binary sequence. interrupt.

example:
player A throws a coin more than a thousand times. player B is included in the game without having information about the previous results and making bets absolutely randomly, despite the statistics of the eagle or tails.
having played 100 times and having lost all the money Player B goes for money. Player A continues to throw a coin. Player A tossed a coin 200 times and at that moment Player B rejoins the game.
question:
the second connection of player B is 101 (continuation of the first game) or 1 (the beginning of a new game)? player B plays at random without paying attention to statistics.
PS if you can answer please prove by mathematics! if you can not, tell me which mathematical forum will help me.

This sounds like the start of an argument, but the key word in your scenario is random.

If it is truly random, then it does not know about it's past, it doesn't know about it's future.

Each random trial is an independent event, which means the odds of occurence of the NEXT event does not depend on the results of any previous event, whether or not there were any previous events, however many results of previous events were recorded, or the distribution of previous results.

https://en.m.wikipedia.org/wiki/Randomness

That page also has a section on "Misconceptions and Logical Fallacies" that are good to know when considering the independence of random events.

I'm not sure what math you're looking for since the problem is based on a concept. Similarly when one counts cards at blackjack they play tens of thousands of hands, if not hundreds of thousands. Of course you could not play all of those in one sitting, so you have to leave and come back. The hands you directly play (or observe since technically it's a hand with a 0 bet) are the only hands that count towards your EV/bankroll.

Thus, if you leave and come back, it doesn't start anything anew... it is a continuation of your previous hands/EV/etc. In your example player B is now playing bet 101 when he comes back...

quote from the mathematical forum: That doesn't seem like a math question at all. If I play poker with the same group each week is it a new game each time or just one long standing game? Seems purely a matter of semantics.What does this have to do with math or probability? I guess the answer depends on what you have agreed with the other people. Most likely it's a new game each time, don't you think?

Quote: evs

...I guess the answer depends on what you have agreed with the other people. Most likely it's a new game each time, don't you think?

It depends on what you're looking at. In your poker example if the players play a cash game, then it's just a continuation of the same game. If they play a tournament then each week is a "new game" but it's still a continuation of their individual EV/bankrolls.

Basically every time you play any game with money, if you leave and come back, your EV continues to add up (even if it's negative from a house edge). Your bankroll doesn't care about "time" only number of hands played.

Example
Here's how you calculate your Expected Value (EV) in blackjack for X number of hands:

EV(x hands) = (X*AvgBet)*(HouseEdge)

You tell me in there where "time" or leaving and coming back has anything to do with your expected value in the game. Re: it doesn't. If you play 100 hands, then your EV at 100 hands is (100*AvgBet)*(HouseEdge). If you play another 100 hands, then your EV is (200*AvgBet)*(HouseEdge)... time/leaving doesn't matter. It's simply a continuation of your total number of hands.

if I play a coin 1000 times without a break without raising or lowering the bet and at the end I lose then starting again in a couple of days, according to the law of large numbers for 1000 shots, I have to recoup ???

Quote: evs

if I play a coin 1000 times without a break without raising or lowering the bet and at the end I lose then starting again in a couple of days, according to the law of large numbers for 1000 shots, I have to recoup ???

Incorrect. The law of large numbers more/less says in the long run (aka a large enough sampling size) things will work out to their mathematical reality. However, another 1,000 flips does not mean you'll necessarily be in the long run. Have you heard of N0 (n-zero)?

Each and every flip, assuming a fair coin, is an independent trial. So for example if 10 heads in a row come up, what's the probability of heads coming up on the 11th flip? Well, it's still just 50%. The previous 10 heads have no baring whatsoever on the next flip. Similarly the previous 1,000 flips where you lost all your money does not mean if you play another 1,000 flips you'll get all of your money back. Mathematically you will end up being in the zone of your EV +/- 3SD (SD = Standard Deviations).

For example, with blackjack... if you play 1,000 hands, WITH AN EDGE FROM CARD COUNTING, with an average bet of $100, your "range" you could end up in is the following:

AvgBet = $100
AvgAdvantage = 1.5%
OriginalSD = VarianceOfBJ * AvgBet = 1.15*100 = $115

SD(x hands) = Sqrt(x) * OriginalSD

EV(1,000 hands) = (1000*100)*(.015) = +$1,500
SD(1,000 hands) = Sqrt(1,000) * 115 = $3,636.62 ... 3SD = $10,909.86

This means with 99% confidence (read up on wikipedia about SD if you haven't heard of it) after 1,000 hands of blackjack, while counting cards with an advantage, with that average bet you can expect to make $1500... +/- $10,909.86.

Thus, you could be UP $12k or DOWN $9k and you still would be completely in the realm of mathematical possibility. It isn't until you reach the long run (N0) for the law of large numbers to show how it works. Let's look at one more example, this time after the card counter plays 100,000 hands of blackjack with an advantage...

EV(100,000) = (100,000*100)*(.015) = $150,000
SD(100,000) = Sqrt(100,000) * 115 = $36,336.19 ... 3SD = $109,098.58

So you see now, after a large enough sampling size and hitting "the long run" the card counter will be +$150,000 +/- $110,000... thus if they get as unlucky as they possibly can, they still mathematically CAN'T be down any money. If they have the worst variance possible, they will still be up $40,000. Of course if they run the best possibly they'll be up $260,000... but I'm hoping you can see here how the larger sampling size and leveraging the law of large numbers helped the card counter realize their advantage over the long run.

Again - all throughout those 100,000 hands being played the counter left, didn't play for a week, whatever, and it didn't matter. Time doesn't matter, only the number of betting events.

Quote: evs

the second connection of player B is 101 (continuation of the first game) or 1 (the beginning of a new game)?

What does that mean? What are the implications if it's a "new game" that would distinguish it from a "continuation"?

I agree with whoever said it sounds like a semantics question.

gentlemen, I have something to say. but I'm already asleep. I do not know about N0. till tomorrow!

quotation: Determine for what process you expect - for a new one (the old one is not taken into account) or the process continues with the same initial moment. (as we see there is a difference!)

to which I replied: In practice, such a characteristic as the mean square deviation of a random sequence (+ 1-1) at a certain moment (the number of 100 tests) is (+1) 4 sigma (√ (100 × 0.5 × 0.5) = 5; 5 * 4 = 20 total (+1) = mathematical expectation (50) + 20 = 70)! the observer understands that further deviation towards (+1) is extremely unlikely and, on the basis of this, assumes that (+1) (-1) a certain time will go nostril to the nostril or will predominate (-1). for the new observer not put in the course of the case the probability of deviation into one sigma and in (+1) and B (-1) will be equally probable! hence we draw conclusions that colleagues are extremely careless and do not keep a test log! but how does the random sequence (+1; -1) refer to the change of the observer? in my opinion she just will not notice.

an experienced mathematician who has repeatedly brought to such an issue grumbles over his shoulder "most likely the coin is not perfect and God knows how many times she submitted before the test and which side and generally she has no memory!" and will gallop to continue to solve their favorite equations. and the coin is new, perfect and before the tests on it and the fly did not sit! so it is logical to assume that (+1) (-1) a certain time will go nostril to the nostril or will prevail (-1). hence the conclusion is scary. the random sequence (+1; -1) is continuous! I did not get an answer!

citation: The coin does not possess memory, therefore the test log of some new information will not give.

to which I replied: a random sequence with a large number of steps approaches a Gaussian (which I use). There is no memory, but by virtue of that the probability is 0/1 = 50% the result is predictable? 1 and 0 will appear with a predictable frequency. I did not get an answer!
citation: Many players mis-apply the concept of 'the law of large numbers' and this may be tripping you up too. It is true that an anomaly like 67% heads tossed in small numbers will almost certainly disappear in large numbers, but the way this happens is not by the coins developing a 'will to correct' - being inanimate objects. The anomaly disappears because the larger set of numbers will cause that anomaly to become insignificant.

Explain with the help of formulas the development of events!