
Help me
could anyone help me with this question? In an experiment, you have four attempts to throw 3 sixes. First all 3 dice are thrown and any sixes are put aside. Any dice without a six (if any) are thrown a second time and once again any sixes are put aside. Once again any sixes are put aside and the remaining dice(if any) are thrown. The remaining dice (if any) are thrown. What is the probability of having a six on each of the 3 dice after the four throws have been completed?
 Ricky
 Moderator
Re: Help me
Seems complicated, so let's just do it for once dice first:
The probability of rolling a 6 is 1/6. So there is a 1/6 chance that you will be done your first time:
1/6
But there is a 5/6 chance that you will have to roll again. Then there is a 1/6 what when you roll again, it will be a 6, and you are done:
1/6 + 5/6*1/6
Or an 11/36 chance that you are done after two rolls. But there is a 25/36 chance that you won't be, and then a 1/6 chance that you get a 6 when you roll again.
1/6 + 5/6*1/6 + 25/36*1/6 = 11/36 + 25/216 = 91/216
Finally, you have a 125/216 chance that you will have to roll again, and again, a 1/6 chance that you will get a 6:
1/6 + 5/6*1/6 + 25/36*1/6 + 125/216*1/6 = 91/216 + 125/1296 = 546/1296
Or a 42% chance for 1 die. You will find that using this method, the probability will approach, but never reach, 1.
But you need this for 3 dice. So that's:
546/1296 * 546/1296 * 546/1296 = 162,771,336 / 2,176,782,336
Or a 7.478% chance.
Someone please check this. At least it sounds right...
"In the real world, this would be a problem. But in mathematics, we can just define a place where this problem doesn't exist. So we'll go ahead and do that now..."
Re: Help me
Ricky, you absolute genius. I was getting suffocated by stacks of dependent events, but your method just separates out all of the dice and makes working out the answer much simpler. And it's right, as far as I can tell.
Why did the vector cross the road? It wanted to be normal.
 Ricky
 Moderator
Re: Help me
Took a few minutes and made a C++ program to run a little numerical analysis:
Runs: 1000 Hits: 64 Precent: 0.064
Runs: 10000 Hits: 785 Precent: 0.0785
Runs: 100000 Hits: 7566 Precent: 0.07566
Runs: 100000000 Hits: 7478819 Precent: 0.0747882
Source code:
#include <iostream> #include <time.h> #include <stdlib.h>
using namespace std;
int main() { srand(time(NULL));
const long NUM_RUNS = 100000; long numHits = 0;
for (long x = 0; x < NUM_RUNS; x++) { int die1 = 0, die2 = 0, die3 = 0; for (int y = 0; y < 3; y++) { if (die1 != 6) die1 = rand() % 6 + 1; if (die2 != 6) die2 = rand() % 6 + 1; if (die3 != 6) die3 = rand() % 6 + 1; } if (die1 == 6 && die2 == 6 && die3 == 6) numHits++; }
cout << "Runs: " << NUM_RUNS << endl; cout << "Hits: " << numHits << endl; cout << "Precent: " << (double)numHits / NUM_RUNS << endl;
return 0; }
Last edited by Ricky (20051215 07:55:18)
"In the real world, this would be a problem. But in mathematics, we can just define a place where this problem doesn't exist. So we'll go ahead and do that now..."
 ryos
 Power Member
Re: Help me
Something about that just doesn't feel quite right, but I'm not good enough with probabilities to figure it out. So, I'll toss out my thinking for review.
We can lump the three dice into one theoretical 18sided die (with 3 sixes). The probability of hitting a six on the first roll is 3/18 = 1/6. Should we wind up removing one die for the next roll, our chances are 2/12 = 1/6. For one die, of course, it's 1/6. So, as we roll along, our chances of hitting a six on one of the dice doesn't decrease.
Making final answer Ricky's first bit, or 546/1296.
El que pega primero pega dos veces.
 Ricky
 Moderator
Re: Help me
ryos, my c++ program agreed with my result to the 4th decimal place, and I'm pretty sure that's no cosmic coincidence. That being said:
"So, as we roll along, our chances of hitting a six on one of the dice doesn't decrease."
This got me at first too. But then I thought about this example:
Keep flipping a coin till you get heads.
Well, the first time I filp it, I have a 1/2 chance of getting heads. The second time I flip it, I have a 1/2 chance of getting heads. So this would be 1/2 + 1/2, no? Does this mean that I have a 100% chance of getting heads if I flip a coin twice? Obviously, that's wrong. Heck, what happends if I flip it three times? I get a 150% chance of getting heads.
The thing is, you _only_ flip a second time when the first time was a bust. How often will that happen? 1/2 the time. So it's 1/2*1/2, or 1/4.
Think about it this way. You are asked to flip a coin two times. What are the possible outcomes?
Flip 1 Flip 2
h h h t t h t t
And all of these are equally weighted. So what's the chance of flipping tails, then heads? 25%.
If you flip a coin till you get heads, what's the chance of flipping heads, and then heads? 0%
Since these are the only two possibilities of getting heads on the 2nd flip, its 25% + 0%, which is 25%.
Last edited by Ricky (20051215 21:13:56)
"In the real world, this would be a problem. But in mathematics, we can just define a place where this problem doesn't exist. So we'll go ahead and do that now..."
 ryos
 Power Member
Re: Help me
Agreed. I was talking more about this:
But you need this for 3 dice. So that's:
546/1296 * 546/1296 * 546/1296 = 162,771,336 / 2,176,782,336
Or a 7.478% chance.
It's really counterintuitive that the probability goes down when you play with three dice. But what I missed is that it's the probability of finishing with all of them having turned up sixes, which makes sense. 8O)
El que pega primero pega dos veces.
 Ricky
 Moderator
Re: Help me
Ah, I see now. I mostly invented this method on the fly, and the part which I was talking about was the part which I had the most trouble understanding. So I guess I just thought that you would have the same.
"In the real world, this would be a problem. But in mathematics, we can just define a place where this problem doesn't exist. So we'll go ahead and do that now..."
