simple question

[BruceZ]

Correct, pudley4 and Bozeman. The 1st roll is most likely to produce the first 6, with probability 5/36. All of the other rolls have lower probability because you have to NOT get a 6 on all the rolls before it AND then get a 6. That's harder. The probability for getting the first 6 on the Nth roll is (5/36)*(31/36)^(N-1) <= 5/36, with = for N=1.

Now, next question. What is the average number of rolls necessary to throw a 6? There are 2 ways to solve this problem that I know of. One requires a little knowledge of calculus. The other requires only logic.

[crockpot]

36/5.

[Wake up CALL]

[ QUOTE ]
Correct, pudley4 and Bozeman. The 1st roll is most likely to produce the first 6, with probability 5/36. All of the other rolls have lower probability because you have to NOT get a 6 on all the rolls before it AND then get a 6. That's harder. The probability for getting the first 6 on the Nth roll is (5/36)*(31/36)^(N-1) <= 5/36, with = for N=1.

Now, next question. What is the average number of rolls necessary to throw a 6? There are 2 ways to solve this problem that I know of. One requires a little knowledge of calculus. The other requires only logic.

[/ QUOTE ]

Sorry BruceZ, I know disagreeing with you can be hazardous to one's health but if you are going to ask this exact question "Now, if you roll until you get a 6, on which roll are you most likely to get it? Isn't the correct answer the last roll since you will then stop rolling?

[Copernicus]

36/5 is correct, but what is the calculus approach? As a discrete function the only approach I know is to sum the infinite series.

I also agree that "last" is a better answer the way the question was originally phrased. While the probability of rolling the first 6 is highest on the first roll, you roll it on the last roll 100% of the time, including when the first roll is the last roll. [img]/images/graemlins/crazy.gif[/img]

[BruceZ]

Summing that particular infinite series requires calculus if you don't already know how to sum it.

How about showing some work for either method?

[BruceZ]

I'll give you the same answer my old abstract algebra professors once gave me after I had submitted some dumbass answer. He said "if all I had found was a trivial answer like that one, I certainly would have looked around to see if I could have found a better answer". In other words, taking the easy way out may work in real life, but it doesn't get you very far in mathematics.

[Wake up CALL]

[ QUOTE ]
I'll give you the same answer my old abstract algebra professors once gave me after I had submitted some dumbass answer. He said "if all I had found was a trivial answer like that one, I certainly would have looked around to see if I could have found a better answer". In other words, taking the easy way out may work in real life, but it doesn't get you very far in mathematics.

[/ QUOTE ]

I don't disagree with your professor BruceZ but I imagine he worded his questions better.

[squiffy]

Wait, the probability of rolling 6 is the same for every roll. So how can the probability be greater on one roll than another?

[BruceZ]

Nothing wrong with the wording of the question in either case, other than the fact that they afforded both an insightful answer and an unthinking answer. Those who chose to answer it insightfully hopefully learned something. Those who took the easy way out, as you did, only cheated themselves.

The roll you get the 6 is obviously the last roll. That was stated in the problem when we said that we roll until we get the 6. Your answer amounted to telling me what was already stated. Here's your F, now go stand in the corner. [img]/images/graemlins/grin.gif[/img]

[Nottom]

Because on the second roll and beyond, you have to have already not rolled a 6.