Today’s Brain Game was inspired by an episode of the original Star Trek that I watched again recently – “The Corbomite Maneuver.” It’s the one where Captain Kirk and his crew are meanaced by first a colorful cube, then by a huge golden chandelier, and finally (gasp) by Clint Howard. Enjoy:
You’re holding two six-sided dice decorated with colors on their sides instead of pips. One die has 2 blue sides and 4 red sides, while the other has 1 blue side and 5 red sides. With this in mind:
How many times would you have to
roll both dice to exceed a 50% likelihood
that you’ve thrown at least one “double-blue” roll?
Comments (17)
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Actually you only have to roll 13 times to get a better than 50-50 chance. On the first roll, the chances not getting blue-blue are 17/18 or 94%. The chances of rolling twice without a blue-blue are 17/18 times 17/18 or 89%. After the thirteenth roll, your chances of not getting a b-b for all 13 rolls are 47.5%, meaning that you have a 52.5% chance of having had at least one b-b roll in that string.
This is similar to the question of how many people do you need to gather in a room before the probability that two of them share a birthday is better than 50-50. The answer, surprisingly, is 24 people.
posted by Joel on 10-1-2009 at 8:41 am
Sandy, I am an actuary and I can attest that the first comment’s solution is correct. You are correct in that the chances of success are 1/18 for each roll. So the chances that two blue sides aren’t rolled is 17/18.
The probability that a blue hasn’t been rolled is equal to (17/18)^x where x equals the number of rolls. So we want to find a scenario where (17/18)^x ln(17/18)/ln(.5). Cleaning that up x > 12.12. So the answer to your question is 13.
posted by Thomas on 10-1-2009 at 9:14 am
Sandy, I have to agree with the other math nerds. The best way to think about these problems is by looking at it \backwards\. Basically, the probability of throwing at least ONE double blue is the complement of not throwing any. In mathemagic language:
P(at least one double blue) = 1 – P(NO double blue). This is easier to deal with because P(NO double blue) is just 17/18 for one roll. And then (17/18)^2 for two rolls. In general, then, the probability of at least one double blue after N rolls is 1 – (17/18)^N.
That goes above 50% after 13 rolls (N=13).
posted by Brad H on 10-1-2009 at 10:14 am
Very interesting… thanks for the explanations, both of you. The mathematics of probabilities has always fascinated me (beck to the days when I learned what factorials were and used to play with them on my old calculator).
I’ve corrected the answer to “13″ to match Joel’s response. Thanks for teaching me something new!
posted by Sandy Wood on 10-1-2009 at 10:17 am
I got 13, via a slightly more laborious route, though similar to what’s described in answer 1. Thomas, I’m curious how you derived the equation that you give with the lns.
posted by Alex on 10-1-2009 at 10:21 am
I used to make my Statistic teacher crazy by telling her that you always have a 50% chance when you are doing anything…you either roll a double blue or you do not.
I would have answered that question in high school like this, “there is no such thing as more than a 50% probability that such an event could occur”.
If you guessed that my grade in that class was a C, you’d be correct.
posted by graham on 10-1-2009 at 10:23 am
Graham, that response made me so happy. I am the exact same way. My wife is obsessed with the weather channel and I always tell her there is a 50% chance of rain everyday. It either will rain or it won’t.
posted by John on 10-1-2009 at 12:29 pm
my head asplode.
posted by emaneff on 10-1-2009 at 12:42 pm
Wait, Joel, can you explain that last comment? I can put 24 random people in the room and someone the probability that 2 of them share the same birthday (and I assume we’re talking same month/date but not year) is 50-50? I can’t think of a way that is possible.
posted by Knobber on 10-1-2009 at 1:48 pm
being the tabletop game nerd that I am, I converted the colors to pips (numbers help me out better, lol). Doing that, I determined that one blue would be 6 on both dice, and the extra blue on the second would be a 5. With that in place, I decided that I needed to figure out how many times I would have to roll to generate a sum of at least 11. Those odds are 2/36, or 1/13. So, I found that I have the chance to roll that sum at least once in 13 times. Granted, I have a feeling that 13 rolls only MEETS 50%, not exceeds it. I think the 14th roll will EXCEED the 50%.
posted by Steven on 10-1-2009 at 3:22 pm
Wait a minute on mine…rolling 11 would be 3/36, down to a 1/13. My mistake, I pressed the wrong key.
posted by Steven on 10-1-2009 at 3:26 pm
HA! I see where I made my mistake. The sum of 11 can only be achieved with those odds if the 5 on the other die exists as well, but it doesn’t. So the odds would tighten to 1/18.
I’m still confused as to how I answered correctly with my twisted math.
(sorry for triple post)
posted by Steven on 10-1-2009 at 3:31 pm
To Knobber:
Solve it backwards – what is the probability that no one in the room shares a birthday?
For two people, P(2) = 365/366.
For three people, P(3) = P(2)*364/366 (i.e. probability that the 1st and second do not share a birthday times that probability that the third does not share a birthday with either of the first 2)
Continuing, P(N) = P(N-1)*(366-(N-1))/366
P(22) = 0.4936, so when you arr e in a room with 22 other people (23 total), the odds are better than 50-50 that 2 share a birthday.
This is a great sucker bet when in a room with more than 25 people — ask if anyone will give you odds that at least two people in the room share a birthday. If you can get better than even odds, you will make money more times than not! If there are 40 people, the odds are 10 to 1 in your favor and with 60 they are more than 99 to 1!
posted by Ric on 10-1-2009 at 4:27 pm
Now I’m dying to know what the original answer was.
(I got it right, but I have a BA in statistics. They’d take my degree back if I missed it.)
posted by Boofus on 10-1-2009 at 5:03 pm
To Knobber:
To follow on Ric’s excellent mathematical explanation, a way to visualize this for the non-statistically inclined is as follows:
With two people in the room, the second person’s birtday can be 1 day to be the same as someone else’s.
With three people in the room, the second person’s birtday can be 2 days to be the same as someone else’s.
This trend continues all the way up to the 23rd and 24th people, who individually can have their birthdays on any of 22 and 23 days, respectively, to match someone else in the room. Once you get numbers this high, for each person you add, you have roughly a 1/15 chance they match someone else for each person you add. Enough 1/15 chances, and you’re likely to hit a match. 24 happens to be that golden number.
posted by Matt on 10-1-2009 at 6:20 pm
Alex,
To answer your questions about use of the logarithims -
You want to solve the following equation for x:
(17/18)^x = .5
That is, the probability of not rolling blue on ‘x’ independent rolls is equal to 50%.
The way to do this algebraically, is to take the log of both sides; and naturally, natural logs are the easiest to use.
ln(a^b) = b * ln(a)
Thus:
(17/18)^x = .5
x * ln(17/18) = ln(.5)
x = ln(.5) / ln (17/18) = 12.13
posted by Ben on 10-1-2009 at 6:24 pm
That natural log statement just reminds me of the horrible joke from BC Calculus.
What’s the integral of 1/[cabin] d[cabin]?
Natural log cabin.
posted by Matt on 10-1-2009 at 6:52 pm