D'Nalor Posted June 21, 2010 Share Posted June 21, 2010 Here is a question I have always wondered how the answer is correct, because I can't manage to make it seem right. A chap at the fair is tring to find a prize that is hidden in one of three caskets. He chooses a casket, whereupon the man running the stall opens one of the other cases, which is empty. He then asks the chap if he wants to stick with the casket he chose originally, or switch to the remaining one. which casket will give the chap the greatest odds of winning? The answer in the back of the books states that if he switches, his chances of winning are doubled. so the chap should take the remaining casket. This always seems quite odd to me. It seems to work on the principle that at the beggining, he has a 2/3 chance of being wrong. but once one casket is opened, that should reduce the chances of being wrong to 1/2, shouldn't it? that there were originally three shouldn't make any difference, should it? Am I right, or is the book? Link to comment Share on other sites More sharing options...

swansont Posted June 21, 2010 Share Posted June 21, 2010 (edited) The Monty Hall problem. Yes, the answer is correct — adding information changes the odds. What if there were 100 caskets, and after you chose, the barker opened up 98 of them? Edited June 21, 2010 by swansont typo Link to comment Share on other sites More sharing options...

ewmon Posted June 21, 2010 Share Posted June 21, 2010 The player begins with a 1/3 chance of having picked the winning casket, and this reality never changes despite what the barker reveals (and the barker can always reveal an empty casket), so the other unopened casket has a 2/3 chance of winning. Link to comment Share on other sites More sharing options...

D'Nalor Posted June 28, 2010 Author Share Posted June 28, 2010 Ok, sorry it's taken me a while to get back to this, but here we go... I've provided a spreadsheet showing why I think the answer is wrong. I have put it in more detail that I probably need to, but I want to know the flaw in my reasoning, and the more detail I provide, the easier it is to find. I haven't managed to find a way to show that you are 2x more likely to find it in the second case. Looking at what swansont said, I would have thought that the main info that changes the situation would be that you have one case less than you began with, which you could basically ignore the existence of. after all, 3-1=2 So I'd say the question given by swansont would have the same answer as this question. Base chances.doc Link to comment Share on other sites More sharing options...

Mr Skeptic Posted June 28, 2010 Share Posted June 28, 2010 I think the example given by swansont was particularly instructive, since the numbers are so very different (rather than 3 and 2). Picking 1 out of 100 gives a 1/100 chance you got the right one, revealing 98 empty ones won't change that fact. However, if you know that the host was only going to reveal empty ones, then you know that the prize is in one of the two left and that your original choice has only 1 in 100 chance of being right, so the other has 99 in 100 of being the right choice. It would be a different story if the host had shown you 98 empty ones before you made your choice. Link to comment Share on other sites More sharing options...

crashonly Posted July 4, 2010 Share Posted July 4, 2010 It has everything to do with the man who is running the stall knowing where the price is, and always revealing an empty one that you did not pick. Your original pick is going to be right 1/3 of the time, and 2/3 of the time the price will be in one of the remaining caskets. So if we make the assumptions that the guy running the show knows where the price is and always reveals an empty one, 2/3 of the time the price is going to be in the casket you did not pick originally, and he did not reveal. Link to comment Share on other sites More sharing options...

J.C.MacSwell Posted July 4, 2010 Share Posted July 4, 2010 It has everything to do with the man who is running the stall knowing where the price is, and always revealing an empty one that you did not pick. Your original pick is going to be right 1/3 of the time, and 2/3 of the time the price will be in one of the remaining caskets. So if we make the assumptions that the guy running the show knows where the price is and always reveals an empty one, 2/3 of the time the price is going to be in the casket you did not pick originally, and he did not reveal. My first thought is that he will reveal your empty casket, if it is empty, right away, thus saving himself giving up a prize, and only open a remaining empty one if you have the prize in your casket, in hopes you may switch.. Link to comment Share on other sites More sharing options...

crashonly Posted July 4, 2010 Share Posted July 4, 2010 My first thought is that he will reveal your empty casket, if it is empty, right away, thus saving himself giving up a prize, and only open a remaining empty one if you have the prize in your casket, in hopes you may switch.. That's not an accurate depiction of the Monty Hall problem, which is what (I think) we're talking about here. If the rules were as you state, that: 1) the guy knows where the price is 2) can reveal -any- casket after you've made your pick, and you won't get to make another pick after he reveals your casket. Then indeed he would always reveal your chosen casket right away if you had picked an empty one, so as not to give you any additional chance of winning. You'd always have 1/3 chance of winning, and you should never switch. Link to comment Share on other sites More sharing options...

jcarlson Posted August 2, 2010 Share Posted August 2, 2010 Perhaps think of it this way: When you initially pick a door out of 3, there is a 1/3 probability that your door holds the prize, and a 2/3 probability that the two other doors hold the prize. When one of the two other doors is revealed, it doesn't change the fact that there is a 2/3 probability that the other 2 doors hold the prize, but since you know which of those other 2 doors doesn't have the prize, it means the unopened of the other 2 doors now has a 2/3 probability of holding the prize. Your original door also still has only a 1/3 chance of holding the prize, therefore, you should always switch to the other door, if given the option after the host opens one. Link to comment Share on other sites More sharing options...

lavoisier Posted August 3, 2010 Share Posted August 3, 2010 Very simple explanation of this problem, I think it was given on a BBC programme about mathematics: - when you make your initial choice, you have 2 possible outcomes: right casket (1/3) or wrong casket (2/3) - next step: the barker reveals one empty casket, and you're asked if you want to switch Case A - you don't switch --> then your chances of having chosen the right one remain 1/3 Case B - you switch --> then if you had originally chosen the right one (1/3), you lose; if you had chosen the wrong one (2/3), you win, because both empty caskets are now excluded, and you can't pick but the right one by switching. In practice you swap your winning chances with your losing chances. So: - if you don't switch, the chances are 1/3 win, 2/3 lose as in the initial choice. - if you switch, the chances are 2/3 win, 1/3 lose, which means you've doubled your chances of winning. The presenter on the programme said there's no shame in not getting this the first time round - apparently the problem puzzled even some professional mathematicians. Link to comment Share on other sites More sharing options...

vordhosbn Posted October 13, 2010 Share Posted October 13, 2010 I had big problem understanding why switching is beneficial, and though I read the wikipedia article, along with its discussion page + the very informative posts from forum members here, from the reasoning behind the solution, what I had biggest issue was this particular statement (in red): I think the example given by swansont was particularly instructive, since the numbers are so very different (rather than 3 and 2). Picking 1 out of 100 gives a 1/100 chance you got the right one, revealing 98 empty ones won't change that fact. However, if you know that the host was only going to reveal empty ones, then you know that the prize is in one of the two left and that your original choice has only 1 in 100 chance of being right, so the other has 99 in 100 of being the right choice. It would be a different story if the host had shown you 98 empty ones before you made your choice. My mind resisted to assert this so furiously that I had to write a short computer simulation in C to prove myself wrong... Here's the code if anyone is interested and is still disbelieving the "switching is beneficial" solution: // This program demonstrates the Monty Hall paradox: // // Suppose you're on a game show, and you're given // the choice of three doors: Behind one door is a car; // behind the others, goats. You pick a door, say No. 1, // and the host, who knows what's behind the doors, // opens another door, say No. 3, which has a goat. // He then says to you, "Do you want to pick door No. 2?" // Is it to your advantage to switch your choice? #include <stdlib.h> #include <stdio.h> #include <time.h> // Defined constants #define cNbValidArg 2 #define cArgDoorsPos 1 #define cArgCyclesPos 2 // Function prototypes void PrintHelp(void); void Simulate(void); // Local data static unsigned int u32NbDoors, u32NbCycles; // Function takes 2 arguments // #1 - number of "doors" // #2 - number of cycles int main(int argc, char *argv[]) { if(argc != cNbValidArg + 1) { PrintHelp(); return(1); } else { //Read Input parameters sscanf(argv[cArgDoorsPos],"%d", &u32NbDoors); sscanf(argv[cArgCyclesPos], "%d", &u32NbCycles); Simulate(); return(0); } } // This function prints help message void PrintHelp(void) { printf("This program illustrates the Monty Hall problem.\n"); printf("It takes %d arguments:\n", cNbValidArg); printf(" #%d - number of doors\n", cArgDoorsPos); printf(" #%d - number of cycles to repeat the simulation\n", cArgCyclesPos); } // Performs actual simulation void Simulate(void) { unsigned int u32Index; unsigned int u32Choice, u32Car, u32UnrevealedDoor; unsigned int u32SwitchingWins, u32TotalWins; double dfWinPct; // Seeds random number generator with system time srand((unsigned int)time(NULL)); u32TotalWins = 0; printf("|Car behind:|Chosen: |Left closed:|Switching wins:\n"); printf("--------------------------------------------------\n"); for(u32Index = 0; u32Index <= u32NbCycles; u32Index++) { // Random door number with car behind it u32Car = rand() % u32NbDoors; // Random door number that player has chosen u32Choice = rand() % u32NbDoors; // Here is the "magic" part // If the player has chosen the door with the car (low probability), // the host can open any door he wants and the player does not benefit from switching if(u32Choice == u32Car) { // Random door number(except chosen one) that remains closed u32UnrevealedDoor = u32Choice; while((u32UnrevealedDoor == u32Choice)) { u32UnrevealedDoor = rand() % u32NbDoors; } } //but if the player has chosen an empty door (much higher probability), // The only choice of the host is to open all doors except the one with the car!!! else { u32UnrevealedDoor = u32Car; } // If switching your choice wins - update variable if(u32UnrevealedDoor == u32Car) { u32SwitchingWins = 1; u32TotalWins++; } else { u32SwitchingWins = 0; } printf("|%10d |%7d |%11d |%14d |\n", u32Car+1, u32Choice+1, u32UnrevealedDoor+1, u32SwitchingWins); } dfWinPct = (double)u32TotalWins / (double)u32NbCycles; printf("--------------------------------------------------\n"); printf("Total win percent when switching: %3.2f%%\n", dfWinPct * 100); } montyh.zip 2 Link to comment Share on other sites More sharing options...

Mr Skeptic Posted October 13, 2010 Share Posted October 13, 2010 Well at least you're thorough. If I recall correctly, several well-educated mathematicians had trouble believing it too. Link to comment Share on other sites More sharing options...

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