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Computer Science 18 Online
OpenStudy (anonymous):

Who here knows how to graph on a computer the rolls of 2 dice and how many rolls go by until a 7? And the dice rolls are determined randomly by the computer. Does anyone know how to do this? I graphed it by hand which is time consuming to get an accurate graph and I don't accurately remember how the graph went but it was something like this: 1st roll: 9%, 2nd roll 8% third roll 7percent 4th roll 6% 5th roll 5% 6th roll 6% 7th roll 7% and from the 8th roll to the 40the it proceeds to be an exponential curve such that the 30th to 50th roll approach 99.9% probability of a 7.

OpenStudy (anonymous):

For one roll, probability(P) of rolling a seven is 1/6, and the P of rolling a non-seven is 5/6. For success on the second roll, you must roll one non-seven, followed by one seven: (5/6)*(1/6). For success on any subsequent single roll “n”: (5/6)^(n-1) * (1/6). To compute how many rolls to hit a seven, you sum the P of all single rolls, from roll 1 to roll n. This gives you your limit function, approaching 1. e.g. Chance of a seven on exactly the 12th roll: 0.02243 – Chance of a seven in twelve rolls: 0.88784 You can build that table out easily in a spreadsheet, then chart it to see the function.

OpenStudy (anonymous):

Impossible, it would be interesting to see the graph. I think you can start counting on the roll following a 7, another 7 could be the first.

OpenStudy (anonymous):

Do I get a Nobel prize?

OpenStudy (anonymous):

I had to draw it out to figure it out, here's a pic of what mine looks like:

OpenStudy (anonymous):

Impossible.

OpenStudy (anonymous):

Break the bank at the casino before everyone else does.

OpenStudy (anonymous):

How can past dice rolls affect the present? Quantum insanity.

OpenStudy (anonymous):

No. Nothing ever affects your current throw except loaded dice. Every throw, by itself, is a 1 in 6 chance. (six combinations that total 7, out of a set of 36 possible combinations.) The chance that this set of rolls succeeds becomes more probable as you make each throw. I didn't say what I meant to, in my first post. Disregard that muck I said about "chance of the current roll." You do still need to sum the series (5/6)^(n-1)*(1/6) for n = 1 to #rolls, in order to compute your percentage, as far as I can tell. Chance of throwing a single roll, hitting seven = 1 in 6. (16.6%) (not sure where 9% came from?...) Chance of throwing two rolls, and odds of a seven are 1 in 3.27. (30.6%) Three rolls, and odds of a seven are 1 in 2.37. etc... Running my chart out to 50 rolls does give you about a 1 in 1.0001 chance (99.989%) BUT, remember, that's how many rolls it takes to hit a *single* seven, not a run of sevens. You're taking multiple tries, for a single result. -And rolling a seven is good odds, to begin with. Four tries at a 1-in-6 shot, and you're right at even money. You hit 99.126% probability at 26 tries. A run of sevens would be totally different. Rolling twice, and hitting two sevens, would be a 1 in 36 chance (2.78%) and it just gets worse from there. It's not amazing that people don't break the bank. Chances are 2 in 3 that you'll shoot a point, and then seven's a loser. Ouch. Sorry, First post, I didn't know about the Medal system on this website. Thought you were joking about a Nobel Prize.

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