Solving the imaginary puzzle and winning millions
March 2, 2011 9:42 AM Subscribe
How can I determine the optimal scenario to get through imaginary radio station contest hell and earn the fictional prize.
All right, so I have no idea how I got myself thinking about this, but neither I nor friends I've asked seem to have the processing power to figure it out.
Here's the situation:
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There is a fictional radio contest that runs from Monday-Friday, where one must call in to win. The odds of getting through are 1 in 5.
On Monday, if you get in, you get in. But each subsequent day, someone new will get in as well, and you have to face-off in a coin flip. Whoever calls it correctly gets to stay. So each day, you have a 1 in 2 chance of going home.
If you are eliminated, you can not call back in, and if you are left standing Friday afternoon, you get $100 million and a lifetime supply of white fudge Oreos.
So what is the optimal day to call in?
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Shooting from the hip, I feel like Friday. But then it's so all-or-nothing on getting in. You might not even get to play. If you start trying on Monday, you'll have a better shot of even getting in in the first place.
What say you, MeFites?
All right, so I have no idea how I got myself thinking about this, but neither I nor friends I've asked seem to have the processing power to figure it out.
Here's the situation:
---
There is a fictional radio contest that runs from Monday-Friday, where one must call in to win. The odds of getting through are 1 in 5.
On Monday, if you get in, you get in. But each subsequent day, someone new will get in as well, and you have to face-off in a coin flip. Whoever calls it correctly gets to stay. So each day, you have a 1 in 2 chance of going home.
If you are eliminated, you can not call back in, and if you are left standing Friday afternoon, you get $100 million and a lifetime supply of white fudge Oreos.
So what is the optimal day to call in?
---
Shooting from the hip, I feel like Friday. But then it's so all-or-nothing on getting in. You might not even get to play. If you start trying on Monday, you'll have a better shot of even getting in in the first place.
What say you, MeFites?
On preview... dammit, DevilsAdvocate beat me to it. Explained it better than I did as well (since I started on Monday and worked forward).
posted by Etrigan at 9:59 AM on March 2, 2011
posted by Etrigan at 9:59 AM on March 2, 2011
Note: the above assumes that if you get through and lose a coin flip (either the same day or a later day) you are not permitted to try calling in again later in the week.
posted by DevilsAdvocate at 10:00 AM on March 2, 2011
posted by DevilsAdvocate at 10:00 AM on March 2, 2011
...which, on re-read, you explicitly said in your question.
posted by DevilsAdvocate at 10:05 AM on March 2, 2011
posted by DevilsAdvocate at 10:05 AM on March 2, 2011
Response by poster: Hmm…I love the thinking. Can you add probability like that, however? If you could, wouldn't you eventually get over 1.0?
Other than that, this makes great sense to me.
posted by jpcody at 10:15 AM on March 2, 2011
Other than that, this makes great sense to me.
posted by jpcody at 10:15 AM on March 2, 2011
Can you add probability like that, however?
You can add probabilities like that — P(A or B) = P(A) + P(B) if and only if the two events are mutually exclusive. "Getting through on Wednesday and surviving to the end of the week" and "not getting through on Wednesday, but getting through on a later day and surviving to the end of the week" are mutually exclusive, so they can be added.
If you could, wouldn't you eventually get over 1.0?
No, because the general form is P($DAY) = [P(getting through on $DAY) * P(surviving all coin flips starting on $DAY)] + [P(not getting through on $DAY) * P($DAY+1)]
Since P(getting through on $DAY) + P(not getting through on $DAY) = 1, and P(surviving all coin flips starting on $DAY) and P($DAY+1) are both less than 1, the sum is always less than 1.
posted by DevilsAdvocate at 10:22 AM on March 2, 2011
You can add probabilities like that — P(A or B) = P(A) + P(B) if and only if the two events are mutually exclusive. "Getting through on Wednesday and surviving to the end of the week" and "not getting through on Wednesday, but getting through on a later day and surviving to the end of the week" are mutually exclusive, so they can be added.
If you could, wouldn't you eventually get over 1.0?
No, because the general form is P($DAY) = [P(getting through on $DAY) * P(surviving all coin flips starting on $DAY)] + [P(not getting through on $DAY) * P($DAY+1)]
Since P(getting through on $DAY) + P(not getting through on $DAY) = 1, and P(surviving all coin flips starting on $DAY) and P($DAY+1) are both less than 1, the sum is always less than 1.
posted by DevilsAdvocate at 10:22 AM on March 2, 2011
Or a more formal proof of same:
Given:
0≤P(A)≤1
0≤P(B)≤1
0≤P(C)≤1
0≤P(D)≤1
P(A)+P(C)=1
We can deduce:
P(B)≤1
P(A)*P(B)≤P(A) [multilply both sides of previous equation by P(A); it is important we know P(A) is non-negative]
P(D)≤1
P(C)*P(D)≤P(C)
P(A)*P(B) + P(C)*P(D) ≤ P(A)+P(C)
P(A)*P(B) + P(C)*P(D) ≤ 1
posted by DevilsAdvocate at 10:33 AM on March 2, 2011 [1 favorite]
Given:
0≤P(A)≤1
0≤P(B)≤1
0≤P(C)≤1
0≤P(D)≤1
P(A)+P(C)=1
We can deduce:
P(B)≤1
P(A)*P(B)≤P(A) [multilply both sides of previous equation by P(A); it is important we know P(A) is non-negative]
P(D)≤1
P(C)*P(D)≤P(C)
P(A)*P(B) + P(C)*P(D) ≤ P(A)+P(C)
P(A)*P(B) + P(C)*P(D) ≤ 1
posted by DevilsAdvocate at 10:33 AM on March 2, 2011 [1 favorite]
Response by poster: Amazing work, DevilsAdvocate, thanks so much for your help on this. I'm going to spread the good word to all those who have tried and failed with me thus far :)
posted by jpcody at 10:42 AM on March 2, 2011
posted by jpcody at 10:42 AM on March 2, 2011
Just for fun I wrote a quick Python program to simulate it with 100000 trials for each day. Here's the results:
Monday: 10.461%
Tuesday: 11.451%
Wednesday: 12.991%
Thursday: 13.075%
Friday: 9.975%
Probability works!
posted by burnmp3s at 10:58 AM on March 2, 2011 [4 favorites]
Monday: 10.461%
Tuesday: 11.451%
Wednesday: 12.991%
Thursday: 13.075%
Friday: 9.975%
Probability works!
posted by burnmp3s at 10:58 AM on March 2, 2011 [4 favorites]
Or, re-route the phone lines to the radio station, ensuring you'll be the correct caller (as Kevin Mitnik and friends supposedly did).
posted by filthy light thief at 11:02 AM on March 2, 2011 [1 favorite]
posted by filthy light thief at 11:02 AM on March 2, 2011 [1 favorite]
This thread is closed to new comments.
If you call on Friday only, you have a 1/5 chance of getting in, and if you do, a 1/2 chance of surviving, so your chance of winning is 1/10=0.1
If you start calling on Thursday (we will assume once you start calling, if you don't get through you will continue calling every day thereafter), you have a 1/5 chance of getting through, and a 1/4 chance of surviving (you must survive two coin flips); plus, on the 4/5 chance you don't get through, you have the same 1/10 chance of winning on Friday. So your overall probability is (1/5 * 1/4) + (4/5 * 1/10) = 13/100 = 0.13
Wednesday: (1/5 * 1/8) + (4/5 * 13/100) = 129/1000 = 0.129
Tuesday: (1/5 * 1/16) + (4/5 * 129/1000) = 1157/10000 = 0.1157
Monday [note the chance of surviving if you get through is still 1/16, since there's no Monday coin flip] (1/5 * 1/16) + (4/5 * 1157/10000) = 0.10506
So the highest probability of winning, 0.13, is achieved by not calling Monday, Tuesday, or Wednesday; then calling Thursday, and also Friday if you don't get through Thursday.
(Another way to look at it: if you call and get through on Monday, Tuesday, or Wednesday, your chance of surviving is 1/16 (for Mon. or Tue.) or 1/8 (for Wed.), which are less than 0.13, the probability of winning if you start on Thursday. But if you get through on Thursday, your chance of surviving is 1/4, which is greater than the 0.1 probability of winning if you start Friday.)
posted by DevilsAdvocate at 9:57 AM on March 2, 2011 [4 favorites]