yellow concept car???
- Thread starter Ceilrux1
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SportWagon
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In my personal experience, winning the yellow Concept car was a lot rarer than winning the purple one.
- 10,544
- Columbia, MD.
- Parnelli_Bones
Yea, I never saw the yellow one, and I won the UK vs. US races several times.
SportWagon
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- Punkeydoodles Corners, Ontario, Canada. (aka GT2t
In one game it took me 12 tries to win any Concept Car, and it was yellow. The next 2 races won me two more yellow Concept Cars, and then the next one finally won the purple Concept Car, completing the set of four prizes for the UKvsUS series in that game.
It seemed to me that that game would get stuck in ruts of winning some particular prize for some event. But considering all my games overall, the purple Concept Car seemed more common than the yellow overall. And it sort of seemed like the "rare" prizes were more common early in the games than later.
Viewing a collection of results from different games, I got...
12 green RX-7
9 purple Concept Car
8 blue RX-7
3 yellow Concept Car
It seemed to me that that game would get stuck in ruts of winning some particular prize for some event. But considering all my games overall, the purple Concept Car seemed more common than the yellow overall. And it sort of seemed like the "rare" prizes were more common early in the games than later.
Code:
80-85 [R]Cerbera UKvsUS sw45to24,20,19,9,8 blueRX7(1/4)i
218-223 [R]RT/10 UKvsUS sw45to24,21,19,11,5 blueRX7(dup)a
224-229 [R]RT/10 UKvsUS q6SS11 1st45to25 blueRX7(dup)a
230-235 [R]RT/10 UKvsUS sw45to23,20,18,13,6 blueRX7(dup)a
236-241 [R]RT/10 UKvsUS sw45to28 blueRX7(dup)a???
248-253 [R]CerberaLM UKvsUS sw45to22,22,20,11,5 blueRX7(dup)i
272-277 [R]CerberaNOT UKvsUS sw45to26 greenRX7(2/4)i+a
278-283 [R]CerberaNOT UKvsUS sw45to28 greenRX7(dup)a
296-301 [R]RT/10 UKvsUS sw45to24,24,14,13,5 greenRX7(dup)a
309-314 [R]RT/10 UKvsUS sw45to25,19,19,11,6 blueRX7(dup)a
315-320 [R]RT/10 UKvsUS sw45to24,24,14,13,6 greenRX7(dup)a
321-326 [R]RT/10 UKvsUS sw45to26,22,14,12,6 yellowConcept(3/4)a
327-332 [R]RT/10 UKvsUS sw45to28 yellowConcept(dup)a
333-338 [R]RT/10 UKvsUS sw45to26,19,18,12,5 yellowConcept(dup)a
347-352 [R]RT/10 UKvsUS sw45to22,20,17,14,7 purpleConcept(4/4*)a!Viewing a collection of results from different games, I got...
12 green RX-7
9 purple Concept Car
8 blue RX-7
3 yellow Concept Car
LeGeNd-1
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When you toss a coin, contrary to your intuition, you won't get HTHTHT (H=heads, T=tails). You'll get something like 5H, then 7T, and so on. Just the weird way random probability works I guess.
And I thought the A-Spec RX7 LM is green and purple? Your TV's colour setting might need some adjusting there
SportWagon
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- Punkeydoodles Corners, Ontario, Canada. (aka GT2t
Actual experiment. Reasonably equivalent.
I flipped 5 American quarters together, and noted the number of heads.
I kept going until I got 5 of a kind. (0 heads in this case).
And then stopped.
These are the actual results (number of heads) I observed.
1 1
2 1
3 3
4 3
5 3
6 3
7 1
8 2
9 3
10 4
11 2
12 2
13 4
14 3
15 2
16 2
17 2
18 2
19 0
Slightly more than the "expected" 16.
P(0H) = (1/2)^5 = 1/32
P(0T) = (1/2)^5 = 1/32
The probability of either is the sum, or 1/16.
If considered as sequential throws, each sequence will have 1/32 probability.
HHHHH
HHHHT
HHHTH
HHHTT
HHTHH
HHTHT
HHTTH
HHTTT
HTHHH
HTHHT
HTHTH
HTHTT
HTTHH
HTTHT
HTTTH
HTTTT
THHHH
THHHT
THHTH
THHTT
THTHH
THTHT
THTTH
THTTT
TTHHH
TTHHT
TTHTH
TTHTT
TTTHH
TTTHT
TTTTH
TTTTT
Or that is to say, that equivalently diagrams the total number of possible positions (T or H up) of my five quarters, and each position is assumed to be equally likely. You can count from the above that there are...
0H - 1 time
1H - 5 times
2H - 10 times
3H - 10 times
4H - 5 times
5H - 1 time
(At this point, by counting without regard to order, we change possibly order-significant results into results where order is not significant--as in my actual experiment).
So the most likely result is 2 or 3 heads, each with probability 10/32.
Or about 1/3 for each, or 2/3 for either.
5/32 or about 1/6 for 1 or 4 heads.
1/32 for 0 or 5 heads.
In fact, I observed.
7 results of 2 - 7/19 or 11.79/32 - versus 10/32
6 results of 3 - 6/19 or 10.10/32 - versus 10/32
3 results of 1 - 3/19 or 5.05/32 - versus 5/32
2 results of 4 - 2/19 or 3.37/32 - versus 5/32
1 results of 0 - 1/19 or 1.68/32 - versus 1/32
0 results of 6 - 0/19 or 0/32 - versus 1/32
which is not far off, especially considering I stopped the experiment on the first 5-of-a-kind. (Note you do not expect your observed results to exactly match the perfect probability distribution, but you do not expect them to be consistently off, either).
I flipped 5 American quarters together, and noted the number of heads.
I kept going until I got 5 of a kind. (0 heads in this case).
And then stopped.
These are the actual results (number of heads) I observed.
1 1
2 1
3 3
4 3
5 3
6 3
7 1
8 2
9 3
10 4
11 2
12 2
13 4
14 3
15 2
16 2
17 2
18 2
19 0
Slightly more than the "expected" 16.
P(0H) = (1/2)^5 = 1/32
P(0T) = (1/2)^5 = 1/32
The probability of either is the sum, or 1/16.
If considered as sequential throws, each sequence will have 1/32 probability.
HHHHH
HHHHT
HHHTH
HHHTT
HHTHH
HHTHT
HHTTH
HHTTT
HTHHH
HTHHT
HTHTH
HTHTT
HTTHH
HTTHT
HTTTH
HTTTT
THHHH
THHHT
THHTH
THHTT
THTHH
THTHT
THTTH
THTTT
TTHHH
TTHHT
TTHTH
TTHTT
TTTHH
TTTHT
TTTTH
TTTTT
Or that is to say, that equivalently diagrams the total number of possible positions (T or H up) of my five quarters, and each position is assumed to be equally likely. You can count from the above that there are...
0H - 1 time
1H - 5 times
2H - 10 times
3H - 10 times
4H - 5 times
5H - 1 time
(At this point, by counting without regard to order, we change possibly order-significant results into results where order is not significant--as in my actual experiment).
So the most likely result is 2 or 3 heads, each with probability 10/32.
Or about 1/3 for each, or 2/3 for either.
5/32 or about 1/6 for 1 or 4 heads.
1/32 for 0 or 5 heads.
In fact, I observed.
7 results of 2 - 7/19 or 11.79/32 - versus 10/32
6 results of 3 - 6/19 or 10.10/32 - versus 10/32
3 results of 1 - 3/19 or 5.05/32 - versus 5/32
2 results of 4 - 2/19 or 3.37/32 - versus 5/32
1 results of 0 - 1/19 or 1.68/32 - versus 1/32
0 results of 6 - 0/19 or 0/32 - versus 1/32
which is not far off, especially considering I stopped the experiment on the first 5-of-a-kind. (Note you do not expect your observed results to exactly match the perfect probability distribution, but you do not expect them to be consistently off, either).
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- 6,694
- Dallas, TX
I remember winning this car years back. It was one of the best cars I have ever driven in the Gran Turismo series. The yellow paint is remarkably stunning to my eyes that my eyes see. If Dodge made the Concept Car into a street car, then I would definitely purchase it no matter how hard the procedure of obtaining the car will possibly be.
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Fine. You'll be sent to prison for 70 years, and then you only get to drive on roads with a limit of 25M/H for 10 minutes.
Doesn"t sound fun now, huh?
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"If" Dodge made the Concept car into a street car.
LeGeNd-1
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Wow. I didn't expect you to actually try it
You should get a medal for that 👍 And I know that eventually the distribution will come close to the probability distribution. I'm merely saying that during the actual run itself you might not necessarily get HTHTHT as some might expect if a fair coin is tossed. 5T and 7H might be too exaggerated though. HHHTTTHHTTTTHHH might be a more realistic sequence.
SportWagon
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HHHTTTHHTTTTHHH
as a precise order-dependent event, is, strictly-speaking, just as likely or unlikely as
HHHHHHHHHHHHHHH or TTTTTTTTTTTTTTT
But characterized only as having 8H and 7T in 15 throws, it represents a lot more likely event than the other two. (I.e. [I claim] there is no characterization of 15H/15 or 15T/15 which is a likely event).
Similarly HTHTHTHTHTHTHTH, THTHTHTHTHTHTHT are each just as unlikely as
HHHHHHHHHHHHHHH or TTTTTTTTTTTTTTT
(But if characterized as 8H and 7T or 7H and 8H they are part of that relatively likely set).
But similarly, as the length of a run of a single result, e.g. HHHH gets longer, that becomes more unlikely. But, as pairs, HH HT TH and TT are all equally likely. Reflecting the fact that subsequent results are not (in a truly random situation) affected by previous events. When you go to three results, a mixed result does become more likely, however. (6 mixed to 2 with the same).
HHH,HHT,HTH,HTT,THH,THT,TTH,TTT
A quick scan of my Tuned series results suggested that the Trueno is a more likely prize than the Skyline. (The Trueno occurs more often, and has longer runs).(That's the only series I repeated and recorded a great number of times, I believe). I might post more details later.
as a precise order-dependent event, is, strictly-speaking, just as likely or unlikely as
HHHHHHHHHHHHHHH or TTTTTTTTTTTTTTT
But characterized only as having 8H and 7T in 15 throws, it represents a lot more likely event than the other two. (I.e. [I claim] there is no characterization of 15H/15 or 15T/15 which is a likely event).
Similarly HTHTHTHTHTHTHTH, THTHTHTHTHTHTHT are each just as unlikely as
HHHHHHHHHHHHHHH or TTTTTTTTTTTTTTT
(But if characterized as 8H and 7T or 7H and 8H they are part of that relatively likely set).
But similarly, as the length of a run of a single result, e.g. HHHH gets longer, that becomes more unlikely. But, as pairs, HH HT TH and TT are all equally likely. Reflecting the fact that subsequent results are not (in a truly random situation) affected by previous events. When you go to three results, a mixed result does become more likely, however. (6 mixed to 2 with the same).
HHH,HHT,HTH,HTT,THH,THT,TTH,TTT
A quick scan of my Tuned series results suggested that the Trueno is a more likely prize than the Skyline. (The Trueno occurs more often, and has longer runs).(That's the only series I repeated and recorded a great number of times, I believe). I might post more details later.
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SportWagon
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If I recorded the AI fields, it was not in the computer files. I do have accompanying paper records of this stuff in 3-ring binders.
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Well can you post these papers?
LeGeNd-1
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Yup. I get this alright. It's kinda hard to explain what I mean lol. I think you get the gist though. Basically it's more unlikely (practically) to get perfectly alternating results, and you're more likely to get straight repetitions when getting something randomly. Which is the case with both the coin and GT's prize cars.
I think we're taking this thread off topic though. So we'll stop here
SportWagon
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Our discussion relates tangentially to any attempt to prove that the yellow Concept car is actually rarer than any of the other 3 prizes for the USvsUK series.
I.e. it relates to how difficult, in practice, it is to get the yellow Concept Car from where you get it.
I.e. it relates to how difficult, in practice, it is to get the yellow Concept Car from where you get it.
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Something was relayed to me with regards to AI fields and the prize cars awarded. I haven't had time to test it properly, but suffice to say that, on the one occasion the grid was as supposed for a certain prize car, I actually did get that prize car and at no other time.
It's difficult to test though, since the variables are hard to standardise. I doubt it was more than a fluke coincidence, but I'd like to be able to test it. If you have the AI fields, it may be possible.
LeGeNd-1
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That's true in a sense. But I think the OP just wants an answer on where you could get it without delving into the world of statistics and whatnot
In my experience though, both of the RX7 A-Spec LMs are much much rarer than both Copperheads (something like 1:25 ratio). The yellow and purple are awarded almost equally in my game. I'm only reciting this from memory though, so it definitely isn't as reliable compared to your notes.
- 2,583
- Kelowna, Canada
- Antonisbob
- No...
Its rare? Last night I decided to pop in GT1 in my PS3, started my new-ish save and traded a Viper GTS-R or w/e its called into my garage, did the UK vs US and won the yellow version first try.👍
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I think I may have uncovered something alarmingly simple about GT1's prize cars. It doesn't seem to tally with what SportWagon noted in his UK vs. US datasheet, but I've now observed it in three different series in my NTSC-J version.
Put simply, if you enter the series when your game days display an odd number day, you win one of the two cars. If you enter the series when your game days display an even number day, you win the other one of the two cars.
I don't have enough data to draw a solid conclusion from it yet, but I have observed it across a number of series (FF, 4WD, Lightweight). I'll be testing the hypothesis quite thoroughly in the next few days but here's the results of me testing the Lightweight Challenge:
Only one variable that I could control changed - game days.
There are other factors that I cannot rule out at this point - notably the "enemy car" theory. Note that the appearance of the Civic SiR, Civic Type-R, Integra Type-R, Levin AE86 and Trueno BZG all correspond with a CTR prize car instead of a Eunos - however, no one of them is ever-present in the line-ups. In theory there could be two "enemy cars" - no line-up without either of the Civic SiR or Integra Type-R has yet resulted in a CTR prize car. There could also be a combination of these factors.
What I'm going to attempt, therefore, is a play-through on even game days and a play-through on odd game days - using licences to skip where necessary - to determine if there's any validity. It is worth noting though that almost every series you start on an odd date will end with you starting the next series on an odd date (same with even to even), which may explain how people repeatedly do the same series back-to-back and end up with the same car every time...
Note: This is all done on the NTSC-J version so far. Please feel free to test on other versions and compare your results to mine. Results for me so far are:
FR Challenge - Odd = SilEighty; Even = S13 Silvia
FR Challenge - Odd = Alcyone SVX; Even = Lancer Evo 4
Lightweight Challenge - Odd = Eunos Roadster; Even = Civic Type-R
US vs UK - Odd = Dodge Concept; Even = RX-7 LM
Edit: Hurr. FR Challenge. Start on day 399 - SilEighty; Start on day 404 - S13 Silvia Qs...
Edit edit: Hurr hurr. US vs UK. Start on day 408 - RX-7 LM; Start on day 415 - Concept Race
Edit edit edit: Hurr hurr hurr. 4W Challenge. Start on day 421 - Alcyone SVX; Start on day 426 - Lancer Evo IV
Incidentally, the odds of this combination (ABABABAB) are 1 in 16, the same as any other combination of prize cars. The odds of me predicting it, however, are 1 in 256 (1 in 2, eight times). But I only have to be wrong once to invalidate the theory. Still, those are very, very long odds...
Put simply, if you enter the series when your game days display an odd number day, you win one of the two cars. If you enter the series when your game days display an even number day, you win the other one of the two cars.
I don't have enough data to draw a solid conclusion from it yet, but I have observed it across a number of series (FF, 4WD, Lightweight). I'll be testing the hypothesis quite thoroughly in the next few days but here's the results of me testing the Lightweight Challenge:
Code:
Start day: 357
Prize: Eunos Roadster
Colour: Lilac
User: Eunos V-Spec (27pt)
CR-X EF8 (14)
Asti RX (10)
Cyborg R (10)
Trueno AE86 (8)
Demio GL-X (6)
Start day: 362
Prize: EK Civic Type-R
Colour: Yellow
User: Eunos V-Spec (27pt)
CTR (18)
Trueno BZG (10)
Civic SiR-II (9)
Asti RX (8)
Demio GL-X (3)
Start day: 366
Prize: EK Civic Type-R
Colour: Yellow
User: Eunos V-Spec (27pt)
CR-X EF8 (18)
Civic SiR-II (10)
Asti RX (7)
Demio GL-X (7)
Trueno AE86 (6)
Start day: 370
Prize: EK Civic Type-R
Colour: Blue
User: Eunos V-Spec (27pt)
ITR (18)
Cyborg (10)
Levin AE86 (8)
CR-X EF8 (7)
Demio GL-X (5)
Start day: 374
Prize: EK Civic Type-R
Colour: Pink
User: Eunos V-Spec (27pt)
CTR (18)
Civic SiR-II (10)
Asti RX (10)
Trueno BZG (7)
Demio GL-X (3)
Start day: 390
Prize: EK Civic Type-R
Colour: Pink
User: Eunos V-Spec (27pt)
ITR (18)
CTR (11)
Trueno AE86 (9)
Asti RX (7)
Demio GL-X (3)
Start day: 395
Prize: Eunos Roadster
Colour: Gold
User: Eunos V-Spec (27pt)
CR-X EF8 (16)
Levin BZG (13)
Asti RX (10)
Starlet (6)
Demio GL-X (3)Only one variable that I could control changed - game days.
There are other factors that I cannot rule out at this point - notably the "enemy car" theory. Note that the appearance of the Civic SiR, Civic Type-R, Integra Type-R, Levin AE86 and Trueno BZG all correspond with a CTR prize car instead of a Eunos - however, no one of them is ever-present in the line-ups. In theory there could be two "enemy cars" - no line-up without either of the Civic SiR or Integra Type-R has yet resulted in a CTR prize car. There could also be a combination of these factors.
What I'm going to attempt, therefore, is a play-through on even game days and a play-through on odd game days - using licences to skip where necessary - to determine if there's any validity. It is worth noting though that almost every series you start on an odd date will end with you starting the next series on an odd date (same with even to even), which may explain how people repeatedly do the same series back-to-back and end up with the same car every time...
Note: This is all done on the NTSC-J version so far. Please feel free to test on other versions and compare your results to mine. Results for me so far are:
FR Challenge - Odd = SilEighty; Even = S13 Silvia
FR Challenge - Odd = Alcyone SVX; Even = Lancer Evo 4
Lightweight Challenge - Odd = Eunos Roadster; Even = Civic Type-R
US vs UK - Odd = Dodge Concept; Even = RX-7 LM
Edit: Hurr. FR Challenge. Start on day 399 - SilEighty; Start on day 404 - S13 Silvia Qs...
Edit edit: Hurr hurr. US vs UK. Start on day 408 - RX-7 LM; Start on day 415 - Concept Race
Edit edit edit: Hurr hurr hurr. 4W Challenge. Start on day 421 - Alcyone SVX; Start on day 426 - Lancer Evo IV
Incidentally, the odds of this combination (ABABABAB) are 1 in 16, the same as any other combination of prize cars. The odds of me predicting it, however, are 1 in 256 (1 in 2, eight times). But I only have to be wrong once to invalidate the theory. Still, those are very, very long odds...
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- 2,583
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- Antonisbob
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That is very plausable, today yet agian I raced the US vs. UK and won the Yelow concept car in hopes of getting the purple, Now I have 2 Yellows and the copper xD
SportWagon
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A question is, though, Famine whether that is always the pattern or whether other unknown factors can change it.
Can you elaborate on that. Especially your first assertion about 1 in 16.Famine
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As I say, all the variables I can control I am controlling. There is no reason to assume other "unknown" factors at this point, because that requires further entities I cannot know about - entia non sunt multiplicanda praeter necessitatem.
As for the numbers... For 4 series with two possible prize cars each, there are four combinations from each event - AA, AB, BA or BB - and four possible predictions. Since we're asserting that the way prize cars are awarded is random, each set of four combinations isn't dependant on the previous one(s), it's simply 1 in 4 for a given combination for each of the four series - 1 in 16. Just like coin tosses.
Since my predictions are dependant on each other, however, it's more like a sequence of 8 coin tosses - and indeed the odds of a given car are 1 in 2 and each prediction is founded on the last. 1 in 2 for each of eight individual predictions is 1 in 256 (and there are 256 possible combinations of prize cars for a pair in each of four series).
It's simply the difference between four discrete pairs of data and one set of eight - what car I get for US vs. UK doesn't (at least according to what I propose) depend on what car I get for FR Challenge, but what car I predict I will win for US vs. UK does depend on what car I predict I will win for FR Challenge.
At present, I've not been wrong. However I only need to be wrong once to invalidate the theory. Note that this, so far, is only valid for the NTSC-J version since that's the only one I've tested. Start an event when your game days are odd-numbered and you get one car, start an even when your game days are even-numbered and you get the other one.
As for the numbers... For 4 series with two possible prize cars each, there are four combinations from each event - AA, AB, BA or BB - and four possible predictions. Since we're asserting that the way prize cars are awarded is random, each set of four combinations isn't dependant on the previous one(s), it's simply 1 in 4 for a given combination for each of the four series - 1 in 16. Just like coin tosses.
Since my predictions are dependant on each other, however, it's more like a sequence of 8 coin tosses - and indeed the odds of a given car are 1 in 2 and each prediction is founded on the last. 1 in 2 for each of eight individual predictions is 1 in 256 (and there are 256 possible combinations of prize cars for a pair in each of four series).
It's simply the difference between four discrete pairs of data and one set of eight - what car I get for US vs. UK doesn't (at least according to what I propose) depend on what car I get for FR Challenge, but what car I predict I will win for US vs. UK does depend on what car I predict I will win for FR Challenge.
At present, I've not been wrong. However I only need to be wrong once to invalidate the theory. Note that this, so far, is only valid for the NTSC-J version since that's the only one I've tested. Start an event when your game days are odd-numbered and you get one car, start an even when your game days are even-numbered and you get the other one.
SportWagon
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Did you do all your experiments in one continuous "powered on" session?
Try saving powering off and reloading between series.
Try doing one series, exiting to Arcade Mode, doing a race (or maybe not), and returning to do the next series.
What state is your Arcade Mode in, in general?
Try losing a race now-and-then.
You need to try varying variables as well as keeping them the same.
Though true, you seem to have proven that, under some conditions, the results are not purely random.
Each single prediction AA|AB|BA|BB might seem to have a 1/16 chance of being a correct prediction, independently. The result would be one of the four, and your guess would be one of the four. Thinking hastily, your chance of guessing one pair correctly might therefore seem to be 1/4 * 1/4 = 1/16. But, in fact, if you enumerate, there are 16 result/prediction combinations, of which four represent you being correct. Which is an elaborate way to confirm intuition. (If there are four equally likely results and you guess, you have a 1 in 4 chance of being correct).
However, the four sets of two would be assumed to be independent, so 256 (4*4*4*4) is the correct number of combinations (permutations), not 16 (4+4+4+4?). Since each result is from an identifiably different event, you can't really view it as an order-independent situation. (The only thing which could be considered order-independently is the combinations of prizes won from the same event, but you don't want to do that; you'd have AA 1/4, AB 1/2 and BB 1/4, and if you're enumerating possible combinations not considering probability, you get 3 combinations, not 4, so perhaps 3*3*3*3 = 81 possible prize results in your garage, counted without regard to order).
Unless I misunderstand, you should really say ABCDEFGH not ABABABAB. There are certainly more than 16 combinations of (A|B,A|B),(C|D,C|D),(E|F,E|F),(G|H,G|H). Just start enumerating them. 4*4*4*4, as I say. As do you, too, actually.
I actually created a file...
AACCEEGG
AACCEEGH
AACCEEHG
...
BBDDFFHG
BBDDFFHH
I sorted each line (so both AACCEEGH and AACCEEHG became AACCEEGH) and got 81 uniq(ue) combinations.
AAAAAAAA
AAAAAAAB
AAAAAABA
...
BBBBBBBA
BBBBBBBB
(Also 256 combinations) gives only 9 unique order independent combinations. (Corresponding to the possible number of A's (or B's), including zero).
Try saving powering off and reloading between series.
Try doing one series, exiting to Arcade Mode, doing a race (or maybe not), and returning to do the next series.
What state is your Arcade Mode in, in general?
Try losing a race now-and-then.
You need to try varying variables as well as keeping them the same.
Though true, you seem to have proven that, under some conditions, the results are not purely random.
Each single prediction AA|AB|BA|BB might seem to have a 1/16 chance of being a correct prediction, independently. The result would be one of the four, and your guess would be one of the four. Thinking hastily, your chance of guessing one pair correctly might therefore seem to be 1/4 * 1/4 = 1/16. But, in fact, if you enumerate, there are 16 result/prediction combinations, of which four represent you being correct. Which is an elaborate way to confirm intuition. (If there are four equally likely results and you guess, you have a 1 in 4 chance of being correct).
However, the four sets of two would be assumed to be independent, so 256 (4*4*4*4) is the correct number of combinations (permutations), not 16 (4+4+4+4?). Since each result is from an identifiably different event, you can't really view it as an order-independent situation. (The only thing which could be considered order-independently is the combinations of prizes won from the same event, but you don't want to do that; you'd have AA 1/4, AB 1/2 and BB 1/4, and if you're enumerating possible combinations not considering probability, you get 3 combinations, not 4, so perhaps 3*3*3*3 = 81 possible prize results in your garage, counted without regard to order).
Unless I misunderstand, you should really say ABCDEFGH not ABABABAB. There are certainly more than 16 combinations of (A|B,A|B),(C|D,C|D),(E|F,E|F),(G|H,G|H). Just start enumerating them. 4*4*4*4, as I say. As do you, too, actually.
I actually created a file...
AACCEEGG
AACCEEGH
AACCEEHG
...
BBDDFFHG
BBDDFFHH
I sorted each line (so both AACCEEGH and AACCEEHG became AACCEEGH) and got 81 uniq(ue) combinations.
AAAAAAAA
AAAAAAAB
AAAAAABA
...
BBBBBBBA
BBBBBBBB
(Also 256 combinations) gives only 9 unique order independent combinations. (Corresponding to the possible number of A's (or B's), including zero).
Code:
#!/usr/bin/perl -w
#
# Sort the characters in each input line.
# That is, "ABGHEEC" will become "ABCEEGH".
# Blanks lines will be ignored and removed.
# (Makes manual generation of input easier).
# It is also assumed spaces will never be significant in the input.
#
# This was for counting order-independent combinations given lists
# of order-dependent permutations.
#
while (<>) {
$s = $_;
next if $s =~ /^ *$/;
$s =~ s/\n//g;
$s =~ s/(.)/$1 /g;
@a = split(/ /,$s);
# print("@a\n");
@a = sort(@a);
$s = "@a";
$s =~ s/ //g;
print $s, "\n";
}
Last edited:
- 84,553
- Rule 12
- GTP_Famine
No.
That's the difference between the independence of the events (chance) and the dependence of my predictions (odds).
The purely random odds of one of the two cars in one event are 1 in 2. Competing in the event twice successively thus gives four possible results AA, AB, BA, BB - 1 in 4 chance of any given result. Entering a new event resets the odds - the prize car in the new event is independent of the previous one and its purely random odds are 1 in 2. Competing in the event twice successively thus gives four possible results AA, AB, BA, BB - 1 in 4 chance of any given result. Because the two events are independent, the odds are summative, not multiplicative. The chance of a given sequence in each of four events is 1 in 4 * 4.
However each of my predictions depends upon the previous one. The purely random odds of my prediction of a prize car in one event are 1 in 2. Competing in the event twice successively thus gives four possible results AA, AB, BA, BB - 1 in 4 chance of any given prediction. Entering a new event doesn't reset the odds - the prediction in the new event is dependent on the previous one and its purely random odds are 1 in 4 * 1 in 2. Competing in the event twice successively thus gives four possible results CC, CD, DC, DD - 1 in 4 chance of any given result. Because the two events are dependent, the odds are multiplicative, not summative. The odds of a predicted sequence in each of four events are 1 in 4 * 1 in 4.
The number of unique combinations isn't really relevant, since each "prize car award event" is unique. AB may be the same as BA on a random basis, but the order they occur is important for the predictions - in the same that flipping heads then tails is the same net result as tails then heads, but not if you predicted it one way round.
However, any way you wish to look at it, I've changed one controllable variable, predicted the result and observed the prediction correctly in eight independent 50:50 situations which, currently, gives a 99.61% confidence of that variable being the correct one. And about as simple as you could get
It's going to be a while before I pull out my PAL console, and as yet this is all only valid for NTSC-J. If anyone else fancies testing it for other regions, please do so.
Last edited:
- 1,274
- Chicagoland, IL USA
- oldmodelt
- 144
Me too! Except I popped my GT1 into the PS3 about a month ago.
I was surfing the web trying to find some decent suspension settings for the car to get it to run, and discovered that I had, on the first try (well, not counting when I played the game back in the day on the original PS) gotten the "rare" yellow car.
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