Martingale, calculation of probability - Markov chain needed?

[kravky]

I am trying to estimate probability of loosing (probability of bankrupt $Pb$) using Martingale system in betting.
I will ilustrate my problem on the following example:

Let:
$p$ = probability of NOT getting a draw (in some match)

We will use following system for betting:
1) We will bet only on draws (probability of draw in a match = $1-p$)
2) If we loose our bet, next bet will be doubled.
3) We have limited amount of money
Lets assume:
$k$ = max. number of matches in a row without a draw (streak of $k$ matches without a single draw = no draw in $k$ matches), that we can handle with our money.
So $k+1$ matches in a row without draw would mean bankrupt. $k$ represent something like a limit for us.
$n$ = number of matches we will bet on. usually $n >k$.

My question is, what is the probability of bankrupt ($Pb$) for given $k, n$ - it means - what is probability of getting streak of at least $k$ matches in a row without a draw (there will be no draw in $k$ matches), supposing betting on $n$ matches.

Lets consider, for exapmle, following values:
$p = 0.75$ (probability of draw is $1-p = 0.25$)
$k = 15$ (maximum number of "NO" draws streak is $k =15$)
$n = 200$ (we will consider a sample of 200 matches)

i tried to calculate it, but i am not sure my solution is correct:

i derived following formula:

$Pb = 1-(1-p^k)^{ (n/k)}$ .

My solution aprroach:
Probability of not getting a draw in k matches in a row is $p^k$.
probability of getting at least 1 draw in $k$ matches is $1 - p^k$.
Probability of getting at least 1 draw in every $n/k$ attempt is $(1-p^k)^{ (n/k)}$.
Probability of getting no draw (streak without a draw) at least in one of the $n/k$ attempt is $Pb = 1-(1-p^k)^{ (n/k)}$

It works at least when $n/k$ is integer. Value $n/k$ is like number of "attempts" in my calculation which i think does not have to represent reality very well.

For given values
$p = 0.75$
$k = 15$
$n = 200$
i found that $Pb = 0.16$.
It seems quite low for me.

Is my solution correct or it is more complex problem ? Is streak an independent event like I assumed or it should be treated as not the independent event - maybe something analogous to random walk process ?
If it is not correct, what is correct approach?
Should we use e.g. Markov chain (discrete time Markov chain like here https://wizardofodds.com/image/ask-the-wizard/streaks.pdf) ?

i would be very grateful if someone could present solution using discrete time Markov chain.

 

[tnich]

As I understand the Martingale system, you bet on the same outcome, doubling the amount bet each time, until you win or go bankrupt. If you don't go bankrupt first, you double your initial stake. The problem is that if you continue to bet using this strategy, you will eventually go bankrupt.

I do not follow your logic above. Is $p$ the probability of winning, or of losing? You say that you are doubling your bet when you lose. Are you also ceasing to bet when you win?

 

[kravky]

As I understand the Martingale system, you bet on the same outcome, doubling the amount bet each time, until you win or go bankrupt. If you don't go bankrupt first, you double your initial stake. The problem is that if you continue to bet using this strategy, you will eventually go bankrupt.

I do not follow your logic above. Is $p$ the probability of winning, or of losing? You say that you are doubling your bet when you lose. Are you also ceasing to bet when you win?

$p$ is probability of loosing

 

[tnich]

Suppose you bet amount $2^{k-1}b$ each time. If you stop betting after the 15th bet, then you lose a total of $(2^{15}-1)b = 32767b$, that is 32767 times your initial bet $b$. On the other hand, if you win at the $k$th bet for $k<15$, your total winnings are exactly $b$.

Supposing that you only have enough money for 15 bets, the probability of losing 15 times in a row is $p^{15}$, which is a very small number. At that point you are out of money and cannot start a new trial, so it does not make sense to look at $n>1$ trials.

By the way, the expected value of your winnings is zero when $p=0.5$. When $p>0.5$, the expected value is negative.