1.  12/13/2004, 07:25 AM Originally Posted by skydivewags sxtg, all you have to do, is this. If you want to subtract 5% from something, just multiply it by .95 so this should work: 3,500 * .95 / 2 = 1,662.5 How do your formula's differ? Are the percentages different? if so..... to subtract 10%, just multiply by .9 to subtract 15%, just multiply by .85 and so on...... Usually the first # is the variable. Yes I know I can work around it. But it sucks. Even my Vx was able to perform these functions.
2.  12/13/2004, 07:46 AM Sorry to nerd out on the caculator thing, but I bought this calculator back in the 80's, and it has every mode the 650's has except finance. and it also can convert between hex, oct, binary, etc. It's about 3/8" thick when closed, and the buttons on the right side are realy built into the cover. It's amazing it took so long to incorporate these functions into the Palm OS. The internet is a very amazing (odd) place. I actually found a web page with this calc. on it. http://www.voidware.com/calcs/fx451m.htm Attached Images Casio.jpg (76.8 KB, 39 views) Blue Skies, Wags
3.  12/13/2004, 08:12 AM Funny, I've never used the % key this way on a calculator. I guess I was never comfortable with something like 100 - 20% being a real mathematical expression.
4.  12/13/2004, 09:45 AM Originally Posted by Jeff DLB I probably shouldn't add to this thread, but... IMHO, both answers are wrong, but the 650's is "less wrong." Normally, entering a number and pressing % divides the number by 100. Thus, the correct answer is 100-20% = 100-0.2 = 99.8. The 650 is inserting parentheses and saying 100-20% = (100-20)% = 80% = 0.8. The 600 is ignoring the % entirely and saying 100-20% = 100-20 = 80. Calc v4.0 on my aging Clie SJ33 does the same thing as the T600. It ignores the % sign if you press it after the second number in an addition or subtraction. If you press it after the 3rd number (or thereafter), it turns the +/- you just typed into a multiplication by (3rd number)%. Example: "100-20-10%" is treated as (100 - 20)*10% = 8. Moral: don't use the ambiguous percent key. Divide by 100 with the calculator or in your head. BTW: Google gets it right. Type: 100-20% into Google and you get: 100 - (20%) = 99.8 as the results page...
5.  12/13/2004, 11:20 AM midmofan, I think that if you do the Lets Make A Deal thing, over time, switching will get you ahead... even if the transaction is new... no?
6.  12/13/2004, 12:43 PM For a basic calculator, the best one I've found is SnapCalc. It can be set to come up with a hard key press on the T600 or 650.
7.  12/13/2004, 03:49 PM Originally Posted by 300MileTrip In response to midmofan, I would say that you dont have to switch your choice to get the better odds. Your odds have improved to 50/50, but you can stay with door #1 or change to door #2, either way it's still 50/50. I won't get into the math of it and all but actually the Monty Door Problem is actually one of those test case to show that common sense is not always right. It isn't a brand new problem your odds are actually better if you switch doors. Here is a site to explain it better. If you just do the problem yourself and play it out, there are plenty of sites on the internet that do it, you will see that switching improves your odds. http://www.uvm.edu/~dhowell/StatPage...ThreeDoor.html As for the bellboy problem it isn't a brand new situation it's just a continuation They paid out \$30 at first \$2 went to the bellboy leaving \$28 dollars \$25 dollars went for the room leaving \$3 \$3 were split among the men. The question is confusing because of the way it is phrased.
8.  12/13/2004, 05:41 PM Originally Posted by silverado Funny, I've never used the % key this way on a calculator. I guess I was never comfortable with something like 100 - 20% being a real mathematical expression. I'm with you. i never understood the point of the % sign on calculators. it just makes more sense to me to use decimals. I'd take a 1/x button over % anyday.
9.  12/14/2004, 10:46 AM I suspect that the result from the T600 is "correct" and that of the T650 is incorrect. The issue is just what does 100-20%= mean to the user since % is not a normal operater. I suspect that to most people it means 100 minus 20% of 100 as in a 20% discount on a \$100 purchase. There arer two ways that this can be calculated: 100 - (push 100 on the stack and store subtraction operator) 20 % ( push 100 on the stack multiply by .2 leaving 20 in the visible register) = ( subtract 20 from 100 yielding 80 in the visible register) You can check to see if this is what the T600 is doing by doing the following 160 - 20%= You should get the following 160 - 20 % 32 = 128 The other route is 100 - (push 100 on the stack and store subtraction operator) 20 % (push 1 on to stack and change change subtaction operator to multiplication orperator subtract 0.2 from the 1 on the stack leaving .8 in the visible register) = (multiply 100 on the stack by .8 leaving the result 80 in the visible register) This can be checked by entering the sequence 160 - 20% = and you should get the following results 160 - 20 % 0.8 = 128 Any other result would be in error if the assumption that the intention is to calculate a discounted value. The same calculation is much easier to use and understand in reverse polish notation suppose for example a dealer offers us a \$160 item with 20% discount and we must pay a 9% sales tax. Our operations would be as follow with the results we would see on the screen. 160 enter (push 160 on stack) enter (push another copy of 160 on stack) 20 % (take 20% of the lowest operand (160) on the stack) 32 - (subtract 32 from the 160 on the stack) 128 enter (push a copy of 128 on the stack) 9 % (take 9% of lowest operand (128) on the stack) 11.52 + (add 11.52 to 128 on the stack) 139.52 Note that in rpn there is no need for parens and give enough registers on the stack that any algebraic expression can be solve either by inside out or left to right reading of the expression. Also note that there is no need for equal signs you simply follow the "Alice in Wonder Land Rule) "Begin at the beginning, keep going until you get to the end, then quit". Deputy-Dawg Originally Posted by mgauss In the 650's calculator, Calc, if you type 100-20%= you get .8 In the 600 you get 80
10.  12/14/2004, 11:36 AM Originally Posted by anoetic I won't get into the math of it and all but actually the Monty Door Problem is actually one of those test case to show that common sense is not always right. It isn't a brand new problem your odds are actually better if you switch doors. Here is a site to explain it better. If you just do the problem yourself and play it out, there are plenty of sites on the internet that do it, you will see that switching improves your odds. http://www.uvm.edu/~dhowell/StatPage...ThreeDoor.html The problem with all of the math "proofs" of the three-door problem is that they do not recognize that the decision NOT to switch doors is, in fact, a new, seperate, affirmative, decision to choose the that door (that you had already picked). As in my example, say that, after one door is tossed out of the mix, Monty Hall then said: "We were just kidding you, you don't have to choose one of three doors, just one of two. Your first pick doesnt count. One of the two doors has a prize. You need to start over and pick one of them, which one do you want?" In that case, there is a 50% chance of being right whichever door you pick. In the classic three-door problem, the person is given the choice of remaining with their first pick or switching to the other door. The boolean math proofs (and the computer programs that have fun with this) dont recognize that if the person keeps their original door choice that, then, is really a brand new independant choice of the door just llike in my example where Monty was kidding. To really do a computer program to test it you would first need to decide which door had the prize (truely randomly, not with a pc random number generator). Save that decision separatly. Then make a choice of doors. Then, turn off/on the computer and start over. Input the "correct" door choice, then show the results of a computer run that either stayed with the original choice or made the switch, showing the results between staying with your first door and switching. That is what is actually going on on "Lets Make a Deal" The second computer program can be as simple as this Input A (the correct door choice from previous true random generation) Input B (the original contestant door choice) "Do you want to change doors?" (y/n) If y then input C (new door choice) if n then B=C If A=C then "win" Then just keep track of wins and losses in relation to if a change was made. Will come out to be close to 50-50 if you do a long enough run. Of course, the "contestant" cannot know what the original door choice was (input A). FWIW, however, I have seen pretty good explinations that, on the quatum level, the three-door choice actually does work as described in the three-door problem, so if Monty has you choosing between three different spinning bosons, then, by all means, change your pick! Last edited by midmofan; 12/14/2004 at 12:07 PM.
11.  12/14/2004, 12:05 PM BTW, this is a lot of fun, sure beats sweating over the patch for the 650 sound problem! Anybody want to debate the flaw in the famous Schrodinger's Cat problem??? Gets real metaphysical. Fun to freak out your local physics prof!
12.  12/14/2004, 12:15 PM On my Handspring Treo 600 (with latest patch) this is what happens in Calc: Selecting "Finance" then entering 100 then - then % then 20 then = the result is 80 Selecting "Finance" then entering 100 then - then 20 then % then = the result is 96
13.  12/14/2004, 01:05 PM Originally Posted by midmofan The problem with all of the math "proofs" of the three-door problem is that they do not recognize that the decision NOT to switch doors is, in fact, a new, seperate, affirmative, decision to choose the that door (that you had already picked). The thing is it is not a new situation it is a continuation. Look at it this way. In choosing three doors with one door being a winner there are two possibilities: 1. 33.3% chance you choose the winning door 2. 66.7% chance you pick the wrong door So lets play out the game with those two choices: - With possibility 1 if you switch you will lose because you chose the right door in the first place and then switched to the wrong door. Therefore you will lose all 33.3% of the time you chose the winning door first - NOW, possibility 2 is the complete oposite. Since Monty can't open the door you chose he has to open one of the other two doors. Since he can't open the door that is the winner that only leaves him to open the door that is wrong leaving the winning door. Therefore, if you chose the wrong door on the first try you will only be left with the winning door on the second try. You win 66.7% of the time. You keep thinking it is a new decision that is 50-50 but it isn't. The thing has a lot to do with your initial decision and the rules of the game. Monty can't open the door you picked or the winning door. If you switich you will win 66.7 % of the time. If you just choose randomly you will win 50% of the time on the second try. 33.3% of time first choice is right --> switch is always wrong. 66.7% of time first choice is wrong --> switch is always right. Go ahead, play it out. You'll see a switch is always right. Last edited by anoetic; 12/14/2004 at 01:28 PM.
14.  12/14/2004, 01:12 PM The real question is Did the bellboy use a Treo (600 or 650) to figure what he should pay, or was it a Pocket PC device?
15.  12/14/2004, 01:17 PM Okay, so I don't read the manual... ... looking at the calc, my initial impression was yuck, so I install Parens, a neat calculator I have used for years on my pre Treo Palms. Parens needs MathLib so I installed that also. Turns out the built-in calc has many of the same capabilities as Parens, so I delete Parens (its no longer supported anyway). Now the question is...do I need MathLib???
16.  12/14/2004, 01:21 PM Originally Posted by anoetic Tthe second try. 33.3% of time frst choice is right --> switch is always means loss. 66.7% of time first choice is wrong --> switch is always right. Go ahead, play it out. You'll see a switch is always right. That is the problem with the math. It carries over the initial % value to the next decision, which is actually a new independant decision to choose the same door. The decision not to change doors is actually a new choice of that door.
17.  12/14/2004, 02:07 PM Originally Posted by midmofan That is the problem with the math. It carries over the initial % value to the next decision, which is actually a new independant decision to choose the same door. The decision not to change doors is actually a new choice of that door. But your first choice does have an impact. You are forgetting the rules of the game. Monty does not remove a door at random, that would create an independant event for the second round because it wouldn't matter if your first choice was right or wrong. But the rules of the game say he can't open your dorr and he can't open the right door. THEREFORE, whether you chose the correct door the first try has an impact on what door is removed. Think of it this way: When you start with three doors you have a 33.3% chance of winning and Monty has a 66.7% chance of winning. If you could choose Monty's two doors you would. So you have your door and in the second round monty says do you want to switch your one door with my two doors, one door that I opened and the one door I didn't open. You are effectively getting Monty's two doors when you make the switch. He just shows you that one of his doors doesn't have a prize. Why should that make any difference in the percentage?
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