Xero wrote:In a gambling game of chance, let's say roulette. There are 38 numbers that may win. 1-36, 0, 00. the odds a single number will hit are 1:38. After playing the game for 3 rounds, 0 hit twice and 36 has only hit once. A history of the last 100 games reveals that black has occurred 60 times, red 35 and green (0, 00) 5.
what is the probability the next winning number will be 0? What is the probability it will be black?
I have flipped a coin 10 times, 8 times it landed on heads, twice on tails. What is the probability it will be heads again?
If you flipped a coin 10 times, 8 times it landed on heads, twice on tails, what is the probability it landed on heads 8 times? 100%. If you flip the coin another 10 times, what is the probability it will land on heads 8 times again? This is a two-outcome experiment repeated multiple times (ignoring any outcomes where the coin does not land on heads or tails, such as it lands on its side, or it is sucked into a dimensional vortex, or any other outcome that is neither heads nor tails). Let's consider all possible ways that it can land on heads 8 times and tails twice. I will write all possible sequences of ten characters where the characters are either H or T (for heads or tails respectively) that satisfy the condition that 8 of the characters are H and two of the characters are T:
I put it in spoiler tags because there are 10C2=10C8=45 different ways of writing that sequence (10C2 is the evaluation of the probability expression nCr where n=10 and r=2 or r=8). What is the probability that if we flip a coin ten times, we get one of those sequences? Let's look at the first one: HHHHHHHHTT. All we do is multiply the probabilities: (0.5)*(0.5)*(0.5)*...*(0.5)=(0.5)^10=0.0009765625. It is not possible that any flip of a coin will be simultaneously heads and tails (since the experiment already stated it ignores any outcome that is not either heads or tails, repeating the experiment as many times as is necessary to obtain 10 trials where the outcome is either heads or tails). Thus, these probabilities are disjoint. So, by the sum principle of probabilities, if we have multiple disjoint events, the probability that one of them will occur is the sum of their probabilities. We have 45 disjoint events, all with the same probability. So, the probability that we flip a coin ten times and we have exactly eight flips that are heads and exactly two flips that are tails is 45*0.0009765625=0.0439453125. So, what are all of the possibilities:
The following table is the number of tails flipped and the probability of achieving that result:
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Number of Tails Probability Formula
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0 0.0009765625 (10C0)*(0.5)^0*(0.5)^10
1 0.009765625 (10C1)*(0.5)^1*(0.5)^9
2 0.0439453125 (10C2)*(0.5)^2*(0.5)^8
3 0.1171875 (10C3)*(0.5)^3*(0.5)^7
4 0.205078125 (10C4)*(0.5)^4*(0.5)^6
5 0.24609375 (10C5)*(0.5)^5*(0.5)^5
6 0.205078125 (10C6)*(0.5)^6*(0.5)^4
7 0.1171875 (10C7)*(0.5)^7*(0.5)^3
8 0.0439453125 (10C8)*(0.5)^8*(0.5)^2
9 0.009765625 (10C9)*(0.5)^9*(0.5)^1
10 0.0009765625 (10C10)*(0.5)^10*(0.5)^0
If we add up the probabilities, we get a total of 1 (in other words, we exhausted all possibilities, and the number of tails we get must be a number from 0 to 10). Now, what is the probability that in those 10 flips, we get at least 1 tails? That is the probability that we don't get heads every flip, or 1-0.0009765625=0.9990234375. So, there is a 99.9% chance that if we flip a coin 10 times, we will get at least one heads.
We can create a similar table for success or failure of receiving legendary items during 10 kills:
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Number of Items Probability Formula
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0 0.99900044988002099748020998800045 (10C0)*(0.0001)^0*(0.9999)^10
1 0.00099910035991601259874008399640 (10C1)*(0.0001)^1*(0.9999)^9
2 0.00000044964012597480314974801260 (10C2)*(0.0001)^2*(0.9999)^8
3 0.00000000011991602519580041997480 (10C3)*(0.0001)^3*(0.9999)^7
4 0.00000000000002098740314958003150 (10C4)*(0.0001)^4*(0.9999)^6
5 0.00000000000000000251874025197480 (10C5)*(0.0001)^5*(0.9999)^5
6 0.00000000000000000000020991601260 (10C6)*(0.0001)^6*(0.9999)^4
7 0.00000000000000000000000001199640 (10C7)*(0.0001)^7*(0.9999)^3
8 0.00000000000000000000000000000045 (10C8)*(0.0001)^8*(0.9999)^2
9 0.00000000000000000000000000000000 (10C9)*(0.0001)^9*(0.9999)^1
10 0.00000000000000000000000000000000 (10C10)*(0.0001)^10*(0.9999)^0
(Note: The last two probabilities are so small, their first significant digit is too far to the right to be accurately represented).
Now, if we want to calculate the probability of getting at least one item, we can add up the probability of getting exactly 1, exactly 2, exactly 3, ..., exactly 10, or we can just subtract the probability of getting 0 from the total probability of 1: 1-0.99900044988002099748020998800045, which is just less than 0.1%.
Let's say we calculate this same table for x kills (omitting the probability, keeping only the formula):
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Number of Items Formula
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0 (xC0)*(0.0001)^0*(0.9999)^x
Now, the probability of finding an item after x kills is 1-(xC0)*(0.0001)^0*(0.9999)^x=1-(0.9999)^x, which is the formula I used in my original post.
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