 48 - 45 = 3 and so on

You can say, this is the Martingale system and it doesn't work because math https://en.m.wikipedia.org/wiki/Gambler's_ruin

You can switch the bets on Even / Odd after you win. Yet even shouldn't you switch them, you can win anyway

Say you put on Even and you had 15 coins

1 - you've either lost 1 - or you win 2 - 1 = 1
2 - you've either lost 3 - or you win 4 - 3 = 1
4 - you've either lost 7 - or you win = 8 - 7 = 1
8 - you've either lost 15 - or you win 16 - 15 = 1 and so on

At this moment you have won 1 coin, and you have 16 of them

You may keep betting on Even

1 - you've either lost 1 - or you win 2 - 1 = 1
2 - you've either lost 3 - or you win 4 - 3 = 1
4 - you've either lost 7 - or you win = 8 - 7 = 1
8 - you've either lost 15 - or you win 16 - 15 = 1 and so on

And after 4 rounds you have won another coin, you have 17 now and so on

I mean, you should start from 1 coin after you win, or 2 - whatever is the number of coins which you plan to win. In other words, you should increase the bets until you win more than you've lost and next start from the difference

Your chances to win on Even / Odd are 50 / 50. If you have \$1,000 dollars and write an automated script to gamble on dollar only, most probably you’ll be fine. There is a very small chance that you may loose all your money, but this is a really small chance

If each time you start from the dollar, the question is will the odds hit you before you win more or not

Sincerely,
Aenne Lee

Edited by Aenne Lee

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Not sure if this is serious, but giving it the benefit of the doubt. Have you considered that there's a zero and sometimes a double zero on a roulette wheel which doesn't pay out odd or even?

2 hours ago, Aenne Lee said:

Your chances to win on Even / Odd are 50 / 50.

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3 hours ago, Tidus said:

Not sure if this is serious, but giving it the benefit of the doubt. Have you considered that there's a zero and sometimes a double zero on a roulette wheel which doesn't pay out odd or even?

I have edited the letter aftewards

In my opinion, you should calculate the chances first, I simply don't know the math. It may be, that your chances to loose this way are smaller than your chances to win. The Gamblers' Ruin theory had been found on supposition that bets all the time increase. I say start from the difference

Intuitively, the chances to hit 4 odds in raw are 1 / 16. In other words, until you don't surpass 4 rounds only, if you stop increasing the bets, you risk a much smaller money which you eventually may get back

Furthermore, if you know the previous results, perhaps you can still win, like put on even should it be many odds. And if we talk about bet's in million dollars, perhaps you can win a million dollars

You should just play for long enough and put the bets after it have been many in raw results only

I'd limit myself to 3 rounds only as 7 is a much smaller risk and easy to make a bid number and move on with up to 1 - 2 dollars only, since the winning strategy appears to count on repeating results and put the opposite bets. Like wait for 3 evens in raw and start putting on odd. Anyway, somebody should calculate it first

Thus let me formulate the question. If you wait for 3 evens and start doubling the bets on odd will you win, if any time you start from the difference between the won and lost money only. It may be, you don't even have to increase the bets for this

And may I offer to you even better strategy - to start increasing the bets after 3 - 4 evens in raw only. Like you may all the time put on odds and keep winning and loosing about the same money, yet should it be 3 - 4 evens in raw, to start increasing the bets

And to make it better, you may even increase the bets by 1 dollar only

Let's say you start putting on odds

1 - you may either loose 1 - or win 1
1 - you may either loose 1 - or win 1
1 - you may either loose 1 - or win 1
1 - you may either loose 1 - or win 1

And by the time you could loose 4, you start increasing the bets by 2

2 - you may either loose 6 - or loose 2
4 - you may either loose 10 - or loose 2
8 - you may either loose 18 - or loose 2
16 - you may either loose 34 - or loose 2

Nay, this don't work. You should make no bids until it have been 3 - 4 evens

Yet wait. Let's say you start putting on odds

1 - you may either loose 1 - or win 1
1 - you may either loose 1 - or win 1
1 - you may either loose 1 - or win 1
1 - you may either loose 1 - or win 1

And by the time you could loose 4, you start increasing the bid to 10

10 - you may either loose 14 - or win 6
10 - you may either loose 24 - or loose 4
10 - you may either loose 34 - or loose 14
10 - you may either loose 44 - or loose 24

Nay. You should definitely make no bids until it have been 3 - 4 evens. And then likely you should put the same amount, because to double the bids may ruin you. You simply bet on much larger chances of odds and thus some average income

Yet let say by the time you could loose 4, you start increasing the bid by 10

10 - you may either loose 14 - or win 6
20 - you may either loose 34 - or win 6
30 - you may either loose 64 - or loose 4
40 - you may either loose 104 - or loose 24

Since you should double the bit, you may get ruined. Yet should you make no bids, you can win

even, the odd chances have been 1 / 2
even, the odd chances have been 3 / 4
even, the odd chances have been 7 / 8
even, the odd chances have been 15 / 16

Intuitively, you've come to 31 / 32 chances of odd. You definitely should win

Thus I recommend to make no bids until it have been 3 - 4 evens in raw and start making \$1 bids  only or any other constant amount of money you prefer. On average you should win. At once, should you increase the bids by certain number, at some point you start loosing the money and should you double or triple it, you may get ruined since you'll run out of money

In this way, your chances to win a dollar begin from 31 / 32 and your chances to loose a dollar start from 1 / 128

Edited by Aenne Lee

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Final edition

Let me write this step by step. If you bid on the even \ odd numbers only and constantly double the rates, in the end you'll win the starting sum

1 - you've either lost 1 - or you win 2 - 1 = 1
2 - you've either lost 3 - or you win 4 - 3 = 1
4 - you've either lost 7 - or you win = 8 - 7 = 1
8 - you've either lost 15 - or you win 16 - 15 = 1 and so on

2 - you've either lost 2 - or you win 4 - 2 = 2
4 - you've either lost 6 - or you win 8 - 6 = 2
8 - you've either lost 14 - or you win 16 - 14 = 2
16 - you've either lost 30 - or you win 32 - 30 = 2 and so on

You can say, this is the Martingale system and it doesn't work because math <a href="https://en.m.wikipedia.org/wiki/Gambler's_ruin">https://en.m.wikipedia.org/wiki/Gambler's_ruin</a>

Yet we talk here about the difference, just \$1 or \$2 - you should start bidding over on amount which you have won only. I mean, you should start from 1 coin after you win, or 2 - whatever is the number of coins which you plan to win

At once this method is fairly risky since you can loose all your money or reach the casino limit

Thus let's play differently. If you make no bids until it was 3 - 4 evens in raw and next start bidding on odds, you start from 31 \ 32 chances to win. In fact, you can wait for just 2 evens only, since your chances to win if you play on odds at once start from 7 / 8

Next let me formulate the question regarding the Gambler's Ruin. If you wait for 3 - 4 evens and start doubling the bets on odd will you win, if any time you start from the difference between the won and lost money only

Just make sure to place the same bids. If the bids grow arithmetically, at some point you start loosing the money and if the bids grow geometrically, you can run out of it

10 - you may either loose 10 - or win 10
20 - you may either loose 30 - or win 10
30 - you may either loose 60 - or win nothing
40 - you may either loose 100 - or loose 20

Intuitively, if you play this way 8 times, you can win 7 bets and loose just 1

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It doesn't work for the same reason as the martingale system doesn't work. You'd need infinite money to be able to withstand the effects of picking the wrong number an improbable but not impossibly high number of times.

I don't fully get your reasoning behind why you say it's different, but I can assure you there's no guaranteed system to profitably playing roulette over the long term. Also, be careful with relying on intuition with probability problems, the human brain doesn't always intuitively get them right (e.g. the monty hall problem).

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18 minutes ago, Tidus said:

It doesn't work for the same reason as the martingale system doesn't work. You'd need infinite money to be able to withstand the effects of picking the wrong number an improbable but not impossibly high number of times.

I don't fully get your reasoning behind why you say it's different, but I can assure you there's no guaranteed system to profitably playing roulette over the long term. Also, be careful with relying on intuition with probability problems, the human brain doesn't always intuitively get them right (e.g. the monty hall problem).

Yes, but if you have made no bids and it was 2 evens in raw, are chances to win if you play on odd 7 / 8

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Final final edition

Intuitively, if you play this way 8 times, you can win 7 bets and loose just 1. Yet of course, the chances may be lower if it was many odds before and at the same time, it all depends on the time span which you investigate

Thus let me rise a new question. Are chances to win if you play on odds after it was 2 evens always 7 / 8

Edited by Aenne Lee

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Errr, nope? They're a bit under 50%.

Unless I'm misunderstanding what you're saying.

Edited by Tidus

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37 minutes ago, Tidus said:

Errr, nope? They're a bit under 50%.

Unless I'm misunderstanding what you're saying.

I believe the chances you get even should be 50 / 50 at first time only. Next its 3 / 4 and 7 / 8 only afterwards. It just won't start from the same point. Nobody knows exactly why it won't start from the same point, yet it just won't because of a different experiment conditions

Edited by Aenne Lee

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Ok, last post here. You can take my word for it or not, but it's always 50/50, or actually just under because of 0 and 00, that it will land on even. Previous spins don't have any effect on future ones. The events are independent of one another.

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On 8/7/2016 at 3:34 PM, Aenne Lee said:

Thus let me rise a new question. Are chances to win if you play on odds after it was 2 evens always 7 / 8

You're kidding? I strongly suggest you stay out of casinos.

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On 8/7/2016 at 2:20 PM, Aenne Lee said:

Thus let's play differently. If you make no bids until it was 3 - 4 evens in raw and next start bidding on odds, you start from 31 \ 32 chances to win. In fact, you can wait for just 2 evens only, since your chances to win if you play on odds at once start from 7 / 8

I'm not sure I understand exactly what the strategy you're proposing is, but this doesn't seem to make any sense. Are you saying that if you wait until 3 or 4 evens in a row come up, you will have probability 31/32 of winning on the next round? This just isn't how probability works. The rounds are all independent of each other.

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