But he didn't know it was a losing booster when he made his choice. Because of that, his odds do not improve. The first guy had a 1/3 chance of chosing the booster that ended up being revealed, too.

If you choose any booster at random, there is a 2/3 chance the right booster is among the others. That choice was made before knowledge of the contents of any boosters is revealed. That chance is 100% random and does not change as we reveal boosters. Even once they are all open, it is still true that there was a 2/3 chance of being wrong.

Bring it back to 1,000,000 boosters. You pick first. I'll take whatever's last. If 999,998 crazy fools open theirs first to find nothing, does that make your first choice have better chances? Nope. You made that choice uninformed. It was truly random. It stays at 1 in 1,000,000.

I'll stick with the last booster.

You made your choice uniformed too. You did not know. it was a 1 in 1000000 chance. What you are doing is applying the changes to one side and not the other for no good reason. Think about it this way. I have a booster. I choe first. You second guy also have a booster. You chose second. then we opened the third and found it empty. That means that one of the two boosters won. I have 1/2 of a chance, you have 1/2 of a chance. The biggest flaw in your theory is this. Lets take the 1000000 thing. Except I am the 500000th person. Acording to you I have a 499998/1000000 and you have 999997/999998 chance to win. This is obviously wrong. This is not the Monty Hall Syndrome. They are different things. For me to believe you A. Show me how this problem and the first one are related. B. Show me how my
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is wrong. C. Show how my error in your theory is wrong. I am expecting perfect without flawed answers or in my mind you are wrong. I am sorry. I do not see how what you are saying works.

And this, ladies and gentlemen, is why Vegas is growing so fast, and there is no downturn in sight.

The numbers part is easy. Knowing how to apply the numbers is not.

Game on! And happy holidays to you all!

PS--If you get one present, and your brother gets two, and the first one he opens is socks, should you offer to exchange your present for his remaining, unopened, present?

ChaimWitz just returned from a poker tournament this time, and is not hungover, but his tiny little Dinosaur brain hurts. LMAO, after reading all of this, it hurts a little worse!

Here is the scenario laid out by DBlizzard. This is the only information we have/had to work with. There is no other information available, and no other assumptions should be able to be made.

Taking it as a whole or piece by piece, it is what it is, and nothing changes it.

NOTE: "Someone" will always refer to the person who provided the boosters and any opportunity after providing the boosters.

"Someone pulls out three Sinister boosters and tells you that one has War in it and he knows which one it is."

Without any information regarding the "someone's" motivation, intentions, bias, or personality traits, I must assume that it is what it is and nothing more or less. There is no sociology to apply to "someone", and is not part of the puzzle at this point.

I know that "someone" pulled out 3 Sinister boosters.

I know that "someone" told me that one of the Sinister boosters has War in it.

I know that "he" (someone) knows which Sinister booster is the one with War in it.

You are allowed to have one of the boosters. Even if you don't want the figure, you know you can always sell it for a bundle, so you pick your booster.

I know that I am allowed to have one of the Sinister boosters.

I know that even if I don't want the War figure, I can always sell it for a bundle.

I know now that I pick my booster.

**At the moment I pick my booster, there are 3 unopened Sinister boosters for me to choose from. At this moment, the probability or odds that I pick the booster with War in it is 1 in 3, or a 33.33% chance to pick the booster with War in it. Conversely, at this moment, the probabillity or odds that I do not pick the booster with War in it is 2 in 3, or a 66.67% chance that I do not pick the booster with War in it.**

Up until this point, I have no idea about anything else. There is no other information available to me.

In summary up until this moment, I now have picked a Sinister booster, which was one of three I could pick from. "Someone" told me that I could have the one of the Sinister boosters, and "Someone" told me that one of those 3 boosters has War in it, and that "Someone" knows which booster it is.

The person who made this offer then opens one of the boosters you didn't pick to show you the horseman isn't in it.

I know that "Someone" opens one of the two Sinister boosters I did not pick.

I know that the Sinister booster "Someone" he has selected to open does not have "the horseman" in it.

I know that "Someone" shows me that "the horseman" isn't in the Sinister booster "Someone" opened.

**At the moment this occurs, this is all I know. In addition to the previous information I had available, I now know that the one booster "Someone" opened, of the two I did not pick out of the original three, does not contain War.**

To summarize, up until this moment, there is nothing here to indicate bias, intent, or manipulation on the part of "Someone". I already was made aware that "Someone" knows that one of the three boosters has War in it and that "Someone" knows which booster that is.

This does nothing to alter the odds/probablity I had in picking the booster with War in it, at the time I selected it. I still had a 1 in 3 or 33.33% chance to pick the booster with War in it, and a 2 in 3 or 66.67% chance to not pick the booster with War in it. The opened booster does nothing to change this, even with the booster opened to show War is not in it.

He then offers to allow you to switch your booster for the remaining unopened booster.

Now, along with the other information provided to me, "Someone" offers to allow me to switch the booster I picked in exchange for the remaining booster which is unopened. At this point, either booster could contain War, and THE IS THE FIRST MOMENT THE ODDS/PROBABILITIES WOULD COME INTO PLAY.

I know that when I picked my Sinister booster, it was random with a 1 in 3 or 33.33% chance to be the booster with War in it, and that there was a 2 in 3 or 66.67% chance that I did not pick the booster with War in it. NOTHING CHANGES THIS FACT.

I know that either the booster I picked or the remaining unopened booster have an equal chance to contain War. This does not mean anything at this point. Call it a 50/50 chance or an equal 1 in 3 or 33.33% chance that the booster I picked has War in it.

This in NO MANNER changes the fact that when I picked my booster, that I only had a 1 in 3 or 33.33% chance of it containing War. This in no manner changes the fact that when I picked my booster, that I had a 2 in 3 or 66.67% chance of it not containing War.

Therefore, the odds/probability that War is in one of the two boosters I did not pick is still 2 in 3 or 66.67%.

This is the tricky part, and can be easily confused:

It is true that now one of the 2 boosters I did not pick has been opened to demonstrate that it does not contain War.

It is also true that I now have an equal chance that my booster has War in it, when compared to the chance the remaining unopened booster has War in it.

HOWEVER

The odds/probability of the booster I picked containing War WILL NEVER be greater than 1 in 3 or a 33.33% chance. There is no way around this, because at the time I chose it, that was the highest possible odds/probability. It will never be a true 50/50 or 50% chance.

THEREFORE, when making my decision as to whether I want to switch the booster I picked for the remaining unopened booster, the odds/probability at the time I selected my booster were still 1 in 3, as opposed to 2 in 3. The odds are better that I DID NOT pick the correct booster at the time I picked one.

This is the basis of why I should switch boosters, when calculating the odds and probabilities on whether to switch or not.

Should you trade boosters, keep the one you have, or is it a tossup? Why do you make that choice?

This is the end of the puzzle.

The answer has been stated. There is no other information available.

Never was there any indication of bias on the part of "Someone". It never states anything about intent or motivation of "Someone". It never gives any indication of "Someone" wanting to keep War for himself. There is no mention of buying anything, other people involved, or any other factor.

The only way that anybody can try to explain an alternative explanation is by making assumptions that do not exist in any part of the puzzle.

If there were a bunch of "if's" to be analyzed, then its a different story. What if instead of assuming that "someone" wants to keep War, we assume that he is just a nice guy giving me an opportunity to get War. You can twist and turn this around many ways, but if you stick to what is laid out in the puzzle, based on probability and/or odds, switching the boosters is still the correct answer.

A couple other items of note:

Calculating probability is considered an experiment, and if you create the experiment in a way that it is impossible for something to happen, then there is no need for it.

As far as the mention of rolling a 1d6 to come up with a result of 53, and being able to prove that it won't happen, this is not correct. First, you cannot prove that it will happen that way. You can only work the experiment to demonstrate that it will not happen. You cannot submit the data until the experiment has been conducted.

Example: You cannot prove to me that a 1d6, when rolled, will add up to a 53. (of course I know it wont'...lol) You have to work to demonstrate the probability that by rolling a 1d6 you will not come up with the result of 53.

The die is a random roll. Once you roll it 1,000,000 times and it still doesn't come up a 53 doesn't "prove" anything. However, you can demonstrate that when rolling a 1d6 you will not come up with a 53, or that the probability of it happening is almost 100% against it. You can do this with data showing that in those million die rolls, it does not come up. However, you cannot say that because it didn't happen in those 1,000,000 die rolls, that it never will...you can't prove it.

Also, if my earlier example of the sun rising were taken literally, then the probability of the sun rising would be almost 100% against it happening. That would be because the real test would be "What is the probability that the Earth will not rotate on its axis tomorrow?":p

Once again, I cannot prove that the Earth will rotate on its access tomorrow. However, I can work to calculate the probability that it will not rotate on its access tomorrow.

Bleh.....I'm tired of writing.

Bottom-line: In the puzzle, there is no behavioral info on the part of "Someone", nor is there any motivation, intent or bias on the part of "Someone" mentioned.

The puzzle is what it is, and in the absence of any info to the contrary, all assumptions must be in line with the concept of fairness and all things being equal.

OK.....I'm really tired of writing, but extremely tired of thinking....lol:D

PS: Spudkin gets a stamp!;)

Not true. The 500,001st person to chose has a 1/500,000 chance times the chance that someone else has already picked the right one (500,000 out of 1,000,000, or 1/2). 1/2 of 1/500,000 is 1/1,000,000. Before anything is revealed, their chances are the same. However, once one in revealed, which group was it revealed from? The 500,000 that the second person was forced to choose from? If so, then it improves that part of their odds. And it also improves the odds of all the remaining choices the first person did not choose. It does not affect the odds of the first choice...

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Kel, explain to me the difference between this scenario and the Monty Hall Syndrome.

A person chooses at random from N choices. N-2 of those not chosen are revealed as losers. What are the odds that the last unchosen option is the winner?

Is that the Monty Hall Syndrome? If not, what is?

-----------------

There are three choices. They are all indistinguishable and have no relevant order.
With three choices there are three possibilities:

1) The chosen one is the winner, and the remaining two are losers.

2) The chosen one is a loser, and one of the remaining ones is the winner.

3) The chosen one is a loser, and one of the remaining ones is the winner.

Now we reveal one choice. That choice is a loser. We know that the first option above wasn't eliminated, since the winner is not yet known. But which of the other two was eliminated? Since the choices are indistinguishable and order is irrelevant, there's no way to know. So the odds for the first choice don't change. But the odds for the unchosen one get better.

See, your example of A, B and C labels the choices. But the labels are irrelevant. It is either the one winner or one of many losers. If you reveal a loser, it doesn't matter which loser it was. It only matters that it was not the winner.

----------------

If you don't see it, I don't know how to make it clearer. I don't want to be annoying, so I won't argue the point any further.

One thing is certain. Probability makes for interesting discussion...

Happy Holidays all. Have fun choosing which presents to open first. :)

Monty Hall Syndrome is different. The way it works is completley different. It does not matter the order. I do not get your choices. The first two guys both have 1 of two boosters. Okay. They could have not taken the 3rd loser booster. I get what you are trying to say. You are trying to say that the last two guys are like monty hall, the first guy is like the contestant. That is wrong. Let us think about it this way. The second guy has a booster. He know's that 66% of the time either the 1st or 3rd guy won. The 3rd guy opens his booster. It is empty. He assumes that the 1st guy stills has 66% chance of winning. No. This is what you are trying to say through the eyes of the first guy. This is not right because I just showed that it works in the opposite direction. The 3rd guy perspective should also be shown. He knows he has a 33% chance to win. But he lost. The remaining 66% chance is split among the 1st and 2nd guy 50/50. So it is 50/50. Because each person has only one booster that is howit works. Monty hall syndrome is diffrent in many different ways.
1. Monty Hall knows which one is a winner. Here noone does. If the second guy knew which one was a winner and he said for the 3rd guy to open the booster does that increase 2's chances? no. I can as easily group booster 3 with 1 than with 2.
2. One guy has two boosters. So what you ask. Here is why this makes a big diffrence. You know that one is empty. Just because he reveals the empty one does not make him any less of a chance to win. Here with one booster everyone has 1/3 chance to win. They could have picked and one of 3 boostes. However when the 3rd guy shows the empty one you know this. Neither 1st or 2nd guy picked. So I ask you, why include it in the probability.


Here are your choices

1) The chosen one is the winner, and the remaining two are losers.
Possible
2) The chosen one is a loser, and one of the remaining ones is the winner.
Possible.
3) The chosen one is a loser, and one of the remaining ones is the winner
Not Possible. You see that is the booster guy three chose. So you can eliminate this.

Now you have 2 choices. They both only have 1 outcome. 1st thew 1st guy wins. 2nd the 2nd guy picks the winning booster. There is the key diffrence. You knew that the empty booster existed in the first problem, here you knew nothing. Therefore the guy losing changes the probabilities. In problem number 1 you essentially got 2 boosters. You got shown a loser which you knew about. That means all of his chances of winning are in the other one, so he basically gave you two boosters but he opened one before hand. You should take the deal. That is not this problem. You can only do it with two people, here is 3. It will not work.

I just wanted to say thanks, especially to WarPlayer and Chaim, for making a dull morning at work a lot more interesting. I love math, but statistics has always been my weak spot. (Calculus is so much more fun and aplicable!)

My first reaction was to say that it's a 50/50 chance at the end, but now I see that given the choice of swapping presents with my brother, I would. Besides, who needs socks when you live in Florida?

Happy Holidays! :D

Well, I thought the answer was pretty obvious : find a blunt object, apply it a couple of times on the guy making the offer, and take off with all the boosters...

There you go 100% chances of getting War :D

OK, I stand corrected, sort of. I missed or forgot the line that said (and he knows which one it is). That makes his choice an informed one, and the odds are better if you switch if puzzle logic requires that he intends to offer you the choice to switch boosters after he opens one.

So, if the logic of puzzles doesn't require this, then the restricted choice probability arguments hold. But, if the logic of puzzles does not require this, then the probability argument is inaccurate, as I believe I have shown. The math teacher in me has designed a simple, 3 card (a black Ace and a pair of red Twos) to show the situations, but it is Christmass break, and I don't want to assign homework.

One last thing:There are only 6 possibilities on a fair six-sided die. They are 1, 2, 3, 4, 5 and 6. Each has a 1 in 6 chance of appearing on any die roll. Adding these together gives 6/6 or 100%. And absolute. Therefore, on a fair 6 sided die numbered 1-6, there is absolutely zero possibility of rolling a 53, or a 79, or any other number than 1, 2, 3, 4, 5 or 6. This is a PROBABILITY argument, not a STATISTICS argument.

So, by rolling a die, say, a million times, you can't statistically prove it won't roll a 53 on the next roll, but you can, by analyzing the universe of possibilities, show that the result (53) is impossible.

Again, it is subtle, but it is important. Statistical arguments involve predicting the future based on observations in the past. Probability arguments involve an analysis of the entire universe of possibilities, and calculating theoretical probabilities from that universe. It may involve math that looks similar to statistics, but they are very different.

All this reminds me of a Garfield cartoon from some years ago. Jon and Garfield are watching the news and hear that someone broke his legs jumping off a building trying to fly. Jon asks Garfield what he thinks the guys problem was. Garfield, with towel for a cape, runs to the roof with the thought bubble “He didn’t believe!” Hope always springs eternal. And Vegas keeps growing.

Well, if anyone wants a homework assignment, set the scenario up with three cards, a friend who can make the “informed” choice of which booster to open, and see if you can set up the scenario.

Finally, as for why someone would set this up and, as I suggested, only offer to switch if you have the wining booster, well, we have all probably seen the “grab bag” bins or auctions, say, of magic cards. The sign on the barrel (or whatever) says you get 4 cards for a dollar, and that one of the 1000 bags has a Black Lotus (a very valuable M:tG card). If one of the first people to pick, say #11, opens the Black Lotus, does the guy selling the cards stop selling, or change the sign? But, then again, this goes to motive, and puzzle logic may rule motive out in our original example.

Perhaps I made it more complicated than necessary, but then again, I’ve seen the winner at some of these grab-bags (a former student of mine had luck that was nothing short of amazing) and the people running the sale never changed the sign when the “prize” was already pulled.

If anyone really wants the math homework, I’ll write the lesson next year.
For now, I’m out of here.

Happy Holidays!

Ok, I admit I've skimmed a lot of the discussion. But in short, all you can go off of is what IS at THIS POINT. What happened in the past has no influence on what may be, pretty much. And I admit, I'm not a statistician, so my discussion is not "mathematically based". My strength is Logic (due to my basis in computer science).

Basically, at this point, there are TWO boosters left. One has War, one does not.

Prior to the "third" booster being opened, the chance that you had War is 33% - 1 in 3. Once it was opened, and shown to NOT have War, your chance that you now have the correct one is 1 in 2. This is because of the ORIGINAL boosters, each has a 33% chance. One of those 33% is shown to be 0%. Thus, there are two options, each that ORIGINALLY had a 33% chance. Based on the new information, each now has a 50% chance of being correct.

Thus, barring any psychological mind games that the other guy may be playing, it's a toss up as to whether you should switch. It's either in the booster you have, or the other booster. The fact that it was NOT in the third booster does not make it more likely to be in either of the two remaining in specific - in essence, it's "odds" were divided among those that are left.


From my brief studies in combination theory, one concept I do remember. If you have X unique items, your chance of pulling a specific item on the first try is 1/X. If one of the items (not the one you want) is removed, your chance goes to 1/(X-1). For every element in X that is removed that isn't the one you want (let's say Y), then your chance of pulling it is 1/(X-Y). I don't remember what this is called, but I remember there being some mathematical proof of it :)


Anyway, just my two cents, from the non-mathematically based sigma guy :)

Nope, your forgetting that the person opening the booster knows which booster has War in it. That changes things.

Note the information I failed to include in the original post (which I should go back and edit in). You know he's going to show you a "non-War" booster from the two remaining boosters.

There are 3 cases, we'll call the boosters A, B & C. The booster you choose is A.

1) Booster A has War. Your friend will open B half the time and C half the time. You should keep the booster.

2) Booster B has War. Your friend will always open booster C. You should trade for booster B.

3) Booster C has War. Your friend will always open booster B. You should trade for booster C.

So, in 2 of 3 cases, you win if you trade. In 1 of 3 cases, you win if you keep it. I'll take the odds of trading any day at a casino.

It shouldn't matter whether he knows or not. If he opens one you didn't pick and it happens to turn out a loser, the situation is the same.

Well, my brain works better on Computer Program stuff than math. So I did what any computer geek would do:

I wrote a program to simulate it.

Here's the "pseduocode"

- Put the WAR in a random box (numbered 1 to 3). The other two are empty.

- Always pick the first box (this is just as likely to pick it as picking it randomly, but it's easier to program this way).

- If Box 1 HAS War, the computer randomly picks one of the boxes left.
- If Box 2 has War, the computer picks box 2 (representing the one that he has left)
- If Box 3 has War, the computer picks box 3 (representing the one that he has left)

Then, it "swaps" boosters, making box 1 have whatever was in the computer's box.

Then it checks to see if box 1 has War (after the Swap).



I ran 50,000 test cases. 33,250 of those times, box 1 has War AFTER the swap. That's 66.50%.

Thus, based on that, I guess there is something mathematically I must be missing :) I'm not a mathemetician, and this is why, I guess ;) But I am convinced, now, unless someone can point out a flaw in the program logic :)

Well, I thought of a way that it fits in my brain "logically", if not mathematically. Though DBlizzards explanation seems to fit, here's how it helped me.


You pick a booster. You have a 33% chance of it being right. Your friend has a 66% of still having it.

In essense, your friend is kind of offering you the chance to swap your ONE booster for his TWO. Why? Well, because you already KNOW that one of the two boosters he has will NOT have it. Thus, his opening a non-War booster is not really a surprise, since he KNOWS it doesn't have it. Thus, nothing "new" is being revealed. This is no different than if you traded your ONE for his TWO - you KNOW one of them will NOT have War. But, you will have a 2/3 chance of having ONE of them with War, and you only have a 1/3 chance of yours having it.

I think I'm starting to see where DBlizzard is coming from - the fact that he KNOWS where War is makes the difference - his opening the empty booster is just smoke and mirrors - and it should be no surprise that it's empty, since he'll obviously pick the one that's not to "keep" (for the purposes of this anyway).

I still maintain it doesn't matter whether he knows or not. What matters is that you now know. Whether Monty is the one offering boosters with his clairvoyance, or the bottom falls out of one of them revealing the contents or some smart-aleck crushes one under his bootheel to see the offeree's reaction, the situation remains the same. The 2/3 odds you had to lose are still true, since your choice was random. But there's only one booster left in the pile you didn't choose.