Take a dice (die ;)) and roll it
Are the chances of rolling a 6 50/50 ?
Try betting on the results
I’ll send you a for every time you roll a 6 if you send a to me for every time you don’t 
The two outcomes will be equally probable when there are only two possibilities, and nothing favors one over the other. For example, the chance that a coin will land heads or tails is 50%, provided you’re willing to ignore the possibility of it landing on its edge. Another good example is that of a die landing on an even number or an odd number: There are six possible outcomes, but these three are no more or less likely than those three. Your lottery ticket will have a 50% chance of winning ONLY if there are only two numbers that can possibly win, and you have one of them.
If you do win, is there then a 50% chance that you will give all your winnings to me? Either you will, or you won’t.
You will. You will. You will. You will. You will. You will. You will. You will. You will. You will. You will. You will. You will. You will. You will. You will. You will. You will. You will. You will…
I meant to add:
The problem with your original question is that although you are correct about there being only two outcomes (the ticket wins, or doesn’t win), one of those outcomes is highly favored by the circumstances. There is only one possible way for you to win, but there are many many ways to lose.
“either it will happen, or not” is different from “It’s 50/50”.
Either you win, or you don’t. BUT, there are a hundred million ways to lose and only one way to win. So you count up the possibilities: a hundred million on one side, and one on the other. So your odds of winning are one in a hundred million, not 50/50.
chowching, if you still don’t understand what we’re trying to explain, then please try to answer my request from before: Can you describe any sort of thing where according to you the chance would be something other than 50-50? What sort of event would have a 1-in-3 chance of happening?
Look at what the “50/50” represents. It represents a scenario where you tried the scenario 100 times, and 50 times it went one way, and 50 times it went another way.
But of course that’s not how lotteries work. If there are 100 names in a jar, you don’t have a 50/50 chance of winning, you have a 1/99 chance of winning. If you could add your name twice, you’ve have a 2/98 chance of winning. It’s only when you add your name to the jar 50 times that you’ve got a 50/50 chance of winning.
Just to extend the OPs logic here:
Fair 6-sided die:
You either roll a 1 or not. Prob(Roll 1) = 50%
You either roll a 2 or not. Prob(Roll 2) = 50%
You either roll a 3 or not. Prob(Roll 3) = 50%
You either roll a 4 or not. Prob(Roll 4) = 50%
You either roll a 5 or not. Prob(Roll 5) = 50%
You either roll a 6 or not. Prob(Roll 6) = 50%
So, the probability of rolling any number at all on a single die is 300%. I like them odds.
I don’t know about you guys, but I want the OP to be in charge of the next mega millions drawing.
There is only one certainty here and someone has to state it: “You can’t win if you don’t play!”.
duh dumdum, cymbal crash.
When you buy up half the lottery tickets in circulation.
Your don’t have a 1:2 chances of winning the lottery because there aren’t 2 possible outcomes. There are (lets say) one million possible outcomes, of which only one (the numbers you picked) is favorable. Ergo 1:1,000,000
Er, not quite. It’s when you buy lottery tickets for half the possible results. It’s always possible that no one has the winning numbers, and then you get a rollover (Your Lottery May Vary).
Actually I think he’s using the understanding most people have. It’s not a mathematical calculation of odds, he’s considering the number of choices available, in this case 2, winnning and not winning. Using dog math, that means there’s a 100% chance of 1 of 2 things happening, and 100 divided by 2 is 50, therefore the odds would be 50/50 because 50 divided by 50 is 1.
That isn’t meant to explain odds, but how people get to this misunderstanding.
If there are 20,000,000 number combos, then you need to buy 10,000,000 tickets (no dupes).
Then you have 50/50 chance, because you have 50% of the combos, and there are 50% you don’t have.
This is why playing tabletop RPGs is so important. Not that it will protect you from dice superstitions, just that you’ll end up with a pretty good grasp of likelihood.
No. Winning and Not-Winning are not the two possible outcomes.
You have two actions: Buy-a-Ticket and Not-Buy-a-Ticket.
If you chose Not-Buy-a-Ticket there is only one outcome: You don’t win anything.
If you chose Buy-a-Ticket there is – assuming a drawn lottery with hundreds of thousands of participants – hundreds of thousand outcomes, because each person that bought a ticket has a chance to win.
So, if you bought a ticket, you have one in hundreds of thousand odds of winning. Not an even 50/50.
I think dice are a good way to intuitively understand how the whole 50/50 thing is wrong. Because on fair dice, all sides are equally likely. So I start with flipping a fair coin, it has 2 sides, each side is equally likely, I have a 1/2 chance of guessing correctly. Now I roll a d6, each side is equally likely, I have a 1/6 chance of guessing correctly. Now I roll a d20, each side is equally likely, I have a 1/20 chance of guessing correctly.
So now we can simulate a lottery drawing by putting each possible ticket combo on a side of a die. Thus, using the pattern we see above, my chance of guessing correctly is one divided by the number of sides. Yes, it just so happens in a lottery that all outcomes are theoretically equally likely, the problem is that there’s usually a very large number of possible out comes. It’s not just I win vs. I lose, it’s I win vs. you win vs. Alice wins vs. Bob wins vs. … etc.
If you’re still seeing the I win vs. I lose, we can try to intuitively understand it with unequal probabilities. Let’s go shoot some hoops. First, take some shots from the foul line. Do you expect to make 50% of those shots? I suppose it depends on how good you are, you might make 50%, you might also make 70% or 30%. But let’s say you expect to make 50% there, now step back to the 3-point line. It’s a harder shot, do you expect to make the same percentage? They’re both I make vs. I miss, but it should also intuitively follow that the latter will be lower than the former because it’s a harder shot. That is, if you succeed at the exact same rate, then it doesn’t really make sense to say it’s a harder shot. Still having trouble? Try half-court shots or even full-court shots. Even the best shooters in the NBA won’t make those shots with anything approaching 50% accuracy. At some point you’re going to be far enough back that you’re missing virtually all of them.
Yes, I know exactly how you feel.
But how can we be so sure that the two outcomes will be equally probable? How can we be so sure that landing a head is equally probable with landing a tail? The characters on the tails side is different from the one in the heads. That might give one side a slight advantage over the other. Even if both sides are mirror image of each other, I will be dismissing some small things if I am to say that they are equally probable to happen.