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Old probability exercise

  1. Apr 16, 2008 #1
    In another website (RHP) forum the following problem was posted:

    A man with a hat shows you 3 cards, one completely gold, one completely silver, and one gold on one side and silver on the other. He puts them in his hat, and picks one at random. He then shows you one side of the card he picked, which happens to be silver. Now he says "I'll bet you even money that the other side of this card is silver too...whaddaya say, partner?"

    Is this bet a fair one?


    Now there are a few users that think it's really a fair bet and it's 50/50 that it's either gold or silver on the other side. They have problems with a few other probability exercises also. The thread there is a quite long debate (fistfight) and I was curious whether someone here could provide a clearly explained solution.

    I know it's not a fair bet, it's actually 66% that the man would win this bet and the problem is quite similar to the Monty Hall problem, but I was wondering if really there is another way to explain the problem than done in the thread given.

    Thank you.
     
  2. jcsd
  3. Apr 16, 2008 #2
    The way I look at it is: You have three cards here: (1,1);(1,2);(2,2). The question is whether the opposite side is the same as the front.

    I therefor assume that if he shows gold, he will offer the same bet. BUT, we can easily see that of the three cards, TWO HAVE THE SAME BACK! Thus the probability is 2/3 that the other side is the same.
     
  4. Apr 16, 2008 #3

    cristo

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    It is indeed quite similar to the Monty Hall problem. My solution is this:

    You've been shown a silver side; this could be side A of the silver/silver card, side B of the silver/silver card, or the silver side of the silver/gold card (i.e. three options). For two of these options the other side of the card is silver and for one it is not, thus the probability of the other side being silver is 2/3.
     
  5. Apr 16, 2008 #4
    Not quite like that. Two have the same back, but one of those two can be excluded.

    3 cards SS (silver/silver), SG, GG. In the given occasion, GG is automatically excluded. 2 cards left - SS and SG. You see a S side. Together there are 3 S sides. To one of them, the other side is G, to other two, the other side is S, therefore, the bet 2/3 in the man's favor.
     
    Last edited: Apr 16, 2008
  6. Apr 16, 2008 #5
    Exactly, but what I'm looking for is an explanation for kids. And explanation from which it would be clear to anyone (I believe in such an explanation).

    Try reading the thread. Some of the posters', that think it's 50/50, posts are actually quite hilarious, if not pathetic.
     
  7. Apr 16, 2008 #6

    cristo

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    Some people just can't see the answer to problems like that. I'm not sure that it's possible to make the solution any easier to understand; I think you should just give up on trying to explain it to them!
     
  8. Apr 16, 2008 #7
    What if he had shown a gold card?
     
  9. Apr 16, 2008 #8
    Well that's actually not in the original problem but if you ask, then:

    If he showed a gold side (not card), then the man would say - I bet ya the other side is also gold - thus the answer would be the same: it isn't a fair bet.
     
  10. Apr 16, 2008 #9
    Maybe you're right. However, it's still a funny thread. One, particular poster is very persistent on his solution.
     
  11. Apr 16, 2008 #10
    kbaumen: Some people just can't see the answer to problems like that. I'm not sure that it's possible to make the solution any easier to understand; I think you should just give up on trying to explain it to them!

    Frankly, I think the reason for that is because they think in terms of CARDS and NOT SIDES. However, that is why I prefer my solution.

    The way I look at it is: You have three cards here: (1,1);(1,2);(2,2). The question is whether the opposite side is the same as the front. I therefor assume that if he shows gold, he will offer the same bet. BUT, we can easily see that of the three cards, TWO HAVE THE SAME BACK! Thus the probability is 2/3 that the other side is the same.

    They way I presented it, the cards are still in the hat when we arrive at our answer. We do not have to break the problem down and consider sides instead of cards--which seems to be the reason for confusion.
     
    Last edited: Apr 16, 2008
  12. Apr 16, 2008 #11
    I suppose your initial thought is, well it is not the gold-gold card obviously,
    so it must be either the silver-silver or silver-gold so it is 50/50.
    That seems perfectly reasonable logic to me, or at least it did.

    I could see the two arguements and both seemed correct, I was just starting
    to write a computer program to simulate it when I though about it in a different
    way.
    What if he had said "I bet you it has the same colour on the back?"
    Then you would think, well 2 cards have the same colour on the back and one
    does not, so it is 2 out of 3 it is the same colour, ie 66%.

    So it is the Monty Hall prob, the twist is that in the MH prob he is 'forced'
    to open a particular door (he can't open the door with the prize) whereas here he
    is forced to say a particular colour, he is 'forced' to say the colour he sees.

    Perhaps the simplest way to say it is he will only pick the mixed card 1 in 3 times.
    so 2 out of 3 times it will be the same colour.

    But I need to go back and disprove my initial logic, where I said
    "well it is not the gold-gold card obviously so it is 50/50".

    Now, I should be able to explain why this is wrong now, *starting* from that
    point of view, but to be honest, I have tried and I am struggling!!!

    I can explain the MH problem by saying that MH is forced in his choice of door to open
    but I am struggling to explain this.
    I guess it is that he is forced in his choice of colour to say? Put I can't quite put
    my finger on it, I might be able to come up with 75% though!!

    Or is it that 1/3 of the time you won't be able to rule out the gold card?
    Because it will be gold both sides!!

    That sounds on the right lines. So 2/3 times it is 50/50 and the other time it
    is guaranteed?
    I think that brings out the right numbers (somehow)

    Anyway I give up!!

    Here is a simulation program and the results!!
    Code (Text):

    #include <stdio.h>
    #include <stdlib.h> // need for srand() function
    #include <time.h>   // need for time() function
    #include <unistd.h> // need for getpid() function
    // #include <iostream>  // need for printing to screen
    int i;
    int x;
    int r;
    int cd[3][2];
    int same, different;
    main(){
    cd[0][0]=0;
    cd[0][1]=0;
    cd[1][0]=0;
    cd[1][1]=1;
    cd[2][0]=1;
    cd[2][1]=1;


      srand( time(NULL) );
       for( i = 0; i < 10000000; i++ ){
            i, x=rand();    
            r=(int)(3*(double)x/(double)RAND_MAX);
            if ( cd[r][0]==cd[r][1]) same++;
            if ( cd[r][0]!=cd[r][1]) different++;

        }
        printf("\nSame %d different %d", same, different);

    }


     
    Drumroll!!!!!!!!!!! Results!!!!!!!!!


    Code (Text):


    Same 6667647 different 3332353

     
    Probably a bug or 3 in there and could be optimised but it is fairly convincing if you are
    a doubter of the 2/3 ratio!!

    Gonna win some money off my grandma now!!! :rofl:
     
  13. Apr 16, 2008 #12
    Yes I can see how you can come up with the solution if you start of thinking in the
    "right" way, but I would like to see a solution when you start think about it in the "wrong" way.

    What i think it does show is that how easy it is to make disasterous errors!!
    And I know even maths professors have got the Monty Hall problem wrong,
    I think it shows how you can be fooled by things even if you have a solid grasp
    of maths and are used to doing certain types of problems, as with the Monty Hall
    problem if felt uneasy about the 'obvious' answer, but 'basic' maths seemed
    to show it was correct.

    I think it is one of those occasions where it is better to go with your 'gut' feeling
    but I think this is trickier than Monty Hall on the gut feeling side of things.
     
  14. Apr 17, 2008 #13
    With the Monty Hall, I just look at three cases (1,1), (1,2), (1,3).

    We get these choices from finding pairs of random numbers from 1 to 3. HOWEVER, for simplicity, we can say that the Contestant always choses 1, and the correct answer is the second member of the pair.

    PRESTO! It is clearly true that if the contestant choses 1, and sticks with it; he will be right only 1/3 of the time. THUS WE CAN IMMEDIATELY CONCLUDE to switch is to be right 2/3 of the time! We can alway put in more detail and see these things on a more individual bases, but I won't bother for our purposes here.

    Now the matter of thinking this thing out on an analytical basis might go: "Well, the original situation is 1/3, but after a door is open and a goat shown, the probability will change to ........?

    BUT THE DIAGRAM SHOWS THAT THE PROBABILITY WILL NOT CHANGE IF THE PERSON DOES NOT SWITCH! Thus the new information is not incorporated into our situation, if we retain our original choice.

    So forgetting about the diagram and trying to think about it, what seems a very convention way of approaching the final probability step by step CAN BE DECEPTIVE.

    The major reason seems to be because the Master of Ceremonies knows what door the treasure is behind, and never opens that! Thus, possibly, he adds to the problem a certain kind of "cunning knowledge" that we may not ordinarly see in probability problems.

    esbo: I think it is one of those occasions where it is better to go with your 'gut' feeling
    but I think this is trickier than Monty Hall on the gut feeling side of things.


    I think that a difficulity is that a person ordinarily assumes he can tell the back of a card from the front. But that is not true here.
     
    Last edited: Apr 17, 2008
  15. Apr 17, 2008 #14
    With the Monty Hall, I just look at three cases (1,1), (1,2), (1,3).

    We get these choices from finding pairs of random numbers from 1 to 3. HOWEVER, for simplicity, we can say that the Contestant always choses 1, and the correct answer is the second member of the pair.

    PRESTO! It is clearly true that if the contestant choses 1, and sticks with it; he will be right only 1/3 of the time. THUS WE CAN IMMEDIATELY CONCLUDE to switch is to be right 2/3 of the time! We can alway put in more detail and see these things on a more individual bases, but I won't bother for our purposes here.

    Now the matter of thinking this thing out on an analytical basis might go: "Well, the original situation is 1/3, but after a door is open and a goat shown, the probability will change to ........?

    BUT THE DIAGRAM SHOWS THAT THE PROBABILITY WILL NOT CHANGE IF THE PERSON DOES NOT SWITCH! Thus the new information is not incorporated into our situation when we retain our original choice.

    So forgetting about the diagram and trying to think about it, what seems a very convention way of approaching the final probability step by step can be deceptive. Like this is a tree type of problem.

    The major reason seems to be because the Master of Ceremonies knows what door the treasure is behind, and never opens that! Thus, possibly, he adds to the problem a certain kind of "inside knowledge" that we may not ordinarly see in probability problems.

    esbo: I think it is one of those occasions where it is better to go with your 'gut' feeling
    but I think this is trickier than Monty Hall on the gut feeling side of things.


    I think that a difficulity is that a person ordinarily assumes he can tell the back of a card from the front. But that is not true here
     
  16. Apr 17, 2008 #15
    That way you get the intended answer but the reasoning is incorrect. Once the man draws a card from the hat (let it be silver on the side he shows), the third card (gold/gold) is automatically excluded. We have to look at 2 cards, 4 sides (from which at 1 we are definitely not looking).

    Actually, this draw is a one-time event so, technically, we don't have to consider the case where we get shown the golden side.

    Anyway, I think now it's impossible to prove it there, in that forum. There is an incredibly persistent poster, who thinks that the solution I provided is wrong, witless, bogus, etc. Everything I post Irrelevant, Ad Hominem (which he equals to fallacy - amazing, isn't it) or Red Herring.

    Check the thread - funny read.
     
  17. Apr 17, 2008 #16

    uart

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    Hi kbaumen, you might find it useful to look at a solution based notions of conditional probability using Bayes theorem. http://en.wikipedia.org/wiki/Bayes'_theorem

    I know that there are probably more straight forward solutions (robert ihnot's for example) but the conditional probability solution has the advantage of breaking the problem into a number of very simple sub-problems with well defined events (and hence well defined assumptions) that really anyone should be able to follow.


    Denote events A and B as follows.

    A : That the randomly picked card is silver-silver.

    B : That a randomly picked side of the randomly picked card is silver.

    First consider the two events in isolation (that is,without any prior knowledge of partial outcomes).

    It is easy to show that P(A) = 1/3 and P(B) = 1/2.

    Now consider the probability that outcomes A and B both occur, we denote this as P(AB). Of course the events A and B are not independent so we cant just multiple P(A) and P(B) together to find this. We can however notice that if event A is true then event B is a certainty, and that tells us that P(AB) = P(A), so therefore we also have P(AB) = 1/3.

    That's all the data needed to apply Bayes theorem and get the solution we seek, the conditional probability of event A occurring given that event B has already occurred, that is P(A|B).

    P(A|B) = P(AB) / P(B) = 1/3 divide 1/2 = 2/3.

    So yes the guy with the hat has a 2/3 chance of winning if you decide to play. I hope this explanation helps.
     
    Last edited: Apr 17, 2008
  18. Apr 17, 2008 #17
    firstly i dont think its fair to call ppl having difficulty with this pathetic or hilarious!![mainly because it would have to include me!]


    it does seem very tricky because when the bet is offered the gold-gold is eliminated
    leaving you with 2 possible cards, 1 of which wins so you think 50/50

    but the thing is he dosnt offer the bet everytime, only when it suits him!

    to explain to kids you could simply draw out all the possible scenarios

    [1] S-G card picked
    side 1 is presented [silver]
    bet is offered. You Win.....................................[GOOD FOR YOU]

    [2] S-G card picked
    side 2 is presented [gold]
    NO BET IS OFFERED

    [3] S-S Card picked
    side 1 is presented
    bet is offered. You loose...................................[BAD FOR YOU]

    [4] S-S card is picked
    side 2 is presented
    bet is offered. You loose...................................[BAD FOR YOU]

    [5] G-G card is picked
    side 1 is presented
    NO BET OFFERED

    [6] G-G card is picked
    side 2 is presented
    NO BET OFFERED

    SO You see the bet is only offered half the time. THE HALF THE TIME THE BET IS NOT OFFERED THE DEALER KNOWS IT WOULD BE BAD FOR HIM!!
    also you could leave out the gold-gold card altogetherand it would not affect the outcome,

    i supoose the slight of hand is that even tho the S-S card looks the same both sides, there are two distint faces of it, each of which being presented is a seperate event!!

    i also wrote a quick prog for this in VB
    50% of the time a silver face is presented and the game is allowed to continue
    66% of the time the other face is silver also!
    actually it ran 10,000,000 games
    4999886 times the dealer offered to bet
    3332669 times the dealer won (66.6548997317139%)
     
    Last edited: Apr 17, 2008
  19. Apr 17, 2008 #18
    Hmmm. I see it now, I should have read the replies more carefully!
    The key is cards have two sides, that is what I was overlooking.

    So although you can rule out the all gold card, the mistake I was
    making was to think it was 50/50 because only two cards remained one
    with a silver back and one with a gold back so therfore 50/50.

    However the key is you could be seeing either side of the silver-silver
    card or just the silver side of the silver gold card.

    So there are 3 silver sides, 2 have silver on the reverse, one had gold,
    hence the odds are 2:1 or 66% of a siliver.

    Incidently I simulated this with a pack of blue and red backed cards by
    sticking them togeather face to face to get a red red, red blue, and a blue blue.
    However I just shuffled them! Hence I would only ever see one side of each
    pair!! In which case it is 50/50!!
     
  20. Apr 17, 2008 #19
    It's not gold-gold, obviously, so its either the silver side of the silver-gold card, or it could be either side of the silver-silver card. So it is 2:1 that is has silver on the back.

    Thats about as simple as I could put it.
     
    Last edited: Apr 17, 2008
  21. Apr 18, 2008 #20

    cristo

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    Aren't these what I said in post #3?
    I don't get it; why do you need to write a program to simulate this if the solution can be shown exactly using probability?
     
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