Simple probability calculation problem from GS.There are 4 boxes, labeled 1,2,3,4. In box n4 there is 1M USD. You randomly pick a box, another person looks in it. He says only "bigger" if the number of the box is bigger then previous, or "less" in the other case. You can look into the box or put it away. If you look into box the game stops. Put it away- the box dissapears and you continue playing. What is the optimal strategy to get the money if you are lucky?

I don't understand this problem. Can you word it a bit more clearly?

Ok, I forgotten to put another stop condition- the game stops when another person says "less". One scenario is you point to a box, another person sees it is N2 (but as it is the first time he is obliged to say "bigger"). You may say stop, but there is no money in N2. Ok, you say "continue", the box N2 is thrown away. If you next pick- box N1 the game terminated. If it is N4 it continues. You may say "stop" and then you have money in our pocket. If you say- "continue" box N4 is thrown. You pick one more (it can be N1 and N3). In either case you loose. Yea, and you do not know what number you are picking, you know just- the money in box N4, another person says only "bigger" or "less", If "less" you loose, if bigger you carry on, but the box is thrown away. What is the strategy you follow? Really more explanation then the solution, but it can be asked on an interview.

Solving this in a backward recursive way like an american option pricing problem,opening on the first go gives you a probability of winning of 1/4 and this is also the continuation value so the optimal strategy is to open the box on the first go or to continue to the next stage and then open the box

I agree with FV. It is clear that receiving the million by opening on the first run is 1/4. The probability to get the right one by open on the second run is 3/4*1/3=1/4. The probability to get the right one on the third run is 1/4*1/2=1/8. Finally, the probability to get it right on the fourth and final run is clearly 1/4*1/3*1/2=1/24. This implies that it is best to open on the first or second guess.

yup, I agree. 1/4 for the first and for the second - I would personaly choose the first go just in order to save time and keep on solving other brainteasers.If the "less stop" condition had not been imposed, the probability would be independent of the draw (and equal to 1/4.)

you could also take a combinatorial approach:There are 4X3X2X1 = 24 possible arrangements of the boxes1X3X2X1 = 6 arrangements where the box is in the first position 3X1X2X1 = 6 arrangements where the box is in the second positionwhen the box is in the third position and you can win buy waiting to that point there are 3 possibilities1->2->41->3->4&2->3->4and only one arrangement where you can wait and win when the box is in the fourth position. In accordance with the previous answer you should look in the box after the first or second instance and your chance at a million is 24%

oops 25%

QuoteOriginally posted by: dahlgrenI agree with FV. It is clear that receiving the million by opening on the first run is 1/4. The probability to get the right one by open on the second run is 3/4*1/3=1/4. The probability to get the right one on the third run is 1/4*1/2=1/8. Finally, the probability to get it right on the fourth and final run is clearly 1/4*1/3*1/2=1/24. This implies that it is best to open on the first or second guess.ok, could somebody enlighten me why "The probability to get the right one on the third run is 1/4*1/2=1/8"??

QuoteOriginally posted by: FedorEQuoteOriginally posted by: dahlgrenI agree with FV. It is clear that receiving the million by opening on the first run is 1/4. The probability to get the right one by open on the second run is 3/4*1/3=1/4. The probability to get the right one on the third run is 1/4*1/2=1/8. Finally, the probability to get it right on the fourth and final run is clearly 1/4*1/3*1/2=1/24. This implies that it is best to open on the first or second guess.ok, could somebody enlighten me why "The probability to get the right one on the third run is 1/4*1/2=1/8"??the only three possibilities for the sequence of box selections where the game hasn't already ended are 124, 134, 234---3 out of 24 possibilities.

ok, as I expected, I misunderstood the problem...thanks for clarifying!