I lost 28 hands in a row
- Thread starter JohnCrover
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So 31 hands consisting of 28 losses and 3 pushes. You don't say what the rules and conditions were, but for a 6 deck S17 DAS SP3 noRSA game, where a win/loss/push is defined as an overall positive/negative/zero result (even after splitting), the probabilities are roughly Win: 43.3%, Lose: 47.9%: Push: 8.8%.
So the probability of 28 losses and 3 pushes is 0.479^28 x 0.088^3. However, there are COMBIN(31, 3) = 4495 ways that the pushes could be mixed within the losses. So the overall probability is 0.479^28 x 0.088^3 x 4495 = 0.00000034% or 1 in 291,358,647. Remember that this is for exactly 28 losses and 3 pushes.
For 28 losses mixed with any number of pushes (i.e. 28 losses without a win) the probability is (0.479 / (0.433 + 0.479)) ^ 28 = 0.0000015% or 1 in 67,680,116, which is still rare but a lot less rare.
My personal record is 23 losses without a win which was a 1 in 2,704,983 event.
So the probability of 28 losses and 3 pushes is 0.479^28 x 0.088^3. However, there are COMBIN(31, 3) = 4495 ways that the pushes could be mixed within the losses. So the overall probability is 0.479^28 x 0.088^3 x 4495 = 0.00000034% or 1 in 291,358,647. Remember that this is for exactly 28 losses and 3 pushes.
For 28 losses mixed with any number of pushes (i.e. 28 losses without a win) the probability is (0.479 / (0.433 + 0.479)) ^ 28 = 0.0000015% or 1 in 67,680,116, which is still rare but a lot less rare.
My personal record is 23 losses without a win which was a 1 in 2,704,983 event.
While I enjoyed AceofSpade's trip reports, especially the photography of Vegas skylines, I never found them to be very credible because of the detailed memory of each and every round played. I mean, unless he has some sort of photographic memory or is some sort of "rainman savant", who recalls every hand played? 
And if he did have that sort of memory or ability, he was wasting it.
And if he did have that sort of memory or ability, he was wasting it. The problem with calculations like this is that they are valid only from the perspective of: "I am now going to play exactly 31 hands; what is the probability that I will lose 28 and push 3?" But, of course, that isn't at all what happened. The OP has been playing blackjack all his life. He has started thousands upon thousands of such 31-hand streaks. And every single one has/had the potential to start the epic streak now being discussed. As such, everyone has to understand how much more likely one is to encounter something like this, given that you don't play just one such 31-hand segment, but perhaps, over one's lifetime, hundreds upon hundreds of thousands of them, if not millions.
As an example, consider the simple calculation of a streak of getting heads five times in a row with a coin flip. You would all answer, it's a 1 in 32 probability. But now, flip a coin 100 times, giving you 96 shots at starting a streak of five such consecutive heads, and now tell me what the probability is of at LEAST one streak of five in the whole 100-toss sample. The answer turns out to be 81%!! In other words, it would be very unusual NOT to have encountered a streak of five, if you toss frequently enough.
So, if you give yourself enough shots at starting this legendary losing streak (surely millions for the average player), it becomes a great deal less surprising that such an event actually took place.
Doesn't change the fact that it sucks!!
Don
As an example, consider the simple calculation of a streak of getting heads five times in a row with a coin flip. You would all answer, it's a 1 in 32 probability. But now, flip a coin 100 times, giving you 96 shots at starting a streak of five such consecutive heads, and now tell me what the probability is of at LEAST one streak of five in the whole 100-toss sample. The answer turns out to be 81%!! In other words, it would be very unusual NOT to have encountered a streak of five, if you toss frequently enough.
So, if you give yourself enough shots at starting this legendary losing streak (surely millions for the average player), it becomes a great deal less surprising that such an event actually took place.
Doesn't change the fact that it sucks!!
Don
Good point and practical insight, as always, Don. I get a different answer for the coin toss example though.
For "at least one" kinds of problems, I usually calculate from the opposite direction. The probability of not starting a streak of 5 heads in 96 sequences is (31/32)^96. So the probability of at least one such sequence would be 1 - (31/32)^96 ~= 95%. Where did I go wrong?
For "at least one" kinds of problems, I usually calculate from the opposite direction. The probability of not starting a streak of 5 heads in 96 sequences is (31/32)^96. So the probability of at least one such sequence would be 1 - (31/32)^96 ~= 95%. Where did I go wrong?
@1forme, you can learn something from posts he does not respond to (post by those he considers trolls) and by noting that he almost always responds to the OP and not make any post about a poster. Note that he does not copy what a poster wrote on another forum and bring it to this forum or vice versa. I am 3 forums that he is on as well and like you, I read and learn from his posts, both content and style.
Try, please.
Try, please.
OK, here's the mail I sent to Gronbog.
First, let me give you the sites for streak calculators that I have used over the years. They don't all agree, unfortunately, so it's hard to know which are the more accurate. (Strangely, two give one answer and two give another, a little more than 1% apart.) Then, I'll describe how to get a rough idea of the right answer, which isn't possible with straightforward numerical calculations. You can come close, but you can't be exact.
http://www.beatingbonuses.com/calc_streak.htm
http://completecurrencytrader.com/streak-probability-calculator.html#book_wrap
http://maxgriffin.net/CalcStreaks.shtml
https://sites.google.com/view/krapstuff/home
Finally, an interesting discussion can be found here:
https://wizardofvegas.com/forum/off...ak-calculator-and-calculating-streaks-runs/2/
Mike and I had numerous discussion of how to do this algebraically, and we agreed on a method that gets you close, but is just too cumbersome to do without a calculator for the more complicated scenarios. For this example, I'm going to go with 100 tosses of a fair coin, trying to get a streak of 5 heads. Now, right away, you have to understand that there can be confusion as to what that means. The first confusion is that you don't mean just exactly one streak of five heads. It's all right to have two or three. Limiting it to exactly one would be somewhat silly. So, it's at least one streak of five. But it's more than that. It's at least one streak of five or more. That is, not only can you have several streaks of five, you can also have a streak of six or seven, and that counts towards your answer of at least one streak of five OR MORE. Again, it would be silly to have a streak of seven and not count it because it wasn't exactly five!
So now, let's get to the math. Visualize a horizontal number line from 1 to 100. We want a streak of five or more, but we're going to calculate NO streak of five or more and then do the usual 1 - routine to get the probability that it WILL occur. Actually, we're going to calculate will, won't, and then will. I'll explain.
The first way to get a streak of five heads is to start with five in a row, from 1-5, and then get a tail, to end the streak. That's a straightforward (.5)^6 = 0.015625. So, the probability that this WON'T happen is 0.984375. We also use this value at the end, to get tails on #95 and then heads from 96-100. So, the probability that this won't happen at the beginning or the end is the product, or 0.96899.
For all the other streaks of five, beginning with 2-6, we have to precede and follow the streak with a tail, to define the beginning and the end. So, each streak of five heads starts and ends with a tail, and we have (.5)^7 = 0.0078125. And for it NOT to happen is 0.9921875. How many times? Well, starting with beginning with #1 a tail, then five heads, and #7 a tail, and ending with beginning with #94 a tail, the next five heads, and then #100 a tail. In all, 94 of them. And (0.9921875)^94 = 0.4784. Don't forget the two end streaks, above. So we multiply: (0.4784)(0.96899) = 0.46357. So that's the probability of not getting ANY streak of EXACTLY five. One minus that would yield 0.53643. But, clearly, that's too low and doesn't jibe with the calculators' values of 81% or 82%. Why? Because we haven't included any and all streaks of six, seven, eight, or more. They add more probability!
So, you see how tedious it becomes. We now need to repeat the whole process I just did for five in a row and expand it to six in a row. And when you do that (try it, for fun) and get how much more to add to the above, you get a great deal closer to the 81% or 82% given by the calculators. There's so little left for streaks of seven or higher, that you're almost all the way there by just doing streaks of five and six. Do you understand the process?
Anyway, calculating streaks is NEVER what people think it is, and there is no straightforward algebraic way to do it.
Enjoy!
Don
First, let me give you the sites for streak calculators that I have used over the years. They don't all agree, unfortunately, so it's hard to know which are the more accurate. (Strangely, two give one answer and two give another, a little more than 1% apart.) Then, I'll describe how to get a rough idea of the right answer, which isn't possible with straightforward numerical calculations. You can come close, but you can't be exact.
http://www.beatingbonuses.com/calc_streak.htm
http://completecurrencytrader.com/streak-probability-calculator.html#book_wrap
http://maxgriffin.net/CalcStreaks.shtml
https://sites.google.com/view/krapstuff/home
Finally, an interesting discussion can be found here:
https://wizardofvegas.com/forum/off...ak-calculator-and-calculating-streaks-runs/2/
Mike and I had numerous discussion of how to do this algebraically, and we agreed on a method that gets you close, but is just too cumbersome to do without a calculator for the more complicated scenarios. For this example, I'm going to go with 100 tosses of a fair coin, trying to get a streak of 5 heads. Now, right away, you have to understand that there can be confusion as to what that means. The first confusion is that you don't mean just exactly one streak of five heads. It's all right to have two or three. Limiting it to exactly one would be somewhat silly. So, it's at least one streak of five. But it's more than that. It's at least one streak of five or more. That is, not only can you have several streaks of five, you can also have a streak of six or seven, and that counts towards your answer of at least one streak of five OR MORE. Again, it would be silly to have a streak of seven and not count it because it wasn't exactly five!
So now, let's get to the math. Visualize a horizontal number line from 1 to 100. We want a streak of five or more, but we're going to calculate NO streak of five or more and then do the usual 1 - routine to get the probability that it WILL occur. Actually, we're going to calculate will, won't, and then will. I'll explain.
The first way to get a streak of five heads is to start with five in a row, from 1-5, and then get a tail, to end the streak. That's a straightforward (.5)^6 = 0.015625. So, the probability that this WON'T happen is 0.984375. We also use this value at the end, to get tails on #95 and then heads from 96-100. So, the probability that this won't happen at the beginning or the end is the product, or 0.96899.
For all the other streaks of five, beginning with 2-6, we have to precede and follow the streak with a tail, to define the beginning and the end. So, each streak of five heads starts and ends with a tail, and we have (.5)^7 = 0.0078125. And for it NOT to happen is 0.9921875. How many times? Well, starting with beginning with #1 a tail, then five heads, and #7 a tail, and ending with beginning with #94 a tail, the next five heads, and then #100 a tail. In all, 94 of them. And (0.9921875)^94 = 0.4784. Don't forget the two end streaks, above. So we multiply: (0.4784)(0.96899) = 0.46357. So that's the probability of not getting ANY streak of EXACTLY five. One minus that would yield 0.53643. But, clearly, that's too low and doesn't jibe with the calculators' values of 81% or 82%. Why? Because we haven't included any and all streaks of six, seven, eight, or more. They add more probability!
So, you see how tedious it becomes. We now need to repeat the whole process I just did for five in a row and expand it to six in a row. And when you do that (try it, for fun) and get how much more to add to the above, you get a great deal closer to the 81% or 82% given by the calculators. There's so little left for streaks of seven or higher, that you're almost all the way there by just doing streaks of five and six. Do you understand the process?
Anyway, calculating streaks is NEVER what people think it is, and there is no straightforward algebraic way to do it.
Enjoy!
Don