Basic probability question

[Guest]

Hi.

I'm trying to solve an embarrassingly simple problem, drawn from real life. It's not about betting, but it's a fitting way to illustrate it.

Picture that you will win 49 times out of 50. The 50th time you'll lose.
You are not limited to only betting 50 times, you will do it
indefinitely.

Question 1:
What is the probability that you will get two losses after each other?

Question 2:
If you get one loss, what is the probability that the next bet also would be a loss?

Question 3: If you get one loss, is there a higher probability that the bet after the next would be a loss, than the next?

Thanks in advance!

[romsek]

Hi.

I'm trying to solve an embarrassingly simple problem, drawn from real life. It's not about betting, but it's a fitting way to illustrate it.

Picture that you will win 49 times out of 50. The 50th time you'll lose.
You are not limited to only betting 50 times, you will do it
indefinitely.

Question 1:
What is the probability that you will get two losses after each other?

Question 2:
If you get one loss, what is the probability that the next bet also would be a loss?

Question 3: If you get one loss, is there a higher probability that the bet after the next would be a loss, than the next?

Thanks in advance!

Google Gambler's Fallacy. In general trials of a game will be independent of one another so you may have just won the last 50 games and the probability of losing the next one is still 1/50 as always.
This isn't true if the game trials aren't independent. If they aren't you'll need to specify a distribution that captures this dependence somehow.

[Guest]

Thank you! I will read up on that.

I guess that answer mostly covers Q3 and part of Q2. If someone would have an idea of also Q1/Q2 it would be appreciated.

[romsek]

Thank you! I will read up on that.

I guess that answer mostly covers Q3 and part of Q2. If someone would have an idea of also Q1/Q2 it would be appreciated.

Assuming independent games

let [tex]P[\text{win}]= p =\dfrac{49}{50}[/tex]

[tex]P[\text{two consecutive losses}] = (1-p)^2 = \dfrac{1}{2500}[/tex]

2) Since we have independent games the result of the previous game has no effect on the result of the current game.

[tex]P[\text{lose game $n$ | lost game $n-1$}] = P[\text{lose game $n$}] = (1-p) = \dfrac{1}{50}[/tex]

[Guest]

[tex]p=\frac{1}{50}[/tex]
[tex]P(A)=P(B)=P(C)=p[/tex]
[tex]P(A \cap B)=P(A) \cdot P(B) =p \cdot p=\frac{1}{50} \cdot\frac{1}{50}=\frac{1}{2500}[/tex]
[tex]P(B|A)= \frac{P(A \cap B)}{P(A)} =\frac{p \cdot p}{p}=p=\frac{1}{50}[/tex]
[tex]P(C|A)= \frac{P(A \cap C)}{P(A)} =\frac{p \cdot p}{p}=p=\frac{1}{50}[/tex]

[AngleWyrm]

I guess that answer mostly covers Q3 and part of Q2. If someone would have an idea of also Q1/Q2 it would be appreciated.

Independent random variables (as in NOT dependent) have no historical context; they are effectively simultaneous. They're called sampling with replacement. Examples are flipping coins and rolling dice.

Dependent random variables on the other hand do have a before/after relationship between them, where one changes the environment in which the other is calculated. They're called sampling without replacement. Examples are drawing cards from a deck, or lottery balls from a jar.

The statement P(b|a) "Probability of b, given a" refers to a dependent arrangement where the variables a and b have a before/after relationship; a comes before b, and then b is calculated afterward.

[apprentus]

Answer 1:
The probability of losing a single bet is 1/50. Since each bet is independent, the probability of losing twice in a row is the product of the probabilities of losing in each bet, which is (1/50) * (1/50) = 1/2500.

Answer 2:
The probability of losing the first bet is 1/50. If you lose the first bet, there are only 49 possible outcomes for the second bet, since you've already lost one. Therefore, the probability of losing the second bet given that you lost the first is 1/49.

Answer 3:
No, there is no higher probability that the bet after the next would be a loss, compared to the next bet. Each bet is independent of the previous outcomes, and the probability of losing any given bet is always 1/50. The outcome of the previous bet does not affect the probability of the next bet being a loss.

[Guest]

Hi.

I'm trying to solve an embarrassingly simple problem, drawn from real life. It's not about betting, but it's a fitting way to illustrate it.

Picture that you will win 49 times out of 50. The 50th time you'll lose.
You are not limited to only betting 50 times, you will do it
indefinitely.

Question 1:
What is the probability that you will get two losses after each other?

Question 2:
If you get one loss, what is the probability that the next bet also would be a loss?

Question 3: If you get one loss, is there a higher probability that the bet after the next would be a loss, than the next?

Thanks in advance!

Q1: The likelihood that two consecutive losses will finally equal one.
Q2: 1/50 is the next loss following a loss.
Q3: (independent) Next or after-next loss equals both 1/50.

[Guest]

1. Two losses in a row: 0.02*0.02=0.0004(0.04%)
2. Next loss after one: 0.02(2%)
3. Both have the same chance - events are independent