There are three doors, two of the doors have a man-eating tiger waiting behind them and one has an ice cream sundae.You’re asked to guess which door has the ice cream. You make your choice. Of the two remaining doors, the host opens one with a tiger behind it.You’re left with two remaining doors: the door you’ve chosen and the other unopened door.Do you stick to your original choice or pick the other door?

Remember your life is on the line and this may have already been discussed at a Harvey Reunion.

Workout your answer before looking at the comments to avoid spoilers.

Hey no googling, the host at the doors does not provide internet access.

This is a staple. I think I last saw it in the newspaper under the column by Ms. Savant or whatever her name is (you know the one who advertises herself as the highest IQ in the galaxy)

First of all since you "are asked to guess which door has the ice cream" the correct solution is to say "No thank you." It's all about maximizing your returns and ice cream just doesn't justify choosing.

But if your interrogator realizes his/her mistake and threatens your family, you pick the other door. It's just like Monty Hall—he was going to show you the wrong door anyway so now you have a 50% chance instead of a 33.3% chance.

I'm taking a probability class right now, and we already did talk about this at a family reunion you were just playing basketbal at the time, I wont spoil it for anyone, but the chances that the other door is right is 2/3 while your original door is only 1/3 chance of being right.... ....

Sorry, I just threw those numbers out there. I said "now you have a 50% chance" which doesn't include the first part of the problem. So, yes, considering the whole problem 2/3 and 1/3 is correct.

However, most riddles, like life, do not have a single answer. There is almost always alternatives with some truth. (Yes, even in the Prop 8 question:)

Here is the answer I thought of second: The riddle is presented with the doors in sequencial order, 1, 2, 3. I would take the last one. If the 'Monty Hall' impersonator then reveals door #1 as hiding a tiger, I would switch, but if he reveals door #2, I would stay pat.

I will put my reasoning as a psychological experiment. You know, the kind that psychology majors have to do, and James would always be volunteering for. Have someone actually perform the riddle and you record the results. I suspect that the person performing the riddle will almost always reveal the 1st sequential door that has a tiger. In other words if doors 1 and 2 have tigers, he/she will simply reveal door #1. It will not be a 'random' or 'tricky' decision, but the one easiest for the riddler—reveal the 1st available tiger.

My major assumption is that the person presenting the riddle/doors is not Monty Hall or any professional riddler. This is the first and only time he/she has done this.

I could bring into play the Edgar Allan Poe solution for riddles, at this point but I'll save that for another time.

Jim, you are incorrect in your reasoning (like that with prop 8-please if you can, provide me with more than one answer).

Your reasoning is just trying to justify sticking with your gut instinct. Psychological bias will never trump statistics. Your reasoning for door choice, host for tiger placement, and door choosing randomizes the experiment that statistics should be trust and not reasoning. As a scientist, I didn't trust mathematicians and conducted an experiment with two other people I know well and thought I could read. Also at no point in the riddle did the guesser have a 50% chance (unless if one had a memory lapse and forgot which door they had pick).

Experimental Protocol:

Three cups were set up, one with a coin under and two empty. The cups were numbered 1, 2, and 3. Which cup gets the coin is decided by randomly picking a number written on three pieces of paper. There are three players involved. Player 1 has to try to guess which cup has the coin. Player 2 mixes the numbered pieces of paper for Player 3 to sample. Player 3 blindly picks one of the pieces of paper and puts the coin under the corresponding numbered cup. The cups are aligned in a row in order and their position does not change. Player 1 is asked to guess which cup has the coin underneath. From the two remaining cups, Player 3 reveals the empty one. Player 1 then has a choice: stick to his first selection or choose the other cup. 100 trials are conducted where Player 1 sticks to his first choice and 100 trials are conducted where Player 1 chooses the other cup. The three subjects alternate roles after several trials.

Results:

Does changing your mind after one of the wrong cups is eliminated improve your chances of picking the right cup? If outcome is not influenced by cup choice, we would expect that whether or not you change your mind, the proportion of choosing the right cup is 50% or p1 = p2, where p1 is the correct proportion if Player 1 changes his mind and p2 is the correct proportion if Player 1 stays with the same choice. Of the 100 trials in which Player 1 sticks to his first choice, the observed proportion of correct selections is 35 out of 100 or 33.65%. The observed proportion of wrong selections is 65 out of 100 or 66.71%. However, of the 100 trials in which Player 1 changes his mind and picks the other door, the observed proportion of correct selections is 69 out of 100 or 66.45% and the proportion of wrong selections is 31 out of 100 or 32.29%. Therefore we reject the null hypothesis p0 = 0.5 in favor of the alternative hypothesis that the probability of making the correct choice is larger than 50% when Player 1 changes his mind since χ2 = 23.16, p-value < 0.0001 (Pearson). Fisher’s Exact Test verifies a p-value < 0.0001 in favor of the alternative hypothesis. That is, the probability of getting a positive outcome is higher when Player 1 changes his mind.

Outcome Group Wrong Correct Total Change 31 69 100 Same 65 35 100 Total 100 100 200

Oops, reviewing what I wrote, I am the one who was not clear. As a matter of fact, what I wrote was completely wrong. Peter's explanation covers the problem admirably and, as he points out, it is the solution "If the outcome is not influenced by cup(door) choice."

I was trying to (semi humorously) switch the problem from a statistical one to one where the choice is, in fact, influenced by human nature. What I was trying to say was: Assume you have originally chosen door 3 1. If there is a tiger behind door 1 and ice cream behind door 2, door 1 will be revealed. 2. If there is ice cream behind door 1 and tiger behind 2, then door 2 will be revealed. 3. If there are 2 tigers left, then—here is my (completely unproven) hypothesis—door 1 will be revealed simply because the host is a. not thinking about randomizing his choice, b. being lazy, or c. not being tricky.

So, when door no. 2 is revealed, you definitely want to make the switch and take door no. 1. Otherwise, well, you may want to stare long and hard into the host's eyes. I said the exact opposite above and I apologize.

Actually this doesn't really change your answer but rather makes you more confident when door no. 2 is revealed.

## 9 comments:

This is a staple. I think I last saw it in the newspaper under the column by Ms. Savant or whatever her name is (you know the one who advertises herself as the highest IQ in the galaxy)

First of all since you "are asked to guess which door has the ice cream" the correct solution is to say "No thank you." It's all about maximizing your returns and ice cream just doesn't justify choosing.

But if your interrogator realizes his/her mistake and threatens your family, you pick the other door. It's just like Monty Hall—he was going to show you the wrong door anyway so now you have a 50% chance instead of a 33.3% chance.

Close but wrong

by "wrong" I mean your numbers

I'm taking a probability class right now, and we already did talk about this at a family reunion you were just playing basketbal at the time, I wont spoil it for anyone, but the chances that the other door is right is 2/3 while your original door is only 1/3 chance of being right.... ....

Sorry, I just threw those numbers out there. I said "now you have a 50% chance" which doesn't include the first part of the problem. So, yes, considering the whole problem 2/3 and 1/3 is correct.

However, most riddles, like life, do not have a single answer. There is almost always alternatives with some truth. (Yes, even in the Prop 8 question:)

Here is the answer I thought of second:

The riddle is presented with the doors in sequencial order, 1, 2, 3. I would take the last one. If the 'Monty Hall' impersonator then reveals door #1 as hiding a tiger, I would switch, but if he reveals door #2, I would stay pat.

I will put my reasoning as a psychological experiment. You know, the kind that psychology majors have to do, and James would always be volunteering for. Have someone actually perform the riddle and you record the results. I suspect that the person performing the riddle will almost always reveal the 1st sequential door that has a tiger. In other words if doors 1 and 2 have tigers, he/she will simply reveal door #1. It will not be a 'random' or 'tricky' decision, but the one easiest for the riddler—reveal the 1st available tiger.

My major assumption is that the person presenting the riddle/doors is not Monty Hall or any professional riddler. This is the first and only time he/she has done this.

I could bring into play the Edgar Allan Poe solution for riddles, at this point but I'll save that for another time.

"They tell me about the agony of men who can't answer questions like yours."

Jim, you are incorrect in your reasoning (like that with prop 8-please if you can, provide me with more than one answer).

Your reasoning is just trying to justify sticking with your gut instinct. Psychological bias will never trump statistics. Your reasoning for door choice, host for tiger placement, and door choosing randomizes the experiment that statistics should be trust and not reasoning. As a scientist, I didn't trust mathematicians and conducted an experiment with two other people I know well and thought I could read. Also at no point in the riddle did the guesser have a 50% chance (unless if one had a memory lapse and forgot which door they had pick).

Experimental Protocol:

Three cups were set up, one with a coin under and two empty. The cups were numbered 1, 2, and 3. Which cup gets the coin is decided by randomly picking a number written on three pieces of paper. There are three players involved. Player 1 has to try to guess which cup has the coin. Player 2 mixes the numbered pieces of paper for Player 3 to sample. Player 3 blindly picks one of the pieces of paper and puts the coin under the corresponding numbered cup. The cups are aligned in a row in order and their position does not change. Player 1 is asked to guess which cup has the coin underneath. From the two remaining cups, Player 3 reveals the empty one. Player 1 then has a choice: stick to his first selection or choose the other cup. 100 trials are conducted where Player 1 sticks to his first choice and 100 trials are conducted where Player 1 chooses the other cup. The three subjects alternate roles after several trials.

Results:

Does changing your mind after one of the wrong cups is eliminated improve your chances of picking the right cup? If outcome is not influenced by cup choice, we would expect that whether or not you change your mind, the proportion of choosing the right cup is 50% or p1 = p2, where p1 is the correct proportion if Player 1 changes his mind and p2 is the correct proportion if Player 1 stays with the same choice. Of the 100 trials in which Player 1 sticks to his first choice, the observed proportion of correct selections is 35 out of 100 or 33.65%. The observed proportion of wrong selections is 65 out of 100 or 66.71%. However, of the 100 trials in which Player 1 changes his mind and picks the other door, the observed proportion of correct selections is 69 out of 100 or 66.45% and the proportion of wrong selections is 31 out of 100 or 32.29%. Therefore we reject the null hypothesis p0 = 0.5 in favor of the alternative hypothesis that the probability of making the correct choice is larger than 50% when Player 1 changes his mind since χ2 = 23.16, p-value < 0.0001 (Pearson). Fisher’s Exact Test verifies a p-value < 0.0001 in favor of the alternative hypothesis. That is, the probability of getting a positive outcome is higher when Player 1 changes his mind.

Outcome

Group Wrong Correct Total

Change 31 69 100

Same 65 35 100

Total 100 100 200

Hmm.... either I didn't make myself clear or I'm not reading you correctly.

Oops, reviewing what I wrote, I am the one who was not clear. As a matter of fact, what I wrote was completely wrong. Peter's explanation covers the problem admirably and, as he points out, it is the solution "If the outcome is not influenced by cup(door) choice."

I was trying to (semi humorously) switch the problem from a statistical one to one where the choice is, in fact, influenced by human nature. What I was trying to say was:

Assume you have originally chosen door 3

1. If there is a tiger behind door 1 and ice cream behind door 2, door 1 will be revealed.

2. If there is ice cream behind door 1 and tiger behind 2, then door 2 will be revealed.

3. If there are 2 tigers left, then—here is my (completely unproven) hypothesis—door 1 will be revealed simply because the host is a. not thinking about randomizing his choice, b. being lazy, or c. not being tricky.

So, when door no. 2 is revealed, you definitely want to make the switch and take door no. 1. Otherwise, well, you may want to stare long and hard into the host's eyes. I said the exact opposite above and I apologize.

Actually this doesn't really change your answer but rather makes you more confident when door no. 2 is revealed.

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