Ah, part of this I agree with. The first half.

The odds of rolling a particular number are 1/6 (excluding the unlikely event of a cocked die).

The odds of rolling that same number on a subsequent roll are again 1/6.

The odds of rolling the same number twice in a row are 1/36.

This second result is exactly equivalent to a pair of dice and any combination.

So let's take a dodecahedron die from a D&D game. Each side has exactly the same probability of selection = 1/12. Getting the same number a second time is still 1/12 but the combined probability of getting the same number in any two rolls is 1/12 * 1/12 = 1/144.

If we had 5 dodecahedrons the combined probability would be a simple extension of that multiplication.

So now let's go to lottery balls. Assume there are 500 balls placed in bins of 100, each ball then numbered 1-100.

Each ball in each separate bin has an equiprobable chance of being pulled = 1/100.

So what is the chance of pulling say 55 from bin 1? = 1/100.

What is the chance of pulling 55 from Bin 2? = 1/100.

Etc.

What is the combined probability of drawing 55 from each bin?

1/100 * 1/100 * 1/100 * 1/100 * 1/100 = 1/10,000,000,000

Pulling a ball from Bin 1 does not affect the probability of pulling any number from Bin 2, Bin 3, etc. This is exactly equivalent to the coin toss and die toss.

B

Let me nitpick you a bit, Bot.

The odds of throwing a given number twice in a row is 1 in 36. The odds of throwing the same number twice in a row are significantly better.

My argument is based on the fact that the outcomes of the lottery are finite, and thus enumerable. Once you make an order of them, making any lottery draw is exactly the same problem as choosing a random number in the proper range (i.e., between 1 and about 130 million). The fact that you are choosing without replacement affects what will be on the list, but not the probabilities of those which remain.

The odds of rolling any number twice are the same as rolling a selected number once-1/6. We're slipping into rocket scientry here.

This is correct. But this is not the way the lotteries I have seen are conducted. Are we still talking lottery?

Chas, what I can't figure out is why you think it matters whether you replace the balls or not. There are still a finite number of equiprobable outcomes.

It matters just like balancing a check book matters.
Say we start with 50 balls. The odds of picking the desired number are 1/50. The odds of picking the second number are 1/49 and so forth.
If the ball is returned then the odds of picking a number would be 1/50 every time. The odds of winning are substantially different between the 2 methods.
The balls in lotteries I have watched are always kept in a tray and never returned. If you have ever played the lottery you know that the winning numbers are always all different.

I've never run a lottery and I don't know whether they have a set of rules. Good points.

I assumed that they draw without replacement but from separate bins (that's how I'd do it if I ran the damned thing!). If they draw without replacement from the same bin then the n-1, n-2 ... certainly holds.

I am not disputing that the balls are drawn without replacement.

I am disputing that this makes it somehow different from a random number generator. It is quite clear that this is exactly what it is.

It would be the same as if you had a fair die with 130 million (or so) faces, each printed with a set of ball numbers. Or a single integer.

Some lotteries are done exactly how you envision. These typically have very few bins and very few balls (meaning a very small universe of possible outcomes), but you do have to choose the balls in order. That is, 1-2-3 is considered different than 2-1-3. These lotteries have good odds of winning, and thus small prizes. But the good odds of winning have absolutely nothing to do with the sample method.

The big games are chosen without replacement, but this doesn't affect the fact that all lotteries have a finite universe (possible outcomes) with equiprobable entries. Since you also can't choose an invalid entry for the big games (like 1-1-2), there is no change in the odds.

You could just as well have 7 bins with balls 1 through 10, and an eighth bin with balls 1 through 13. This is also a one in 130 million odds. No better, no worse than a choice without replacement with 130 million possible outcomes.

I think he is making a mistake and suggesting that the lottery draws from one drum, instead of one drum for each spot. All he's saying is if you draw from one drum of 50 and don't replace the ball, then redraw, then the odds for each number change because there is one less ball in the drum.

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1984 300TD

No, he is making a completely different mistake. Many lotteries do draw without replacement.

I will agree that at any point in the game, the chance for a particular ball appearing next is not fixed. It goes up as the game is played (unless already drawn).

Where the fallacy lies is thinking that this makes the lottery any harder or easier to win than a sequence of coin tosses. Assuming a fair coin and lottery, these are equivalent activities.

Gosh, I've completely switched sides!

Going into this discussion I thought that the lottery and coin-tossing were equivalent because both were probabilities computed from independent events. But that is not the case--sampling without replacement is a dependent system. So instead of equiprobability that you have with coin toss or die toss, you have conditional probability and that kind of probability is a whole 'nuther kettle of fish.

We deal with problems in conditional probabilities more often than with problems with equiprobabilities. But we usually simplify them because conditional probability math makes our heads hurt.

Take 5 card draw poker. Only the first card pulled from the stack has a probability of 1/52 of being an Ace of spades. If that card is not pulled the probability of pulling the ace increases to 1/51. And so forth until the last card.

This is an entirely different class of problems from coin tossing in which each event is independent of every previous event.

If I say what is the probability of drawing the ace of spades on the next draw given that 30 cards already drawn are NOT the ace of spades, you would probably say 1/(52-30) = 1/22.

If I say what is the probability of calling a heads given that the previous 30 tosses were all heads, you'd say 1/2.

Right?

B

If the number ball is drawn from different bins or the ball replaced after drawing, a drawn number can be drawn again. Have any of you seen a number appear more than once in a lottery drawing? I never have.

Oh cripes. Realistically, from the lottery player's perspective, 1 in 380,000,000 is the same as 1 in 200,000,000.

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You don't need a weatherman to know which way the wind blows - Robert A. Zimmerman

That's correct. But you have to look at a lottery draw as the entire set of numbers, instead of each one. Each number means nothing by itself.

If you could bet on each number as it came up, there would be a point to this discussion. But alas, there is not.

Since you can enumerate the draws and each potential set is exactly as likely as any other, it's just a random number generator.