Martingale, calculation of probability - Markov chain needed?

In summary, the Martingale betting system is a strategy used in gambling and other scenarios where the player doubles their bet after every loss in order to eventually make a profit. It is based on the assumption of a 50% probability of winning, but in reality, this is not always the case. Markov chains are used to calculate probabilities in the system, but there are limitations and drawbacks to using this strategy, such as assuming an unlimited bankroll and not accounting for long losing streaks. Overall, the Martingale system is not a reliable or recommended strategy for decision making.
  • #1
kravky
22
3
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.
 
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  • #2
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?
 
  • #3
tnich said:
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
 
  • #4
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.
 

1. What is the Martingale betting strategy?

The Martingale betting strategy is a gambling system in which the player doubles their bet after each loss, with the goal of eventually breaking even and making a profit. It is based on the belief that a win is inevitable after a series of losses.

2. How does the Martingale strategy relate to probability?

The Martingale strategy is based on the assumption that the probability of winning is always 50%, regardless of previous outcomes. However, this is not always the case and the strategy can lead to significant losses in the long run.

3. What is the role of Markov chains in the calculation of probability in the Martingale strategy?

Markov chains are used to model the probability of future outcomes in the Martingale strategy. This is because the strategy is based on the assumption that future outcomes are independent of past outcomes, which is a key concept in Markov chain theory.

4. Can the Martingale strategy be used to guarantee a profit?

No, the Martingale strategy cannot guarantee a profit. While it may result in short-term wins, in the long run it is mathematically proven that the strategy will lead to losses due to the finite amount of money a player has and the potential for a long streak of losses.

5. Are there any alternatives to the Martingale strategy for calculating probability in gambling?

Yes, there are many alternative strategies for calculating probability in gambling. Some examples include the Kelly Criterion, the Paroli system, and the D'Alembert system. These strategies may also have their own drawbacks, so it is important for players to understand the risks involved and to gamble responsibly.

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