tautology
Joined: Jan 30, 2004
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 Posted:
Jan 04, 2006 - 09:41 |
 
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| Quote: |
I did it like this...
Chance of throwing an accurate pass that is caught (no reroll needed) PLUS
chance of throwing an inaccurate pass that is caught (no reroll needed) PLUS
chance of throwing inaccur. pass that is rerolled to an accurrate pass and caught PLUS
chance of throwing accurate pass that is dropped and rerolled to be caught PLUS
chance of throwing inaccur. pass that is dropped and then rerolled to be caught PLUS
chance of fumbling then rerolling to an accurate pass that is caught PLUS
chance of fumbling then rerolling to an inaccurate pass that is caught
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I suspect that you added in a few too many factors.
For instance,"chance of throwing inaccur. pass that is dropped and then rerolled to be caught" should actually be "chance of throwing inaccur. pass that is REROLLED AND STILL INACCURATE (and then) dropped"
You should exclude any chance of "throwing inaccur. pass that is dropped and then rerolled to be caught" as it is mutually exclusive with the possibility of throwing and Inaccurate pass and rerolling it, which is ALWAYS the best chance of success in this case.
A full possibilty tree will need to include all possibilities of failure as well as success, and is quite unwieldy in this case.
Best to simply calculate success chances, but remember that a reroll can only be used once, and assume that it will be used at the first chance of failure. |
Last edited by tautology on %b %04, %2006 - %09:%Jan; edited 1 time in total |
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tobardin
Joined: Apr 08, 2005
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| Posted:
Jan 04, 2006 - 09:44 |
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Arrrg, was nonsense.....
Correction: You should forget the 50% for RR.
Mistake is because i used a probability tree to calculate it.
Sorry for this Postflood |
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Vicimus
Joined: Nov 16, 2005
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| Posted:
Jan 04, 2006 - 09:53 |
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Hmmm. To my understanding probability is worked out by boolean ANDs being represented as multiplication and boolean ORs being represented as sums.
So the chance of rolling a 4 on 2 dice is:-
1/6 (chance of a 3) * 1/6 (chance of a 1) PLUS
1/6 (chance of a 2) * 1/6 (chance of a 2) PLUS
1/6 (chance of a 1) * 1/6 (chance of a 3)
= ~0.278 + 0.278 + 0.278 = ~ 8% chance. |
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Vicimus
Joined: Nov 16, 2005
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| Posted:
Jan 04, 2006 - 10:02 |
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| tautology wrote: | | Quote: |
I did it like this...
Chance of throwing an accurate pass that is caught (no reroll needed) PLUS
chance of throwing an inaccurate pass that is caught (no reroll needed) PLUS
chance of throwing inaccur. pass that is rerolled to an accurrate pass and caught PLUS
chance of throwing accurate pass that is dropped and rerolled to be caught PLUS
chance of throwing inaccur. pass that is dropped and then rerolled to be caught PLUS
chance of fumbling then rerolling to an accurate pass that is caught PLUS
chance of fumbling then rerolling to an inaccurate pass that is caught
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I suspect that you added in a few too many factors.
For instance,"chance of throwing inaccur. pass that is dropped and then rerolled to be caught" should actually be "chance of throwing inaccur. pass that is REROLLED AND STILL INACCURATE (and then) dropped"
You should exclude any chance of "throwing inaccur. pass that is dropped and then rerolled to be caught" as it is mutually exclusive with the possibility of throwing and Inaccurate pass and rerolling it, which is ALWAYS the best chance of success in this case.
A full possibilty tree will need to include all possibilities of failure as well as success, and is quite unwieldy in this case.
Best to simply calculate success chances, but remember that a reroll can only be used once, and assume that it will be used at the first chance of failure. |
You might be onto something here ... I dunno. My brain is exploding. The problem with assuming that it will be used at the first chance of failure is that an innaccurate pass is not failure. The result can still succeed, the probability is just very low, but it still adds up.
| Quote: | | For instance,"chance of throwing inaccur. pass that is dropped and then rerolled to be caught" should actually be "chance of throwing inaccur. pass that is REROLLED AND STILL INACCURATE (and then) dropped" |
You can't do that. The aim is to add the probability off all possible avenues of success. Obviously, that is failure so you can't put it on the "success" pile. |
Last edited by Vicimus on %b %04, %2006 - %10:%Jan; edited 1 time in total |
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tautology
Joined: Jan 30, 2004
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| Posted:
Jan 04, 2006 - 10:04 |
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| Quote: |
So the chance of rolling a 4 on 2 dice is:-
1/6 (chance of a 3) * 1/6 (chance of a 1) PLUS
1/6 (chance of a 2) * 1/6 (chance of a 2) PLUS
1/6 (chance of a 1) * 1/6 (chance of a 3)
= ~0.278 + 0.278 + 0.278 = ~ 8% chance.
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Quite correct, however the chance of rolling 4 or worse would need to include 1,1 1,2 and 2,1 as well, for a total of 16.67% |
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tobardin
Joined: Apr 08, 2005
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| Posted:
Jan 04, 2006 - 10:05 |
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It is as tautology stated. He used boolean ANDs and ORs and is completely right. But that's the problem in probabilities. You may make mistakes or overlook something very fast.
By the way, in no calculation we considered nuffle!!!
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tautology
Joined: Jan 30, 2004
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| Posted:
Jan 04, 2006 - 10:08 |
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| Quote: |
You might be onto something here ... I dunno. My brain is exploding. The problem with assuming that it will be used at the first chance of failure is that an innaccurate pass is not failure. The result can still succeed, the probability is just very low, but it still adds up.
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If you do not reroll an inaccurate pass, then you have NO CHANCE of the result " throw inccurate pass, reroll it to become accurate, catch the pass"
You cannot add the possibilities of both rerolling and inaccurate pass and not rerolling an inaccurate pass, as they are mutually exclusive. You have to choose a path
Alternatively, you could treat all decisions as random, in which case you need to apply the full weight of all preceding probabilites at EACH branch in the tree...this gets messy in a hurry. |
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Vicimus
Joined: Nov 16, 2005
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| Posted:
Jan 04, 2006 - 10:09 |
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There's one thing here's the hurting my brain.
In that example I gave, the result works out to 8.34% using that method.
If you work out the percentages by saying there is 1 chance in 12 of getting a 4 = (1/12)*100 you get 8.34% too, so it seems to go together perfectly. The problem is that it's not 1 chance in 12, it's 1 chance in 11. You can't roll a 1 with 2 dice. Can someone please explain why this math works so well when it shouldn't? Keep in mind I'm not good at math. |
Last edited by Vicimus on %b %04, %2006 - %10:%Jan; edited 1 time in total |
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Vicimus
Joined: Nov 16, 2005
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| Posted:
Jan 04, 2006 - 10:11 |
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| tautology wrote: | | Quote: |
You might be onto something here ... I dunno. My brain is exploding. The problem with assuming that it will be used at the first chance of failure is that an innaccurate pass is not failure. The result can still succeed, the probability is just very low, but it still adds up.
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If you do not reroll an inaccurate pass, then you have NO CHANCE of the result " throw inccurate pass, reroll it to become accurate, catch the pass"
You cannot add the possibilities of both rerolling and inaccurate pass and not rerolling an inaccurate pass, as they are mutually exclusive. You have to choose a path
Alternatively, you could treat all decisions as random, in which case you need to apply the full weight of all preceding probabilites at EACH branch in the tree...this gets messy in a hurry. |
Woohoo! it clicked! Thankyou! I will reconfigure that statistic before it leaves my brain. |
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macike
Joined: Jun 25, 2005
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| Posted:
Jan 04, 2006 - 10:26 |
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| Vicimus wrote: | | Second, there are some interesting statistics I figured out using probability mathematics, check it out... |
Once again. You mess up (pseudo)randomness with statistics.
A number generator must pass some statistic test to be a random number generator. So with your page you try to prove something that was proved a long time ago.
Your page will not change one basic thing - it is pseudorandom number generator. What does it mean in practice was explained by Rynkky. |
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Vicimus
Joined: Nov 16, 2005
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| Posted:
Jan 04, 2006 - 10:34 |
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Okay so here's my new way of thinking...
Throw accurate pass, catch it, plus
(3/6 * 4/6) +
throw accurate pass, drop it, reroll and catch it, plus
(3/6 * 2/6 * 4/6)
throw innaccurate pass, reroll to become accurate, catch it, plus
(2/6 * 3/6 * 4/6)+
throw innacurate pass, reroll fails, ball scatters nicely, catch it, plus
(2/6 * 2/6 * 0.625 * 3/6)
throw fumble, reroll to become accurate, and catch it, plus
(1/6 * 3/6 * 4/6)+
throw fumble, reroll it to become inaccurate, ball scatters nicely, catch it
(1/6 * 2/6 * 0.625 * 3/6)
= (0.5 * 0.67) +
(0.5 * 0.34 * 0.67) +
(0.34 * 0.5 * 0.67) +
(0.34 * 0.34 * 0.625 * 0.5) +
(0.17 * 0.5 * 0.67) +
(0.17 * 0.34 * 0.625 * 0.5)
= .335 + .114 + .114 + .036 + .057 + .018
= 67.4% chance of success.
Phew! |
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Vicimus
Joined: Nov 16, 2005
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| Posted:
Jan 04, 2006 - 10:39 |
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| macike wrote: | | Vicimus wrote: | | Second, there are some interesting statistics I figured out using probability mathematics, check it out... |
Once again. You mess up (pseudo)randomness with statistics.
A number generator must pass some statistic test to be a random number generator. So with your page you try to prove something that was proved a long time ago.
Your page will not change one basic thing - it is pseudorandom number generator. What does it mean in practice was explained by Rynkky. |
what!? The only thing here that is not making sense is you.
The only thing I ever tried to prove is that the Java Random Number generator is not faulty and will not cheat you in a game on Fumbbl. I am having a lot of trouble understanding your what is you're even trying to disagree with and your logic to why you disagree.
The little bit about probability statistics is just general interest. It's not an attempt to prove anything about psuedo random numbers. It's just a bunch of facts (assuming my math is correct). |
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koadah
Joined: Mar 30, 2005
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Jan 04, 2006 - 11:13 |
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I've got to ask you, does javascript use the same RNG as java?
I have always believed that javascript has nothing at all to do with java other than the name.
Is this meaningless unless you put your code into an applet?
Please excuse me if I am talking from outdated knowledge.
Is the luck meter meaningless. Yes probably.
Four POWs on the first 4 blocks of the game taking four players off the field is bad luck. Especially against Dwarves who are going to take their full 8 turns to batter your remaining players before you can get any reserves on.
The luck meter will probably still say that you had good luck at the end of the game because of all the dodges you made trying to keep your players alive.
The luck meter is just there to give us something to moan about.
Is anyone really suggesting SkiJunkie gives up any more of his free time to rip out tha java RNG and build a better one from scratch? |
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Buur
Joined: Apr 29, 2004
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| Posted:
Jan 04, 2006 - 11:15 |
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Pie.... Pie pie pie..... Want pie now..... Pie ... PIE! PIE! PPPPPPPIIIIIIIIEEEEEEE!!!!!!!!
-Buur |
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For most people, reason is nothing but their own believes. |
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Vicimus
Joined: Nov 16, 2005
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| Posted:
Jan 04, 2006 - 11:28 |
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| koadah wrote: | | I've got to ask you, does javascript use the same RNG as java? |
I have no idea. JavaScript was invented by the same people who made Java, although they are two different things really. JavaScript is interpreted by whatever browser you are using. The best way to really test the Java RNG would be to look at that page through a java browser. But that's just getting silly. I can guarantee now there will not be any discernable difference in the results. Any random number generator from any programming language is going to be absolutely fine and unfaultable for our purposes. Hell, even for online gambling purposes.
| Quote: | | Is anyone really suggesting SkiJunkie gives up any more of his free time to rip out tha java RNG and build a better one from scratch? |
Why on earth? With a set of 500,000 numbers I think I've shown that any variation in what is generated randomly is non-existent or statistcly insignificant to the nth degree.
If a set of half a million simulated dice rolls is not enough to show any sort of pattern whatsoever, who could properly suggest that a few hundred in a game of fummbl is going to? It's absurd. |
Last edited by Vicimus on %b %04, %2006 - %11:%Jan; edited 1 time in total |
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