As if…lol
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use the force!
I just wear my lucky socks, never fails
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Place a bet on every single number
That little 0 always gets you, and two 00 in the states, changes the odds somehow, 
There is one… Do not play! 
No failsafe because of the 0 and 00. But you would have odds.
Say you bet RED non stop, after every miss you double + profit. Something like 1, 4, 11, 26, 57, etc. You want to win $1 for every spin. When red comes up, you restart at 1. The odds of missing on the 10th roll are 0.17%. It’s not a 0 and will never reach a 0.
Right, all you need is infinity dollars and you’ll be guaranteed to win a small amount!
Not true. The odds of betting red and missing on the 10th spin are 10 in 19 or 52.63%.
talking about missing 10 bets on RED in a row. It’s simple math.
I know exactly what you are talking about, but your math is wrong because you are not betting on 10 fails in a row, you are only betting on ONE fail. If the house even let you bet on 10 fails in a row, they would be very happy to pay you out only 2 to 1 if it somehow came in. ; )
I like when you use the word “exactly”.
AI Overview
To calculate the probability of getting at least one red in 10 roulette spins, you need to consider the probability of NOT getting a red in each spin and then use the concept of complementary probability
.
1. Probability of NOT getting red in a single spin:
- In American roulette, there are 18 red, 18 black, and 2 green slots (0 and 00), for a total of 38 slots.
- The slots that are not red are the 18 black and 2 green slots, so there are 20 non-red slots.
- The probability of not getting red on a single spin is 20/38.
2. Probability of NOT getting red in 10 spins:
- Since each spin is an independent event, the probability of not getting red in 10 consecutive spins is the product of the probabilities for each spin: (20/38) ^ 10.
3. Probability of getting AT LEAST one red in 10 spins:
- The event “at least one red in 10 spins” is the complement of the event “no red in 10 spins”.
- Therefore, the probability of getting at least one red in 10 spins is 1 - (probability of no red in 10 spins).
- Probability (at least one red in 10 spins) = 1 - (20/38)^10.
Calculation:
- (20/38)^10 ≈ 0.0033
- 1 - 0.0033 ≈ 0.9967
Therefore, the odds of catching at least one red in 10 tries in American roulette are approximately 99.67%.
Yes, the odds of whiffing red 10 times in a row are 0.16%. I did not dispute that.
The error in your calculation is that you are using the wrong formula for your bet. If you were betting that red would not come up 10 times in a row, then yes your math would be spot on, but you are NOT betting missing red 10 times in a row. You are betting missing red just ONCE ten separate times, and the percentage of missing red each time is 10 in 19 (20 in 38) or 52.6%.
Just to make this extra clear, you are not telling the house, “Spin this thing 10 times and if red does not come up once in those 10 spins I win.” You are betting against red at 52.6% just once. And then again just once, and then again just once … until you’ve bet it ten separate times.
If you are arguing that if you see red fail 9 times, there is a 99.8% chance that red will come up on the 10th, you are mistaken. It is still just 9 in 19 or 47.4%.
Actually, they are. The increase in bet size does mean that as long as a red comes up eventually, we win. The problem is we have to keep risking more and more money for proportionally smaller rewards.
Think about this in terms of EV and it becomes obvious how terrible this strategy is:
| Round | Bet Size | Total EV |
|---|---|---|
| 1 | 1 | -0.053 |
| 2 | 4 | -1.211 |
| 3 | 11 | -5.579 |
| 4 | 26 | -17.368 |
| 5 | 57 | -45.000 |
It’s basically exponentially bad.
To be fair, your original post said:
I know what you meant, but if you take that literally it’s both incorrect and very misleading.
you are right, the odds of missing Red (in American roulette) 10 times in a row is:
- (20/38)^10 ≈ 0.0033
he is saying 47% which sounds extremely stupid.
The actual percentage wasn’t the misleading part, you said “on” the tenth roll, when you meant “up to and including”
I see, I get it.
The original question of 100% failsafe is just impossible to achieve.
99.xx% is best possible.
Well, that’s misleading too, because you’re risking a lot of money to win a little. If you care about how much money you win/lose overall, as opposed to how often you walk away with something, then the “best possible” thing is to put as little money in as possible every single time.
Nope. You do not understand the math. I’ve explained it. You refuse to understand. That’s on you. Now go play roulette and lose your shirt, your bank account, your house …
Actually, please don’t do that. My advice to you is to stay well clear of any casino or gambling room. Seriously. I know people who have lost everything (home, business, family) using precisely your math. I’m not kidding. I’ve had someone show me boxes of paper calculations trying to prove EXACTLY your theory/system. He was homeless, lost his family, everything because he literally thought that if he bet black 4 times in a row he couldn’t lose, yet all he did year after year after year was lose. That is why I refute your math. It is fatally flawed.
So please, do not use that math. It is not 100% failsafe, it is not 99% failsafe, it is actually 99.99999999% destined to fail … disastrously. Do NOT use that math, and absolutely do NOT tell anyone else to use that math. It is fatally flawed.