Brain Teasers

[folsoml]

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

Now you just knocked me back to 50:50. Before, I was not taking into account the fact that you could take no action at all. If I look above, I see six possible outcomes, three of which win.

[kevin c]

See # 49 on the analogy with rolling dice.

To calculate odds, you must include all the possible combinations and factor in the frequency at which they occur.

With the dice, you have an eleven consisting of a five and a six. There are two ways of gettng a five and a six with two dice, so you must include both ways to calculate the odds correctly. On average, there will therefore be two elevens in every 36 rolls. Both elevens look the same, but they still represent two, not one of the 36 possible combos. So the frequency of an eleven is twice that of a twelve.

With the poppers when the shooter picks the winning popper (call it "A"), the RO has two nonwinning poppers to pick from ("B" and "C"). Each of them can be removed, and the other picked or not by the shooter. To calculate the true odds, you have to include all the combinations, even though they look the same to the shooter - you have to include the possible outcomes for B and C both. That's what I did in post #40, and that is why there are eight listed outcomes, half wins and half losses, regardless of switching strategy.

Noting that the shooter picks the winner and the RO picks a nonwinner as one of three possible scenarios underestimates the frequency of that particular scenario, since there are two ways it can happen (RO picks B or RO picks C), and so there are twice as many outcomes compared to what is shown. That changes the ratio of wins and losses.

If the poppers were labelled A, B and C, or were different colors, it would be easier to see.

The diagram in #50 should be modified to show:

Winner (A) - B removed - pick A and win

Winner (A) - B removed - switch to C and lose

Winner (A) - C removed - pick A and win

Winner (A) - C removed - switch to B and lose

The rest is the same.

Now add up the switches, and you will see that you have 50% wins, 50% losses, same as not switching.

Edited for clarity

KC

Edited July 8, 2006 by kevin c

[Waltermitty]

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)

There are a number of restrictions placed on the RO that are not explicit in the description of the problem here that are fleshed out in detail elsewhere. I missed questioning a couple of them completely. This is what makes the math assumptions work.

I assumed the RO wouldn't knock down the winning popper (as was stipulated in another post). This seemed logical and appears to be correct. I did not assume that the RO could not knock down the popper I had chosen if it was wrong and I did not assume the RO was required to allow me to change my choice only after revealing the other loser. These stipulations are what connects the two games into one set of probablilities. The game will not work out any other way with any other assumptions.

As you pointed out in your post Kevin, the RO could queer the game by only offering you the opportunity to switch when you chose the correct popper in the first place, or visa versa.

The RO would also screw up the assumption if he knocked down the popper you selected if it was wrong.

So it's a little bit more complicated than just "the RO knows where the prize is". Without these rules restricting the poppers the RO can eliminate and when, no logical person up to a PhD in Mathematics can state the popular conclusion, which is to always switch.

Unless you can ask the RO these questions and have the correct answers returned, the final choice remains 50:50 on the last guess.

So this appears to be one of those little anomalies that requires (somewhat) tortured stipulations (that aren't fully revealed) to engender an incorrect response. I would guess that somewhere nearing 99.99% of the people hearing this riddle the first time guess wrong, particularly if they know anything about statistics or have trained themselves to understand random systems and processes. That's probably part of what makes it fun for each successive storyteller.

I'd also hazard to guess that an important percentage of those that do give the "right" answer (now) do so only because they have seen it before; and haven't a clue as to why or how it really works.

[boo radley]

So it's a little bit more complicated than just "the RO knows where the prize is". Without these rules restricting the poppers the RO can eliminate and when, no logical person up to a PhD in Mathematics can state the popular conclusion, which is to always switch.

I agree, mostly. It's hard to phrase the question in such a manner that it's both interesting to try to solve AND doesn't descend into a mere equation with a set of rules an agent must follow, etc.

It's a one-time shot -- who's to say the RO/Monty Hall/whomever, will offer this specific challenge again? But once you start assigning rules this party MUST follow, it becomes a non-puzzle, though still somewhat interesting, I think.

[Waltermitty]

So it's a little bit more complicated than just "the RO knows where the prize is". Without these rules restricting the poppers the RO can eliminate and when, no logical person up to a PhD in Mathematics can state the popular conclusion, which is to always switch.

I agree, mostly. It's hard to phrase the question in such a manner that it's both interesting to try to solve AND doesn't descend into a mere equation with a set of rules an agent must follow, etc.

It's a one-time shot -- who's to say the RO/Monty Hall/whomever, will offer this specific challenge again? But once you start assigning rules this party MUST follow, it becomes a non-puzzle, though still somewhat interesting, I think.

Well said.

[JD45]

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

This description changed my mind.

What I see is if you stay 3 times you win once. If you switch 3 times you win twice. If those are all of the combinations I'll switch every time.

[shred]

To calculate the true odds, you have to include all the combinations, even though they look the same to the shooter - you have to include the possible outcomes for B and C both. That's what I did in post #40, and that is why there are eight listed outcomes, half wins and half losses, regardless of switching strategy.

You have 3 choices to start with. You get one binary action after that. How many combinations can that be? 3 * 2 = 6.

What song the RO decides to sing between picking poppers has no bearing on the outcome, yet adds infinite combinations, same as the RO picking loser 1 or loser 2 to knock down. No effect on the outcome.

[kevin c]

By golly, Shred, after lookng over my diagrams and line of thinking, I think you were right all along. My errors in logic:

WM and I are right in thinking that removing one popper before or after didn't influence the odds of 50/50 for the INITIAL pick, and the odds of picking at random a second time are still 50/50. But my argument afterwards there being no way to improve on 50/50 was off because the RO's actions afterwards are, as you point out, not random - he can only remove a nonwinning popper, and because the shooter has a SECOND, nonrandom, pick to make after the RO's actions.

I did diagram the eight possible outcomes, and that is mostly what made me sure it was 50/50 at the end, but I didn't take into account what I had actually mentioned in another post - likelihood. There may actually be eight permutations, four of which include the shooter picking the winner and the RO picking one of the two losers, but these four occur only at half the frequency of the other possibilities where the shooter picks a loser and the RO takes the remaining loser; half the probability because the winner is first picked by the shooter only once for every two times one of the two losers is first picked. That means the number of losses by switching from the winner are only half the number I counted up, and that works out to odds of one in three.

My dice analogy doesn't fly because each of the two dice comes up with a number independent of the other, but there aren't "independent trials" here because RO's actions are constrained by what the shooter picks first, altering what is left for him to pick, and that he cannot pick randomly (can't remove the winning popper if it is available).

Succinctly put, there is a two out of three chance that the popper left by the RO is the winner. Another way of looking at it is that your last (not first) choice is between one popper (your first choice) and the other two poppers together, and, yes, indeed, the odds are twice as high that the two will have the winner vs the one (66.6% vs 33.3%). The remaining two may not be there after the RO removes one, but the one remaining popper still represents two of the three possible locations of the prize, and higher probability location at that. The option of a second choice makes all the difference because it allows you to get to this scenario.

A more extreme (but clearer, to me) example: one prize in a hundred poppers, grouped into one group of one popper and one group of 99 poppers. 1% chance with the first, 99% chance with the second. Eliminate 98 poppers from the second. Now you have two groups of one each. For somebody walking onto the scene at this point and making a pick, the odds are 50/50 that they will pick the winning popper, which nonetheless has a 99% chance of being in the group that started with 99 and had 98 non winners removed from it. The shooter, however, was not blind to which group had more chance of having the winner in it.

All this assumes the RO, as WM pointed out, cannot take away either the shooter's first popper or the winning popper if it remains among the poppers not picked.

Blinded by my own certainty and misapplied reasoning

Hat's off to you, Shred.

Kevin C.