Dealing with choices in disease treatment
Back in law school, I wrote a column for the University of Chicago Phoenix on the "Let's Make a Deal" problem of accepting Monty Hall's offer to switch doors. The correct answer is to switch doors, and you will win 2/3 of the time.
In the October 1998 issue of Intellectual Property Today, I reviewed the matter:
BACKGROUND
Of my columns for my law school newspaper, the most discussed was the presentation of the "Let's Make a Deal" problem. On seeing it, most people get it wrong, although on discussion, most people get it right. Some never do. On August 6, I was explaining it to a former colleague of mine as we rode the PATH to the World Trade Center. He got it wrong. More interestingly, a stranger who was listening to us, who claimed to be a Ph.D. in statistics from Harvard, intervened to argue for the wrong position. At that point, I thought it worthwhile to resurrect the material.
LET'S MAKE A DEAL
The set-up of the problem arises from a game show, begun in the 1960's and hosted by Monty Hall, entitled "Let's Make a Deal." There were three doors, behind only one of which was a valuable prize. After some preliminaries, a contestant would choose one door from among the three. The selected door remained closed. At this point, Monty would sometimes open one of the two unchosen doors, behind which was no prize. Monty would then offer the contestant the opportunity to switch choice from the initially-chosen door to the remaining unopened door. The question is simply: should the contestant switch?
Most people see the situation as one of equal probability. There are two unopened doors, and behind one of them is the prize. In fact, by switching choice, one wins two times out of three. The key point is that there are three states of the system, not two.
To work through the situation, let's assume the contestant picks door number 1. One-third of the time, the prize will be behind door number 1, and the contestant will win if does not change and he will lose if he changes. Two thirds of the time, the prize will not be behind door number 1 (one third of the time behind door number 2; one third of the time behind door number 3). The contestant will lose if he does not change and he will win if changes. One could stop here.
However, the difficulty arises when one confronts the opening of the door without a prize. There are two doors left, and one prize, ergo a seeming probability of one-half. But, because Monty cannot open the initially-selected door of the contestant, he cannot change the probability as to that door. It will remain at one third. Some people do not find this correct.
Instead of the "three door" problem, visualize a lottery with one million tickets, only one of which is a winner. You buy one ticket and put it in your pocket. Monty, who knows which ticket is a winner, tears up 999,998 losing tickets and offers to trade you his ticket for yours. There are only two tickets left and one is a winner. Do you think the probability is 50/50?
There are many messages from this problem. I'll consider two.
KNOWLEDGE IS GOOD
If you understand the problem, you will switch doors and you will win 2/3 of the time. The elimination of the one door gave added knowledge. The probabilities were not 50/50.
Did you ever wonder why appellate panels have more than one member? Assuming a district judge and an appellate judge have the same probability of "getting it right", and that the probability is greater than 50%, a panel will give the right answer more frequently than the single judge.
To take a simple case, let us assume that a given judge has a probability of getting the right answer (2/3) of the time, and that this probability is independent of the analysis of the other judges. Consider a three judge panel which requires at least a 2-1 vote for a decision. The joint probability of getting it right is 20/27 (=74.1%), which comprises the 3-0 vote (8/27=(2/3)<3>) and three 2-1 votes (each of joint probability 4/27). This is 11.1% higher than the probability the single judge will get it right ((74.1-66.7)/66.7).
An interesting twist on this involves the choice to employ a drug.
Imagine that you have a certain disease that is not life threatening but greatly diminishes quality of life. There is a treatment that will be effective in removing symptoms 2/3 of the time, but ineffective 1/3 of the time. Good idea to try it.
HOWEVER, there is an additional factor. In the 1/3 of the time the treatment does not work, you develop an additional problem.
If the problem were death, likely this would be a bad idea. But suppose the problem is the development of another disease (similar to the first), so that 1/3 of the time you don't solve the initial problem AND you develop a new problem, making your situation "twice" as bad as when you started. What do you do?
***
See also
"Let's Make a Deal" problem
In the October 1998 issue of Intellectual Property Today, I reviewed the matter:
BACKGROUND
Of my columns for my law school newspaper, the most discussed was the presentation of the "Let's Make a Deal" problem. On seeing it, most people get it wrong, although on discussion, most people get it right. Some never do. On August 6, I was explaining it to a former colleague of mine as we rode the PATH to the World Trade Center. He got it wrong. More interestingly, a stranger who was listening to us, who claimed to be a Ph.D. in statistics from Harvard, intervened to argue for the wrong position. At that point, I thought it worthwhile to resurrect the material.
LET'S MAKE A DEAL
The set-up of the problem arises from a game show, begun in the 1960's and hosted by Monty Hall, entitled "Let's Make a Deal." There were three doors, behind only one of which was a valuable prize. After some preliminaries, a contestant would choose one door from among the three. The selected door remained closed. At this point, Monty would sometimes open one of the two unchosen doors, behind which was no prize. Monty would then offer the contestant the opportunity to switch choice from the initially-chosen door to the remaining unopened door. The question is simply: should the contestant switch?
Most people see the situation as one of equal probability. There are two unopened doors, and behind one of them is the prize. In fact, by switching choice, one wins two times out of three. The key point is that there are three states of the system, not two.
To work through the situation, let's assume the contestant picks door number 1. One-third of the time, the prize will be behind door number 1, and the contestant will win if does not change and he will lose if he changes. Two thirds of the time, the prize will not be behind door number 1 (one third of the time behind door number 2; one third of the time behind door number 3). The contestant will lose if he does not change and he will win if changes. One could stop here.
However, the difficulty arises when one confronts the opening of the door without a prize. There are two doors left, and one prize, ergo a seeming probability of one-half. But, because Monty cannot open the initially-selected door of the contestant, he cannot change the probability as to that door. It will remain at one third. Some people do not find this correct.
Instead of the "three door" problem, visualize a lottery with one million tickets, only one of which is a winner. You buy one ticket and put it in your pocket. Monty, who knows which ticket is a winner, tears up 999,998 losing tickets and offers to trade you his ticket for yours. There are only two tickets left and one is a winner. Do you think the probability is 50/50?
There are many messages from this problem. I'll consider two.
KNOWLEDGE IS GOOD
If you understand the problem, you will switch doors and you will win 2/3 of the time. The elimination of the one door gave added knowledge. The probabilities were not 50/50.
Did you ever wonder why appellate panels have more than one member? Assuming a district judge and an appellate judge have the same probability of "getting it right", and that the probability is greater than 50%, a panel will give the right answer more frequently than the single judge.
To take a simple case, let us assume that a given judge has a probability of getting the right answer (2/3) of the time, and that this probability is independent of the analysis of the other judges. Consider a three judge panel which requires at least a 2-1 vote for a decision. The joint probability of getting it right is 20/27 (=74.1%), which comprises the 3-0 vote (8/27=(2/3)<3>) and three 2-1 votes (each of joint probability 4/27). This is 11.1% higher than the probability the single judge will get it right ((74.1-66.7)/66.7).
An interesting twist on this involves the choice to employ a drug.
Imagine that you have a certain disease that is not life threatening but greatly diminishes quality of life. There is a treatment that will be effective in removing symptoms 2/3 of the time, but ineffective 1/3 of the time. Good idea to try it.
HOWEVER, there is an additional factor. In the 1/3 of the time the treatment does not work, you develop an additional problem.
If the problem were death, likely this would be a bad idea. But suppose the problem is the development of another disease (similar to the first), so that 1/3 of the time you don't solve the initial problem AND you develop a new problem, making your situation "twice" as bad as when you started. What do you do?
***
See also
"Let's Make a Deal" problem
posted by Lawrence B. Ebert at 4:31 AM
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