Robert:
Check your e-mail :) I sent you an Excel (97) spreadsheet with a complete breakdown, board-by-board, up to board 136. I used an average fruit calculation of 2,700 points from board 9 to 136.
Hope this helps.
Robert:
Check your e-mail :) I sent you an Excel (97) spreadsheet with a complete breakdown, board-by-board, up to board 136. I used an average fruit calculation of 2,700 points from board 9 to 136.
Hope this helps.
Hello Ron:
Got the enclosure...thanks !! Now then...here is what you must count on for a 1M game on Ms Pacman. I will break this out into several scenarios. Keep in mind, this is based on achieving a consistent threshold throughout the game, regardless of how the game's distribution actually turns out to be.
The point here is that we all know that the game awards unfavourable average fruit distribution...that is a given. This thread is to assess what, under normal law of average circumstances, is the necessary thresholds to hope for in order to achieve a 1M kill screen game.
Here goes...
1st - The "fixed" points assume a perfect game with respect to eating all dots and ghosts. That is a given. Based on the data that Ron provided, here is the breakdowns...
-> Sum of all dots & energizers for 136 stages is 345,660
-> Sum of all fixed fruits thru stage 7 is 19,000
-> Sum of all ghosts eaten thru stage 18 is 204,000
-> Total fixed points thru kill screen is 568,660
Thus, you must achieve 431,340 points from the remaining fruits, of which there are 129 mazes with random fruits, times two per maze, is 258 random fruits. Onto step 2
2nd - We have 431,340 points for 258 fruits to achieve for a threshold. This equates to an expectation of 1,671.860465 per fruit...or 16,718 per ten (10) fruits (i.e. 5 complete stages). In other words, for sake of simplicity, let's focus on what every block of ten (1) fruits needs to be in order to meet or surpass this threshold average
CASE ONE would be the expectation of getting no more than ONE (1) banana per ten (1) fruits/five (5) screens
-> That would leave nine (9) fruits to attain the remaining 11,718 points, or 1,302 each
-> If you assumed getting one (1) pear in the remaining nine (9) fruits, that would require the remaining eight (8) to make up the balance of 9,718 points, or more than just apples only, so this is not feasible
-> If you assumed getting two (2) pears in the remaining nine (9) fruits, that would require the remaining seven (7) to make up the balance of 7,718 points, or about an apple each...still not feasible as this is too top heavy
-> Three (3) pears of the remaining nine (9) would leave the remaining six (6) to absorb the balance of 5,718 points, or 953 each...almost an apple each so this is stil ltoo top-heavy
-> Four (4) pears means the remaining five (5) fruits must absorb the balance of 3,718 points, or 743.6 points each.
This would require the remaining five (5) to be better than one of each from the 1,000 apple down to the 100 cherry...even with two (2) apples in the mix, the remaining three would need to yield 1,718...and with the pretzel being worth 700, this is still too top heavy.
Conclusion - one (1) banana is not enough...case closed.
CASE TWO - two (2) bananas
-> The remaining eight (8) fruits would have to pick up the balance of 6,718 points, or 839.75 points each. Not bad. However...
-> If so, 1 of each remaining fruit plus one extra each of the pretzel and orange, for average purposes, is only 5,700 points...still not enough, so we could hope for one (1) pear, two (2) apple, two (2) pretzel, two (2) orange, one (1) strawberry and one (1) cherry for 6,700 points...almost, but not quite.
Thus, a top-heavy distribution is a given...of the seven fruits, thus far, in every ten (10) collected, we would have to hope for only two (2) to be worth less than 500 points...not very practical, averages aside, so two (2) bananas and one (1) pear by themselves still is not enough
-> Let's try two (2) pears, then. This means the remaining six (6) fruits must pick up the balance of 2,718 points...that's just 453 points each, and very do-able. Of the remaining 5 fruits, let's say the average is the 500 point orange. Since 500 exceeds 453, we can, on the average, make the threshold with some slack to spare.
CONCLUSION - in an admittedly top-heavy game, per every ten (10) fruits, you NEED two (2) bananas AND two (2) pears to make up 14K of the required 16,718 per ten (10) fruits collected.
This does not imply that you will get this distribution, again. It only provides the statistical necessities in order to achieve this pre-set threshold.
Obviously such a game would be rare indeed, which is an understatement as no one has achieved a million yet. But, this is a good way to reach that total. And no matter what, you cannot achieve the total without bananas...period, as to make the balance of 431,340 points using only pears and apples would average 1,500 apiece, and 258 fruits times 1,500 points is only 387K
Okay...the statistics have spoken. And once again, this is NOT taking into consideration what pac-masters already know, that the average fruit is closer to strawberry/orange than even orange/pretzel...it only attempts to give you a better guide as to how incredibly favourable the average distribution must be in order to get a 1M game. And this is for every run of ten fruits, so that means 25.8 consecutive runs averaging 16,718 in fruits per run, from stages 8 thru kill screen.
Hope that shed some light on the likelihood of a 1M game. Thanks again, Ron, for the data.
Robert
How does one get 136 screens for Ms. Pac-Man? Prior to The Chase there are five mazes. After the chase there are eight cycles of sixteen mazes; a cycle consists of four orange Chase screens, four blue Junior screens, four pink Chase screens, and four brownish Junior screens. Now, 5 + 8*16 = 133, so there are 133 mazes that may reliably be played on Ms. Pac-Man.
If the machine is "in a good mood", you may be able to play an additional eight mazes - four upside-down Chase mazes, and four normal-looking blue Junior mazes; this makes the grand total 141 mazes, which evidently is the absolute maximum number of mazes one can play. I have never heard of anyone playing more than 141 mazes and it probably is not possible.
I too have created a spreadsheet on the matter; I finished the spreadsheet almost a year ago, and it is in Excel 2000. My spreadsheet considers the fact that there are 2,580 points in dots and energizers on Chase mazes and 2,540 such points on Junior mazes. It also considers the probability of a lower-valued fruit (cherries through pretzels) as being 5/32 and the probability of a higher-valued fruit as being 4/32 = 1/8.
After 133 mazes, there is a maximum of 204,000 points that can be scored from eating monsters. The total points from eating dots and energizers is 340,280. Of course the fixed fruit value through the 7th maze is 19,000, so the total number of "fixed" points through 133 mazes is 563,280.
(After 136 mazes, the total points from dots and energizers is 348,020.)
After 141 mazes, the total points from dots and energizers is 360,760, so the total number of "fixed" points becomes 583,760.
Now, Robert went through more math regarding the difficulty of reaching a million points; I won't repeat Robert's math here. I'll just state that the average score after 133 mazes of perfect play, based on my spreadsheet, is exactly 874,342.5 points. (Bear in mind that after 133 mazes, the last two digits of one's score is *always* "80".) After 141 mazes of perfect play, one's score should average 914,572.5 points, although the last two digits of the score will always be "60".
I made a calculation based on standard deviations the better part of a year ago, and based on that I guesstimate that one perfect 141-maze game in about 5,000 would produce a million points. The odds against a 133-maze game producing such a score is around a million to one. (These again are guesstimates; I'll have to re-do my math.)
Robert, I'll be happy to mail you my spreadsheet if you want; again it is Excel 2000.
-- Richard
Richard:
If you wouldn't mind, I too would like to copy.
Ron,
No problem. :-)
-- Richard
cuz the kill screen can first appear in the middle of the next cycle of 16 as you have defined them above.Originally Posted by rwmarsh741
In MAME I just rack advanced through it. Using your definition of a "cycle" above, you have the first 5 boards, then I see 8 complete cycles...8*16==128...that's 133 boards so far.
Then I saw 4 red chase boards and 4 blue junior boards, then the kill screen....so that was really 141 boards...I guess it was in a good mood for this rack advancing...hehe although on those final 4 red-dot chase boards I didn't see 2 of them upside down. Everytime I have ever rack advanced a mspac I get the ultimate kill screen at the same spot. Now it depends on what you call a kill screen. I call it where it is impossible to get past it cuz there are no dots to clear...just you and the monsters and the monster box...no dots..no maze.
I remember the same thing in the arcade game....the 4 blue junior boards being the last "set" prior to the kill screen...not the brown junior boards...so it's really 8.5 cycles plus the first 5 boards.
I guess it's possible the behavior can be different if you actually play out all the boards. If so then that means other things happen from playing it...eating so many fruits etc. that end up contributing to what form etc. your board is in for #133-141.
Hello Richard:
Ron sent me a spreadsheet indicating 115 mazes after stage 21...thus I based my summations on 136 mazes. As I am not an expert, I would not know for sure what the number was.
However, adjusting for 134 mazes is not a problem...I will do the math later this evening and post the results.
To give you a sample of what it would take for 136 mazes (below) should give you an idea as to what I will approximately come up with later this evening for 134 mazes. Here goes for now...and this covers ONLY the random fruits...
51 bananas = 255,000
51 pears - 102,000
25 apples = 25,000
26 pretzels = 18,200
51 oranges = 25,500
26 strawberry = 5,200
26 cherry = 2,600
This adds to 433,500...just enough to pass 1M assuming there were 136 stages, and adding this to the fixed points (dots, energizers, all ghosts, fixed fruits). And to reiterate my previous posting, this is a realistic sample (136 stages) of what it takes to maintain the pace necessary to get 1M...not based on the game's propensity for less-than-favourable fruit distribution.
I will restructure my calculations for a 134 stage scenario and will re-provide later this evening after normal business hours. Regards.
Robert
man, that's a lot of pears and bananas. I find it interesting in that scenario you include twice as many peaches(your "oranges" hehe) cuz I do find myself getting lots and lots of peaches...more so than cherries and strawberries.
Perhaps that's just a coincidence...or was that based on the fact you personally have also gotten more peaches/oranges versus other fruits?
I really can't say that I've noticed oranges appearing more than cherries. (In fact, when I really want a banana I usually seem to end up with cherries.) In fact, from my experience the fruit distribution over the long run has been pretty equal. The truth is, until about a year ago I naively assumed that they all appeared with 1/7 probability.
-- Richard
yeah, I was going to ask if you saw that article about mspac fruit at
http://www.ionpool.net/arcade/pac_man/MsPacFruit.txt
Cherry 100 pts @ 5/32 = 15.625 %
Strawberry 200 pts @ 5/32 = 15.625 %
Orange 500 pts @ 5/32 = 15.625 %
Pretzel 700 pts @ 5/32 = 15.625 %
Apple 1000 pts @ 4/32 = 12.5 %
Pear 2000 pts @ 4/32 = 12.5 %
Banana 5000 pts @ 4/32 = 12.5 %
In short do you agree with this assessment from that article? Using these probabilities instead of equal probablities how much does that affect the score over the course of all the fruit? Given you have 258 "random" fruit that roughly would make it so you get a about 4 less apples, pears, and bananas versus an equal distribution game but get extra cherries-pretzels instead of those higher scoring fruit.
Adding the above table up to get the statistical average fruit value gets 1234.375 points. 2 fruits per board that's 2468 pts.
When you assume equal 14.3% of each fruit that comes to 2714 points per board.
That means over the course of 129 slow boards, that's means a difference of 31k points.
It means those 3 top scores for mspac are already beating the odds by quite a bit as the most probable score if the above odds of fruit are correct eating all monsters and all fruit is really just in the 880-885k range.
However, if you assume all 3 of those games they played 141 boards total instead, so 134 random fruit boards, that's an extra 5 boards versus your above which gives back about 25k points getting you statistically up to around 905-910k....right where they are.
If ou was truly equal add that 31k back which would have you up around 940k which no one has reached...supporting the above fruit odds.
Whatcha think? That makes your above scenarios even harder.