Statistics and probability: bike raffles

[joosttx]

What is the probability ?

No. It is the same.

[vvv321]

What is the probability ?

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

[joosttx]

No. It is the same.

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.

[vvv321]

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

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.

[vvv321]

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.

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.

[Lionel]

1 chance out of a thousand. 10^3.

10: number of combination
3: number of drawing

[caleb]

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.)

[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.

[caleb]

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.

How would you model the non-replacement situation?

[Mabouya]

This is my favorite probability problem:

Monty Hall problem - Wikipedia

[j44ke]

Never mind. Missed the second page.

[joosttx]

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.

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.

[taz]

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.

[zachateseverything]

How would you model the non-replacement situation?

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.