Brain Teasers

[gm iprod]

Bollocks.

Shoot them both and make up your own mind and live with the decision.

[kevin c]

Bollocks.

Shoot them both and make up your own mind and live with the decision.

[folsoml]

Ask either genie this one question, "Which path will that genie (pointing to the other genie) tell me is the path to the village?" Whichever path he points to, take the opposite path and you will be right.

If you are asking the question to the lying genie, he knows what the truthful genie will say is the correct path, but will lie about it and tell you the other path.

If you are asking the truthful genie, he knows the lying genie will tell you the wrong path, and he will tell you the truth about it.

Either way, the genie will point to the wrong path. Take the other path.

[boo radley]

This is an old and classic one, but it came up today in a work conversation, and thought I'd throw it out for those that have never heard of it...

At a large area match, there's a side-contest: if your name is drawn, you get to stand in front of 3 poppers. You get to shoot at 1 of the 3 poppers. Behind 1 of the poppers is a Rolex watch. Behind the other two is nothing.

Your name is drawn, and you stand at the line, and pick a popper to shoot at. The RO asks, "are you SURE?". You say you are, then the RO says, "I'm going to make it more interesting -- I'm going to knock down one of the two poppers that isn't hiding the prize," and he proceeds to do so.

Now there are just two poppers standing. Behind one of which is a prize. The RO says, "Your final chance -- are you SURE you want to shoot the same popper you picked." You feel good with the choice and don't switch. You shoot it, and get nothing.

One the ride back, your shooting buddy says, "You moron -- if you had switched that second time, you would have *doubled* your chances at being right!"

Is he correct?

[Condition 1]

Mongo confused....me wike cake!!!!

[shred]

Always switch if the RO knows where the rolex is and isn't just knocking down any old popper.

Think of it this way: Your first choice has a 33% chance you picked the right one, 66% chance it's behind one of the other two. If the RO knocks down one of the ones you didn't pick, the odds are still 66% it was behind one of those two, but now there's only one left to shoot at.

I swear we had this thread a long time ago.

[kevin c]

Hmmm, mebbe I'm missing something here:

My chances were one in three when there were three poppers. One is now gone, so my chances are 50/50 now, but that applies to either of the two remaining poppers, no?

[Bubber]

Hmmm, mebbe I'm missing something here:

My chances were one in three when there were three poppers. One is now gone, so my chances are 50/50 now, but that applies to either of the two remaining poppers, no?

Hindsight is 100%. After you had shot and pick the wrong one if you had changed your mind it would of doubled your chances form 50% to 100%. But it only works after you made the wrong choice.

[Brazos SC Shooter]

This is an old and classic one, but it came up today in a work conversation, and thought I'd throw it out for those that have never heard of it...

At a large area match, there's a side-contest: if your name is drawn, you get to stand in front of 3 poppers. You get to shoot at 1 of the 3 poppers. Behind 1 of the poppers is a Rolex watch. Behind the other two is nothing.

Your name is drawn, and you stand at the line, and pick a popper to shoot at. The RO asks, "are you SURE?". You say you are, then the RO says, "I'm going to make it more interesting -- I'm going to knock down one of the two poppers that isn't hiding the prize," and he proceeds to do so.

Now there are just two poppers standing. Behind one of which is a prize. The RO says, "Your final chance -- are you SURE you want to shoot the same popper you picked." You feel good with the choice and don't switch. You shoot it, and get nothing.

One the ride back, your shooting buddy says, "You moron -- if you had switched that second time, you would have *doubled* your chances at being right!"

Is he correct?

This was on an episode of that show "Numbers". The idea is you have twice as many chances to be wrong than right with three choices. So while you may feel confident in your original choice, the probability that it was wrong is higher and so you choose to shoot the other target. The math guru on the show had a much better explanation but this is the basic idea.

[Waltermitty]

This is an old and classic one, but it came up today in a work conversation, and thought I'd throw it out for those that have never heard of it...

At a large area match, there's a side-contest: if your name is drawn, you get to stand in front of 3 poppers. You get to shoot at 1 of the 3 poppers. Behind 1 of the poppers is a Rolex watch. Behind the other two is nothing.

Your name is drawn, and you stand at the line, and pick a popper to shoot at. The RO asks, "are you SURE?". You say you are, then the RO says, "I'm going to make it more interesting -- I'm going to knock down one of the two poppers that isn't hiding the prize," and he proceeds to do so.

Now there are just two poppers standing. Behind one of which is a prize. The RO says, "Your final chance -- are you SURE you want to shoot the same popper you picked." You feel good with the choice and don't switch. You shoot it, and get nothing.

One the ride back, your shooting buddy says, "You moron -- if you had switched that second time, you would have *doubled* your chances at being right!"

Is he correct?

The odds do not change after the third popper is knocked down. You have a 50/50 chance of guessing the right one, no matter how many times you're asked or change your selection.

This is a simple coin toss with drama. You can call Heads or Tails before the toss, do a dance, kiss your girl for luck, etc, but the process is the same. Some people guess right after some of these procedures and learn the wrong lesson. That's why Casino's and Lotteries make money; but that's another thread.

So if he had changed his guess the second time, he would have increased the *accuracy* of the guess, but unless the RO was trying to help him win (doncha just hate it when the RO says " IF you are finished...long pause..." ) the chances cannot change.

[p99shooter]

Definitely switch poppers.

The only time you'll lose when you switch is when you picked the right popper initially; a 1-in-three chance. All other times, you will win.

Another way to look at this is to run through all of the permutations of the game ([nerd]which is pretty easy to look at with SQL and some cross joins[/nerd]). Overall, you have a 1-in-three chance of winning if you don't switch, and a 1-in-2 chance if you do take the strategy of switching.

[shred]

Anybody that thinks the odds are 50-50 after the RO knocks down a popper, well.. I need them nearby the next time we run a side match, I could use the $.

It's not 50-50 because the RO knows and never exposes the winning popper.

If it were random and the RO didn't know, then it would be 50-50, but then 1/3rd of the time the winning popper would get knocked down first and there wouldn't be any point to the second round.

Try it with some playing cards or coins. Take some cash from the neighbor kids..

[p99shooter]

I didn't mean 50/50 in a single permutation of the game. I meant overall for all possible combinations of the game. Like I said before, the only time you don't win if you switch is when you picked the right one to begin with. So you have a 1-in-3 chance of losing, and a 2-in-3 chance of winning.

[Waltermitty]

The third popper is a diversion. There are only two poppers in the game. Sorry guys. Pick Heads or Tails, Door #1 or Door #2. 50/50.

BTW p99shooter, a "1-in-2" chance *is* a 50/50 chance.

The RO's knowledge, or lack thereof, after the game changes to a two popper game of chance cannot effect the outcome of an individuals' choice. If you play the game enough times you'll win about half the time.

Line up 100 people and give them coins. Tell the people your goal is to produce "Heads". Everybody flip, about half get Heads. Fire everybody else and have the ~50 flip again. Continue till you get down to the last 2-5. Ask them how they got so good at flipping Heads. They might give you an answer, but it won't mean anything. The only way to get worse results is to let the people try to predict THE NEXT TOSS. That's the trap of this example.

Some will think they have a "system" but it will be wrong. The error here is to include the 1 in 3 chance with the 1 in 2 chance and then trying to drive the probabilities as if the two games have an effect on each other. They do not.

As far as anybody wanting to bet me money on side matches, I choose my bets very carefully. I use Casino games (like Craps, Cards, coin flipping, etc) to teach statistics in manufacturing plants. In the classes, we run the games and I *prove* that the systems are stable and give predictable results. But even if we have just rolled a pair of dice 100 times, established a Mean, 3 sigma limits and the normal distribution, NO ONE can tell you what the next roll will be. Because each roll is separate and has its' own probabilities.

The House then sets the rules so you have to bet against the process and then counts on three basic facts: 1) most people are bad at math, 2) many people believe in "Luck" and 3) most people won't quit when they are ahead.

[kevin c]

The third popper is a diversion. There are only two poppers in the game... The error here is to include the 1 in 3 chance with the 1 in 2 chance and then trying to drive the probabilities as if the two games have an effect on each other. They do not.

That's my understanding too.

Think of it this way, the game was ALWAYS about two poppers, because the RO, in this example, will always remove one of the nonprize poppers, had always intended to, so that you will always be left with one prize and one nonprize choice. If he didn't remove one, then you are playing a different game, with odds of one in three.

Actually, there are few enough permutations here to put them all down:

You pick popper A, which has the prize. RO removes popper B, and you keep A - you win.

You pick popper A, which has the prize. RO removes popper B, and you go to C - you lose.

You pick popper A, which has the prize. RO removes popper C, and you keep A - you win.

You pick popper A, which has the prize. RO removes popper C, and you go to B - you lose.

You pick popper B, which has no prize. RO removes popper C, and you keep B - you lose.

You pick popper B, which has no prize. RO removes popper C, and you go to A - you win.

(Note that the RO cannot remove the prize containing popper A).

You pick popper C, which has no prize. RO removes popper B, and you keep C - you lose.

You pick popper C, which has no prize. RO removes popper B, and you go to A - you win.

(Note that the RO cannot remove the prize containing popper A).

Moving the prize to B or C generates identical sets with identical results in terms of the pattern and number of wins and losses.

Check it out: Odds overall are 50/50 on wins/losses. Always switching (the second "go to" option in each pair) is also 50/50.

What would be truly evil (cheating actually) is if the RO only offered you the choice to switch after you picked the winning popper. You'd go from a 100% down to a 50% chance of winning. Of course, if you knew that he'd do that only in that situation, then you'd always win by refusing. Then again, if he truly did want to give you a chance when he knew you picked wrong, and you refused to change, you'd always lose. Converse applies as well. The irony is, without knowing the RO's mind as far as strategy goes, you odds are still the same 50/50 no matter what you think.

If the RO removes a popper at random after you pick, then you'd get additional outcomes for initially picking B or C, all four of which lose, since these apply when the RO removes the winning popper in those two sets of outcomes. The odds of winning go down to one in three overall. The odds of winning if the winning popper is still in the game are still 50/50 regardless.

If the RO doesn't remove a popper at all, your odds are a straight one in three.

Edited July 6, 2006 by kevin c

[boo radley]

The third popper is a diversion. There are only two poppers in the game... The error here is to include the 1 in 3 chance with the 1 in 2 chance and then trying to drive the probabilities as if the two games have an effect on each other. They do not.

That's my understanding too.

Think of it this way, the game was ALWAYS about two poppers...

No, you both are wrong.

Look at it like this:

A

1/3

B

1/3

C

1/3

You would agree when you intially get to pick a single popper the chances of winning the prize are 1/3 no? Say you pick "a". And if you could pick TWO poppers, the chances of winning would be 2/3, right? For example, if you were able to pick both "b" and "c".

So the diagram would look like:

A

1/3

B

1/3

(+) 2/3

C

1/3

Now, what happens is the RO knocks down a NON-prize concealing popper, every time, from the pair that you didn't pick ("b" and "c") and because of that, the 2/3 chance of being right "shifts" to the remaining standing popper you didn't pick. Like this:

A

1/3

B

2/3

X

Make sense? Suppose the RO didn't knock down any poppers, but offered you a choice to switch and pick BOTH of the two poppers you didn't select. That, in effect, is what he's doing.

[Waltermitty]

The third popper is a diversion. There are only two poppers in the game... The error here is to include the 1 in 3 chance with the 1 in 2 chance and then trying to drive the probabilities as if the two games have an effect on each other. They do not.

That's my understanding too.

Think of it this way, the game was ALWAYS about two poppers...

No, you both are wrong.

Look at it like this:

A

1/3

B

1/3

C

1/3

You would agree when you intially get to pick a single popper the chances of winning the prize are 1/3 no? Say you pick "a". And if you could pick TWO poppers, the chances of winning would be 2/3, right? For example, if you were able to pick both "b" and "c".

So the diagram would look like:

A

1/3

B

1/3

(+) 2/3

C

1/3

Now, what happens is the RO knocks down a NON-prize concealing popper, every time, from the pair that you didn't pick ("b" and "c") and because of that, the 2/3 chance of being right "shifts" to the remaining standing popper you didn't pick. Like this:

A

1/3

B

2/3

X

Make sense? Suppose the RO didn't knock down any poppers, but offered you a choice to switch and pick BOTH of the two poppers you didn't select. That, in effect, is what he's doing.

Makes no sense to me at all. The three popper part of the story is still a diversion. You're never given the option of picking two poppers and checking the result. The probablility of winning does increase by going from 3 poppers to 2, it would also increase if one more was knocked down before you choose .

The question about picking two poppers is also not material unless those are the rules of the new game.

You're engaging in a practice that humans are hard-wired to perform, that is, you view events as related on a single continuous time-line. The artificial rules of a game of chance negate the value of this practice. This is how casino's and con-men make their living.

If I flipped a coin 5 times and saw Heads every time, you would be highly tempted to bet that Tails would come up next; and I wouldn't blame you. But the odds of Heads coming up a 6th time is exactly the same as it was the first time the coin was flipped, unless something is wrong with the coin.

The probablilities cannot "shift" from events (or elements) that are no longer in the game. Any gambling addict will tell you (and believes) that he's lost so many games that it's "time" for him to win. But the probabilities of the game(s) reset each time the game is run or if the rules of the game are changed.

Do you suggest that if the doodling with a third popper had never occurred that the chances of picking the winning popper from two choices is anything other than 50:50? What scientifically verifiable process alters that fact by having a third popper shown, then removed, before the game begins?

I really want to learn this thing if it is true, but so far, the evidence presented is always based on the third popper having an effect on the selection between popper 1 and 2 alone. Its removal simply improved your chances of picking the right one to the maximum possible of 1 in 2 or 50:50.

[shred]

Once more, the RO KNOWS where the prize is. This is the key. Without that it is 50-50. With that, it isn't.

You have a 66% chance of picking the wrong one first, right?

Those odds then never change.

What changes is one of the two you didn't pick (66% chance it's behind either of those two, right?), and never the winning one goes away.

The odds are still 66% you picked the wrong one, but there's only one other one left now.

It's the same as if the RO said "You can either keep your popper, or take what's behind both of the other two". No brainer, right?

Here's a variant that might help: There are a hundred poppers and one Rolex. You pick one. One in a hundred chance you got the right one. The RO carefully knocks down ninety-eight of them with no watch behind them and leaves one up. You can switch or stay.

Googling "Monty Hall" will turn up a thousand hits on this very problem.

[Waltermitty]

A decent explanation of the "Monty Hall Problem"

*IF* the RO must ALWAYS knock down an incorrect popper and it CAN'T be the one you picked:

*IF* the RO must ALWAYS offer you the chance to change your original choice.

If one or both of the requirements above are at the discretion of the RO based on what he/she knows about your choice, this proof is not valid.

Those details are pertinent to the evaluation of the problem.

Marilyn is tricked by a game show host

Edited July 7, 2006 by Waltermitty

[kevin c]

All the possible permutations for the game as described are put down in my 7/5/05 post. I count up to best odds of only 50/50.

If you want to satisfy yourself, put a mark on one side of one of three quarters, get a friend to play the all powerful RO, and go to it! (but do make sure you are using the same rules as originally described, otherwise it's a different game)

Edited July 7, 2006 by kevin c

[shred]

You did try it, right? I did and it still works out just the same-- twice the odds of winning if you switch (taking into account the clarifications waltermitty posted)

Here's the permutations. Leave the marked coins face up:

Put your finger on the winning one. Take away either one of the losers. Switching loses.

Put your finger on the first losing one. Take away the other loser. Switching wins.

Put your finger on the other loser. Take away the first loser. Switching wins.

Edited July 7, 2006 by shred

[kevin c]

I don't think you put down all the possible combinations. And to play the game the way it was described, you need to be blinded to the choices. To calculate the odds, though, you can look, but need to include all the different choices with all the poppers involved.

You put your finger on the winner and take away one of the losers. Switching loses.

You put your finger on the winner and take away the OTHER loser. Switching loses.

Now you have an additional loser. The total # of permutations in your diagram becomes four with two wins and two losses. 50/50. (In reality the total # of possible combinations of prize location and choices is 24, not 8 as in my diagram or 4 as in yours, but the odds are still 50/50).

Let's turn it around. You start with only two poppers: one winner and one loser. Your odds are 50/50. The RO adds another popper (or another 98 - it doesn't matter). Your odds of winning are still 50/50 if you keep the same choice or pick the other original popper, because the other poppers obviously don't matter since you know the winner has to be in the first two.

Now what if the RO takes the other original popper and mixes it into the other added poppers so that you can't tell them apart, then removes all but one of these poppers, showing, of course that none of those removed was the winner. He then gives you a choice of keeping your original choice or switching. The odds are still 50/50, switching or not, because either your first choice has the winner (in which case it doesn't matter what the second popper is, since all the others are losers), or the second original popper does, in which case the second original popper is the still other choice, because the RO can't remove it from the game if it is the winner. All he did was add extra poppers, and then take the same ones away again. "Window dressing", as WM put it. There were only two choices all along.

Edited, supposedly for clarity

Edited July 8, 2006 by kevin c

[folsoml]

Put your finger on the winning one. Take away either one of the losers. Switching loses.

Put your finger on the first losing one. Take away the other loser. Switching wins.

Put your finger on the other loser. Take away the first loser. Switching wins.

I was convinced it was 50:50 until it was explained this way. I don't see any other permutations.

[kevin c]

Still nope.

It's the same difference as rolling a twelve or an eleven with a pair of dice.

A twelve requires both dice to come up six so there is only one way out of the 36 possible combinations of dice rolls for this to happen.

An eleven can come up two ways: a six on the first die and a five on the second, or a five on the first and a six on the second. The two look the same, but there are two ways to get five and six, so the odds are two out of 36. The dice look the same so the two ways may not be obvious. If the dice were each of different colors, then it would be easier.

Similarly, there are two ways for the RO to eliminate one popper once the shooter picks the winner - the RO can pick either of the two nonwinning poppers, and each remaining popper can be picked by the shooter or not. Shred didn't diagram all the possibilites. If each coin were different, then it'd be clearer.

Edited July 8, 2006 by kevin c

[shred]

I diagrammed all the starting and ending possibilities. There are more intermediate states, but the outcome is win or lose, not which loser you got.

Which of the losers the RO removes if you pick the winner is irrelevant; each of the losers is identical so it you get loser #1 or loser #2, it makes no difference-- the two states collapse into one outcome. There are only 3 possible start positions, and you get one binary action each; thus six possibilities.

Start On --- Action --- Result

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

Winner --- Switch --- Lose

Winner --- Stay ---- Win

Loser 1 --- Switch --- Win

Loser 1 --- Stay ---- Lose

Loser 2 --- Switch --- Win

Loser 2 --- Stay ---- Lose