Real Expensive Pills


You have Some Terminal Condition, which necessitates taking two pills a day:
one Pill A and one Pill B. If you neglect to take either pill, you die; if you take more than one A or more than one B, you die. If you don't take them at exactly the same time, you die.

This morning you are going through your usual routine. You pick up your bottle of A Pills and gently tap one into your palm. Then you pick up your bottle of B Pills and tap it, but two pills accidentally fall into your hand. You now hold three pills (one pill of A and two pills of B), you don't know which are which, and they are completely indistinguishable from each other. The A Pills are the same color as the B Pills, they are the same shape, same size - they are identical in every respect.  Your doctor is charging you $10,000,000 a pill! So you dare not throw any away.

Thus, the puzzle: what can you do to ensure that you take only one A Pill and only one B Pill today, without wasting any pills (either today or in the future)?

Update (04/03/2010): Solution posted by Nikhil Pandey (CSE, IITB Alumnus and working at ASUS Taiwan) in comments!!

Comments

  1. Gold and Iron10:09 PM, March 04, 2010

    you were very foolish to tap out 2 tablets at once

    however do this now
    take another pill A out......and take care not to take out 2 pills

    now u have 2 pills A and 2 B pills

    next aim divide 2 a and 2 b into one a and one b

    powder all the 4 tablets mix them well and have half today half tomorrow

    warning - if u can not mix well, then cut all 4 in half
    have one half today...another tomorrow

    ReplyDelete
  2. Pratik Poddar10:32 PM, March 04, 2010

    @Pandey.. Nice solution..Thanx..

    ReplyDelete
  3. Earthling7:59 PM, March 05, 2010

    If you follow the powdering approach, you will die with probability 1 :)
    That is because to not die, the mixture should have exactly equal proportions of A and B which is a possible but extremely improbable event(prob. = 0).
    The deterministic solution is -
    Take another A out. Now you have 2 As and 2 Bs. Cut each in half and keep the left halves and right halves separate. Eat the collection of left halves today and right halves tomorrow.

    Now since I have brought extreme probabilities into picture, I should emphasize that please use an infinite precision scale to chop the tablets in perfect halves otherwise you are still probably dead :)

    ReplyDelete
  4. Pratik Poddar1:22 PM, September 08, 2026

    That is exactly the intended solution - take one more A, halve all four pills, eat one half-set today and the other tomorrow: each serving is guaranteed to contain exactly one A and one B. And your objection to powdering is mathematically spot on: exact 50-50 proportions is a measure-zero event. Infinite-precision scale duly noted. :) (replied using AI)

    ReplyDelete

Post a Comment

Popular posts from this blog

Polya's Urn Problem: Expected Balls in the Smaller Urn

Puzzle: There are two urns with one ball each. Each of subsequent n-2 balls is placed into one of these urns, with probability proportional to the number of balls already in that urn. What is the expected number of balls in the smaller sized urn? Source: P. Winkler's Puzzles book. (Chapter: Probability). Solution: Highlight the part between the * symbols for the answer. * This problem can be reformulated as the following problem. Suppose I have a stack of black cards and one red card. Initially I take red card in my hand. Now I add black cards randomly between any two cards (so, initially its either above or below red). Note that the probability that I add the card above the red card, when x-1 is the number of cards above red and y-1 is the number of cards below red is x/(x+y). Let the problem be if red card is dividing the black cards into two sets, what is the expected number of black cards in the smaller section. So, we see that the two problems are equivalent. No...
Read more

Lion and Man in a Circular Cage: Can the Lion Catch the Tamer?

Source: Asked to me by Pramod Ganapathi (PhD Student at Stony Brook University) Problem: A lion and a lion tamer are enclosed within a circular cage. If they move at the same speed but are both restricted by the cage, can the lion catch the lion tamer? (Represent the cage by a circle, and the lion and lion tamer as two point masses within it.) Solution: Highlight the part between the * symbols for the answer. * Answer: surprisingly, no - the tamer can avoid being caught forever. This is the famous Lion and Man problem, posed in the 1930s and settled by Besicovitch; the proof appears in Littlewood's "A Mathematician's Miscellany" (and opens Bollobas's "The Art of Mathematics"). What the lion CAN do (Rado's strategy): move to the center, then always stay on the radius through the tamer, using any spare speed to move outward. Since the lion runs on a smaller circle, it can match the tamer's angular position while steadily increasing its radiu...
Read more

Expected Tosses for Consecutive Heads: 2 Heads vs 3 Heads Puzzle

Let's say A keep tossing a fair coin, until he get 2 consecutive heads, define X to be the number of tosses for this process; B keep tossing another fair coin, until he get 3 consecutive heads, define Y to be the number of the tosses for this process. 1) Calculate P{X>Y} 2) What's the expected value of X 3) What's the expected value of Y This is probably the hardest puzzle I have ever put on my blog. Hence, I will post its solution in the post directly rather than on comment. Solution: 1) (Solved by me finally after 13 months :)) Make a state diagram. Let the state be (a,b) where a is the number of consecutive heads streak "A" is on currently and b is the number of consecutive heads streak "B" is on currently. So, (0,3) (1,3) are final accepted states and (2,0) (2,1) (2,2) (2,3) are failure states. If you get tails, your contribution to the state reaches to "0" f(State) = P(X>Y | "State" configuration initially) ...
Read more