I think the basic approach to this is that in the Monty Hall model, you are basically switching one booster for two boosters when you swap. That's why your odds go up to 2/3 when you switch.

Effectively, you get both of the boosters that the other guy had - the one that is empty (well, you don't get it, but you might as well have) and the other.

It's not simply the increase in information - it is that fact that you have information about the 2/3 winning side.

The teacher in me can't help it.

First, let me congratulate Simazero13 for DOING something. His results proved that if the person making the offer knows which is the prize, and is bound to open a loser, then you are, statistically, more likely to win if you switch. And my little card example was going to suggest only 50 tirals. 50,000 is much better. A+

And, actually, showing that if he selected one at random is different than if he selected one with knowlegde and aforethought is tricky without doing the experiment.

So, anyone who want to try it (sigma, it should be easy to modify your code for this--bonus points are being offered here, and the semester ends January 21) can set it up as follows.

1) Set up a data table with three entries:Win, Lose, No-Game

2) Get three cards, an ace and two others. The Ace is the prize, the others are losers.

3) Shuffle or randomely select one. Since it is hard to randomly shuffle only three cards for yourself, I recomend using a fair die.

4) Flip over one of the remaining 2 cards. If it is the Ace, record No Game and go back to step 3. This should happen about 1/3 of the time.

5) If it is not the Ace, your card for the remaining card. Then, flip it and record Win (if the Ace) or Lose Iif not the Ace). Then go back to step 3.

Do the loop 99 times.

I claim you are expected to have 33 No Games, 33 Wins, and 33 Loses. This is because the flip was made without knowledge of whether it was a win or not.

Spudkin, who sees all, claims that since knowledge is not relevant, that your wins should be 44 to 22 loses.

Of course, it is unlikely that the numbers will be exact, but 50/50 is enough different than 2/1 that it should be obvious which is right.

Or, if Sigma will oblige us, he can modify his code a touch and run it 50,000 times. (bonus points!)

I leave the rest, as they say, an exercise for the reader.

I didn't claim that knowledge is not relevant. I claim that it doesn't matter whether Monty is revealing a loser because he knows, or if a loser is revealed by chance. If a loser is revealed for any reason, _your_ knowledge is increased from when you randomly chose. Therefore you would be better off switching.

It's your knowledge that matters, not how you get it.

Obviously, I'm not going to write anymore long posts trying to demonstrate or justify my conclusion for this puzzle. However I have a couple of questions/comments about the "homework" you described, and especially the parts I've quoted.

Setting up the experiment the way you did is fine with me, as I haven't noticed anything that would give me pause, except for the "No Game" part. Please bare with me.

-Wouldn't counting the "No Games" as part of the results skew the results? I understand that it could show that one-third of the time, the Ace would pop up, as would the others (in a perfect world) However, counting these in the results would eliminate the decision aspect of switching the boosters.

Basically, including "No Game" results would not be representative of the puzzle, since you go no further once the Ace is shown.

-Wouldn't it be more representative if the "No Games" were excluded from the 99 or 50,000 times the loop is run? The reason I ask this, is that it would indicate that "War" was in one of the two boosters that weren't chosen.

-If you take out the "switch" aspect, isn't it just as easy to test it by picking one card and leaving the other two? Then, just see if the Ace is the card you picked, or one of the two you didn't pick?

-This would give results as to where the Ace is while running all the loops, and the switch isn't necessary to test this.

Basically, it will show that either you picked the Ace or didn't, and how many times you picked it compared to how many times it still was one of the two cards you did not pick. This would demonstrate how close the probability is in a realistic test....I think...lol

Perhaps I'm missing something here, but I believe the only valid results would be from "Win" or "Lose", and "No Games" should not be counted in the total desired sample amount.

:confused:

Important Point I left out:

If the results of the 3 possible results (Win, Lose, No Game) are true to form and all end up occurring 1/3 of the time, then taking out No Game results would definitely mislead the student.

The reason for this is that results of Win = 33 times, Lose = 33 times, and No Game = 33 times would be correct with each possiblity occurring 1/3 of the time. If you were to eliminate the "No Game" results, it would give the impression that Win or Lose would have an equal chance of being the result.

This is not a pure result, as one must remember that every "No Game" that occurred is representative of the Ace being one of the two cards you did not choose.

This would definitely work to prove that the odds are still 2 in 3 that you did not pick the Ace (War booster).

The reason I write this is that now that I think about it, even eliminating the "No Game" result will skew the results more toward 50%.......That is not a fair test.....You would basically eliminate the 3rd option.....

This is what the end of the previous post was supposed to say....sorry

The reason I write this is that now that I think about it, even eliminating the "No Game" result will skew the results more toward 50%.......That is not a fair test.....You would basically eliminate the 3rd possibility.

This will give the illusion that Win or Lose would happen about 50% of the time, although both are limited to the fact that they are equally expected to occur 1/3 of the time.

1/3 of the time + 1/3 of the time ....does not equal a 50/50 chance....you are still missing the third possibility that makes up the missing 1/3 of the time.

**ChaimWitz shakes his head to regain focus on silly stuff**

No, you are not missing anything. In fact, your question reveals you are thinking in exactly the direction I wanted you to think. As a math teacher, I am happy for both of us.

Just for fun, let’s not keep track of the "No Game" numbers. They are not used in the calculated results of Wins to Loses anyway.

In that case, modify step 4 to read:
4) Flip over one of the remaining 2 cards. If it is the Ace, ignore record nothing and go back to step 3.

And my analysis at the end to:
Do the loop until you have recorded results 66 times. [This will take about 99 total shuffles.]

I claim you are expected to have 33 Wins, and 33 Loses. This is because the flip was made without knowledge of whether it was going to be an Ace or not.

Spudkin, who sees all, claims that since knowledge [b]on the part of the person flipping of the two remaining cards[b] is not relevant, that your wins should be 44 to 22 loses.

*****
Short side note: Spudkin, if it wasn’t clear, I have the highest respect for you as a Warlord and promoter of the game. I was not trying to be flip, or rude, by referring to your seeing all, rather, I was obliquely referencing the fact that the potato sees all, ‘cause it has so many eyes. And the knowledge I am referring to is the knowledge on the part of the person flipping the second card (after you have selected yours), not knowledge in general. That specific knowledge is relevant precisely because of the argument to follow. I am certain that you value knowledge in general. In fact, after you do the experiment, you might even appreciate the fact that I have helped one eye of the potato see a little more clearly.
*****

Why does the knowledge of the person flipping the second card (opening the booster) matter. Well, it matters for exactly the reason that there will be about 33% of the shuffles that lead to a non-game. Follow me here ChaimWitz, you are 95% of the way to understanding this theoretically. In the case where the person knows which card (or booster) is the winner, he will never (by the rules of the puzzle) pick that card. Thus, those 33% of the possible outcomes are NON-RANDOMLY removed. That is, if you have picked one of the two loser cards, he will always (in this puzzle/game) pick the non-winner. That way leads, as Sigma’s computer simulation shows, to a 2/1 statistical advantage in switching.

However, if the person flipping the second card is doing so RANDOMLY, then, if you have picked a loser card, half of the time he will reveal the prize. The other half of the time, you would win. That is, in a random pick situation, one-half of the wins you would have gotten by switching (which gives you the 2-1 advantage in switching) are removed. Half of 2/3 is 1/3. The same odds as you having picked correctly in the first place.

One last thought experiment. Spudkin, ChaimWitz, and WarPlayer are told that one three boosters contains War. We all want that piece (assume we do for this example). Spudkin gets to pick first. Whichever booster he picks we will label Booster A. ChaimWitz picks second, and whichever booster he picks we will label Booster B. I pick last, the remaining booster, Booster C.

Now, at start, I claim we all have a 1/3 chance of having the winning booster. I think everyone will agree. (Well, I have terrible luck at these things, but theoretically we do.) Spudkin wants one of us to open our booster first. ChaimWitz does, and, sorry to say, he pulls all *figures and Domains he already has. No relic. No item. Tough luck.

Spudkin now wants to trade with me, because he thinks I have a 2/3 chance of having the winning booster. If I believe Spudkin’s logic, I say sure. You see, using his logic, the odds of my having the right booster was only 1/3 originally, with there being a 2/3 chance that one of you two had the prize. Since ChaimWitz doesn’t have it, that means that there is a 2/3 chance of either he or ChaimWitz having it originally, and ChaimWitz doesn’t have it, so my odds double if I switch boosters with him.

As it is, I’d still agree to switch with him, because I know it’s 50/50.

I hope you see now why I had you record the No Games, even though we never use them in a calculation. They are relevant to understanding the difference between a random reduction of the pool, and a non-random one.

Finally, I highly recommend doing the experiment I described, with or without recording the number of non-games. It shouldn’t take long. Two shuffles a minute is an easy enough pace to keep, once you get the first few done, it become a pattern and goes a bit faster. It is unlikely to take you much over 100 shuffles to complete the experiment. So, for less than an hour’s work you’ll gain an insight into probability and statistics. That is a great investment for a gamer. So, get a 6 sided fair die and three cards. (Shuffling so small a deck is hard.) And, at the end, you will either see that the numbers are about 33 to 33, or about 44 to 22. You will KNOW which of us is right, and, hopefully, you’ll know why.

A matrix, for those who prefer them:

There are three objects, one winner (W), and two losers (L1 and L2).

These three cards lead to six possible combinations for the non-informed revealing of the second card. They can be set up in a simple matrix:
My pick Revealed card Remaining card
W L1 L2
W L2 L1
L1 W L2
L1 L2 W
L2 W L1
L2 L1 W

That’s it. Just 6 possibilities.

Since I am going to switch with the remaining card every time, if the remaining card is the winner, then I win.

Let’s remove the two that don’t fit our story/puzzle, that’s the case where the revealed card is the Winner. This leaves us with
W L1 L2
W L2 L1
L1 L2 W
L2 L1 W

In the random reveal of the cards, that is a 2/4 chance of winning if I switch or not.

Now, let’s bring knowledge of the person revealing the card into play. [note: For those of you who also teach (or taught) probability, I know we need not differentiate between the two losers, but it is easier to show as a matrix this way with both matrix’s being 6, rather than explain why we don’t need to distinguish between L1 and L2. and have a matrix universe of 3.]

This leads to the following matix
My pick Cards in pool Revealed card Remaining card
W L1, L2 L2 L1
W L1, L2 L1 L2
L1 W, L2 L2 W
L1 W, L2 L2 W
L2 W, L1 L1 W
L2 W, L1 L1 W

Because the revealed card was NOT selected randomly, this gives a 2/4 ratio of it being the Winning card. This situation, where the revealed card is NON-RANDOMLY chosen, is the much referenced Monty-Hall syndrome.

Now, suppose that if you are not sure if the situation is one of random selection of the revealed booster, or non-random selection. If it is random, you don’t increase or decrease your odds by switching. If it is non-random, then you double you chances of winning. Since it is never worse to switch, it shouldn’t hurt, in the long run, if you switch every time. But this only holds true if it is determined, before, that one booster will be opened and that you will be offered a chance to switch if the winner is not revealed. Otherwise, you just might be in the miniatures version of a 3Card-Monty game. And, like Vegas, in the long run, the house always wins.

And, once again, i highly recomend that the interested reader perform the experiment himself.

Peace Out!

The forum didn't like the spaces I used to make my beautiful matrix's. I repeat them here, in a more readable form:

My pick Revealed card Remaining card
W ………………. L1 ………………… L2
W ………………. L2 ………………… L1
L1 ……………… W …………………. L2
L1 ……………… L2 …………………. W
L2 …………….. W …………………. L1
L2 …………….. L1 …………………. W

This leaves us with
W ………………. L1 ………………… L2
W ………………. L2 ………………… L1
L1 ……………… L2 …………………. W
L2 …………….. L1 …………………. W

My pick Cards in pool Revealed card Remaining card
W ………… L1, L2 ………… L2 ………… ………… L1
W ………… L1, L2 ………… L1 ………… ………… L2
L1 ………… W, L2 ………… L2 ………… ………… W
L1 ………… W, L2 ………… L2 ………… ………… W
L2 ………… W, L1 ………… L1 ………… ………… W
L2 ………… W, L1 ………… L1 ………… ………… W

ChaimWitz, you must have posted your most recent messages while I was typing my notes.

You got it! A+ for you. If you do the experiment, you can have bonus points too.

Now, if only I could motivate the other half of my (real) classroom to focus as intently as you, Sigma, Spudkin, cpastore, KelMasterP, DBlizzard, and the others who posted seem to be.

PS--Olan and joedj get bonus points for “thinking outside the box”, and ‘cause Olan scares me a little.

Ok, created a new program to test COMPLETE randomness on all parties. The pseudocode is as follows:

- Put WAR in one of 3 boxes randomly.

- Pick a random box for ME.

- Pick a random DUMMY box.

- Pick the last box for the other guy.

- Tally which box has it.

- If it's not in the DUMMY box, swap boxes, and if you have it after the swap, add a tally*


I ran this 50,000 times (well, it was in a loop). The * means that the reason it only swaps if the dummy does NOT have it is because of the nature of the initial problem, it is given the "dummy" booster does NOT have WAR. Thus, only those cases where the dummy booster was empty were counted for the "swapping" part.


The results were as follows:
16534 times, I picked the WAR the first time. (33.07%)
16648 times, the other guy ended up with WAR (33.30%)**
16819 times, the Dummy had WAR (33.64%)

** It is these times where swapping would be beneficial

There were 33182 times that the dummy did NOT have War. Of these, it is beneficial to swap 16648 times. This is 50.17%


Thus, these numbers have me convinced. And as mentioned by ChaimWitz, here is why it works (in my own words).


For the RANDOM case, the dummy not having WAR was random as ever. It's only because we know that it does NOT have WAR that we throw the cases out where it does. Of the times left, half of the time, I have it, half of the time, he has it.

For the times where the other guy KNOWS, the dummy will ALWAYS be empty - there are no samples that need to be thrown out. The other guy basically gets "2" boosters, you get 1. Thus, you have a 1/3 chance of having it, he has a 2/3 chance of having it. He throws away a box that HE knows is empty. However, this is no surprise - one of his boxes MUST be empty. Thus, his chances of still having the War box is still 2/3, because the empty box becomes smoke and mirrors - it is always empty, but serves a purpose as a "placeholder" to take the odds out of your favor.

I'm not very good at explaining it, but I think it's finally settling in my head WHY this works. Basically, it just boils down to the fact that 2/3 of the time, your buddy will have the War booster after you pick. Throwing away the empty booster doesn't detrace from the fact that he had a 2/3 chance of getting it.

whew, my brain hurts :) Time for a vacation!

Whew!:eek:

After all this, can I at least get a stamp? :p

I'm not sure why you threw away the cases when Dummy had War. In this case you would swap with Dummy because "the other guy" would have revealed the empty box.

Basically, you have the opportunity to swap your one box with the other two boxes.

Because, if the revealed box had the prize, then it would no longer fit the conditions mentioned in the puzzle.

Bonus points for sigmazero13. Also, thanks. And your explaination is excellent.

And now, since we aren't in a real classroom, I suggest we break and meet at the local pub for a pint. ("It come in pints!")

Yes, but "dummy" isn't necessarily the revealed box. Only one of them is the revealed box. If Dummy has War, then the other box ("other guy") is the revealed box.

The simplified program (S013's first run) is a more accurate representation of the problem.