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I am editing this question as requested so I am more clear in what I am asking.

Assuming that a coin flipped has a 50% chance of landing heads and a 50% chance of landing tails, I had wondered how many times I would have to flip the coin on average to end up with specifically 7 tails in a row. The answer I think is that after 128 flips I would run the risk of getting 7 tails in a row, and after 254 flips I could be expecting to get 7 tails in a row.

The second part of my question was, if I bet on my coin flips trying to get heads and I started with a 5 dollar bet and used a betting progression that looked like 5/5/5/30/100/300/500 for the first through 7th coin flips in such a way that any time I flipped heads and won I would start the progression over, and any time I flipped tails I would advance to the next step in the progression order and try to get heads, then how many flips can I make before I would be more likely to start losing more than I gain, or after how many flips would it be advisable to keep what I have won so far? (edit: if there was a 60% chance to land tails each flip)

Everything below this point is my original question, above is my reworded question. I took out all the blackjack related aspects as its way too complicated to figure odds without building a specific program that follows my personal blackjack strategy on top of betting progression.

First off I have a more simple problem, and then a more involved complex problem.

The simple problem is, if I flip a coin over and over, how many times do I need to flip it before I'm likely to end up with 7 tails in a row?

From what I calculated by adding up the number of times I could flip heads or tails, is that in every 7 flips there are 254 heads+tails possible. And if only one of those combinations is tails 7 times in a row I took 1 and divided it by 254 and ended up with 0.0039370078740157 (edit=actual number is .0078126) chance of getting 7 tails in a row on any given 7 flips. Or should it be 1 divided by 128 because by the seventh flip there are 128 possibilities and only one of them as a seventh tail?

And I think that if I flipped coins over and over I would have to flip it 128(edited) times before I would be likely to get 7 tails in a row but I'm not sure.

Now the second more involved question is:

If I play blackjack and there are 8 decks and I double down on 11, split on 2(being a pair of aces), 6, 7, 8, or 9 and otherwise always stand on 12 or higher, while betting a progression bet that looks like 5,5,5,30,100,300,500 returning to the initial 5 dollar bet after any win, with an initial bankroll of 1000, what are my chances of losing all my money If I play until I have 7000? While the dealer must hit on 16 or soft 17, push if we both get blackjack, but otherwise the dealer wins if he has blackjack before I can hit. (house rules I found played near me that are helpful to the player: can hit to reach 21 and still get blackjack payout bonus, can hit after splitting aces, can double down after hitting) Unsure if splitting ten value cards is productive, could get 21 or another 20 and seems risky

I would really appreciate help clearing up the myth that betting progression systems do not work, as most people refer to the martingale system where you just double your last bet until you win, my system is more complex because you watch 3 small 5 dollar hands for a string of 3 losses then start ramping up the bet a lot until you win.

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    It's a well known, mathematically proven result (not a myth) that if a game is such that a player has a negative expectation on each indivual bet, there is no betting strategy which will give the player a positive expectation. – 2017-01-03
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    What you mean is that if you have less than 50% chance of winning then no string of bets can be likely to end in 1 win in 7 hands? because if I win on the first hand I win 5, second hand win break even, third hand win I lose 5, fourth hand win I gain 15, fifth hand win I gain 55, sixth hand win I gain 155, seventh hand win I gain 55. I only lose my 1000 bankroll if I lose 7 hands in a row. – 2017-01-03
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    I get $2^7=128\neq 254$ outcomes on 7 coin flips... – 2017-01-03
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    So that means only 1 out of 128 blocks of 7 flips would I end up with tails 7 times in a row correct? Or I would have to flip the coin 128 times before that would begin the 7 flips that would end in 7 tails. – 2017-01-03
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    @Aykens: In your reply when you said "What you mean is ...", you clearly don't understand what I actually said. My claim was about expected value. No betting system can yield a positive expectation for the player if on each indidual bet, the player's expectation is negative. – 2017-01-03
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    Doubling your bet until you win is also a bad blackjack strategy. There is a way to gain an edge in blackjack, and that is defining your bet based on the probability of winning, since the dealer's strategy works better when there are more low-value cards in the deck and worse when there are more high value cards in the deck. This requires counting cards. – 2017-01-03
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    Your question may be better received if you edit it to ask "Does my system work?" rather than "Help me prove that my system works". – 2017-01-03
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    I did point out that I know the simple doubling bet until win doesn't work because when you do win you just win the small initial bet. But my strategy involves betting the minimum until I lose 3 times in a row then ramping the bet up around 3 times what I have bet to get a large win before returning to the small bet. I suppose what I'm wondering is if I'm betting for a limited time, how small a time period should I employ my strategy to gain more often than i lose. – 2017-01-03
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    @M.Aykens the answer, as you have been told several times, is that there is no scheme of this sort that will improve your chances – 2017-01-04
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    I am starting to realize that anything equal to or less than zero can never amount to one. But there is a lot of randomness in card games when you factor in the other players whims to hit or stand. If I can end up gaining in the short runs and set proper limits on myself I can maybe reach a final goal when I could quit for good. I am sorry I dragged this out so much, but I was unable to move it to a chat and it was a learning experience for me, being the first question I asked here. I saw a poker area but not blackjack, yet I could try another place to ask about my blackjack action strategy. – 2017-01-04

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Regarding the coins, the expected time to get $7$ heads in a row is $254.$ The time to get a specific generic length n sequence of heads and tails (like HTHHTHHHTTHTT) is about $2^n$ (=128 in your case) but the time to get strings that have a subsequence at the beginning that is the same as the one at the end is longer. The string of all heads (or all tails) is the worst and takes $2^{n+1}-2.$

Regarding the blackjack strategy, I don't know the edge for blackjack so I can't calculate that. (And I wouldn't want to even if I did.

What I can tell you is:

  1. If you play any progression strategy like that with negative edge and stop when you've lost a fixed amount of money, you will lose money on average. This is a theorem. Progression systems can only work (in the sense that they make money on average) if you can weather infinite downswings.

  2. If you have negative edge and you play smaller bets until you lose or win a fixed amount of money, then you have a worse chance of making money the smaller your bets are. This is not to say you ever make money on average. But you will make money in more individual sessions and you stand a chance of getting lucky. Whereas with small bets the law of large numbers takes over and you lose money all the time.

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    If you take blackjack out of the equation and simply assign positive and negative values to the 7 coin flips like this: flip 1 heads is positive 5, flip 2 heads is 0 change, flip 3 heads is negative 5, flip 4 heads is positive 15, flip 5 heads is positive 55, flip 6 heads is positive 155, and flip 7 heads is positive 55. with 7 tails being negative 945. I am likely to get a positive return if I only flip 200 times then? since 252 times would be an expected 7th tails? – 2017-01-03
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    The "Generically" sentence is unclear. +1 for the rest. – 2017-01-03
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    The reason I asked this question is because I have played blackjack with a starting amount of 400 to 1000 dollars 6 or 7 times and I know that of those times I walked away with over 7000 dollars twice or roughly one third of the time. So I ended up positive 10000 at least, and I'm wondering if this was just random luck or if the strategy is something that can be relied on over time to turn a profit. – 2017-01-03
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    I think you gave me a good answer in that you point out that the chances of tails 7 times in a row is going to be rare in 100 total flips. I know blackjack is not exactly 50 50 odds but if I play 100 hands at a time using my progression betting strategy I think I should gain money even if its a 55% chance on any flip that I will get tails. And I think that playing with at least 2 other players at the table should add more randomness to the chances of winning or losing any given hand. – 2017-01-03
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    The answer is that it was certainly random luck. Again, this is a very general theorem, so if we were to actually do the math in detail it would come out that you have a negative expected value on each session. Eyeballing your numbers, they look statistically consistent with random chance to me, but even if they didn't, it's simply impossible to have positive expected earnings with such a strategy. – 2017-01-03
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    I agree that you should expect to lose 2 out of 3 times, because that is what I did on average. But I won over 6000 dollars each time I won. So if I wager 1000 every time and lose 2 out of 3 times but when I do win I walk away with 7000 dollars I should expect to gain at least 3000 dollars for every 3 times I play. – 2017-01-03
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    7/3>1 is exactly what has happened so far, I start with 1 and stop if I reach 7. So even if I lose 1 a couple times before I hit 7 I end up positive. The more complicated math would be to assign the positive and negative values to the coin flips to see if on average I would gain in 100 flips. Being that I could get 5,0,-5,15,55,155,55 in a heads flip, losing only on the 3rd heads or 7th tails flip 5 and 945 respectively. – 2017-01-03
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    What I'm telling you is that your average winnings have to be negative. On a session where you start with 1000 and walk away when you lose it all or when you reach 7000 we must have: 6000*Prob(you win 6000) - 1000*Prob(you lose 1000) < 0. So for these numbers you should expect to lose more than 6/7 times on average. It's totally possible for that to be the case and still you've observed 2 out of 3 given you've only played 6 or 7 times. You absolutely win more when you win but the odds cannot be enough so that the average comes out positive. This is true no matter how you bet. – 2017-01-03
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    One thing to keep in mind is that if I say start with 1000 and gain 1000 over an hour of play, then lose 1000 in the next 7 hands I still have 1000 left to keep playing and trying to hit my goal of 7000 chips. But if you are confident that playing this way I will lose in the long run mathematically and that being up 10000 dollars after 6 or 7 trips to the casino was just luck, then maybe I will do something else with my money like an investment instead of going to the casino whenever I have an extra 1000 dollars. – 2017-01-03
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    Yeah, that's all taken into account in the theorem... it's designed to address the exact question you're asking (BTW it's called the optional stopping time theorem). This is a common, well-studied problem so if I seem overly confident, it's cause I've thought about it a lot in the past as have many others. – 2017-01-03
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    I read what I could understand of https://en.wikipedia.org/wiki/Optional_stopping_theorem and what I seem to gather from it is that if you employ basic martingale system of bet doubling until you win that you always end up with about the same amount of money you begin with, or taking in house limits you end up losing every single time in the long run. But I don't use basic bet doubling, I test the waters 3 times then on average I triple the bet to gain more when I DO win. So in the short runs I should win more than lose I think? – 2017-01-03
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    The 'martingale' in the wikipedia article does not just refer to the winnings under the martingale betting system. It refers to the winnings under any betting strategy in a fair game (0 edge). A game with house edge you play with a particular strategy is a 'supermartingale'. The relevant case is (c) since your winnings/losses under your stopping strategy (up 6000, down 1000) are bounded. The conclusion is that average value of your winnings in a given session <= 0. This means 'in the long run' playing many sessions, you will lose. Short run, perhaps not. – 2017-01-03
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    By the way, the way the martingale betting system 'gets around' the theorem is that the stopping strategy is to play until you win, no matter how long it takes. Thus you can get arbitrarily deep in the hole, and the conditions of part (c) do not apply (nor (a) and (b)). However, as soon as you put a condition that you stop after some amount of loss (even a ridiculous number), the theorem holds and you can't win on average. If you have a very large loss limit, then you will win almost all the time, but that one rare time you lose up to your (huge) limit will drown out all your gains. – 2017-01-03
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    I realize your system is quite different from the martingale, both in betting and stopping strategies, so maybe that's not super helpful, but it does give a feel for how that strategy obeys the theorem once practical limitations are put on it. And since you have a win/loss limit from the get go, your strategy must obey the theorem. – 2017-01-03
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    If I used a method that was not (+6000 or -1000) but instead tried (+50 or -1000) I would only need one win on the 5th 6th or 7th hand. If I gave the house a win rate of 60% a hand that would be just under a 3% chance of any given sequential 7 hands all losing. I would be much more likely to win at least once on the 5th, 6th, or 7th hand before I lost 7 in a row. Also 4 wins on the 4th hand would net +60 and merit stopping too. From what I have learned here it would be a low risk way to get a quick 50 bucks. Just a slim chance I could lose a lot. One net loss would mean I need 200 net wins. – 2017-01-04
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    I just had a thought to create a wave that begins at zero, arcs up to 6000 then down to -1000 then up to 5000 then down to -2000 then up to 4000 then down to -3000 then up to 3000 then down to -4000 then up to 2000 then down to -5000 then up to 1000 then down to -6000. It could then become my limits: 6k,-1k / 5k,-2k / 4k,-3k / 3k,-4k / 2k,-5k / 1k,-6k could be my upper and lower stopping limits for 6 trips to the casino. Would require 21k$. A trip I met my positive goal I could try for 1k higher on the following trip. I just need to figure out what order my limits should be from the get go. – 2017-01-04
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The second part of my question about how many coin flips I can make, (or hands of blackjack played) before I am likely to lose 7 times in a row betting on the 40% odd heads, (or my blackjack hand) is in my estimate 50 flips or hands. And since I only really lose money if I lose 7 bets in a row, and blackjack shoes run out before 50 hands are played on full tables, I could sit at one blackjack table for many hours slowly accumulating chips.

I personally prefer what is referred to as the 3rd base position, at the right hand of the dealer. Because all of the random actions the 2 or more players to my right ensure breaking up long strings of cards that exist in the shoe after the cards are dealt. And my decision about hitting, standing, splitting, or doubling down is all derived from the 2 cards I am dealt. So whatever 3rd card I get, or leave for the dealer to hit with is random.

Previously I had left the casino only after I lost my initial 1000, or gained 6000 or so. But I think it would be better to try for simply 3000 profit or 1000 loss each trip to the casino. Maybe I will be lucky again in the future.

I read some interesting things about a lot of math topics based off your responses and found the gamblers ruin Wikipedia page and related pages informative.

https://en.wikipedia.org/wiki/Gambler%27s_ruin

For anyone interested in gambling, or who do gamble at times, its worth checking out.

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the chance of getting a head would be 1/2 so 1/2^6 = 1/64, so the chance would be 1/64. the reason I did 1/2^6 instead of 1/2^7 was because it doesn't matter if the first ones a head or a tail.