I don’t think you understand what I posted, I don’t advocate risking anything.
- the OP asks about 100%, I posted saying it cannot be 100%.
- I don’t advocate playing or not playing roulette or any other Casino.org game.
You butted into a conversation you do not understand, the numbers you posted show some kind of impairment. Missing a bet on RED 10 times in a row is NOT 47%. I know I am wasting my time.
I agree. I have never disagreed. You have misread what I wrote.
I don’t think so.
That is not true in the slightest. He is claiming that if we bet red and lose 9 times in a row (that is 9 separate bets on 9 separate spins), on the 10th time we have a 99.83% chance of winning, and since we double up each successive bet, it’s almost impossible to lose, we win back all of our losses. That could not be more WRONG! It’s horribly wrong. It’s waaay off.
He is using the odds if whiffing 10x on a bet to whiff just ONCE. The odds to whiff that once is 52%, not 99+%. His math is correct, but he is using the WRONG formula to calculate his odds.
Like I said, if he had arranged a bet with the house to make one bet that covered 10 straight spins and bet to miss red on all of them, THEN his odds of winning (not losing) would be 0.16%. If he had arranged a bet with the house to cover 10 straight spins and at least ONE of them would be red, his odds to win would be 99.83%, however, the house would laugh at him if he expected 2:1 payout odds. They might pay him 1.01:1, and in order to make that worthwhile he’d need to bet 5 or 6 figures. The house is not going to waste it’s time paying out a dime on a 10 dollar bet, or even spinning 10 times for 10 lousy bucks.
you lack basic understanding of odds, arguing with you is pointless.
I know that’s what they said, but it was also clear to me that’s not what they actually meant (see our exchanges above)
What they’re proposing is just a modified martingale strategy (Martingale (betting system) - Wikipedia), and it would work provided you have infinite money and infinite time.
I just meant that the bets we have to make each time a red doesn’t hit increase much faster than the money we can win. ie On the 10th roll you’d be betting over $2K and would have over $4K invested in total for a potential reward of only $55.
@johnlittle actually nailed it. The only way not to lose is not to play. The more money you put in, the faster you lose, so any strategy that increases bet sizes is worse than just betting a fixed amount.
Then stop. (shrug)
Yes I get that, but he still thinks that if he loses 9 times in a row he can’t miss winning it all back because he thinks he has a 99.8% chance of winning the tenth spin.
I don’t blame you for not understanding the math, but then you patronize like any low IQ would do?
Comical.
You haven’t actually been that clear on what you are saying. Napkin on her very first post questioned:
and I don’t see where you’ve actually attempted to clear that up.
What is wrong with my math? Explain in detail for my ‘low IQ’.
here is a much simpler example: the COIN TOSS.
You bet to catch HEADS, just one time out of as many tries as possible. You have a 50% chance since it’s 2 faces. What you have been saying is, at any number of tosses it is always a 50% chance. That is wrong. There is a math algorithm for this that goes: 1-(1/2)^N . where ^=power. N=number of tries.
for American roulette (38 numbers) it would be 1-(20/38)^N
Don’t know if this is relevant but I found this on Roulette Bets, Odds and Payouts - The Complete Guide
Avoiding Common Misconceptions
…For example, a simple system may be to bet on red but increase your bet size after a loss. The theory is that increasing the bet size will allow you to win back any losses. But the reality is you are simply increasing the amount you wager on the next spin. Each spin has no correlation to the following spin, at least not in the way that such a betting system assumes.
Another example is consider 10 reds spinning in a row. One roulette player says “red is on a streak, I must bet red”. Another player says “black is due to spin next, so I’ll bet black”. So who is right? Neither of them. This is because the odds of red or black spinning are still 50/50 (neglecting the existence of zero for now)…
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Your last replay is still confusingly worded - there’s sentences in there that are incorrect if you take them out of context.
Let me see if I can get you both on the same page, because I’m pretty sure you both understand the math correctly, just not each other.
You flip a coin 9 times, they’re all tails, what are the odds that the tenth flip is a head?
What are the odds of flipping a coin ten times and getting only tails?
So many people have actually tried this!
such as?
odds are 1-(1/2)^10=99.9023%
unlike death & taxes it is not 100%
I will add that if there was a 100% chance to win at gambling, the game would be removed/altered.
No, you are the one that is wrong. That is exactly what I thought you were arguing, and that is exactly why I have argued against it, so I have after all understood you 100% clearly right from the beginning because this is the exact error that I was trying to point out.
The coin toss is always 50%, no matter how many times heads or tails has come up in previous tosses. Always always always.
The reason that the chances of tails coming up 10 straight times is 1 in 1024 (.00098%) is because each successive toss is 50%. .5x.5x.5x.5x.5x.5x.5x.5x.5x.5 = .00098.
What you are trying to do (as I have explained more than once in this thread) is to apply the 10x odds (99.999%) to the individual 10th toss. This is a grave error in your calculation, sorry to say.
If we use your theory of exponential odds in successive flips, then the 2nd toss (after a fail of tails) should be 75% odds in favor of tails, but it’s not because that would put your odds of 2 straight tails at 37.5% (50%x75%), when in reality it is 25% (50%x50%). Your odds of winning are much lower than you think they are, are much lower than your calculations indicate, because you are using the wrong formula. Do you see?
Heads Heads = 25%
Heads Tails = 25%
Tails Heads = 25%
Tails Tails = 25%
Where is the 37.5%? It’s not there! Why? Because you are using the wrong formula.
If we maintain this flawed exponential odds, we get WAY out of whack in a hurry, loooong before the 10th flip. It won’t take long to realize that by your (incorrect) calculations, at the 10th toss we will be nowhere near 99.999%. The reason that we get to 99.999% is because each individual toss remains steadfastly 50%.
I’m not trying to make you look like a fool. I’m trying to save your bank account. And if you are not willing to listen, then I am trying to save the bank account of anyone else who might be reading. Do NOT use this system. Like EVER!
ok, napkin was right, I was wrong, and so are you.
The odds of any flip never change. Yours is a correct answer to my second question, but not the first. The chance of getting a head is always 50/50 no matter how many tails have come up already.
No, I understand him perfectly. I’m pretty sure that he understands me, he just disagrees with me. However, I am right and he is wrong. That’s all. And it’s not about being right and showing him up (as he is trying to do to me with his trolling rude comments). It is simply about setting out the truth.
My answer: 1 in 1024.
His answer: 1 in 1024.
We agree.
My answer: 1 in 2.
His answer:
We disagree, but I am right and he is wrong. This is not opinions on whether a pop song or an art piece is good or not, this is math.
As you can see, I understand him perfectly and have all along.
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@BR001 If you’re thinking - if the odds of ten tails in a row is 1 in 1024, the odds of a tenth flip being tails given 9 tails in row must also be 1 in 1024, the problem is the “given 9 tails in a row” part. We’re assuming an event that only has a 1 in 512 chance of happening has happened. We only need to half those odds to get to the 1 in 1024 answer for the tenth flip, hence the 50/50 probability for a tenth tail still holds.