# Roulette Probabilities

I was thinking about a roulette wheel, and, to keep ths questions simple, and the numbers easy, I'm going to forget about the 0 or 00 so that there are just 36 possible outcomes.

For example, the likehood of 4 blacks coming up in a row is 1/2 x 1/2 x 1/2 x 1/2 = 0.0625, which is low. The probability of 5 blacks coming up in a row is even lower - 0.03125. This can be compared to tossing five coins at once, and calculating the probability of them all coming up Heads.

This type of calculation can be used to predict the probabilty of a losing run when doubling up bets in the Martingale system, the limitations being the the minimum and maximum bet limits imposed.

Consider a table with a min \$1 and max \$512 bet. This would give you 10 chances to double up before you lost:

1,2,4,16,32,64,128,256,512

Question: I calculate the probability of losing on a run like this to be 0.5^10 = 0.000976563 which strikes me as being low. It would mean in 1024 spins, there is one chance that this losing run will occur. Is this correct?

Arthur_Wood

membership
Create an account to see this answer
Signing up is free. No credit card required.

According to that scheme, with say 50 spins, it is unlikely that a losing run will come up. How often would one win \$1? The minimum number of wins would be once every ten spins, and the maximum number of wins would be every spin, or 10/10 spins. So lets say that on average a \$1 win occurs every 5 spins. 50/5 = \$10. I think the standard bell shaped probability curve applies here, with 5 in the center? Is this correct?

If so, the odds would be stacked in favour of the player winning \$10 bucks for 50 spins only. The more spins of course the more you will win, but at the same time the chance of a big lose coming up increases.

I'm neglecting the 0 / 00 for now.

If you take out the 0 and the 00 then the odds become exactly fair and your expected return is \$1 per \$1. Yes, you would only lose 1 time for every 1023 times you win; however, when you win in martindale you gain \$1 (try it) and when you lose you lose 1+2+4+8+16+32+64+128+256+512 which is \$1023. So for every 1024 times you play you win \$1023 and lose \$1023.

Now let's put the 0 back in. Your odds of losing are now 19/37. ^10 gives .001275... You are now losing once out of every 784 runs. So for every 784 runs you lose \$1023 and gain \$783 for a net loss of \$240
Here's the main thing. The spins are independent events so no betting system will do any better than random guessing. Any reasonably intelligent person who took a good statistics class will tell you this.

If there are no 0/00 then you will average getting exactly what you started with. If there is a 0, you average losing 2.7% of all your bets added together. No matter what betting style you use, these numbers  will hold. Guaranteed.

Leaving the zero out, as I specifically requested in the opening question....

If there are no 0/00 then you will average getting exactly what you started with. If there is a 0, you average losing 2.7% of all your bets added together. No matter what betting style you use, these numbers  will hold. Guaranteed.

No. The odds only become even if you play enough times for the 1/1024 statistics to start to show up. But my point is that if you play just a few times, say 50 times max, then the odds are in favour of the player.

The average would only apply after a lot of spins, may be 100,000 or so. Then the pattern of 1/1024 losses would start to show up.

SOLUTION

membership
Create an account to see this answer
Signing up is free. No credit card required.
So, yes. Your chance of walking off with \$50 is pretty high. But you lose A LOT in that 5%. So if you do 50 runs often enough it comes back to even money. What's important is not your chance of a gain but your expected return. Which without the 0/00 is \$1 per \$1. If everyone goes out and plays 50 times, most will gain a small amount, a few will lose a great deal. This is why casinos make a lot of money. (Slots and lotteries work the other way around).

In every case for every casino game, your expected return is less than \$1 for every dollar.
>>But my point is that if you play just a few times, say 50 times max, then the odds are in favour of the player

This is not true. I know it's a hard concept to really grasp, but the small chance of great loss compared to the high chance of small gain in this case (since the loss*loss_chance = gain*gain_chance) make the odds even and of course with the 0/00 the odds are against the player) even for one run of the martindale method (or any other).

Looking at another scenario, again, ignoring the zeros. This time looking at high, medium and low numbers. Betting on one of these groups means you have 2/3 chance of losing, and a losing run of 10 would have the probability of 2/3^10 = 0.017342, or 1/57 spins. Not very favourable.

But what if you bet a chip on two groups? Then you'd have 2/3 chance of winning, and a 1/3 chance of losing. Also it would mean you would need to bet 2 to win 1.

If you lose, then you need to get back the money you lost (2), plus that which you wanted to win (1). So next win needs to be 3, and to win that you'd need to put up 6. Again, if you lost You'd need the 6 back, plus 1, requiring a bet of 14. Losing this requires you to win back 14, plus your one.

The sequence appears to be:
2
2 x (2 + 1) = 6
2 x (6 +1) = 14
2 x (14+1)= 30
2 x (30+1)= 62
2 x (62 +1)= 126
2 x (126 +1) =  254
2 x (254 +1) = 510

With a table limit of 512 it gives a possibility of a losing 8 times in a row.

The probability of that losing run of 8 occurring would be 1/3 ^ 8 = 0.000152 which would mean it would be likely to occur once in every 6578 spins.

Isn't this rather like having a die with 6578 faces, and betting on the chance of a number between 1 and 4386 to come up, but hoping that the number 6578 doesn't come up?

This is not true. I know it's a hard concept to really grasp, but the small chance of great loss compared to the high chance of small gain in this case (since the loss*loss_chance = gain*gain_chance) make the odds even and of course with the 0/00 the odds are against the player) even for one run of the martindale method (or any other).

This is what interests me. Forget the money for a moment, and just think of the probabilities. A 2/3 chance verses 1/6678 chance. If you had \$100 to bet, what would you do?
2
2 x (2 + 1) = 6
2 x (6 +1) = 14
2 x (14+1)= 30
2 x (30+1)= 62
2 x (62 +1)= 126
2 x (126 +1) =  254
2 x (254 +1) = 510

There is an error here: If you win on the third spin you get your 7 chips but you've lost 8 so far, not 6. So you lose \$1
2
2 x (2 + 1) = 6          + 2 = 8
2 x (8 +1) = 18         + 8 = 26
2 x (26+1)= 54         + 26 = 80
2 x (80+1)= 162       + 80 = 242
2 x (242 +1)= 486

So you lose a total of \$728 if you lose and gain \$1 if you win.
Since you can only actually play 6 rounds it's (1/3)^6 = .001372.
1/.001372 =  729
So you are expected to lose 1 out of 729 games. So you should win \$1 728 times and lose \$728 once.

Amazing as it may seem, you are thus expected to even out again. Trust me when I tell you that you will always even out on a roulette wheel with no 0/00.
It takes most people getting math degrees (including me) a while to really believe this, but since the spins are independent evens you will always, without exception, see these results. Of course the 0/00 means you lose not break even.
>>This is what interests me. Forget the money for a moment, and just think of the probabilities. A 2/3 chance verses 1/6678 chance. If you had \$100 to bet, what would you do?

Yes. It does look better one way. But I believe in math and would personally never bet money at all unless my expected return was greater than 100% (which only happens if a casino has a major screw-up).

Actually, the stock market can be seen as gambling with a >100% expected return.

Thanks for pointing out the error. I was only doubling - you have to treble of course. The formula would be:

I trust the math too, but that evening out will only occur in the long haul, ie if you play a lot. But if you only play a few times, the 1/3 wins are likely to come up while 1/728 loss isn't likely. For example, if there were 728 numbers, it would probably be a wasted dollar to bet on one of those numbers, yet millions of people buy lottery tickets when there is something like 1-in-million chance of winning....

I don't see how the stock market guarantees >100 return?
SOLUTION

membership
Create an account to see this answer
Signing up is free. No credit card required.
SOLUTION

membership
Create an account to see this answer
Signing up is free. No credit card required.