Statistics and probability: bike raffles

So I entered a raffle to win a Bingham Built MTB. My number is 131. The drawing (really 3) will randomly pick three numbers 0-9 to get the three digit number (not quite how they are doing it but this makes the question a little more simple). The three drawings will be independent. An example will be:

Drawing 1: 5
Drawing 2: 6
Drawing 3: 7

Winning # 567

So are my odds better for winning having three different numbers like 123, 729, 891 than having a number with at least two numbers the same like 775, 131 955....

If memory serves I have a better chance of winning with 3 different numbers since I think (guessing) there are more three different number combinations than same number (meaning at least 2/3 same number) combinations.

Ok smartie pants lets here your arguements and explainations...

Are they selling exactly 1,000 tickets? (Numbers 000-999)

The repeating, or consecutive numbers may seem rarer, because they are the go-to scenarios for explaining statistics, but the odds are all the same.. For example- the odds of the draw-er pulling 1-1-1 are 0.1% or 1 in 1,000, but so are the odds of them pulling 5-6-7.

I'm assuming the selections are made with replacement based on your explanation.

Your odds would not change. the odds of 1-1-1 getting pulled are the same as 4-7-2 getting pulled. You have a 1 in 10 chance of pulling the #1 out of hat for the first number. With replacement you have the same 1 in 10 odds of pulling a 1 for the 2nd and 3rd place.

The probability would change dramatically. As the probability of of getting the same number is a function of the prior event. So think (1/10) x (1/10) x (1/10)

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Originally Posted by robin3mj

Are they selling exactly 1,000 tickets? (Numbers 000-999)
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Assume they are... for the excercise.

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Originally Posted by joosttx

So I entered a raffle to win a Bingham Built MTB. My number is 131. The drawing (really 3) will randomly pick three numbers 0-9 to get the three digit number (not quite how they are doing it but this makes the question a little more simple). The three drawings will be independent. An example will be:

Drawing 1: 5
Drawing 2: 6
Drawing 3: 7

Winning # 567

So are my odds better for winning having three different numbers like 123, 729, 891 than having a number with at least two numbers the same like 775, 131 955....

If memory serves I have a better chance of winning with 3 different numbers since I think (guessing) there are more three different number combinations than same number (meaning at least 2/3 same number) combinations.

Ok smartie pants lets here your arguements and explainations...

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Well, if they are using an online "random" number generator, a lot of those actually use something, like the temperature of your processor or the temperature of the weather in certain location to base their numbers off of. In the end it all depends on how they getting their numbers.

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Originally Posted by vvv321

Well, if they are using an online "random" number generator, a lot of those actually use something, like the temperature of your processor or the temperature of the weather in certain location to base their numbers off of. In the end it all depends on how they getting their numbers.

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Assume completely random. This is more a theorical discussion than real.

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Originally Posted by joosttx

Assume completely random. This is more a theorical discussion than real.

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If it's actually random, then it's completely random, no number is better than the next.

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Originally Posted by vvv321

If it's actually random, then it's completely random, no number is better than the next.

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Then why when you flip a coin and get heads five times in a row your odds of getting tails is higher than heads on the next flip?

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Originally Posted by joosttx

Then why when you flip a coin and get heads five times in a row your odds of getting tails is higher than heads on the next flip?

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They aren't.

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Originally Posted by joosttx

Then why when you flip a coin and get heads five times in a row your odds of getting tails is higher than heads on the next flip?

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It isn't more likely. You might PERCEIVE it to be more likely but it's not in fact more likely.

Even if tails does come up - it wasn't more likely than heads.

There is no gravitational pull in randomness, by definition.

Wait - have you been gambling behind the A&P and the dealer was selling you increased odds on coin tosses?

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Originally Posted by joosttx

Then why when you flip a coin and get heads five times in a row your odds of getting tails is higher than heads on the next flip?

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You are confusing the terms ‘odds’ and ‘probability’

The odds stay the same every flip. The probability changes because it’s a function of the prior flip.

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Originally Posted by j44ke

Even if tails does come up - it wasn't more likely than heads.

There is no gravitational pull in randomness, by definition.

Wait - have you been gambling behind the A&P and the dealer was selling you increased odds on coin tosses?

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In series h,h,h,h it does increase the probability you will get a tails. History matters. Therefore if you are picking a series of number randomly (ten sided coin 0-9) the prior result has bearing on the future result. Isnt this the bonferroni thingy?

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Originally Posted by vvv321

It isn't more likely. You might PERCEIVE it to be more likely but it's not in fact more likely.

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Flip enough times the series will equal .5 chance. Therefore, a subset (h,h,h....) of the series which
Skews the result say .6 heads .4 tails then the next flip in the series has a greater chance to be tails than heads

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Originally Posted by joosttx

Flip enough times the series will equal .5 chance. Therefore, a subset (h,h,h) of the series which
Skews the result say .6 heads .4 tails then the next flip in the series has a greater chance to be tails than heads

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This isn't true. Unless it's an untrustworthy coin each flip will be random, equal chance of being heads or tails.

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Originally Posted by joosttx

In series h,h,h,h it does increase the probability you will get a tails. History matters. Therefore if you are picking a series of number randomly (ten sided coin 0-9) the prior result has bearing on the future result. Isnt this the bonferroni thingy?

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Each coin flip is an independent event. The coin doesn't care how many times it landed on heads previously.

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Originally Posted by Matthew Strongin

Each coin flip is an independent event. The coin doesn't care how many times it landed on heads previously.

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Got it. 50% chance no matter of history. But what can be predicted the possibility of flipping a series of heads is less than a head tails repeat.

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Originally Posted by joosttx

Got it. 50% chance no matter of history. But what can be predicted the possibility of flipping a series of heads is less than a head tails repeat.

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So would the probability of selecting 2 of the same numbers in a series of 3 numbers be less than selecting 3 different numbers?

.5^3 probability to flip head 3xs.

.1*.9 vs .9*.8 (am I right on this?) I don’t think I am

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Originally Posted by joosttx

So would the probability of selecting 2 of the same numbers in a series of 3 numbers be less than selecting 3 different numbers?

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No. It is the same.

What is the probability ?

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Originally Posted by vvv321

No. It is the same.

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Originally Posted by joosttx

What is the probability ?

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Of getting a number out of a thousand? One in a thousand, 0.10% chance.

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Originally Posted by vvv321

No. It is the same.

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To further simplify what is the probability of drawing the same number in two drawings with a range of 0-9?

If I draw the number 1 from 0-9. The probability of me drawing 1 again (the next time) is .1. The probability of draw another number is .9.

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Originally Posted by vvv321

Of getting a number out of a thousand? One in a thousand, 0.10% chance.

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As long as the way they are picking the numbers is actually random, it will be completely random, again no number has a better chance than the next.

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Originally Posted by joosttx

To further simplify what is the probability of drawing the same number in two drawings with a range of 0-9?

If I draw the number 1 from 0-9. The probability of me drawing 1 again (the next time) is .1. The probability of draw another number is .9.

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You have a 10% chance, 1-10 of getting any number 0-9. Unless you are taking out the number that is picked, it's always going to be 1-10 or a 10% chance of getting that number every time a truly random number is picked. Yes there is a better chance that a different number will be picked, since there are 9 of those and 1 of that same number, this dosen't change the 10% chance of hitting any number in that range each time a truly random number is picked in that range.

You are seriously overthinking this.

1 chance out of a thousand. 10^3.

10: number of combination
3: number of drawing

I'm not a statistician, so everything below might be hooey. That said, here's how I'm thinking about it:

If they're physically "drawing" numbers and not replacing them (ala the Powerball), your odds of winning are lower with a repeating sequence than with three unique numbers. For example, imagine a pool of 90 digits, with 1 through 9 each represented 10 times each. If your ticket were 555, your odds of drawing a first five are 10/90 (11.11%). Since a five is removed from the pool on the second round, your odds of drawing a second five are only 9/89 (10.11%). If you were to then draw a second five and remove it from the pool, your odds become 8/88 (9.09%). The longer the sequence of repeating digits, assuming each draw is removed from the pool, the more your odds decline. In contrast, with a sequence of unique numbers, your odds would slightly increase as the rounds progressed since all desired digits remain in the pool, even as the pool shrinks in size.

However, if you mean that the numbers are replaced (literally or virtually), then the pool for each round is identical, and the odds remain consistent at each step. No prior round reveals any information, so we aren't picking doors on Monty Hall. The odds of 555 would be the same as 123.

My best short answer is that it depends on what you mean by "draw." (Cue the Bill Clinton jokes.)

you should read up on conditional probabilities. in the coin flipping example of getting four heads in a row, you have a 6.25% (50%^4) of flipping four heads in a row. but if you've already flipped three coins as heads, the odds of flipping a fourth coin heads up is 50%.

transferring this to your raffle, as long as the digits are drawn from an even distribution with replacement there isn't any number that will be more advantageous than another.

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Originally Posted by zachateseverything

you should read up on conditional probabilities. in the coin flipping example of getting four heads in a row, you have a 6.25% (50%^4) of flipping four heads in a row. but if you've already flipped three coins as heads, the odds of flipping a fourth coin heads up is 50%.

transferring this to your raffle, as long as the digits are drawn from an even distribution with replacement there isn't any number that will be more advantageous than another.

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How would you model the non-replacement situation?

This is my favorite probability problem:

Monty Hall problem - Wikipedia

https://upload.wikimedia.org/wikiped...n_door.svg.png

Never mind. Missed the second page.

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Originally Posted by zachateseverything

you should read up on conditional probabilities. in the coin flipping example of getting four heads in a row, you have a 6.25% (50%^4) of flipping four heads in a row. but if you've already flipped three coins as heads, the odds of flipping a fourth coin heads up is 50%.

transferring this to your raffle, as long as the digits are drawn from an even distribution with replacement there isn't any number that will be more advantageous than another.

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I understand. What I was trying to do was create two sets, same number, different number. If I drawn a 2. The chances for me to draw another 2 are 1/10 (picking from a set of 0-9) where there is a 9/10 chance I will draw a different number. But that answers or asks a different question.

If the draw is truly random, then the probability of the same number drawn three times is the same as three different numbers, i.e. 1/1000.

However... if this is a real life physical drawing, with (for example) numbered ping-pong balls or a 10 sided die rolled three times, there's a possibility that the object displaying the number is not completely random/neutral. So there's a non-zero probability that a single number is more likely to occur. What that probability is depends on how far from neutral the die or ping-pong ball is and may not be significant, but it's a factor. (like the difference between a loaded die and one that has a microscopic chip off one corner).

So, if you were betting on a flipped coin and the coin came up heads 9 times in a row, probability says that the 10th flip has a 50/50 probability of coming up heads or tail, but the smart bet would be on heads. Why? Because the previous results may be evidence that the coin isn't truly neutral so you should use that to your advantage - but if the coin is neutral, you haven't increased your risk/odds. It's still *at most* .50 tails.

It took me a little while to wrap my head around this - I had trouble letting go of the assumption of true randomness and learn from the prior observed results.

Check out Nicolas Taleb's 'Fat Tony' story in The Black Swan for a better written explanation.

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Originally Posted by caleb

How would you model the non-replacement situation?

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there are a couple ways to model it, the easiest is to turn it into a counting problem.

if the order of the picks matters, the answer is 1 in (10 ways to pick the first number)*(9 ways to pick the second number)*(8 ways to pick the third number) or 1:720.

if the order of the picks doesn't matter (ie. 123 is the same as 132), it's that previous answer divided by the number of combinations of 3 unique digits (ie 6) so 1:120.