That's correct.

Indeed, as I understand it if the month ends at a loss it is prefinanced by the largest stockholder and forwarded to the next month where the large shareholder will first recoup their loss only after which remaining profits will be distributed.

Wasn't it still May when you posted that?  I think it's still May now for another hour maybe, server time.

I'm using "expected" in the statistical sense, and "in a row" in the usual sense. So the derivation is for mean number of losses in a row for a game with a win probability of 'p', loss probability 'q' (q=1-p), and number of bets, 'bets'.

As for ignoring "losses in a row", of course. Expected losses in a row can of course be derived the same way by redefining p and q.

Edit: That should be "expected longest number of wins in a row".

I've confused "expected length of a run" and "expected longest run", and probably confused you too.

I can derive the "expected length of a run":

Expected length of a run

For a series of Bernoulli trials where:

P(win) = p
P(loss) = 1 - p

the probability of k of losses before a win is the probability of k - 1 losses multiplied by the probability of 1 win:

P(X = k) = (1 - p)^(k-1) * p

From this the expected number of losses before a win can be derived.

E(X) = sum(n = 1 to n = k)  [ n * (1 - p)^(n - 1) * p ]

= p * sum(n=1 to n=k) [ n * (1 - p)^(n - 1)]

The derivative of a geometric progression sum(n=1 to k) n * x^ (n - 1) = 1/(1 - x)^2, so substituting (1 - p) for x:

p * sum(n=1 to n=k) [ n * (1 - p)^(n - 1)] = p * 1 (1 - ( 1 - p))^2

 = p p^2 = 1/p

So the expected number of losses before a win is 1 p

This holds true for the number of runs in a win, by redefining

P(loss) = p
P(win) = 1 - p


Expected longest run

I'm pretty sure I didn't derive this one. Maybe it was dooglus.


As far as I can tell, neither converge to the correct answer although they clearly converge on each other. I'll see if I can figure it out.





Edit:  I found a better estimate here: http://gato-docs.its.txstate.edu/mathworks/DistributionOfLongestRun.pdf

Expected longest run = -log(n*p)/log(q) + 0.57722.../log(1/q) -1/2

Note: 0.57722... is the Euler-Mascheroni constant, which is the limit(n -> infinity) { sum(k = 1 to k = n)[1/k] - log(n) }

The fan in my laptop makes a horrible noise every time I generate a new report for this thread.  I guess something's wearing out, and the extra load caused by loading the whole blockchain into RAM drives it too hard.

I've ordered a new laptop which should be here within a week, but until then I won't be posting the reports very regularly.  I'd like this laptop to keep working until I get the new one set up.

Having said that, I'm running a report right now because I'm curious to see how profits have been over the last couple of days, and will post it when it's done.

Something tells me that this month is a good sdice month.

Me too, I can't believe my eyes. I would be very surprised that this is from someone who does not have a lot of money. Makes no sense. It would be like buying a ferrari with your pension fund. Ofcourse exception to the rule always exist. But I think the rule is that rich people play big, not middle class people.

New laptop arrived, how exciting!  I get to spend days getting it set up just the way I like it! 

Money Mayweather bet $5.9 million on Miami Heat -7 over Indiana in game 7.

This level of betting is still small potatoes relative to the past, but does look pretty good so far...  At any moment S.DICE has the potential to have an explosive month of revenue.  I bought BTC after many people gave up on it, around $2, because it had the potential to go back to $33.  Well, apply the same logic to S.DICE...