Re: I'm bored, let's debunk another martingale gambling strategy! New experiment 🏇!

Quote from: stompix on Today at 02:47:29 PM

Quote from: KTChampions on Today at 12:46:14 PM

- I can calculate the exact number of bets, but I will answer empirically: no, I will not answer, maybe I will write when I find the formula (which describes the length of the segment needed to obtain a series of a certain size), I have it saved somewhere haha.

I have two models ranging from 100 to 2400....so god knows, as one is already wrong!
~

Fractional odds are breaking my head. Am I right in assuming that the average odds you bet on are 2.0 (in decimal format)?

Quote

This is a version of Martingale with stop loss then doubling the chain so it goes
1/2/4/8/16 loss, you stop -31
2/4/8/16/32 but only 16 times to recover the previous lost bets 2*16>31, if it happens again
4/8/16/32/64 again
8/16/32/64 wipeout will happen in 6 consecutive losses so a least 6 days, far more likely at least 12!

It is quite difficult to calculate the probability of losing even for one day, since it is unknown how many bets you will make on this day. As I understand it, this is a random value. But I can calculate the probability of getting 5 losses in an infinite game: 2⁽⁵⁺¹⁾-2 = 62 bets. This is provided that we have averaged your odds to 2.

So, if you get a loss (5 in a row) every 62 bets, then the minimum value of the total loss will be 62x6=372 bets. But we need to take into account that you can "return" to the previous day. Hmmm... how to do it?
To get back you need 16 wins, so 32 bets on average? So (if I'm not mistaken haha) 32/62 is the probability of your return (looks good, but we don't take into account the bookmaker's margin, let's continue like this for now).
Then everything is simple: the probability that you will not return is the inverse of the probability that you will return. That is, 1 - 32/62 = 30/62. To lose, you need to not return 5 times (from the first doubling to the base, from the second doubling to the first doubling, etc.), so (30/62)⁵ =~ 0.0265 Like 2.65% Oops, no, this is a random walk problem, there are different calculations, because if you return from the 4th day to the 3rd, you start trying to lose from the 3rd day and not from the very beginning.
Thank God 5 is a small number and the solution is known = 1/63
So, if the bookmaker did not take a margin, you would lose approximately in 63 tries with 62 bets, i.e. 3906 bets.

If we take into account the margin, then 1.9⁽⁵⁺¹⁾-2 = 45 bets to get day lost (5 lost bets in a row).
To lose completely you need 63x45 = 2865 bets.

And this is not the final figure yet, in theory it will be even worse. I have to think about how to correctly estimate the probability of wandering - returning to the previous level/doubling the base rate, obviously it is not 50/50 (all the same because of the bookmaker's margin).

I need to rest Grin