Monty Hall revisited
Suppose you're a contestant on a game show and the host shows you 5 curtains. The host informs you that behind 3 of the curtains lie lumps of coal, but behind the other 2 curtains lie separate halves of the same $10,000 bill. He then asks you to choose 2 curtains; if these 2 curtains are the ones with the 2 bill halves behind them then you win the $10,000, otherwise you go home with nothing.
After you choose your 2 curtains, the host opens one of the remaining 3 curtains that he knows has just a lump of coal behind it. He then gives you the following options:
(0) stick with the curtains you initially chose,
(1) swap either one of your curtains for one of the remaining curtains, or
(2) swap both of your curtains for the remaining 2 curtains.
Let p ( k ) , k = 0 , 1 , 2 , be the respective probabilities of winning the money in scenarios ( k ) as outlined above. Then
p ( 2 ) − p ( 0 ) − p ( 1 ) = b a ,
where a and b are coprime positive integers. Find a + b .
The answer is 21.
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1 solution
Excellent problem Brian!! Had great fun solving it... Tx for posting such gems:)
Did same way awesome problem
Ok so what's 21 have to do with the possibility of choosing the right curtains?
This type of question usually generates some differences of opinion, so don't worry if you disagree with my solution method and my answer. I thought about this question for a while before posting it and my solution, but that of course doesn't guarantee that I'm right. :)
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Great problem, Brian... Loved it! This one really took me a while to figure out... Phew! I can finally rest at night! :D
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Thanks! I'm glad that you enjoyed the problem. :)
What is wrong with the following logic?
Say I want to calculate p ( 1 ) . For us to win in this case, we first have to choose 1 winning door and 1 coal door, the probability of which is 5 3 .
The host then reveals the other coal door and, since this is the case ( 1 ) , we change the door. Which door do we get? Well, there are exactly 2 choices - 1 coal door and 1 winning door. The probability of choosing the winning door is then 2 1 , thus we get 5 ⋅ 2 3 ⋅ 1 = 1 0 3 .
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We also have to change the right door. If we swap the winning door for another door, we don't win. There is 2 1 chance of swapping the coal door, and 1 0 3 , of choosing the winning door (as you calculated). So p ( 1 ) = 1 0 3 × 2 1 = 2 0 3
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ahh, right, he doesn't have to change the coal door, he could've changed the winning door instead. :(
please point out what is wrong with the following reasoning
to win in scenario(1) we have to pick one correct curtain and one false curtain, then choose the false curtain from the chosen ones and swap it with the correct curtain from the remaining two
hence p(1) = (2/5)(3/4)(1/2)(1/2) = (3/40)
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Of the 1 0 possible pairs of curtains initially chosen, 6 have a "winning" curtain paired with a "losing" curtain, i.e., a probability of 1 0 6 = 5 3 . We would then multiply this by 2 1 ∗ 2 1 as in your analysis.
Your method left out the possibility of first choosing one losing curtain and then one winning curtain. So taking this into account, your calculation should be
( 5 2 ∗ 4 3 + 5 3 ∗ 4 2 ) ∗ 2 1 ∗ 2 1 = 2 0 3
as found in my solution.
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ah so the order matters, thanks for pointing out! sigh, stupid mistakes like this
Finding p(1) was definitely the trickiest part of this problem. I originally multiplied my 3/5 probability by 1/5. I somehow counted 5 switching options instead of 4 the first time. I then arrived at the given answer in a roundabout way of finding the losing options using strategy 1 and subtracting from 1.
Nice extension
I loved the problem Brian! I apologise if my doubt is silly or previously explained but after going through all the solutions and comments, I am still getting stuck at the dividing the 3/5 probability into 4 parts. If you reply it will help me sleep better at night. After struggling for some hours on the p(1) bit, I made all the possible 20 (keeping each piece unique) combinations. Now there are 12 cases out of 20 that have coal and a piece of the cheque. Out of these when all the doors were closed there is a 1/3 chance of us getting it right. Now that one door is opened it's probability is divided equally into the other two doors 1/6+1/3=1/2 is the probability of getting it right by choosing a door. So that gives us 12/20*1/2 but there is another 1/2 that rears it head. I don't understand how to account for it. Does it come because of the two pieces? I just can't get my head around it.
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There are just ( 2 5 ) combinations of 2 curtains, 6 of which involve one with coal and one with a piece of the cheque, so I think that the value of 20 you mention refers to permutations rather than combinations, i.e., if the order in which you choose the doors matters then there are 20 ways to do that. That said, there are indeed 12 permutations involving one with coal and one with a piece of the cheque. With the curtains closed, if we decide to change one of our chosen curtains we would first have to choose the right curtain to eliminate, which has a probability of 1 / 2 , and then choose the curtain from the 3 others that has the second piece of the cheque, which has a probability of 1 / 3 . Thus before the host opens a curtain, if we decide to swap precisely one curtain then the probability of winning would be ( 1 / 2 ) × ( 1 / 3 ) = 1 / 6 , and not the 1 / 3 you have mentioned. So in this scenario the 3 / 5 probability would be divided by 6 .
Now when the host reveals a curtain with coal behind it, and our initial curtains include one with coal and one with a piece of the cheque, then of the 2 remaining curtains there is now a 1 / 2 probability of choosing the curtain with the other piece of the cheque. So as there is still a 1 / 2 probability of eliminating the right curtain from our initial pair, a successful swap will occur with probability ( 1 / 2 ) × ( 1 / 2 ) = 1 / 4 . So now the 3 / 5 probability gets divided by 4 , yielding p ( 1 ) = 3 / 2 0 .
Hope that clears things up. I'm glad that you enjoyed the problem. :)
Enjoyed the problem. Many Thanks!
Here's a question. Why can't you solve this problem like this? The initial probability of picking two correctly in a row without replacement is 2/5*1/4=2/20=1/10. The initial probability of picking 1 correctly is 2/5=4/10. Finally, the probability of picking 0 correctly is 1-1/10-4/10=1/2. I've been struggling to figure out under what circumstances to apply each method. Why isn't this like a ball in urn situation? Any thoughts.
excellent problem. loving brilliant due to such great questions by such great people.
There are ( 2 5 ) = 1 0 combinations of 2 curtains, so any particular choice of 2 curtains has a probability of 1 0 1 of winning. Thus p ( 0 ) = 1 0 1 .
There are ( 2 3 ) = 3 combinations of 2 of the 3 curtains you did not initially choose. Thus there is a probability of 1 0 3 that the winning pair of curtains is among these remaining 3 curtains. Once the host opens a curtain he knows has a lump of coal behind it then this probability of 1 0 3 "attaches itself" to the remaining 2 unchosen curtains. Thus p ( 2 ) = 1 0 3 .
Now there is a probability of 1 − 1 0 1 − 1 0 3 = 5 3 that the winning curtains are comprised of one of your chosen curtains and one of the 3 unchosen curtains. After the host opens the curtain with the coal behind it this probability of 5 3 gets distributed equally to the 4 possible combinations involving one of your initially chosen curtains and one of the remaining unchosen curtains. Thus p ( 1 ) = ( 4 1 ) ( 5 3 ) = 2 0 3 .
So finally p ( 2 ) − p ( 0 ) − p ( 1 ) = 1 0 3 − 1 0 1 − 2 0 3 = 2 0 1 . This gives us a = 1 , b = 2 0 and a + b = 2 1 .