In the above 2 links, it's mentioned that if we toss a fair coin ten times then the exact probability that NO consecutive heads come up in succession (i.e. n = 10 and k = 2) is
p(10,2) = 0.14...
So, the probability of at least one pair of heads, or tails, in 10 tosses is approx 1-0.14 ~0.86
Event A: prob. of next toss is always 0.5 and law of large numbers overules everything else (gamblers fallacy applies)
Event B: prob. of having consecutive heads in cluster of 10 tosses is > 0.86
Considering event B only as our universe, can we somehow create a strategy with this probability to have a positive expectation game, maybe related to Penney Ante concept?
The key is Not about increasing bet size after a win Or loss but rather on the condition that "If a coin was tossed 10 x 10 times,
i.e. Ten events B, there would be at least 8 out of 10 events, a consecutive head appearing within an event"...
.i.e. A 86% prob. single event
The short answer is yes. A coin toss is a 51/49 proposition in favor of which ever side is "up" at the start. (Persi Diaconis, Stanford).Quote: algleThe short answer to your question is NO.
A related article but can't find any examples.
Quote: FleaStiffThe short answer is yes. A coin toss is a 51/49 proposition in favor of which ever side is "up" at the start. (Persi Diaconis, Stanford).
At the 'start', meaning......a PAST result? Hmmm, I thought everything was equal? The AP (cough) crew would say its a 50/50 shot.
Wait, wait, let me guess......unless there is a BIAS with the coin? (ROFL)
Ken :)
Quote: FleaStiffThe short answer is yes. A coin toss is a 51/49 proposition in favor of which ever side is "up" at the start. (Persi Diaconis, Stanford).
If heads was up five times and tails the other five at the start, would that even this out?
Quote: Krazycat
Considering event B only as our universe, can we somehow create a strategy with this probability to have a positive expectation game, maybe related to Penney Ante concept?
Try this game (it's easier). Shake up four coins in your hand and lay them down under your palm in order. Two sides to the bet:
Player A) If there are three of either heads or tails in a row , player B pays $3 to $1
Player B) If there are not three in a row player A pays $1 to $1
Take turns shaking the coins so there is no possibility of cheating.
Which would you rather be? Player A or Player B. Or does it matter?
No. There is a bias on each and every toss toward the side that is "up".Quote: mrjjjAt the 'start', meaning......a PAST result? Hmmm, I thought everything was equal? The AP (cough) crew would say its a 50/50 shot.
Quote: FleaStiffNo. There is a bias on each and every toss toward the side that is "up".
Can you give me the short version reason as to WHY, so I dont have to open links.
You do agree?......it is based on a PAST RESULT, correct? One prior toss is the past, correct?
Past result(s) = Gamblers fallacy? You know where I'm heading with all of this so might as well play along.
Ken
Quote: FleaStiffNo. There is a bias on each and every toss toward the side that is "up".
Like the bias toward banker in baccarat?
It is not because the last flip was H or T. It depends on whatever orientation is chosen for the coin before the flip.
Quote: FleaStiffThe short answer is yes. A coin toss is a 51/49 proposition in favor of which ever side is "up" at the start. (Persi Diaconis, Stanford).
No, the short answer is NO.
The long answer *may* be yes, but even after reading all of the theory it is still inconclusive for practical purposes. Hence the short answer is no. That is why a simple coin toss is still widely used, even for some important decisions.
Ken
Only from looking at the face that is up can you possibly squeeze out an edge. If the tosser tosses it 'vigourously' like they do in the academic paper.
Quote: dwheatleyNo. Streaks of length 1 are more common than of length 3 or more. No edge.
This is of course trivially true. Why I asked the question is that since the average is two, some people unconsciously equate the distribution to normal distribution and then the conclude that the mode (the most likely value) has to be two as well! Once I had to go through a lot of effort to convice a player that it isn't the case here, and you simply can't get an edge by observing streak lengths.
Quote: mrjjjMy point being.....can the math fellas please be consistent with definitions !!! Any posts regarding myself is gamblers fallacy BUT........
The Gambler's Fallacy is the mistaken belief that patterns in sequences of independent events can be predictive of future outcomes. The Diaconis paper shows that there is a small bias in favor of the face-up side of a coin to be flipped -- this is not related to the past flip or flips, but related to the pre-flip position of the *current* flip. That's apples and oranges -- there's no inconsistency at all.
Similarly, most roulette systems are based on prior numbers and the flawed assumption that they can predict future numbers. However, a proper roulette analog to the work of Dr. Diaconis would be something like a study that showed that, when a croupier releases the ball into a roulette wheel, it comes to rest in the half-wheel centered around the release point with p=0.51 rather than 0.50. That result would be highly interesting but wouldn't have anything to do with past outcomes.
For the record, I'd love to do a study like that.
If YES is your answer, even ONE of something is still the PAST, correct? The PAST, by definition is gamblers fallacy. This is the same definition that gets jammed down my throat when I post. If I am wrong, please give an example of.....'pre-flip position of the CURRENT flip'. THANKS.
Ken
Quote: mrjjj"but related to the pre-flip position of the *current* flip" >>> Ok, definitions again. Lets say I just flipped and it was heads. It is NOW in the 'PRE-FLIP position', correct? So heads on the NEXT flip has a SMALL edge of being heads again, correct?
If YES is your answer, even ONE of something is still the PAST, correct? The PAST, by definition is gamblers fallacy. This is the same definition that gets jammed down my throat when I post. If I am wrong, please give an example of.....'pre-flip position of the CURRENT flip'. THANKS.
Ken
I'll try:-
A person flips a 'Head' and then places the coin, 'head up' for the next flip - there is a 0.51 probability that the next flip will now be 'Heads'.
Same person flips a 'Tail' and then places the coin, 'head up' for the next flip - there is a 0.51 probability that the next flip will now be 'Heads'.
So, despite the previous outcomes being different, for the 2 flips above, the probability of a 'Head' in the next flip is 0.51.
Therefore, it's not the prior outcome that determines the probability it's the current flip and the way that it is set up.
Quote: SwitchI'll try:-
A person flips a 'Head' and then places the coin, 'head up' for the next flip - there is a 0.51 probability that the next flip will now be 'Heads'.
Same person flips a 'Tail' and then places the coin, 'head up' for the next flip - there is a 0.51 probability that the next flip will now be 'Heads'.
So, despite the previous outcomes being different, for the 2 flips above, the probability of a 'Head' in the next flip is 0.51.
Therefore, it's not the prior outcome that determines the probability it's the current flip and the way that it is set up.[/
Ok but I'm asking WHY it would have the 0.51 edge? I say its a load of s**t, just my opinion. I would never tell my family/friends what you just posted. I would be in a white jacket in no time flat.
Ken
Quote: mrjjj
Ok but I'm asking WHY it would have the 0.51 edge? I say its a load of s**t, just my opinion. I would never tell my family/friends what you just posted. I would be in a white jacket in no time flat.
Ken
I'm using the premise that if you start with a 'Head - Up' then you have slightly more chance (0.51) of a 'Head' appearing on the next flip.
I'm assuming that someone performed an adequate number of trials to come up with this probability within statistical acceptance levels.
If you refute the figure of 0.51 then what would be the figure that you agree to and what evidence could you put forward to counter the figure of 0.51 that has got statistical evidence behind it?
Ken
Quote: Math ExtremistDr. Diaconis would be something like a study that showed that, when a croupier releases the ball into a roulette wheel, it comes to rest in the half-wheel centered around the release point with p=0.51 rather than 0.50. That result would be highly interesting but wouldn't have anything to do with past outcomes.
For the record, I'd love to do a study like that.
For the record, I HAVE done that study.
By the way, if the casino paid you even money less a 5% commission for a coin flip, there would still be no way, even with this knowledge to win in the long term.
Quote: mrjjjOk but I'm asking WHY it would have the 0.51 edge? I say its a load of s**t, just my opinion. I would never tell my family/friends what you just posted. I would be in a white jacket in no time flat.
You're entitled to your opinion, and you're also entitled to disagree with the rigorous results of a widely-respected mathematician who has spent his life studying gambling probabilities. This isn't a graded exam where you fail if you get the answers wrong.
Or you could read the source material and pull the veil from your own eyes:
Article about Persi Diaconis
Dynamical Bias in the Coin Toss, Diaconis, et al.
Quote: KeyserFor the record, I HAVE done that study.
And did you publish it anywhere?
I've tested exactly what you have described, plotted the standard deviation for it as well as the chi square. The testing that I conducted was far more complex than what you might imagine. There's far more to it than just simply plotting the release number and the outcome number. Wheel model used, wheel tilt, wheel speed, ball track health, ball type and size used, coefficient of restitution, etc..., all must be considered.
Quote: Jufo81How about this: In (infinite number of) coin tosses the average streak length of consecutive heads or tails is precisely 2. So since 2 is the average streak length, should you push your bet up after the first head or tail to get an edge?
Like the progression 1,3,2,1? Question here is if we consider 10 tosses as a single event, is it still gambler's fallacy to apply a cancellation scheme just within interval of 10 tosses?