MVee
MVee
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March 12th, 2017 at 11:53:58 AM permalink
A good friend of mine has recently discovered roulette. It has become something of an obsession for him. He believes the martingale approach can't possibly lose in the long run. So I'm working very hard at making him understand the dangers of his newly found strategy.

Basically, his approach is to bet on the last number that has appeared in hopes that it repeats immediately.

So for example, this series of ten numbers would be a winner for him, as number 35 repeats in two successive spins:

34 - 22 - 8 - 5 - 18 - 19 - 3 - 21 - 35 - 35

In discussing roulette odds together, he and I were wondering what are the odds that over a given number of spins, no number will repeat in successive spins. To be more precise, how likely is it that no number will repeat successively over a 300 spin sample?

Thanks very much!
billryan
billryan
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March 12th, 2017 at 12:18:07 PM permalink
How is he betting?
18 comes in. He bets 1 on 18. 23 comes in, he loses one and best what?
If he is betting one unit on the last number and doesn't go to two units until he's lost 38 straight times, it's not a horrible system. Not a winning system and boring as Hell but probably cheaper than doubling up after every loss.
The difference between fiction and reality is that fiction is supposed to make sense.
MVee
MVee
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March 12th, 2017 at 12:44:31 PM permalink
If 18 comes in, he bets 1 on 18. If 23 then comes in (bet lost), he bets 1 on 23. On a European wheel, this continues for 35 spins until a win. On spin 36, if the same number has not appeared in two successive spins,he doubles the bet (2 units) and continues at this amount per bet for another 35 spins. If he still has not seen a successive repeat after the 70th spin, he doubles again (now 4 units - hence the martingale) for another 35 spins. This continues until a new bankroll high is reached, after which he resets to 1 unit.

The martingale can get out of hand, of course, so a stop loss is definitely necessary.

He and I have simulated this with some basic programming using Google Sheets. Our simulation is nearly at 60,000 spins, with a stop loss set at spin 210 (-2204 units - so the martingale is now at 32 times the base bet), and our result is +6151 units.

My guess is we just got lucky. My buddy believes we have just discovered the road to long term riches...

Hence the reason for my original question: What are the expected odds of no two numbers repeating successively over a given number of spins (in our case, 210)?

Thanks for your time and interest.
charliepatrick
charliepatrick
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March 12th, 2017 at 3:43:22 PM permalink
The chance of the last number repeating is 1/37 (assuming UK single zero). The chance of the 2nd spin not winning for you is therefore 36/37. For this to happen 210 times is POWER(36/37,210) = 0.003171 i.e. it will happen one time in about 1 in 315 (for US the figure is more often at 1 in 270).

It is about 40% that you will not see a losing Martingale in 60k spins (60k/211 trials), so you have been lucky.
MVee
MVee
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March 12th, 2017 at 4:16:00 PM permalink
Got it. Thanks very much!
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