- #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.
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.
