I hope there are some good mathematicians here!
What I want to ask is this: "On a single zero wheel, how do you calculate the probabilitily that any one of the last four numbers will repeat in the next, say, five spins?"
Quote from: purple on March 12, 2013, 08:03:31 AM
I hope there are some good mathematicians here!
What I want to ask is this: "On a single zero wheel, how do you calculate the probabilitily that any one of the last four numbers will repeat in the next, say, five spins?"
The probability is 4/37 for each of the 5 spins. That's 10.8%.
Thanks, but I think what you wrote is the probability that after placing a bet on any 4 numbers that any one of them will hit.4/37=0.108 x 100=10.8%
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Quote from: purple on March 12, 2013, 08:03:31 AM
I hope there are some good mathematicians here!
What I want to ask is this: "On a single zero wheel, how do you calculate the probabilitily that any one of the last four numbers will repeat in the next, say, five spins?"
Here's how to calculate that one of the last 4 numbers that hit will hit again once within the next 5 spins:
First, calculate the probability of
none of those 4 numbers hitting... which is 33/37 =
89%. That is the probability that on your next spin,
none of your numbers hit. Now, what is the probability that
none of your numbers hit for
5 consecutive spins? (in other words, that you lose 5 in a row.) Well, that would be 33/37 * 33/37 * 33/37 * 33/37 * 33/37 = 0.558, or
56%. This is telling you that in the next 5 spins, there is a 56% probability that your 4 numbers will
not come up even once.
But, what about the other 44%? That is the
opposite of the probability that none of your numbers show up; in other words, that 44% is the probability that at
least one of your 4 chosen numbers will hit
at least once within the next 5 spins.
If P = [Probability that you will lose 5 in a row] then: If P = 56%, then 1 - P = 1 - 0.56 = 0.44 = 44%.
So the probability that any one of the last four numbers repeats within 5 spins is 44%.
Quote from: nullified on March 12, 2013, 11:56:36 PM
Here's how to calculate that one of the last 4 numbers that hit will hit again once within the next 5 spins:
First, calculate the probability of none of those 4 numbers hitting... which is 33/37 = 89%. That is the probability that on your next spin, none of your numbers hit. Now, what is the probability that none of your numbers hit for 5 consecutive spins? (in other words, that you lose 5 in a row.) Well, that would be 33/37 * 33/37 * 33/37 * 33/37 * 33/37 = 0.558, or 56%. This is telling you that in the next 5 spins, there is a 56% probability that your 4 numbers will not come up even once.
But, what about the other 44%? That is the opposite of the probability that none of your numbers show up; in other words, that 44% is the probability that at least one of your 4 chosen numbers will hit at least once within the next 5 spins. If P = [Probability that you will lose 5 in a row] then: If P = 56%, then 1 - P = 1 - 0.56 = 0.44 = 44%.
So the probability that any one of the last four numbers repeats within 5 spins is 44%.
I hope you're not serious here.It doesn't take a genius to figure out your calculation is flawed.
Last hit 4 numbers is exact same as any 4 numbers. Remember, it will doesn't care what four numbers are ,roulette wheel doesn't care if the 4 numbers are the last four numbers or any 4 numbers
Next spin, probability of last 4 numbers come up is 4/37=10.8%
Next 5 spin，is (4*5)/(37*5)=10.8%
Next 100 spin, is (4*100)/(37*100)=10.8%
I found the answer from a book.
As I said the answer to what's the probability of ANY 4 numbers occurring the answer as I said is 4/37=0.108 x100=10.81%
but to the question:
Question: What is the probability that there will be no repeats within the next 4 spins? What is the probability that there will be at least one repeat within the next 4 spins?
The probability of 4 spins without a repeat is
(36/37) x (35x37) x (34x37)=42840/50653=0.845, or 84.5%
So the probability of one, or more, repeats within 4 spins is 1-0.845=0.155, or 15.5%
So I guess I've answered my own question!