I gave a programming problem in class today. Simulate 1000 six sided die rolls. Count each roll and show a distribution of the outcomes in terms of percent, eg:
How many rolls? 1000
1: 157/1000 = 15%
2: 175/1000 = 17%
3: 168/1000 = 16%
4: 163/1000 = 16%
5: 179/1000 = 17%
6: 158/1000 = 15%
Knowing some students would finish quickly, I threw in a twist. Simulate a weighted dice 20% more likely to roll a one. Not surprisingly, answers start to look like this:
How many times would you like to roll?: 1000
1: 364/1000 36.4%
2: 132/1000 13.200000000000001%
3: 125/1000 12.5%
4: 132/1000 13.200000000000001%
5: 118/1000 11.799999999999999%
6: 129/1000 12.9%
Since the student succeeded in weighing the die, this is acceptable for my purposes. After all, I'm teaching problem solving and programming, not applied probability. That said, while 36% is in the ballpark of 20 more than 13% it appears this die is actually more like three times more likely to land than the other sides.
I would like to present with confidence the correct math used in determining this outcome.
The most closely related QA I found on this site is here: The probabilities of outcomes of throwing a weighted die
I sort of follow the logic and I come up with probabilities:
Chance to roll a …
6 = p
5 = p
4 = p
3 = p
2 = p
1 = 1.2 p
6.2p = 1
p = 1/6.2 = 16.129% (rounded)
Chance to roll a 2-6 = p = 16.129%
Chance to roll a 1 = 1.2p = 19.355% (also rounded)
The math checks out since 19.355% is ~20% more than 16.129%. Did I correctly solve this problem?
Best Answer
"$20\%$ more likely" is ambiguous until you say more likely than what.
You seem to want the answer for "$20\%$ more likely to be $1$ than to be $x$, for $x$ any one of $2, 3, 4, 5, 6.$"
But it could be "$20\%$ more likely to be $1$ than it would have been if the die were fair." For a fair die, $1$ has probability $\frac16,$ and $20\%$ more than that is $1.2 \times \frac16 = 0.2.$
Or it could be "$20\%$ more likely to be $1$ than to be in the set $\{2,3,4,5,6\}.$" Then the probability of $1$ is $\frac6{11},$ which is $20\%$ more than $\frac5{11},$ which is the probability of not-$1.$
Your student seems to have taken the interpretation "$20\%$ more likely than it would have been if the die were fair," but then become confused by percentages of percentages (since they were expressing the probabilities as percentages), and as a result interpreting "$20\%$ more" (in effect) as "add $0.2$" rather than as "multiply by $1.2$."
So since the fair probability is $\frac16 = 16\frac23\%,$ the "$20\%$ more" probability is $\frac16 + 0.2 = 36\frac23\%.$
To make up for adding $0.2$ to one outcome, they then subtracted $0.04$ from each of the other five outcomes, which then each have probability $\frac16 - 0.04 = 12\frac23\%.$