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My friend has this plan where he implies that it's impossible to lose, as long as the odds of winning are 50/50 on each bet. His idea is that basically you keep doubling your bet until you win and then start over again.

So for example, you bet 1 dollar and you lose, your net profit is now -1 dollar. Now you double your bet to 2 dollars and you lose again so your net profit is -3 dollars. Now you double your bet to 4 dollars and you win. This means you gain 4 dollars and now your net profit is 1 dollar. So you've made a profit. Now you start again. The reasoning here being that it is highly unlikely for you to lose a 50/50 toss x number of times in a row.

My counter-argument here is that basically if you go in with 50 dollars with the aim of doubling up to 100 dollars, you have the same odds of winning if you do one bet of 50 dollars or the technique outlined above. I cannot wrap my head around explaining this issue in a clear manner though, so maybe you wonderful folk at Mathematics can help!

Oh and I've pointed out that he uses gamblers fallacy in very obscure way, as he insists you need to go back to betting 1 dollar once you've won. This appears to be an obscure case of gamblers fallacy to me as it implies there is some hidden force which are changing the odds on each individual coin toss.

JaTochNietDan
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    Because you don't have unlimited amount of money. And if you did, you wouldn't be playing the game in the first place! – user2345215 Mar 07 '14 at 23:58
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    You will almost surely (e.g. with probability $1$) make a profit, but that doesn't mean that you're guaranteed to make a profit. Regardless, because of house limits, you can't double your bet indefinitely. –  Mar 07 '14 at 23:58
  • Casinos and lotteries are huge money makers because odds are a bit less favorable than 50/50... – vonbrand Mar 07 '14 at 23:59
  • I'm aware of the fact that casinos and lotteries are not fair games. I wish to assume that the game is in fact 50/50 and therefore the odds of winning via this method are in your favour. What doesn't make sense to me is how it can possibly be 50/50 if you can actually game it and have a more favourable outcome. – JaTochNietDan Mar 08 '14 at 00:01
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    I don't think 50/50 chance is really relevant. You could just bet more to cover your expenses each turn anyway and wait until you win. Also related http://math.stackexchange.com/questions/83904/on-martingale-betting-system You may want to read the answer. – user2345215 Mar 08 '14 at 00:03
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    You might find it interesting to compute the average amount you are up/down during this process of winning. i.e. If the game terminated at a random point, what would be the expected outcome? After all not only are resources limited, time is not unlimited either. – Mark Bennet Mar 08 '14 at 00:06
  • My friend did this too! He actually tried it out in a casino (roulette) , and he made around 1000 dollars using 100. But he did something different (and very important) to what your friend says, he kept on betting on one color (red, didn't change his choice). He said that even when black appeared, it works, because it's highly unlikely that black turns out continuously for more than 3 times or so. I hope you see that you don't need an unlimited amount of money after all. – The Artist Jan 27 '15 at 12:36

4 Answers4

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Let's play out your specific scenario: you start with \$50 and want to stop at \$100. That means you need to play your sequence (double until you win) 50 times without busting.

However, if your sequence begins with six losses, then you are busted, because you just lost $1+2+4+8+16+32=63$ dollars, and your system requires you to have \$64 left to bet. Six losses in a row happens $1/2^6=\frac{1}{64}$ of the time.

To avoid 6 losses in a row, happens $\frac{63}{64}$ of the time, and you need to do this 50 times, so $(\frac{63}{64})^{50}\approx 0.455$. Hence, if all you need to do is avoid six losses, there is about a 45% chance of "winning", i.e. getting to \$100.

However, there are other circumstances when you bust; at the beginning even five losses in a row will bust you, since $1+2+4+8+16=31$ dollars, and you need to bet \$32. Hence until you get to \$63, you need to avoid five losses in a row. Thus, the true answer is $$\left(\frac{31}{32}\right)^{13}\left(\frac{63}{64}\right)^{37}\approx 0.370$$ Thus your friend's system has a 37% chance of getting to \$100, and a 63% chance of getting to \$0. You are much better off just betting the whole \$50 in a single bet.

vadim123
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    Hmm. You seem to suggest that if I lose five times in a row from the start, leaving me with $19, those nineteen dollars are mysteriously sucked into a vortex because I don't know what to do with them. Isn't it simpler to just argue that since every permissible bet has an expected value of zero, any combination of bets will also have an expected value of zero, and so every strategy is equally unhelpful? – Ben Millwood Mar 08 '14 at 00:53
  • Please read my solution again. I'm not treating bets singly, but as sequences of losses followed by a win, as per the strategy laid out. It is not true that every strategy is equally unhelpful, because the game has specific endpoints: 0 and 100. – vadim123 Mar 08 '14 at 02:09
  • Okay, I'll stick to my simpler comment for the time being: your losing scenarios, e.g. losing the first five bets, do not leave you with $0 as you claim. After your first five losses you have $19, at which point it is not specified what you do. I believe this discrepancy is causing you to substantially underestimate the expectancy of the given strategy. – Ben Millwood Mar 08 '14 at 02:24
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    @BenMillwood Upon testing this theory by writing a program in C#, I found that it confirms what I suspected if we add in the clause to go all in when the player does not have enough left to bet double. This will result in the user doubling their money 50% of the time. I tested this over a few hundred thousand games and they won 49.9% of the time. – JaTochNietDan Mar 08 '14 at 17:05
  • That's nice, but that's not the question that was asked. With the strategy given, there is only a 37% chance of getting to $100. – vadim123 Mar 08 '14 at 19:37
  • @vadim123: Please tell me what happens after you lose for the fifth time in a row and have $19 left. Either you do not continue playing, in which case $0 and $100 are not the only possible outcomes, or you do continue playing, in which case it's possible you'll go on to win, and your 37% is an underestimate. – Ben Millwood Mar 08 '14 at 19:51
  • @vadim123: there is a symmetry argument which shows you can play any strategy in reverse (e.g. play with "negative dollars") and so any strategy with a below-average payoff automatically gives rise to one with an above-average payoff. – Ben Millwood Mar 08 '14 at 19:59
  • Fascinating stuff! I had a hard time believing this (not that Math needs faith), so I coded this in Java - and yes, it's about a 37% chance only! -> https://github.com/andrejss88/martingale/blob/master/src/main/java/App.java – Andrejs Mar 29 '18 at 21:11
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Can we change the scenario to a plausible one to make a comfortable living on?

Let's say you had $\$50,000$ and wanted to make $\$100$ every day. What are the chances you could go 20 years making $\$100$/day without busting? Important to note that you increase your wealth by $\$100$ every successful day.

To make it even more realistic, can you decrease the odds from 50/50 to the actual odds of a casino game? From a quick search, I found that you can get 49.3% odds of doubling-up via Craps (http://www.betus.com.pa/sports-betting/guide/craps/best-craps-strategy-for-winning-in-the-short-term-2010-09-18/). At 49.3% chance of winning each turn (doubling your bet), what are the odds of successfully winning \$100 every day for 20 years if you have a $50,000 bankroll to support the losses?

TonyK
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Max Davis
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Thanks for a detailed explanation @vadim123. for the said strategy where the multiplier is 2, if we start from $2 instead of 1 so that we would increase the winning by 2 and decrease the required winning times from 50 to 25 - more risky; we still have 0.938 (15/16) chance of winning PER SEQUENCE. From your example above, we have 0.984 (63/64) chance of winning PER SEQUENCE.

For simple words to answer the question "Help me understand why it's flawed". As per @vadim123; the chance of winning PER SEQUENCE is ~0.9** which is high, but we forget that we must do this ~0.9** many times depending on our goal. Making the probability less and less, as with the example of vadim123 31/32 is high but trying it many times lessen the chance.

It'll be exciting to think when we use multiplier of 3. Alright!! :)

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The simple solution would be to sit and wait until 5 loses come. This will take statistically 32 turns. Then use this betting system (Assuming you lose) £5 £12 £30 £60 £70. At this point you have a 1/1024 chance of losing. £150 £200 £300 £500 £700. At this point you have a seriously low chance of this happening and when it does you'd be extremely likely to be rich already. You've risked £2027 however the chances that your previous winnings will cover this loss are 100.000s in your favour. Always stay with the same strategy of playing the game you're playing. For example: rollette. If you start on black stay on black.

With your friends strategy you will always be playing to gain £1 profit. By his tenth loss he has put the bet on of 512 losing £1023 for that £1 profit WHAT THE HELL IS WRONG WITH HIS LOGIC. It gets worse by his 15th hand he would be placing £16384 and has already lost 16383 so when that fails he has lost 32767 all for chasing the £1 profit. This is just plain stupid. Casinos have minimums and maximum betings for a reason and that is to put them in favour to win.

Now I'm not encouraging ANYONE to gamble but if you're going to do it regardless then please use mine over the original strategy. Casinos and so on are winners for a reason! Remember that. Bye for now and if you want to email me your thoughts please email andytomkins11@gmail.com p.s, my maths maybe slightly out as I done them in my head.

Andy
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