Statistic / Probability Roulette Question

[da_creme]

Red is 47.7% per spin. The chance of 3 reds in a row is about 10% and getting 4 in a row is about 5%. What is probability or any other result of getting a black [47.7%], after 3 reds.

 

[Dr.Peterson]

Red is 47.7% per spin. The chance of 3 reds in a row is about 10% and getting 4 in a row is about 5%.

I suppose you mean that the probability of getting a red on one spin is 47.7%. You are right that this implies the probability of getting RRR is 0.477^3 = 0.108531333, about 11%, and the probability of getting RRRR is 0.477^4 = 0.051769445841, about 5%.

What is probability or any other result of getting a black [47.7%], after 3 reds.

What does this mean?

If you already got 3 reds, that doesn't change the probabilities for the next spin; it is still 47.7% for red, and 47.7% for black. If you're thinking otherwise, search for "gambler's fallacy".

 

[da_creme]

I don’t disagree because I have always heard no matter what flipping a coin is always 50/50, but I then don’t understand why it’s not 47.7% for getting 1 red in a row or 10 in a row? Can you help me understand that? If there is a 5% difference in getting 4 reds in a row versus 3 how would that calculate into the 4th time or are they 2 different probabilities?

 

[Dr.Peterson]

Each time an event happens, you multiply by its probability. One red happens 47.7% of the time; a second will happen 47.7% of that 47.7% of the time, so that two happen 0.477*0.477 = 0.227529 = 22.8% of the time.

The probability of that second red is 47.7% by itself; but the probability of two in a row is smaller, because both have to have happened.

 

[mikegonz00]

I don’t disagree because I have always heard no matter what flipping a coin is always 50/50, but I then don’t understand why it’s not 47.7% for getting 1 red in a row or 10 in a row? Can you help me understand that? If there is a 5% difference in getting 4 reds in a row versus 3 how would that calculate into the 4th time or are they 2 different probabilities?

Your question at the end is related to independence. For example, when flipping a coin, let's say you've flipped the coin four times and gotten heads each time. Assuming a fair coin, the probability that the fifth toss will be heads is still 1/2 (even though you've just flipped heads four times in a row). This is because the event of getting heads is independent of previous throws; the probability is always 1/2 because the coin is fair! We're assuming in your example that landing on red at any spin is independent of previous spins. You could say P(RRRR)=P(RRR)*P(R) if you really want to use previous data since P(RRR)=P(R)*P(R)*P(R), but the more usual solution is just P(RRRR)=(P(R))⁴.

Maybe you've already learned about conditional probabilities, but that's related as well. I'm sure you could think of an example of real-world events that are not independent.