About how impossible mateo's winning were.. I made a few mistakes when
computing the prob and corrected it. Sorry ! Since it was discussed
here and
here, I thought I'd come back.
The approximation actually gives a probability of winning +600 btc like mateo did of about 1 in 1 to 10 millions... (for extreme values like here final result is highly influenced by the computers numerical methods and their accuracy).
See what a hundred thousand simulations of normal 60k bets looks like (histogram=simulations, curve=model), the max profit was 424.
Play with the numbers...
I do not think this is the correct model for computing the probability of mateo. It seems you assume that the player arrives and blindfolded bets for 60K times and at the end he just leaves the website with what is left in the pocket. But this is not like that, the players bets until there is money in the bankroll and for fun (I guess winning the bankroll is a lot of fun). If he just wins the bankroll in the middle of playing he just goes away, without extra rollings.
Also I cannot reproduce the result the 1 to 10 millions estimate, I run a simple script simulating a player betting 60K times at 0.5% of the bankroll (half-kelly as in the website) and on 10000 repetitions I get 4 times a final bankroll smaller than 15%.
Actually, isn't that how you should model someones gambling session? Blindly gambling until they stop when they've run out of money, or when they've hit a certain number of bets? (there is 0% chance in which a player can walk away with 100% of the total casino bankroll) Which is what Joecker's doing in his model.
I'm sorry, but you're using Dubious math. Kelly Betting is actually the amount the House "bets" which will increase their bankroll the quickest. The Player does not enjoy kelly bets as they are betting in a -EV situation. Kelly Betting is only for +EV (which in the case only pertains the House). Thus the formula you are using to determine the chances of bankroll going to a certain % is not correct. The correct % will be much much lower than 2%.
I do not follow your reasoning (maybe you can be clearer?) but I am using a very standard tool. If you run some simulations you can check by yourself that the formula you find in the link is correct.
What i'm saying is that your premise is incorrect, Because you are using the Kelly Formula to predict risk of ruin. But the formula requires the betting situation to be +EV (positive). Dice is -EV (1% house edge remember).
What joecker has done is the correct model.. let me explain in more detail. here is a simple model of how this dice session should go. The Total wagered amount by mateo should be around 50-60K, his profit around 500-600. The Standard Deviation of dice at 49.5% is 1 (other %'s will provide very different standard deviations). His total bet # is 60K. So average bet size is 1BTC. We can then find the probability of this happening by doing a statistical test. I have reposted Joecker's analysis below.
(...) By the way, for fun can someone explain the math of how monstrously improbable Matteo's run of "luck" (fraud) was?
Let X_i be the random variable modeling the profit of each bet.
Assumptions:
1) profits went from 260 to -330, that is manlteo's profit = (+/-) 600btc
2) N=60k bets of 1btc each @ 2x payout
3) math bullshit (iid random variables), q=0.495, P[X=+1] = q, P[X=-1] = 1-q.
Let S= X_1+X_2+ ... + X_60000. We should expect E(S)= 60000 x ( 1x0.495 + (-1)x0.505 ) = -600 (a loss).
We want to know P[S >= 600]. Central limit theorem states that [S - E(S)] / stdev(X)*
\sqrt(N) =: Z
is normally can be approximated by a normal distribution N(0,1).
var(X) = E(X^2) - E(X)^2 = q+1-q - (2q-1)^2 = 4q(1-q) = 0.9999, so stdev ~ 1
So
P[S >= 600]
~= P[Z >= [600 -(-600)] / (1*
\sqrt(N)) ] = P[Z>= 1200/244] = 1-P[Z<=4.89922] =
4.8x10^-7.
To be honest, the average betsize might be smaller than 1BTC since after the bankroll dropped smaller than 200btc, the maxbet was smaller than 1btc. This would make the analysis above on the conservative side. With a Z-score approaching almost 5, the evidence points more and more to something virtually impossible happening. It could happen, but at the probability of 4.8x10^-7.
Joecker, I don't understand why you need to model the stdev as approaching 1. The standard deviation of the winnings of dice played at 49.5% is = 1. you're probably using a different method to calculate that?
i just used a dice simulator over millions of rolls to determine the STDEV of the dice game at different %s, since different % really do provide a different standard deviation.
