These are independent of each other.
You either roll a 1 or not. If you land a 2, then it is on the not part. If you land a 3, then it is on the not part.
Yup. The lottery is a tax on people who are bad at math.
I can’t think of anything that will give 1/3 chance of happening. That would be like dividing by zero.
[QUOTE=Blaster Master;15933400
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.[/QUOTE]
When you try a shot, it will always be either you will make it or not. Although by intuition, you can say that it is more likely you will not make it as you move farther.
I think the easiest way out of the problem is, as others have said, to remember that there are more than two possibilities. It’s not just win or lose, it’s get the one winning ticket or one of the many losing tickets. That’s what odds are. Not whether you win or lose but what is the likelihood that you will win vs the likelihood that you will lose. To get the chances, you have to actually calculate something. The OPs idea is more like a statement about how many outcomes there will be, not a calculation of the chances to arrive at those outcomes.
I’m good at math and I consider it cheep entertainment. I can daydream for 3 or 4 days about what I’d so with the money for only a dollar. Of course more importantly I can afford to throw a dollar away twice a week.
How about this: the outcomes are not “I’ll either win or lose.” It’s “I’ll win with 1-2-3-4-5” or “I’ll lose with 1-2-3-4-6,” “I’ll lose with 1-2-3-4-7,” “I’ll lose with 1-2-3-4-8…” etc.
I have no dog in this fight, nor any issue with the OP, but I think this might be my new sig line.
How would you fill in this table?
Prob(Roll 1) = ______
Prob(Roll 2) = ______
Prob(Roll 3) = ______
Prob(Roll 4) = ______
Prob(Roll 5) = ______
Prob(Roll 6) = ______
Having a bit of a leg pull here? You either are or you are not.
I don’t think it’s GQ nor IMHO. It’s MPSIMS. Seriously, this is so mundane and pointless.
However, I’m just a lousy guest. The moderators are the real experts. So… yeah.
“I’ll win with 1-2-3-4-5” or “I’ll lose with 1-2-3-4-6,” “I’ll lose with 1-2-3-4-7,” “I’ll lose with 1-2-3-4-8…” etc will of course fall under I"ll lose.
yet you were perfectly willing to throw out important information to arrive at your black and white view. if your OP is genuine; you should learn fractions first, then percentages, then a whole lot of stuff before you touch probabilities. and please don’t gamble.
You mean meteors don’t strike twice in the same place?
I think you are confusing two different 50-50s here. You are right to assume that 50% of the possible outcomes are “I win” and 50% are “I lose” as those are the only two possible outcomes. That is completely unrelated to the odds of any one of those outcomes actually occurring, which is patently not 50%.
More generally: the number of events being considered is utterly unrelated to the probability of each of those events occurring, and there’s no reason why those two numbers ought to be connected.
The simpler answer is “because you keep buying tickets and never win”. The longer your losing streak goes on, the plainer it is that your assumption is wrong.
I had asked:
and the response was:
I think this response conclusively demonstrates that chowching’s understanding of these concepts is fundamentally different from the rest of us.
chowching seems to be using terms like “chance”, “odds”, and “50-50”, to simply ilustrate the “either/or” aspect of the situation: either the lottery ticket wins, or it loses. And from that perspective, it is correct.
But the rest of us are using these terms with the specific purpose of clarifying how likely it is for the event to happen. It goes much further than “either/or”. Some things are very likely, some are extremely likely, and some are somewhat unlikely. And if enough details are known, we’re able to put a number on it. For example, suppose I’m playing ordinary 5-card poker, and I am dealt 5 random cards. What are the odds that I’ll be dealt two pair? chowching would say that the odds are 50-50, because I’ll either be dealt the two air, or I won’t, and that the same calculation applies to being dealt three of a kind - either I’ll get them or I won’t.
But Wikipedia says that the odds of being dealt two pair are about 20 to 1, while the odds of being dealt three of a kind are about 46 to 1. This is very useful information! It tells me that if I play poker for a very long time, I will probably be dealt two pair several times, and that I’ll be dealt three of a kind less than half as often. I can use this information for my betting strategy and for other purposes.
chowching, this is why we have these probabilities to begin with. No one cares if something might or might not happen, because that’s true about everything. What we care about is: How likely is it? And if you can actually put a number on it, then you’re really winning.
Everyone have a good night.
There is after all only a 50% chance that the sun will rise tomorrow.
Nice try, but let’s try it a different way. Since fair 6-sided dice exist in the real world, we don’t have to play weird word games. We can try it out in real life.
You either roll a 1 or not, therefore,
Prob(Roll 1) = 50%
Prob(Roll 2 through 6) = 50%
So, let’s make a bet.
You will roll a single fair 6-sided die 1000 times. Every time it comes up 1, I give you $1. Every time it is “not 1”, you give me $1. Since the probability is 50/50, we should come out close to even at the end of the 1000 rolls. I’ll even be nice and say that I’ll admit you’re right if it’s within 45% and 55% via experiment and I’ll start playing the lottery every day. And ff I’m right, this is much more fair to you than playing the lottery.
Care to take me up on the bet?
Alt proposal.
We’ll make it more like a lottery. You get to pick a number between 1 and 6.
Then roll the die. Either you guessed correctly or you didn’t.
If you guessed correctly, I give you $1. If you didn’t guess correctly, you give me $1. We do this 1000 times.
If it’s 50/50, we should be close to break even. If I’m right, I’ll net more than $600.
I’m scared of a terrorist with bomb on my flight. I bring a bomb on my flight, thus doubling the odds of not finding a terrorist with one. I fly safer.