Discussion Forum

New Year's Resolutions (Weekly Brain Potion Discussion)

 
 
Picture of John Lensmire
New Year's Resolutions (Weekly Brain Potion Discussion)
by John Lensmire - Friday, 19 January 2018, 11:58 AM
 

Emily, Emma, Madison, Olivia, and Hannah are five friends in middle school. On January 1st this year, they decided to each make a New Year's resolution and share it with the group. They would then meet every month and share whether they had kept or failed their resolution. (Note, once you have failed your resolution, it is failed for the entire year.)

At the beginning of the year, everyone was excited about their resolution, so each friend had a $90\%$ chance of keeping their resolution during January. (During the month they do not discuss their resolutions, so this $90\%$ is independent for all the friends.)

When they meet each month and share their progress, they are discouraged if another friend fails their resolution, making it less likely they keep their own resolution. Suppose that once a friend fails, it reduces the chance others succeed by $10\%$. For example, if $2$ friends have failed so far during the year, the chance each of the others keep their resolution during the next month is $90\% - 2\times 10\% = 70\%$.

What is the chance that all five friends report failing their resolution during the February meeting? (Meaning they have failed in the first month.)

What is the chance that all five friends report failing their resolution during the March meeting? (Remember that once a resolution is failed it is failed for the entire year.)

Challenge:  In answering the second question, you'll notice that a lot of casework is needed. For those of you with some programming experience, try to write a program to help answer/approximate the chance that all five friends report failing during the April, May, June, etc, meeting. In general try to find the probability that all five friends report failing after $12$ months. Your program can either simulate the problem (to come up with an approximate answer) or calculate the cases directly (to come up with an exact answer).

Please share any thoughts or questions you have below. We'll monitor the responses and give our thoughts as well!

 
Picture of Christopher wu
Re: New Year's Resolutions (Weekly Brain Potion Discussion)
by Christopher wu - Tuesday, 6 February 2018, 5:45 PM
 

For all of the friends to fail during the first month, there is a 1/10^5 or a 0.001% chance. For all 5 to have failed during the second meeting, we have 6 different cases, all of which depend on the first month

Case 1:None fail. This is pretty much the same as part 1, and you get 1/100000, but you multiply by 59049/100000, as that is the probability that none fail the first month. Result:59049/10000000000

Case 2:1 fails. The probability that 1 fails the first month is 32805/100000, and the probability that the other 4 fail during the second month is 1/5^4 or 1/625. Result:6561/12500000

Case 3:2 fail.The probability that 2 fail the first month is 7290/100000, and the probability that the other 3 fail is 3/10^3 Result:19683/10000000

Case 4:3 fail.The probability that 3 fail the first month is 810/100000, and the probability that the last 2 fail is 4/10^2 Result:162/250000

Case 5:4 fail.The probability that 4 fail the first month is 45/100000, and the probability that the last fails is 1/2 Result:45/200000

Case 6:all five fail during the 1st month. We have already calculated this, as 1/100000, then as all 5 have already failed, we are left with the same number.Result:1/100000

Add all of these 6 numbers, to obtain 59049/10000000000+6561/12500000+19683/10000000+162/250000+45/200000+1/100000, which gives our result, 33820849/10000000000.

I hope I didn't make any mistakes, as there were a lot of numbers, and a ton of zeros. Hopefully I didn't add/subtract any extra zeros