Lotto Numbers

  • Thread starter Thread starter ZZII
  • 89 comments
  • 3,314 views

Who has a better chance of winning? (read post)

  • Person A

    Votes: 6 11.8%

  • Person B

    Votes: 13 25.5%

  • both the same

    Votes: 21 41.2%

  • i don't care

    Votes: 11 21.6%
  • Total voters
    51
Nope. The balance of probability is STILL that you are 66% likely to be wrong. That is now distributed evenly over the 1 remaining door.

Put it this way. I'm thinking of a number. It's between 1 and 3, inclusive. Guess what the number is (I'll roll a die to determine it).
 
Okay - so now you've picked a number, you're 33% likely to be right. It's 66% likely to be either 1 or 3.

It's not 3.

Would you like to change your answer to 1 or stick with 2?
 
Then imagine the same thing, but with a much larger number... I'm imagining a number between 1 and 100 (and I do have a 100 sided die to randomise it with). If you were to guess 1, and I told you that the answer was either 1 or 100, would you change your mind, or stick with the 1% chance you were right to start with? Will the other 98/100 have been taken away, as you put it, or will you still be only 1% likely to guess the number from the original sample?
 
if the answer's either 1 or it's 100, then there are 2 choices you would logically choose. Sure, there are 98 other choices, but they've been pulled out of the picture by logic, so you're now dealing with either option a or option b. you have a 50 percent chance of being right either way. You chose option a, you're chosing either 1 or 100 out of 1 or 100, not of 1 or 100 out of 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20, 21, 22, 23, 24, 25, 26, 27, 28, 29, 30, 31, 32, 33, 34, 35, 36, 37, 38, 39, 40, 41, 42, 43, 44, 45, 46, 47, 48, 49, 50, 51, 52, 53, 54, 55, 56, 57, 58, 59, 60, 61, 62, 63, 64, 65, 66, 67, 68, 69, 70, 71, 72, 73, 74, 75, 76, 77, 78, 79, 80, 81, 82, 83, 84, 85, 86, 87, 88, 89, 90, 91, 92, 93, 94, 95, 96, 97, 98, 99, or 100 because only 1 and 100 have a chance at being right, the other 98 have no chance of being right because you've already told me they're wrong.

It works easier this way:
{1, 100}:{1, 100} not {1, 100}:{1, 2, 3...98, 99, 100} because you removed {2, 3...98, 99} when you said it was just {1, 100}.


Now, here's another riddle:

Statement 1: Statement 2 is True
Statement 2: Statement 1 is False
 
OK that probability riddle was...amazing. I'm still going over it in my head. One of the best riddles I've heard in a while. Anyway, here is the answer to the lotto number riddle. Don't read on if you still want to try it.

For all of you who voted that the probabilities would be equal, do you think it would be THAT simple?

The correct answer is Person B has the (slightly) higher probability. There are many ways to think about this. For one, we will do the opposite of the doors riddle and make the large number smaller. Imagine you flip a coin twice. If you were to pick heads both times, you would only have a 3/4 chance of getting it at least one of those times. If instead you flipped the coin only once but picked both heads and tails, you would have to be correct. The same logic applies for lottory number.

Another way to think about it is as follows: When picking cards from a deck, your probability of picking a certain card in five tries is slightly higher if you don't replace the cards each time you pick. Playing 52 sets of lotto numbers is like doing that. If your first set isn't a winner, the chances that your next set is are very slightly higher. Playing a single set 52 times is like replacing the cards every time that you play them.

An interesting spin on this is that the payoff probability, not the win probability IS the same for both. This is because person A has the chance, although small, to win mulitple times and person B does not. This balances out the probability difference so that the money they would win is indirectly proportional to the probability that they have.

Edit: Imagine the coins again. If you won a dollar for each time your were correct, then the payoff probability would be as follows. Person B would be certain to win (by picking heads and tails) but could only win once. The two possibilites are {H} and {T}. In both scenarios, he wins $1. That is $2 for $2 scenarios. Person A has four possible scenarios, {H,H}, {H,T}, {T,H}, {T,T}. In two scenarios, he wins $1. In one scenario, he wins $2. In one scenario, $0. This totals to $4 for 4 scenarios.
 
Riddles that get people to argue vehemently against the correct answer are the best ones. Simplify it:

You have a set of ten numbers to draw from: {0,1,2,3,4,5,6,7,8,9}

You choose the same number on three separate drawings. Your probability of winning at least once is 1/10 + 1/10 + 1/10 = 3/10. You are absolutely correct, it is an independent relationship. But that is doesn't matter.

Now you choose three seperate numbers for one drawing. The probability of your first number being the one drawn is 1/10. Assuming your first number is not the one drawn, the probability of your second number being the one that is drawn is 1/9 (you cannot include your first number in the sample space because if it was chosen, then you would have already won). Assume your first and second number were not drawn. The probability of your third number being drawn is then 1/8 (same reason). The total probability is 1/10 + 1/9 + 1/8 = 121/360 > 3/10.
 
Famine you're wrong man, if one door is rigth and one door is wrong you have a fifty percent chance. Ill elaborate: in the beginnning doors: 1)33%, 2)33%, 3)33% agreed? so then you pick door 1 and have a 33% chance of being right. if you cross door 2 out that doesn mean you were wrong in the first place therefore door 3 has the same chance of being right as door one.
I can prove this mathematically but its stupid u ahve a 1 in 2 chance of being right, just because the middle door was wrong doesn mean door 3 has more chance of being right.
 
Originally posted by ZZII
An interesting spin on this is that the payoff probability, not the win probability IS the same for both. This is because person A has the chance, although small, to win mulitple times and person B does not. This balances out the probability difference so that the money they would win is indirectly proportional to the probability that they have.
Click to expand...

Well its not that intersting considering that that is how odds are supposed to work, I.e the less chance you have of winning the more money you get, mind you in horse racing its not so much on chance but what is the percieved chance which is in fact determined by otehr factors. But just to keep it simple the idea is that if your less likely to win and you do you get more.
 
Originally posted by Crayola
I can prove this mathematically but its stupid u ahve a 1 in 2 chance of being right, just because the middle door was wrong doesn mean door 3 has more chance of being right.
Click to expand...

Bring on the math. Famine proved it correct using math. If you want to prove it incorrect, you have to show a flaw in his math. Saying it's stupid is not a proof. I'm not trying to be a smart-ass, but if you know Famine is wrong and you have the math to prove it, then it would go a long way toward convincing us if you showed us that math.

Originally posted by Crayola
Well its not that intersting considering that that is how odds are supposed to work, I.e the less chance you have of winning the more money you get, mind you in horse racing its not so much on chance but what is the percieved chance which is in fact determined by otehr factors. But just to keep it simple the idea is that if your less likely to win and you do you get more.
Click to expand...

You're quite right. I even said that: "This balances out the probability difference so that the money they would win is indirectly proportional to the probability that they have." Of course in the lottery the payoff is always slightly less than that probability since some of the money goes toward the government budget, exiting the mathematical equation altogether. So it is interesting in terms of the riddle, but absolutely expected in real life.
 
Originally posted by Jpec07
if the answer's either 1 or it's 100, then there are 2 choices you would logically choose. Sure, there are 98 other choices, but they've been pulled out of the picture by logic, so you're now dealing with either option a or option b. you have a 50 percent chance of being right either way. You chose option a, you're chosing either 1 or 100 out of 1 or 100, not of 1 or 100 out of 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20, 21, 22, 23, 24, 25, 26, 27, 28, 29, 30, 31, 32, 33, 34, 35, 36, 37, 38, 39, 40, 41, 42, 43, 44, 45, 46, 47, 48, 49, 50, 51, 52, 53, 54, 55, 56, 57, 58, 59, 60, 61, 62, 63, 64, 65, 66, 67, 68, 69, 70, 71, 72, 73, 74, 75, 76, 77, 78, 79, 80, 81, 82, 83, 84, 85, 86, 87, 88, 89, 90, 91, 92, 93, 94, 95, 96, 97, 98, 99, or 100 because only 1 and 100 have a chance at being right, the other 98 have no chance of being right because you've already told me they're wrong.

It works easier this way:
{1, 100}:{1, 100} not {1, 100}:{1, 2, 3...98, 99, 100} because you removed {2, 3...98, 99} when you said it was just {1, 100}.


Now, here's another riddle:

Statement 1: Statement 2 is True
Statement 2: Statement 1 is False
Click to expand...

Okay. Here's a situation.

You have 1000 GTP$ (non-exchangeable) to stake on this. I am thinking of a number between 1 and 100, inclusive. You have to guess it. After you have guessed it I will remove all the other numbers bar one. You then have the chance to change your number to the other remaining number, or stick with your original guess which, I'll remind you, has a 1% chance of being correct.


You are not gambling on whether you should stick or change - you are gambling on whether you can guess a number correctly out of a sample size of 100.

I suggest you draw a probability flow diagram.
 
Originally posted by ZZII
QUESTION:

Who has the better overall probability of winning?
Click to expand...
He's asking "who has a better chance of winning" Not "Which person picks the winning nubmer"

person B has a better chance of picking a winning number than person A. Because Person B has 52/1,000,000 chances. While person A never gets better odds than 1/1,000,000.

The chances of winning are different from actually winning.
Originally posted by Famine
On a similar subject (probability problems):

You are in a room with a guard and 3 doors. One leads to escape, the other two to... not escape.

You have decided, arbitrarily, to go for the left-hand door. The guard interrupts your train of thought by saying to you "I think you're innocent, but I can't tell you how to escape. I will tell you that it's NOT the middle door though".

If the guard is telling the truth, would you have a better chance with your current door choice, or should you make a new one?
Click to expand...
I reworded it a bit. However this one cuts a bit grayer line. Currently you have a 33% chance of being right and a 66% chance of being wrong. If he's telling you the truth you've got a 66% chance of being right. It's a gray area, that only assumes your making a choice, and not actually opening the door.

In other words, you have a higher chance of picking the escape hatch if you change your mind, but still only a 50% of opening the escape hatch.

Probability is a tough path to calculate.

AO
 
Originally posted by Famine
Okay. Here's a situation.

You have 1000 GTP$ (non-exchangeable) to stake on this. I am thinking of a number between 1 and 100, inclusive. You have to guess it. After you have guessed it I will remove all the other numbers bar one. You then have the chance to change your number to the other remaining number, or stick with your original guess which, I'll remind you, has a 1% chance of being correct.


You are not gambling on whether you should stick or change - you are gambling on whether you can guess a number correctly out of a sample size of 100.

I suggest you draw a probability flow diagram.
Click to expand...

if the answer's either 1 or it's 100, then there are 2 choices you would logically choose. Sure, there are 98 other choices, but they've been pulled out of the picture by logic, so you're now dealing with either option a or option b. you have a 50 percent chance of being right either way. You chose option a, you're chosing either 1 or 100 out of 1 or 100, not of 1 or 100 out of 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20, 21, 22, 23, 24, 25, 26, 27, 28, 29, 30, 31, 32, 33, 34, 35, 36, 37, 38, 39, 40, 41, 42, 43, 44, 45, 46, 47, 48, 49, 50, 51, 52, 53, 54, 55, 56, 57, 58, 59, 60, 61, 62, 63, 64, 65, 66, 67, 68, 69, 70, 71, 72, 73, 74, 75, 76, 77, 78, 79, 80, 81, 82, 83, 84, 85, 86, 87, 88, 89, 90, 91, 92, 93, 94, 95, 96, 97, 98, 99, or 100 because only 1 and 100 have a chance at being right, the other 98 have no chance of being right because you've already told me they're wrong.

It works easier this way:
{1, 100}:{1, 100} not {1, 100}:{1, 2, 3...98, 99, 100} because you removed {2, 3...98, 99} when you said it was just {1, 100}.

Face it, Famine, no one can know everything, and we all make mistakes sometimes (some less frequently than others). I also agree with DA's response about it, you have a 1/3 chance of choosing the right hatch, but a better 1/2 chance of actually opening the right one (my god, I thought I left quantum physics in the 8th grade...)
 
Now repeat it 100 times. I can guarantee you that by sticking with your original choice you will average being right only ONCE on average per 100 times, but by changing you will be right 99 times - that is changing your choice will make you right 99 times as often as sticking with the first. This is patently NOT 50:50.

If you want to stick some money on it, I'd be more than happy to walk off with it. Remember - I HAVE tested this theory out on a double-blinded human subject.
 
Umm, if you've eliminated the intergers 2-99, then you're only left with 2 options to choose from, option a and option b. the other 98 options are gone, poof! byebye! so you're dealing with a 1/2 chance of being correct. Face it Famine, you're wrong...
 
IF the doors were eliminated before you made your initial choice, then it would be 50/50. However since they are eliminated afterward, the probability is better if you change doors. Imagine this:

You have to choose a card from a deck and if you choose the right one you win. You choose one. Then, someone takes a completely new deck and eliminates all the cards except yours and one other one. Do you have a better chance of sticking with your card from the first deck or picking the other card from the second deck? You cannot pick your card from the second deck because that you made the choice when all options were still present, not afterwards.
 
Okay - so now you've picked a number, you're 33% likely to be right. It's 66% likely to be either 1 or 3. It's not 3.

Would you like to change your answer to 1 or stick with 2?
Click to expand...

I enjoyed reading your riddles Famine, but I have one tid-bit to add, and please bear in mind, I sucked at stats at varsity.

You say guess a number between 1 and 3 that you are thinking of. That gives a 1/3 chance. You stipulate that 3 is not the number. Thus there are two numbers to choose, giving you 1/2 chance of getting it right. #2 cant have a 2/3 probability of being correct because, as #3 was excluded, the new set of choices is {1,2} excl {3}. Thus 1 and 2 have a 50% of being correct, no matter the initial choice by the particpant.

Because #3 was excluded outright, it can no longer be included in the probability calculation. Not so :confused:
 
I agree with the lotto numbers thing im not worried about taht its the other one. I cant remeber how i worked it out I've been doin other stuff since then. (Mathematical) But my point is if you have 3 doors and you choose one, then find out that a different one is wrong that doesn mean that you were wrong in the first place.
Here comes the maths:(Well sorta)
Step 1] we have 3 doors 1/3 chance each.
Step 2] We have 3 doors now cross one out BUT just pretend it was still there in terms of a greter picture so all three doors still have the same probability (you just have greater insight into which one is correct)
Step3] Now completely disregard the middle door except for the fact it was there originally and the other two doors still have same probability.

In fact this relates to the answer to the lotto numbers, if you picked 2 doors and were wrong the first time you have a greater chance of being right the second time.

Wait ill conclude i was sorta thinking out loud there:
If you were to choose the left door AND the middle then find out the middle door was wrong you are left with the right door and the left door which both now have a higher chance of being right.

Ok this is my last explaination, do the question in reverse, you have 2 doors ones right ones wrong, now throw in another door that doesnt actually mean anything and you have the origianal 2 with the same chance.

SORRY but you're wrong Famine, if theres 2 to choose from the third doesnt matter.
 
I get what the crayon is trying to say. all the doors have a 2/3 chance of being wrong, changing from your original choice to the other is completely illogical because they both have a 2/3 chance of being the wrong door and a 1/3 chance of being right door. So they're all 1/3 right and 2/3 wrong, no matter which door you choose.
 
I'm willing to stake money on it. Are you?

Let's just pause for a moment, in amongst the "Face it Famine, you're wrong" and "No-one can be right all the time", and think for a minute.

Why is it I was voted the "Most Knowledgeable Member"? It's not because of what's in my head, that's for sure. Could it be, possibly, that I check all my facts first, in books? I wonder where this puzzle came from... Hmmm...

Jpec07 - you at least have my MSNIM. I'd be more than happy to go number guessing with you. Let's make it any number between 1 and 10, to bring the odds down to a little nearer being in your favour, shall we? I'm pretty sure that you can trust me to not cheat - given my extant reputation on the GT3 board - but you don't have to if you don't want to. I can PM you a list of the chosen numbers, with a read receipt so I know you haven't read it, before we start.


Oh, and ZZII has it SPOT on.
 
I don't have MSN. I would gladly face you in a duel of numbers, but I'm not gonna dl another messenger solely for that purpose. I'm gonna print this off and show it to my math teacher (masters' degree in Algebra) and see what he says; but I'm pretty sure that you and ZZII are wrong...
 
Could have sworn I had you on MSN... I'm more reachable on YahooIM, or at a push ICQ.
 
Probably an ICQ, as those are compatible with AIM (the only messenger I have at the moment)
 
Famine isn't wrong. Seriously think about it. Your argument is making no sense. You must remember that the two decisions are not mutually exclusive, and can affect each other. This is very important to remember.
 
Me number's in me profile here.

Try http://www.trillian.cc - it's a chat client, with a perfectly usable free programme as well as a more functional buyable one, which brings ICQ, AIM, MSNIM, Y!IM and IRC all together in the one place.
 

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