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Tagged: character design, for fun, kasnas
 This topic has 17 replies, 9 voices, and was last updated 15 years, 7 months ago by riddles.

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July 23, 2005 at 10:14 am #552318WhiteKnight
 Posts : 878
 Gelatinous Cube
Apologies to those in the room who have heard me state this one before, but it’s been a while since we had “Petals around the Rose”.
You are a contestant on a game show. It is the oldest game in the world, sometimes called the shell game, sometimes cups and ball, but in this instace we have three upturned containers, underneath one of which is the key to the grand prize, a new car.
The compere offers you your choice of the three containers. Once you have made your choice, he removes one of the remaining two containers, in the process revealing it does not have the key underneath it.
Finally he offers you the chance to change your mind about which container you have selected.
Question is, should you change your mind, and does doing so alter your chance of winning?
July 23, 2005 at 10:37 am #564668Dragonkin Posts : 485
 Thrikreen
Ah… I believe I will change my mind, for it increases my chance of winning to the double.
July 23, 2005 at 11:47 am #564669tomlib Posts : 154
 Orc
That is correct. You must switch you choice. It’s simple math although this problem confounds people for some reason.
Beginning of Game:
Choice A: 33% Chance of Being Right
Choice B: 33% Chance of Being Right
Choice C: 33% Chance of Being Right
So, you have a 33% chance of being right with your initial choice.
But after the host takes away one of the wrong choices (there will always be at least one wrong choice after you make your initial decision because there are 2 wrongs and 1 right).
Switch or Not:
Your Choice: 33%
Other Choice: 67%
Simple math and yet a real conversation starter at your next party.
Tomo
July 23, 2005 at 2:28 pm #564670Slag Posts : 12
 Commoner
tomlib wrote:That is correct. You must switch you choice. It’s simple math although this problem confounds people for some reason.Beginning of Game:
Choice A: 33% Chance of Being Right
Choice B: 33% Chance of Being Right
Choice C: 33% Chance of Being Right
So, you have a 33% chance of being right with your initial choice.
But after the host takes away one of the wrong choices (there will always be at least one wrong choice after you make your initial decision because there are 2 wrongs and 1 right).
Switch or Not:
Your Choice: 33%
Other Choice: 67%
Simple math and yet a real conversation starter at your next party.
Tomo
Let me be the one to disagree. The first set of odds I agree with, 33% chance for either of the options. I do have a problem with the second set though, if the game is played as described in the original posters description and the content of the first removed container is revealed then the odds shift to a 50/50 shot for either container to have the key. If you are making a decision to choose one of 2 options and only one of the 2 is correct then you have a 50/50 shot. The previous choice is no longer releveant in the calculation of odds.
July 23, 2005 at 4:54 pm #564671Dragonkin Posts : 485
 Thrikreen
It may look counterintuitive, but you actually have a greater chance of winning if you swap.
The reason some people think you have the same chance whether you choose to swap or not is because there are only two left at that point, ie something that looks like a 50/50 chance. However, look at it this way: What you’re actually choosing is do you keep your own choice, or do you choose all the others – you’re effectively choosing the removed ones as well if you swap.
To make it a little simpler, make it more extreme and make it a hundred choices instead of three:
You choose one, and 98 are then removed leaving you to stay with your one or swap. Since you only have a 1% chance of getting the right choice the first time, the chances of the other one being correct would therefore be 99 (even if 98 of those “percent” have just visually been taken off the table)%.
So, effectively yes – if the second choice is completely isolated, it’s a 50/50 chance… but since you picked which would be one of the options, the previous choice very much affects the outcome.
Or, for mathematic… say you choose B (out of a set of X). If you choose to swap, you’re actually choosing !B, not necessarily A. Just because C and D and E and so on were removed before the second choice doesn’t mean that didn’t affect the first one.
Make sense?
July 23, 2005 at 6:07 pm #564672Wellard Posts : 194
 Orc
Now I remember…I HATE MATHS
July 24, 2005 at 6:17 pm #564673Shug Posts : 118
 Orc
The only real test is to code a computer simulation. It’s a personal opinion that math’s only real purpose is for practical application.
Math may say your chances are greater, but I think it’s arithmetic semantics. When you’re fed the results of the one container, the game ceases to be a compiled set of trials, and becomes two mutually exclusive trials.
In any case, this reminds me of a joke I heard when I was in Uni for Engineering:
Quote:A mathematician, a physicist, an engineer went again to the races and laid their money down. Commiserating in the bar after the race, the engineer says, “I don’t understand why I lost all my money. I measured all the horses and calculated their strength and mechanical advantage and figured out how fast they could run…”The physicist interrupted him: “…but you didn’t take individual variations into account. I did a statistical analysis of their previous performances and bet on the horses with the highest probability of winning…”
“…so if you’re so hot why are you broke?” asked the engineer. But before the argument can grow, the mathematician takes out his pipe and they get a glimpse of his wellfattened wallet. Obviously here was a man who knows something about horses. They both demanded to know his secret.
“Well,” he says, “first I assumed all the horses were identical and spherical…”
July 24, 2005 at 7:49 pm #564674Dragonkin Posts : 485
 Thrikreen
Shug wrote:The only real test is to code a computer simulation. It’s a personal opinion that math’s only real purpose is for practical application.Math may say your chances are greater, but I think it’s arithmetic semantics. When you’re fed the results of the one container, the game ceases to be a compiled set of trials, and becomes two mutually exclusive trials.
Yes and no – looking at it from a standpoint of how to easiest code a computer simulation, it’s quite easy… just for the sake of argument, assume that you always swap, in order to determine how often that’ll win you.
So… assume 3 choices, you choose 1 (which then has a 1/3 chance of being right and a 2/3 chance of being wrong), then it, at the same time, wipes all the others bar one and chooses that one without any choice being made at all. The chance of getting the right one after swap is then 2/3.
Basically, remove the step where it says “remove all but one of the others, which are all false, and ask if you want to swap”. Since all of those are removed, we can just assume that they can as well be chosen in the second bit with an or operator.
So, after this short mental exercise, and (just to humour you and since I’m doing some programming exercises atm anyway) running a computer simulation, the chance of getting the right choice after swapping (with three initial choices) is indeed 66.66666…%.
There are no semantics involved, really…
July 25, 2005 at 3:46 am #564675mchvlichldprdgy Posts : 1168
 Owlbear
YES! Its the Monty Hall problem, with the car and the two goats. Switch the choices. I want a car.
love the wiki
July 25, 2005 at 11:14 am #564676WhiteKnight Posts : 878
 Gelatinous Cube
No shame in being wrong with this one. There was a long running battle within the mathmatical community regarding this one. Several eminent mathematicians nailed their colours to the pole and got it wrong.
However the correct answer, as has been eloquently stated is that you must switch your choice to maximise your chances. 🙂
At your first choice there is a 1 in 3 chance you got it right, and a 2 in 3 chance you got it wrong. The removal of one of the remaining two makes no alteration to the fact that at the point in time you chose your container, it became a 1 in 3 chance of being the right one.
The removal of the third (empty) container leaves you in the situation where you still have a 1 in 3 chance of being right, so the remaining unselected container must have a 2 in 3 chance of being the right one.
I think the difficulty here lies in the fact that the gameshow host has prior knowledge of the setup. He knows he is removing an empty container. If he were to remove a container at random, and not show you its contents then no amount of switching between the two remaining will alter the odds. It is the foreknowledge that the one being removed is a losing selection that makes the difference.
Imagine the situation where having made your selection the host offers you the chance to swap your one choice for the other two, without revealing anything. Here it is clear you are doubling your chances of winning.
I like this problem, the 5050 view is indeed true if you are coming to the two remaining cups and making a choice with no foreknowledge. This is not however the case, and hence having made a choice with a 1 in 3 chance of success, the chance that you made the right choice remains the same, even if one of the incorrect choices is later removed.
July 25, 2005 at 4:50 pm #564677Slag Posts : 12
 Commoner
Im wrong now eh? 🙄
The previous choice is inmaterial it was made inmaterial by revealing that the first choice did not contain the key. This reults in there being one key under two different cups with me making a conscious decision between the two.
Its going to take more than someone on a messageboard saying that choosing between the two different options is anything other than a 50/50 shot so you can stop trying at any point in time. Trying rewording the original post and then you might have something.
July 25, 2005 at 5:50 pm #564678Dragonkin Posts : 485
 Thrikreen
Slag wrote:The previous choice is inmaterial it was made inmaterial by revealing that the first choice did not contain the key.I think you misread the problem… at the time of the second choice, nothing has been revealed about the first choice being correct or not – it merely removes one of the two options that weren’t chosen and show that as being false.
Example: You choose A. B is removed – you still don’t know if your first choice, A, was correct or if it’s in fact C that is correct. All you know is that B isn’t correct… but since you had a 1/3 chance of getting A right in the first place, that means it’s a 2/3 chance of either B or C being correct, which with B removed transfers to C (if B is false, then (B  C) = C).
July 25, 2005 at 11:27 pm #564679Shug Posts : 118
 Orc
Click the Wikipedia link, Slag…
It explains it well enough.
*muttering and grumbling, digging out his Visual Studio install CDs*
I’m trying this out.
July 26, 2005 at 1:24 am #564680mchvlichldprdgy Posts : 1168
 Owlbear
Slag wrote:Im wrong now eh? 🙄The previous choice is inmaterial it was made inmaterial by revealing that the first choice did not contain the key. This reults in there being one key under two different cups with me making a conscious decision between the two.
Its going to take more than someone on a messageboard saying that choosing between the two different options is anything other than a 50/50 shot so you can stop trying at any point in time. Trying rewording the original post and then you might have something.
Like I said: Love the Wiki!
July 27, 2005 at 8:32 pm #564681McMouse Posts : 23
 Commoner
Get a friend and try it yourself. I know we spent some time doing this problem in Stat’s class.
I believe that there’s a similar problem… Goes something like “You have a desk with three drawers. In each drawer there is a coin, two silver, one gold. You open the first drawer to find a silver coin. What is the probability that the second drawer will also contain a silver coin?”
Or something to that effect. It was extra credit, and I got it right for three extra points 😀

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