Double Ball Roulette: Probability of two roultette balls dropped adding up to either 37 or 39
0-28-9-26-30-11-7-20-32-17-5-22-34-15-3-24-36-13-1 - 00-27-10-25-29-12-8-19-31-18-6-21-33-16-4-23-35-14-2 (double 00 roulette wheel)
So here is my math question. What is the probability of two balls being spun/dropped (or two wheel roulette games) where the sum of the two balls (or double wheel) when landing = 37? Same question for 39?
ie: ball 'a' lands on 26 and ball 'b' lands on 11 for total of 37 Q: what is the probability the sum of two balls = 37?
ie: ball 'a' lands on 15 and ball 'b' lands on 24 for total of 39 Q: what is the probability the sum of two balls = 39?
Quote: behronlee15I read on Wizard of Odds a most interesting fact about roulette....Most numbers are part of a pair, with one number between them. These pairs add to either 37 or 39.
0-28-9-26-30-11-7-20-32-17-5-22-34-15-3-24-36-13-1 - 00-27-10-25-29-12-8-19-31-18-6-21-33-16-4-23-35-14-2 (double 00 roulette wheel)
So here is my math question. What is the probability of two balls being spun/dropped (or two wheel roulette games) where the sum of the two balls (or double wheel) when landing = 37? Same question for 39?
ie: ball 'a' lands on 26 and ball 'b' lands on 11 for total of 37 Q: what is the probability the sum of two balls = 37?
ie: ball 'a' lands on 15 and ball 'b' lands on 24 for total of 39 Q: what is the probability the sum of two balls = 39?
I really hope that I am not screwing this up by thinking about it wrong.
It seems to me that you have 36 potential first numbers (1-36) that could then result in a total of 37 given the second number. In this case, the only way that you would absolutely fail to make 37, based only on the first ball, is if the first ball were 0 or 00.
36/38 = .9474 or 94.74% on the number not being 0 or 00.
If the first number is not 0 or 00, then only one number can give you a sum of 37 for the second ball, so 1/38 or .0263 or 2.63%
.9474 * .0263 = .02491 or approximately a 2.49% chance of making a total of 37. 1:40.16
We will apply the same principles to 39, except 0, 00, 1 and 2 are all no good.
34/38 = .8947 or 89.47%
You can still only make 39 with one number.
.8947 * .0263 = .02353 or 2.35%. About 1:42.6
I may have misunderstood your question, if you mean two balls dropped at once. In that case, the two balls cannot land in the same spot, I don't believe, I've admittedly never seen such a wheel. If the two balls cannot land in the same spot, just change the second variable from 1/38 to 1/37 and do the math accordingly.
Back to my question....with the fact that both numbers can be hit at the same time, and/or both balls can land on the same number....does this change your calculation? I'm not sure it does because neither 37 or 39 is an even numbrer and therefore no two balls landing on the same number would matter, correct?
Quote: behronlee15THANKS FOR YOUR REPLY. Yes, the ball CAN fall on the same number. Just as there is now a new double action roulette game whereas there are two wheels and one ball....which would be the same as one wheel shooting out two balls.....that game now exists, too. www.doubleballroulette.com
Back to my question....with the fact that both numbers can be hit at the same time, and/or both balls can land on the same number....does this change your calculation? I'm not sure it does because neither 37 or 39 is an even numbrer and therefore no two balls landing on the same number would matter, correct?
You're welcome. I still await verification that my pursuit of the question posed was correct, but as long as I was approaching the question right, the math is fine.
The second factor does not change my calculations, as they were based simply on two balls being dropped in two wheels separately from one another. If the fact that two balls being dropped at the same time does not make it any less likely for the, "Second ball," to end up in the same spot as the first, then the calculations are unchanged.
However, two balls landing on the same number would matter. It would matter because, if they could not land on the same number, the second step of the equation (in the case of 37 or 39) goes from 1/38 to 1/37 because one number is discluded as impossible. For instance, the probability of 37 becomes:
36/38 * 1/37 = x
.9474 * .0270 = x
.02558 or 2.56% or 1:39.06
You can see that the odds become slightly better when a potential number is eliminated as impossible. The best way of looking at it is from a standpoint of Double-Zero Roulette and betting only individual numbers, straight-up. If you block off any given number as impossible, or make that number a push, you are effectively playing Single-0 Roulette. That's not true from an Even Money, 2:1, or corners bet standpoint unless you are actually blocking 0 or 00, but you see my point.
Quote: buzzpaffI predict a future thread from someone exploiting the bias of double roulette wheels, thereby quadrupling the profit available to AP roulette players.
I wonder if numbers can get entangled even when they don't pair up. It would be worth a Nobel Prize to merely observe Quantum Mechanics at a macroscopic scale...
Speaking of physics, Einstein got it wrong. Another way to make money on roulette is to past-post your bet when it hits. Well, not that Einstein was wrog, but his Theory of Universal Roulette Winnings was incomplete.
If I could persuade a physicist to write a paper about it, including the opticla tricks needed to fool casino security, I'm sure we could get an Ignobel Prize. That has to be worth something, say a dime more than a betting system (meaning it has to be worth a dime).
Quote: DJTeddyBearIf the assumption that the two results are completely independent, and that they can be the same number, then Mission's math is correct.
Yes - if they are independent. However, it seems to be that they are not, since, once a ball lands into a slot, there is a possibility that the other ball would have landed in the same slot and stayed there had the first ball not been there, but the second ball ended up ricocheting off of the first ball and landed on a different (not necessarily adjacent) number.
What effect the presence of a ball has on the other ball can't be calculated accurately with just mathematics; you probably need to resort to a Monte Carlo method (i.e. base the results on some "large" number of actual spins).
Quote: ThatDonGuyYes - if they are independent. However, it seems to be that they are not, since, once a ball lands into a slot, there is a possibility that the other ball would have landed in the same slot and stayed there had the first ball not been there, but the second ball ended up ricocheting off of the first ball and landed on a different (not necessarily adjacent) number.
What effect the presence of a ball has on the other ball can't be calculated accurately with just mathematics; you probably need to resort to a Monte Carlo method (i.e. base the results on some "large" number of actual spins).
I was sort of considering that possibility. I figured if the, "Second Ball," was going just a little too fast it could pop out of there. I also considered the possibility that both balls would not physically fit, but that's not the case.
I really don't see a potential for an AP here. I guess it could happen if the casino pays 40:1 on a total of 37, but that would never happen. I could see them doing 35:1 because that's simple and the player may think, "There are so many ways to make 37, that's a great bet!"
The second value in the equation is simply not 1/38 or 1/37, but something in-between, probably closer to 1/38. In short, it makes very little difference.
Thanks for the answer to my question. Your math makes total sense.
On '37' the probability of winning was 1:40.16, therefore,
if the payout is 35 units for a win on 37, than,
(40.16-35) / (40.16+1) = 5.16 / 41.16 = 12.54% house edge
AND
if the payout is 40 units for a win on 37, than,
(40.16-40) / (40.16+1) = .16 / 41.16 = .388% hourse edge
On '39' the probability of winning was 1:42.60, therefore,
if the payout is 35 units for a win on 39, than,
(42.6-35) / (42.6+1) = 7.6 / 43.6 = 17.42% house edge
AND
if the payout is 40 units for a win on 39, than,
(42.6 - 40) / (42.6+1) = 2.6 / 43.6 = 5.96% house edge
IS THIS CORRECT?
Advantage player.Quote: behronlee15Call me stupid on Tuesdays but what is an AP? Advanced player?
I.E. Someone who has analyzed the game for mathematical advantages, or dealer weaknesses, and plays using that advantage.
Quote: behronlee15Call me stupid on Tuesdays but what is an AP? Advanced player?
What ought we call you on Wednesdays? ;)
Seriously, DJ answered your question. But I wouldn't call you stupid for not knowing what AP means (Associated Press?) All forums slowly but surely accumulate acronyms and abreviations for common, and not so common, terms. Not to mention specialized lingo. You should see what could crop up in Trek boards...
Quote: behronlee15Call me stupid on Tuesdays but what is an AP? Advanced player?
Don't forget to stop back on Tuesdays. I need a day off !
Quote: behronlee15Yes, there is a possibility with the larger of the two spec roulette balls that a ricochet can occur to stop ball #2 from falling in same number but rare. With respect to the double wheel roulette this is not possible because there are two wheels and only one ball.
Thanks for the answer to my question. Your math makes total sense.
You're welcome, my pleasure, glad to know I didn't screw up!
Quote: behronlee15Perhaps I should answer my own question.
On '37' the probability of winning was 1:40.16, therefore,
if the payout is 35 units for a win on 37, than,
(40.16-35) / (40.16+1) = 5.16 / 41.16 = 12.54% house edge
AND
if the payout is 40 units for a win on 37, than,
(40.16-40) / (40.16+1) = .16 / 41.16 = .388% hourse edge
On '39' the probability of winning was 1:42.60, therefore,
if the payout is 35 units for a win on 39, than,
(42.6-35) / (42.6+1) = 7.6 / 43.6 = 17.42% house edge
AND
if the payout is 40 units for a win on 39, than,
(42.6 - 40) / (42.6+1) = 2.6 / 43.6 = 5.96% house edge
IS THIS CORRECT?
Nice work!!!
Don't worry, I've always known what the concept of an advantage play was, but never would have abbreviated it, "AP," nor have taken the abbreviation to mean anything besides Associated Press until stopping here.
Quote: Mission146Nice work!!!
Don't worry, I've always known what the concept of an advantage play was, but never would have abbreviated it, "AP," nor have taken the abbreviation to mean anything besides Associated Press until stopping here.
