So I have a math probablity question.
Lets say there is a pile of money that I can win. I get 15 random numbers assigned to me (i get no say in the matter) and then a number out of 100 is drawn at random. If I hit on that 15% chance I win the money.
Now lets say I get three shots. Get 15 random numbers, number gets drawn out of 100 if I win I win game stops, If i lose we do it again. Get 15 random numbers, number gets drawn out of 100 if I win I win if I lose we do it again. Get 15 random numbers, number gets drawn out of 100 if I win I win if I lose I lose.
So basically I have a 15% chance of winning three times.. what is my overall chances of winning the money?
In: 2
Just some clarification:
Did it say anywhere that the 15 random numbers you draw are always going to be different?
EG you will never have a situation where you are assigned the number 6 twice in your 15 numbers.
If nothing says and you can get duplicates.
Then it doesn’t actually change much.
You’ve received numerous explanations on how to get a 15% chance in 3 tries.
Now you’re dealing with a 1% chance, 45 times.
Just some clarification:
Did it say anywhere that the 15 random numbers you draw are always going to be different?
EG you will never have a situation where you are assigned the number 6 twice in your 15 numbers.
If nothing says and you can get duplicates.
Then it doesn’t actually change much.
You’ve received numerous explanations on how to get a 15% chance in 3 tries.
Now you’re dealing with a 1% chance, 45 times.
Lots of great replies here with the right answer already as well as lots of theory.
You mentioned you aren’t a “math genius” and so I would suggest just using an online calculator in case you ever have a similar math problem in the future.
A quick Google found me this one:
https://stattrek.com/online-calculator/binomial
In your case, probability of success on trial = 0.15, number of trials = 3, and number of successes = 1. Note that it provides you a variety of answers. The one you are looking for here is the one that is P(X>=1) which suggests you win ‘at least once’ which differs from for example the P(x=1) which is winning ‘exactly once’.
Lots of great replies here with the right answer already as well as lots of theory.
You mentioned you aren’t a “math genius” and so I would suggest just using an online calculator in case you ever have a similar math problem in the future.
A quick Google found me this one:
https://stattrek.com/online-calculator/binomial
In your case, probability of success on trial = 0.15, number of trials = 3, and number of successes = 1. Note that it provides you a variety of answers. The one you are looking for here is the one that is P(X>=1) which suggests you win ‘at least once’ which differs from for example the P(x=1) which is winning ‘exactly once’.
This is actually relatively simple to calculate.
To find the probability of winning at least once in three attempts, find the probability of not winning in three attempts, and subtract it from 1.
The probably of not winning a game is 85%, or 0.85.
The probability of not winning three times is 0.85*0.85 *0.85. 0.85^(3)
And then the probability of winning at least once is 1 – 0.85^(3)
Which gives 0.3859. About 39%.
And you can plug in any number of attempts into the same formula to figure it out.
If you draw a number 5 times, it’s 1 – 0.85^(5) = 0.56.
So you only need to draw a number 5 times before you’re more likely to win than lose. That’s nice.
This is all assuming that the different draws are independent, i.e. any number can be picked more than once.
This is actually relatively simple to calculate.
To find the probability of winning at least once in three attempts, find the probability of not winning in three attempts, and subtract it from 1.
The probably of not winning a game is 85%, or 0.85.
The probability of not winning three times is 0.85*0.85 *0.85. 0.85^(3)
And then the probability of winning at least once is 1 – 0.85^(3)
Which gives 0.3859. About 39%.
And you can plug in any number of attempts into the same formula to figure it out.
If you draw a number 5 times, it’s 1 – 0.85^(5) = 0.56.
So you only need to draw a number 5 times before you’re more likely to win than lose. That’s nice.
This is all assuming that the different draws are independent, i.e. any number can be picked more than once.
Start from this:
You have a formula for “what is the chance of a thing happening *every* time”, and that’s to multiply the chances of each trial together.
A chance of a coin flip being heads both times in two flips is 1/4, because each flip had a 1/2 chance, and 1/4 is 1/2 of 1/2.
A chance of a coin flip being heads 3 times in 3 flips is 1/8, because that’s 1/2 of 1/2 of 1/2.
So if you have this rule, then you can invert the odds of it to figure out your chances of FAILING to have a thing happen every time.
If your odds of getting 3 heads in 3 flips is 1/8, from the above, then your odds of getting anything OTHER than 3 heads in those 3 flips is the remaining 7/8. (which is 1 – 1/8).
So you do that here with your example. If success means you just have to hit 15% once in 3 tries, that’s the inverse of saying “to fail you have to hit 85% every time for 3 tries.”
And the odds of hitting 85% three times in a row is 85% of 85% of 85%.
Which is 61.4125%. That’s your chance of NOT hitting 15% 3 times in a row.
So the inverse of that is your chance of hitting 15% at least once out of 3 tries: 100% – 61.4124% = 38.5875%.
Start from this:
You have a formula for “what is the chance of a thing happening *every* time”, and that’s to multiply the chances of each trial together.
A chance of a coin flip being heads both times in two flips is 1/4, because each flip had a 1/2 chance, and 1/4 is 1/2 of 1/2.
A chance of a coin flip being heads 3 times in 3 flips is 1/8, because that’s 1/2 of 1/2 of 1/2.
So if you have this rule, then you can invert the odds of it to figure out your chances of FAILING to have a thing happen every time.
If your odds of getting 3 heads in 3 flips is 1/8, from the above, then your odds of getting anything OTHER than 3 heads in those 3 flips is the remaining 7/8. (which is 1 – 1/8).
So you do that here with your example. If success means you just have to hit 15% once in 3 tries, that’s the inverse of saying “to fail you have to hit 85% every time for 3 tries.”
And the odds of hitting 85% three times in a row is 85% of 85% of 85%.
Which is 61.4125%. That’s your chance of NOT hitting 15% 3 times in a row.
So the inverse of that is your chance of hitting 15% at least once out of 3 tries: 100% – 61.4124% = 38.5875%.
You have two ways of calculating this
1. You can try and calculate the probability that you win the first time *or* you lose the first and win the second *or* you lose the first two and win the third.
2. You can calculate the probability that you won’t win any of them, and subtract that from the whole. (do you see why this works?)
The first approach is complicated. It requires figuring out the probability of three different scenarios and working out how the hell to combine them.
The second approach is simple: You just need to work out the probability of a single scenario: something happening three times in a row (where *something* is “not winning”)
On each draw, there’s an 85% (i.e. 0.85) probability of not winning. To work out the probability you lose all three, you just take 0.85 × 0.85 × 0.85. That’s the probability that you lose all three, so the probability that you win at least one of them is 1 – (probability of losing all three)
You have two ways of calculating this
1. You can try and calculate the probability that you win the first time *or* you lose the first and win the second *or* you lose the first two and win the third.
2. You can calculate the probability that you won’t win any of them, and subtract that from the whole. (do you see why this works?)
The first approach is complicated. It requires figuring out the probability of three different scenarios and working out how the hell to combine them.
The second approach is simple: You just need to work out the probability of a single scenario: something happening three times in a row (where *something* is “not winning”)
On each draw, there’s an 85% (i.e. 0.85) probability of not winning. To work out the probability you lose all three, you just take 0.85 × 0.85 × 0.85. That’s the probability that you lose all three, so the probability that you win at least one of them is 1 – (probability of losing all three)
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