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I was recently given a promotion by an online gambling website for £10 free credit which I could withdraw after having wagered at least £200 in total. What's the best betting strategy for maximising my EV when playing roulette? The non trivial feature of this problem being that, if I lose my initial £10 I cannot continue gambling to achieve the long term expected value of my bets.

I decided at first I was continually betting £0.50 on red. Reasonably quickly I decided to go with a different strategy of betting £0.50 on red and £0.50 black each time meaning I'd break even as long as the ball did not land on 0.

How does one calculate the EV in each scenario, simply betting on a single a colour, or betting on both black and red. It's trivially simple if you have unlimited money, and in both cases the expected value is 36/37~0.973 (since we're playing european roulette with just one green square) however in my case, if I go through a bad streak and lose my £10 I can not continue. Therefore it seems the EV becomes partially dependent on the variance of the strategy.

Secondly what produces a higher EV for both the strategies above, a large amount of small bets, or a smaller amount of larger bets. I.e. one thousand £0.50 bets or one hundred £5?

My gut instinct says that because of this feature that once I'm down a total of £10 I can't continue makes me think that smaller bets are produce a higher expected value but I don't know how to prove it.

Edit: EV stands for expected value, however I may be using the term incorrectly. To explain what I mean here's a simple example of two strategies which are easy to calculate. One where you bet £5 on red and £5 on black every bet, your mean returns is £10*(36/37)^20=~£5.8 i.e. you need to avoid hitting a 0, 20 times.

If instead you bet all the money you have on red every time, you need to win 5 times in a row before you've wagered at least £200, so your mean returns is £320*(18/37)^5=~£8.7 which is greater than £5.8. This is what I meant by expected value. So not the expected value of an individual bet, but the expected value of your winnings after betting at least £200.

Or you could think of it as, if a million people all got the same promotion and then pooled their winnings and shared them out between them, whats the best strategy they should all follow.

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  • $\begingroup$ EV = expected value? Usually the expected value with these games is independent of the strategy you choose. But it seems to me you are asking something else: what strategy maximizes the change of having bet a total of 200 before going bankrupt. Do I understand you correctly? $\endgroup$ – Marc Paul Jan 2 '18 at 20:37
  • $\begingroup$ Almost, I've added an example of what I mean above. $\endgroup$ – generic purple turtle Jan 2 '18 at 21:08
  • $\begingroup$ As with any betting strategy, run a simulation before placing any money. $\endgroup$ – Weather Vane Jan 2 '18 at 21:17
  • $\begingroup$ That was gonna be the next course of action haha :) $\endgroup$ – generic purple turtle Jan 2 '18 at 21:18
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    $\begingroup$ To clarify: if you bet £10 on red and win, will you get £20 back? If so, your bets can be on a winning streak £10, £20, £40, £80, and £50 (with a total of £200). Where did you get £320? $\endgroup$ – karakfa Jan 2 '18 at 21:43
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The expected return on a total bet of $200$ is $200 \cdot \frac {36}{37}\approx 194.59$ That does not change whether you bet one big lot or many small lots, nor whether you bet high odds things like one number or even money things like red/black. What can change is the chance you are down $10$ at some point in the run. If you make many small bets at even odds the dispersion of results will be reduced, so that is what you should do to have the best chance of getting some money out of the deal.

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  • $\begingroup$ So I've included an example in the original question to explain what I mean. I don't want the chance of getting some money back, I want to maximise the average amount of money I'll get back. So in the above example betting £5 on both gives me a 58% chance I'll get some money back, while betting all my money every time only gives me a 2.7% chance of getting any money back, but increases the average amount of money won. So this strategy has a higher EV if you will and would be better if an a large amount of people all got the same promotion and then pooled there winnings. $\endgroup$ – generic purple turtle Jan 2 '18 at 21:14
  • $\begingroup$ The reason the doubling strategy has a more negative expected value is because you bet too much-you bet a total of $310$. If you follow the doubling strategy but on the last bet (when you have bet $150$ so far) you only bet $50$ you will get back to the same expected value. $\endgroup$ – Ross Millikan Jan 2 '18 at 23:49
  • $\begingroup$ The expected value of the doubling method is higher than betting on both black and red at the same time unless I've made a mistake in the maths above, but I don't believe I have. When you use the slightly modified doubling method the average returns becomes the probability that you get the probability that you get 5 reds in a row (18/37)^5 multiplied by the the amount of money you'd win which is 210 plus the probability of 4 reds and one black which is again (18/37)^5 multiplied by the amount of money you'd win, this time 110. (((18/37)^5)*210)+(((18/37)^5)*110)=(((18/37)^5)*320)=~8.7 $\endgroup$ – generic purple turtle Jan 3 '18 at 0:26
  • $\begingroup$ Interesting to note the two modified versions of bet it all scheme, have the same EV. Which makes sense because the odds of the last bet are the same regardless of how much your betting, and it its result doesn't effect your ability to place any future bets, as its the last bet. $\endgroup$ – generic purple turtle Jan 3 '18 at 0:35
  • $\begingroup$ It is much simpler than that. You lose $\frac 1{37}$ of every bet. If you bet exactly $200$ you will have an expected value of $10-\frac {200}{37}$. That is as I said, the EV will be the same for any sequence of bets. The thing that changes is the variance. Many small bets have small variance, a few large ones have large variance. High odds bets, like one number, have high variance. $\endgroup$ – Ross Millikan Jan 3 '18 at 1:43
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The goal is to have hit 200 total bet, with the highest expected payout when you do so

So it actually does make a difference:

You can hit 9.466764061$ expected return like so

Bet 10 on a 1/37 shot (one of the 36 numbers but keeping in mind the 0)

If you win, you've got 360, you just then need to be on it once at: 36/37 (a 10 chip on each number) to hit your 200 requirement.

This has E(V) = 1/37 * 36/37 * 360 = 9.467...

Edit, you don't need to bet your whole remainder on the survival bet so actually:

E(V) = 1/37 * (36/37 * 190 + 170) = 9.590942294

I ran this for lots of other starting bets, the chance of getting them, vs how many 'survival' bets I'd need to hit the 200 total ever bet requirement (which is key) and none did as well as this.

I didn't consider any fancy stuff because we're so close to the maximum possible anyway.

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  • $\begingroup$ Welcome to MSE. Didi you read it before posting it? $\endgroup$ – José Carlos Santos Jan 3 '18 at 10:37

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