Calculate the odds

[CraigD]

Ok let me write them all out.

A is right

Choose A, Monty removes B, stick with A, win.

…

Hmmm, seems like all possible sets have a 50/50 chance of winning if I stick with my original choice.

Cwes, you’ve made a mistake in your systematic counting, due to neglecting to calculate some conditional probabilities.

 

Expanded, here’s your enumeration:

1. A is right, Choose A, Monty removes B, stick with A, win.
   
2. A is right, Choose A, Monty removes C, stick with A, win.
   
3. A is right, Choose B, Monty removes C, stick with B, lose.
   
4. A is right, Choose C, Monty removes B, stick with C, lose.
   
5. B is right, Choose B, Monty removes A, stick with B, win.
   
6. B is right, Choose B, Monty removes C, stick with B, win.
   
7. B is right, Choose A, Monty removes C, stick with A, lose.
   
8. B is right, Choose C, Monty removes A, stick with A, lose.
   
9. C is right, Choose C, Monty removes A, stick with C, win.
   
10. C is right, Choose C, Monty removes B, stick with C, win.
    
11. C is right, Choose A, Monty removes B, stick with B, lose.
    
12. C is right, Choose B, Monty removes C, stick with C, lose.

Your calculation of the chance of winning when you stick with your original choice being 1/2 would be correct if each of these outcomes occurred with equal probability. This is not, however, the case.

 

[3], [4], [7], [8], [11] and [12] have equal probability. Monty has no choice in which door to remove, so the probability is unaffected by his action. [1] and [2], represent 2 possible choices for Monty for a single random door setup and choice by Contestant, so the sum of their probabilities = the probability of [3], as do the probabilities of [5] + [6], and [9] + [10]. It’s convenient, and has no effect on the calculation, to assume that Monty chooses between [1] and [2] randomly, giving [1], [2], [5], [6], [9] and [10] equal probabilities.

 

Totaling and normalizing the enumeration with these weights give the probability for [1] = 1/18, for [3] = 1/9. So the chance of Contestant winning when sticking with his original choice is 6/18, not 6/12.

[cwes99_03]

Cwes, you’ve made a mistake in your systematic counting, due to neglecting to calculate some conditional probabilities.

 

Expanded, here’s your enumeration:

1. A is right, Choose A, Monty removes B, stick with A, win.
   
2. A is right, Choose A, Monty removes C, stick with A, win.
   
3. A is right, Choose B, Monty removes C, stick with B, lose.
   
4. A is right, Choose C, Monty removes B, stick with C, lose.
   
5. B is right, Choose B, Monty removes A, stick with B, win.
   
6. B is right, Choose B, Monty removes C, stick with B, win.
   
7. B is right, Choose A, Monty removes C, stick with A, lose.
   
8. B is right, Choose C, Monty removes A, stick with A, lose.
   
9. C is right, Choose C, Monty removes A, stick with C, win.
   
10. C is right, Choose C, Monty removes B, stick with C, win.
    
11. C is right, Choose A, Monty removes B, stick with B, lose.
    
12. C is right, Choose B, Monty removes C, stick with C, lose.

Your calculation of the chance of winning when you stick with your original choice being 1/2 would be correct if each of these outcomes occurred with equal probability. This is not, however, the case.

 

[3], [4], [7], [8], [11] and [12] have equal probability. Monty has no choice in which door to remove, so the probability is unaffected by his action. [1] and [2], represent 2 possible choices for Monty for a single random door setup and choice by Contestant, so the sum of their probabilities = the probability of [3], as do the probabilities of [5] + [6], and [9] + [10]. It’s convenient, and has no effect on the calculation, to assume that Monty chooses between [1] and [2] randomly, giving [1], [2], [5], [6], [9] and [10] equal probabilities.

 

Totaling and normalizing the enumeration with these weights give the probability for [1] = 1/18, for [3] = 1/9. So the chance of Contestant winning when sticking with his original choice is 6/18, not 6/12.

 

I'm beginning to see some light at the end of this tunnel.

 

My only question is on how to prove that [1] and [2] have less probability independently than [3]. Could not [1] and [2] happen just as often as [3] and [4]? This seems to be where we must carry over the 1/3 2/3 into the second round of choices. The chance that I might pick A (winner) is 1/3. Therefore [1] and [2] initially have a 1/3 chance of even getting into the situation that Monty has to choose B or C. [3] and [4] each have a 1/3 possibility, but Monty can only choose one door to remove in each of these occurences.

 

My stipulation has been that after you have made your initial choice, the odds of winning become 50/50. I say this because after you have made your first choice, there are four possible outcomes as I have enumerated. You are correct in that the odds of each one occuring before you've made your choice are 1/3 for choosing A, B, or C. My only sticking point is in seeing how this affects the outcomes after you have chosen A, B, or C. I contend that at that point your options become [1], [2], [3], and [4], for which if these were your original choices we all agree there would be a 50/50 chance of winning.

 

Now as for the writing of a program that only chooses A and sticks with A, then I would say yes you only have a .33 chance of winning, because the random choice of which door will contain the prize is random and therefore also 0.33. That means that only 33% of the time will it choose door A, and then choosing and stick with door A for every trial in a 100, 1000, or 10000000000000000000 billion loop trial should net a 33% success rate. However, we are not stipulating that you choose only one door and stick with that through all the trials. No we say randomly choose a door and stick with that door for one trial then randomly choose a door and stick with it for that trial, etc.

For this we look back to those results posted on the math.ucsd.edu page. Of course these results could be off if people tried the method of choosing one door and sticking with it for hundreds of trials but given enough different people hitting that website, the results should be pretty random. Given that there was a ~50% success rate for sticking with your first choice, it seems that empirically (using real world participants, not a computer program) the odds are what I believe them to be.