3 Numbers Puzzler


#1

I am not sure I agree with the puzzler answer because the initial problem statement did not seem to allow the use of three known numbers. The problem statement said the numbers had to be any numbers… and how would you increase your odds not knowing the numbers. I believe this Puzzler is was actually put in terms of the Game Show Host Porblem similar to Let’s Make A Deal. I would think the answer is not 50 percent better but 66.67 percent better if you do a switch.


#2

Sort of, yes this is just a twist on an old problem. Made popular from a Marilyn vos Savant column over a decade ago

But that problem is different in this new one in that Monty doesn’t open up a door to reveal a goat. Read the section on the wiki about why the odds are always at least 50%.


#3

Vinnie’s solution did not improve the odds from 1/3. If the first card was the lowest-the second could be the middle(loser) or highest(winner). Similarly if the first picked was the middle, the next could be winner or loser. Picking the highest card first leads to losing
whichever card comes second. So-2 wins out of 6.


#4

Their answer was correct. The answer does not depend on any knowing which three numbers are under the paper. You only need to know that the total papers will be three.
This problem is actually a simplified variation of the general problem. Given any number of pieces of paper, the optimal strategy is to uncover 1/3 of the papers, throwing all away, and then choose the next paper with a number higher than the highest seen so far. Martin Gardner described it in one of his books, I believe.
Although it seems on the surface related to the Let’s Make a Deal problem, it is really a different problem.


#5

Their answer was right, and the odds change to 50-50. It was a mistake for them to mention specific numbers.


#6

@Victor88:

“Vinnie’s solution did not improve the odds from 1/3. If the first card was the lowest-the second could be the middle(loser) or highest(winner). Similarly if the first picked was the middle, the next could be winner or loser. Picking the highest card first leads to losing whichever card comes second. So-2 wins out of 6.”

There’s where you’re missing something: “If the first card was the lowest” You never have this information until you finish applying the solution, which does indeed result in odds of 1/2.