The Game Show Forum
The Game Show Forum => The Big Board => Topic started by: beatlefreak84 on March 14, 2008, 12:37:12 PM
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Hello everyone,
This past week, I had my math students, who are currently learning probability, write a short paper on some probability issues. Essentially, I just wanted to see where their thinking was on some key issues; I didn't care whether they gave right or wrong answers.
However, one of the questions asked them to look at the rules of the car games (on gscentral.net), and, assuming they didn't know anything about the prices of items, which game would they most like to play for the new car, and which game would they least like to play for the new car, and why?
But, because a good number of them weren't that familiar with TPIR, I got a few off-the-wall responses. Thus, I thought I'd ask this group the same question and see what you all say: If you didn't know anything about prices, which game would you like to play for your new car, and which would you NOT like to play for your new car?
Just for the record, my answers:
Most like: Pass the Buck (just for showing up, you have a 1/6 shot of winning the car!)
Least like: Switcheroo (even though you have a 1/5 shot of winning the car, unless you get all 5 prizes right the first time around, which has a 1/120 shot, you can easily second-guess yourself out of a car!)
Anthony
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Let 'em Roll would be a good one. Just roll 5 cars and you win. You don't need to know anything about prices for cars. And even if you don't win the car, you still win some cash.
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[quote name=\'beatlefreak84\' post=\'181312\' date=\'Mar 14 2008, 12:37 PM\']Least like: Switcheroo (even though you have a 1/5 shot of winning the car, unless you get all 5 prizes right the first time around, which has a 1/120 shot, you can easily second-guess yourself out of a car!)[/quote]
If you're a math teacher, and you're talking about random distribution of numbers ("assuming they didn't know anything about the prices"), surely you realize that Switcheroo has a 1/5 chance of winning the car no matter what you do. True, you might switch from a correct first choice to an incorrect second choice, in which case you live with the regret, but you might just as easily switch to the right one. No matter what, on the reveal it's a one-in-five shot. Purely mathematically, that's one of the best games to play. I now wonder where YOUR thinking is on some key issues!
Taking your assignment literally, I think the worst game from a probability level would be Ten Chances. Without having a really good idea about the first and last digit, placing five random numbers into the correct order would seem to have the longest odds of any game. And to take it a step further, if you truly don't have a clue about prices, you may not even get past the first two smaller items to even have a chance at the car!
Meanwhile, the probabilities in Cover Up fascinate me. Has any math type tried to tackle that one?
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How 'bout Double Prices?
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-Let Em Roll
-Pass the Buck
-Spelling Bee
-Card Game...long as I draw the $2000 card. :-P
Temptation switched to cash instead of the prizes, no? I don't need another car, so I'd be perfectly all right with a couple thousand dollars extra in my pocket. However, if I happened to win the car, I have no problem selling it. ;-)
Would hate to have One Away, Any Number, or Cover Up...basically any game that actually requires you to be familiar with the price of a car. I don't watch the show enough to figure out how much a feature adds to the price...
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[quote name=\'fostergray82\' post=\'181325\' date=\'Mar 14 2008, 01:33 PM\']
-Card Game...long as I draw the $2000 card. :-P[/quote]
So if you draw the $5,000 card, you're screwed? ;)
[quote name=\'fostergray82\' post=\'181325\' date=\'Mar 14 2008, 01:33 PM\']Temptation switched to cash instead of the prizes, no?[/quote]
Unless they're converting all the ARP's of the gifts to cash, no. Cash is part of the package sometimes, and there was a $5,000 cash gift in the last aired playing, but there's still other prizes.
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I always thought Lucky 7 was murder even in the days when car prices were four numbers and the first was a gimme.
Now you have to guess 4 numbers 0-9 and only miss by six total, doing it by pure chance; the odds seem astonomical.
The Dice Game seems somwhat simplier if you don't the the price. Only 1-6 available and at least two rolls(three if the number isn't a 1 or 6) get the number without guessing.
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I'd rather have Bonus Game. It's a 50% shot at winning a car (if you're lucky enough to play for one) if you base it on pure luck.
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[quote name=\'Matt Ottinger\' post=\'181315\' date=\'Mar 14 2008, 11:53 AM\']
If you're a math teacher, and you're talking about random distribution of numbers ("assuming they didn't know anything about the prices"), surely you realize that Switcheroo has a 1/5 chance of winning the car no matter what you do. True, you might switch from a correct first choice to an incorrect second choice, in which case you live with the regret, but you might just as easily switch to the right one. No matter what, on the reveal it's a one-in-five shot. Purely mathematically, that's one of the best games to play. I now wonder where YOUR thinking is on some key issues!
Taking your assignment literally, I think the worst game from a probability level would be Ten Chances. Without having a really good idea about the first and last digit, placing five random numbers into the correct order would seem to have the longest odds of any game. And to take it a step further, if you truly don't have a clue about prices, you may not even get past the first two smaller items to even have a chance at the car!
Meanwhile, the probabilities in Cover Up fascinate me. Has any math type tried to tackle that one?
[/quote]
First off, yes; I did acknowledge that the chances of winning a car in Switcheroo were 1/5 (reread my original post!); I said what would make the game undesirable for me would be that the only way I would be sure I'd win the car is if I priced all 5 correctly; the other best case would be if I got 0 right (then, I'd have a 1/4 chance since I'd know the digit I have is wrong). But, anything in the middle, and I'd have no clue how to proceed. The chance that I would switch myself out of a car is not something I'd like to deal with. But, yes; from pure mathematics, it's a very good game to play. It's just that "gray area" in the middle where you don't know for sure that scares me about the game.
You know, a lot of my students said 10 Chances for both directions; the ones that said it would be a really good game to play were those who knew the "ends in 0" rule, as, worst case, you'd give yourself 2 shots at the car. One of my students also (correctly) argued that, if you knew what the first digit of the car was as well, 6 chances for the car guarantees a win. But, without the "ends in 0" rule, you're absolutely right, as that's pretty terrible odds...:)
I don't know if anybody's tackled Cover Up, but we're on spring break now, so I may take a look at it...;)
Anthony
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[quote name=\'WhammyPower\' post=\'181351\' date=\'Mar 14 2008, 02:53 PM\']I'd rather have Bonus Game. It's a 50% shot at winning a car (if you're lucky enough to play for one) if you base it on pure luck.[/quote]
Do you mean Double Prices? And how often is it played for a car, outside of the $1M Spectaculars?
I'll take Pick a Number, and I have to fill in the first digit.
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[quote name=\'MikeK\' post=\'181356\' date=\'Mar 14 2008, 11:59 AM\']
Do you mean Double Prices? And how often is it played for a car, outside of the $1M Spectaculars?
[/quote]
No, he means Bonus Game. If you wild-ass guess, you should win 2 of the 4 screens, which means you have a 50/50 chance of getting the screen with the Bonus in it.
What about Let 'em Roll? Wild-assed guesses should get you one extra roll for a total of 2, so you have a 50/50 chance of getting a car on each die, and two chances to roll each one. In strict terms of probability, assuming a perfect spread of possible outcomes, those feel like pretty good odds.
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Cover Up is actually a 1/720 chance on the first guess, if we take this probablilty exercise literally. However, since the first digit is usually a gimme, and given that most car prices don't have a repeating number, the odds are much better than 1/720. From then on, it varies based on which numbers you get right.
You have pretty good odds on 5 Price Tags as well. Even if you have a brain dead contestant, they have a 1 in 2 shot of getting a guess at the car with the small prizes. Not only that, but then they get up to four tries. From then on, the odds are the number of small prizes right to 5.
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[quote name=\'clemon79\' post=\'181357\' date=\'Mar 14 2008, 03:03 PM\']
[quote name=\'MikeK\' post=\'181356\' date=\'Mar 14 2008, 11:59 AM\']
Do you mean Double Prices? And how often is it played for a car, outside of the $1M Spectaculars?
[/quote]
No, he means Bonus Game. If you wild-ass guess, you should win 2 of the 4 screens, which means you have a 50/50 chance of getting the screen with the Bonus in it.
[/quote]
Yes the Bonus Game(or of couse the Shell Game) in the end is 50/50 even with wild ass guesses. If you want a longer explaination to get to the same answer; my formula for wild ass guesses at Plinko that got me in to so much trouble in the other thread works perfectly here.
There are 16 combinations of t/f guesses. 1 combination of wild ass guesses gets you a win. 4 combos get you a 3/4 chance 6 a 1/2 chance 4 a 1/4 and 1 combo no chance
That adds up to 8/16.
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[quote name=\'clemon79\' post=\'181357\' date=\'Mar 14 2008, 03:03 PM\']
[quote name=\'MikeK\' post=\'181356\' date=\'Mar 14 2008, 11:59 AM\']
Do you mean Double Prices? And how often is it played for a car, outside of the $1M Spectaculars?
[/quote]
No, he means Bonus Game. If you wild-ass guess, you should win 2 of the 4 screens, which means you have a 50/50 chance of getting the screen with the Bonus in it.[/quote]
The key word in your argument is "should". You won't necessarily get a 2 in 4 chance all the time. I read WP's comment as a guaranteed 50/50 shot every time. Hence, Double Prices.
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[quote name=\'Joe Mello\' post=\'181346\' date=\'Mar 14 2008, 02:41 PM\']
[quote name=\'fostergray82\' post=\'181325\' date=\'Mar 14 2008, 01:33 PM\']
-Card Game...long as I draw the $2000 card. :-P[/quote]
So if you draw the $5,000 card, you're screwed? ;)
[/quote]
Can't tell if this is in jest...did they up the high range card?
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[quote name=\'fostergray82\' post=\'181364\' date=\'Mar 14 2008, 03:39 PM\']
[quote name=\'Joe Mello\' post=\'181346\' date=\'Mar 14 2008, 02:41 PM\']
[quote name=\'fostergray82\' post=\'181325\' date=\'Mar 14 2008, 01:33 PM\']
-Card Game...long as I draw the $2000 card. :-P[/quote]
So if you draw the $5,000 card, you're screwed? ;)
[/quote]
Can't tell if this is in jest...did they up the high range card?
[/quote]
They did on May 11, 2005...also made the starting bid $12,000. Really screws with the game, because you could draw the $5,000 card, immediately stop, and win (with the right car, of course).
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[quote name=\'beatlefreak84\' post=\'181353\' date=\'Mar 14 2008, 02:56 PM\']First off, yes; I did acknowledge that the chances of winning a car in Switcheroo were 1/5 (reread my original post!);[/quote]
Please. I didn't miss anything in your original post. I was commenting further that it's one in five no matter what you do.
[quote name=\'beatlefreak84\' post=\'181353\' date=\'Mar 14 2008, 02:56 PM\']It's just that "gray area" in the middle where you don't know for sure that scares me about the game.[/quote]
And I understand that as well. Short of getting a zero on your first try, the second round of Switcheroo never made much sense to me. But for a post that started out being about math and probabilities, it just surprised me that the math teacher marked as his "least favorite" a game that has much, much better odds than most car games.
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I would say Master Key as the game I'd like to play for a car. Assuming you had no problems guessing the actual prices of the small prizes, am I correct in calculating that you will have a 30% chance of picking either the master key or the key that opens the car lock?
3/5 odds of picking a key that doesn't open the car lock
x
2/4 odds of picking the car key or the master key
=
6/20
=
3/10
The game I'd hate to play for a car is Money Game.
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[quote name=\'MikeK\' post=\'181356\' date=\'Mar 14 2008, 02:59 PM\']I'll take Pick a Number, and I have to fill in the first digit.[/quote]
Oooohhhhhhhh, so sorry, Mike...that digit doesn't rise. :-)
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[quote name=\'Steve Gavazzi\' post=\'181372\' date=\'Mar 14 2008, 01:38 PM\']
Oooohhhhhhhh, so sorry, Mike...that digit doesn't rise. :-)
[/quote]
"I'll show you a digit that rises, Trebek!"
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Is it me, or has nobody selected Three Strikes yet? That one's not easy to win even if you know the exact price of the car.
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[quote name=\'Steve Gavazzi\' post=\'181372\' date=\'Mar 14 2008, 04:38 PM\']
[quote name=\'MikeK\' post=\'181356\' date=\'Mar 14 2008, 02:59 PM\']I'll take Pick a Number, and I have to fill in the first digit.[/quote]
Oooohhhhhhhh, so sorry, Mike...that digit doesn't rise. :-)
[/quote]
Then give it some Viagra.
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My apologies if I gave the impression that I wouldn't play Switcheroo strictly based on the probability...Matt's absolutely correct (as if we doubted it...;) ) that, regardless of the situation, theoretically, you have a 1/5 chance of winning the car. So, in that sense, yes; it is a great game to play (and better than my choice of "Pass the Buck!"). What would scare me about it is simply that "second chance;" I'd certainly be apt to switch, but that doesn't necessarily increase my chances of winning unless I had 0 right...:).
So, I'll call a mea culpa on my part and replace my original answer with the one that I actually had in mind when I wrote the assignment, "Lucky Seven." You really don't have all that much leeway, and, without any knowledge of prices, being able to guess four digits 0-9 to save that leeway is very hard (heck; even having an idea how much a car costs doesn't make that game easy, either!). "Switcheroo" was one that I thought of after I had graded the assignment (and, rest assured, they were not graded on whether I agreed with their arguments or not; they got points simply for giving the correct rules of the game and a plausible argument for their viewpoint).
But, since "Shell Game" and "Five Price Tags" have come up, I've thought about something with these games: In each, you have to call four items on essentially a 50/50 shot. However, I've never seen a "Shell Game" playing where all 4 were higher or all 4 were lower (same goes with "Five Price Tags;" just replace those with true or false). So, if I had absolutely no idea about prices, I think I would probably just say the same thing for all four, figuring I'd get at least one right. For those with more familiarity with the show: Have they ever had a playing of one of these where all four were the same answer?
Anthony
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[quote name=\'beatlefreak84\' post=\'181380\' date=\'Mar 14 2008, 05:25 PM\']So, I'll call a mea culpa on my part and replace my original answer with the one that I actually had in mind when I wrote the assignment, "Lucky Seven." [/quote]
As an absolutely random exercise, yeah, Lucky Seven would suck, as would Three Strikes (which is hard enough as it is).
One of the hundreds of things that makes TPIR so fascinating is this very discussion of probabilities, because in most of the games (as we've been discussing on another thread about Plinko) you've got the part that's absolutely mathematical, and you've got the part that's absolutely not. With any of the guess-the-digits car games, you almost always go into it knowing for sure the first digit, which dramatically changes the odds. And with any of the games that have you choose prizes for additional chances, there's no way to say for certain what the odds are on guessing those prizes right.
And those who are more philosophical than mathematical will say that even the Showcase Showdown, which would seem the most pure game of probability in Studio 33, isn't really because of the physicality of the wheel, the occasional effort by a good player to control the spin, and probably a dozen other factors.
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Thought the Card Game rule made it around long enough for most people to get it. Sorry.
While it's doesn't necessarily fall under what the OP asked in his question, I wouldn't mind playing More or Less (http://\"http://gscentral.net/more.htm\").
/Why is this a pricing game?
//And why has it only been won twice?
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I'd like to play Danger Price from the Gametek PC game.
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[quote name=\'beatlefreak84\' post=\'181312\' date=\'Mar 14 2008, 09:37 AM\']However, one of the questions asked them to look at the rules of the car games (on gscentral.net), and, assuming they didn't know anything about the prices of items, which game would they most like to play for the new car, and which game would they least like to play for the new car, and why?[/quote]Pass the Buck, Spelling Bee and Let 'em Roll seem designed to be games where the car is almost a gimme. Lucky Seven is playing roshambo with Roger, and hoping you make the right call (picking middle digits when the car has middle digits, picking outliers when he did...)
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Lucky Seven, it seems to me, was easier to win in the four-digit days for the simple reason that guessing the first digit in the price is easier than guessing the second digit. Often contestants would nail the first digit or miss it by just one, leaving them more leeway for later digits. Now that the contestant has to start with the second digit, he's less likely to have dollars left over by the time he gets to the last digit (if he even gets there). It would be interesting to see the win/loss record of this game over the years to find out if it's had fewer winners on average since going to five digits.
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I'm surprised there has not yet been any mention of Money Game for easiest car game. Assuming there are three possibilities for the first two digits, and six possibilities for the last two digits, and assuming the contestant does not pick a first-two digit combination for the last two digits--and most contestants don't--that's a 50% chance of winning with everything else being random guessing! And if I did my math right, there are 35 out of 126 possible 5-out-of-9 combinations that would win the car, even with the most random of guessing, which is about a 27.8% chance to win. Any sort of knowledge about what cannot realistically be the first two numbers only increases those already pretty good odds.
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Any game with simple odds(Double Prices, Switch?, Bonus Game, Shell Game & 5 Price Tags as well as One Right Price) are the most likely games I'd like to play for a car.
As for the worst, 3 Strikes & Spelling Bee as well as the Range Game based on long odds. Even with 5 numbers & the strike chip being thrown back in after it's pulled, I often wonder how contestants continue to screw up that game by picking the same number & trying that digit AGAIN when they know it wasn't right the first time baffkes me. Spelling Bee requires spelling C-A-R out of 30 cards(11 Cs, 11 As & 6 Rs as well as the 2 CAR cards). Spelling C-A-R & the 2 CAR cards n one felt swoop has the longest odds(something like 11,000+:1). Unless I guess one of the 3 small prizes correctly, this might make an impossible game to win even if you put the $1M tag for getting C-A-R-CAR-CAR in one felt swoop.
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[quote name=\'Fedya\' post=\'181374\' date=\'Mar 14 2008, 02:56 PM\']
Is it me, or has nobody selected Three Strikes yet? That one's not easy to win even if you know the exact price of the car.
[/quote]
Which might have something to do with why no one's selected it.
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I wasn't thinking math - but a cash option when I'd choose "Let Em Roll" or "Pass The Buck". With the latter -you have a good shot if you know nothing about prices and/or can get some cash if you choose.
And with Let Em Roll, $500 is the minimum prize.
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[quote name=\'Craig Karlberg\' post=\'181444\' date=\'Mar 15 2008, 04:53 AM\']
Any game with simple odds(Double Prices, Switch?, Bonus Game, Shell Game & 5 Price Tags as well as One Right Price) are the most likely games I'd like to play for a car.
[/quote]
Yeah wouldn't you love to be the woman who got to play switch for two cars over 19,000??
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[quote name=\'Seth Thrasher\' post=\'181445\' date=\'Mar 15 2008, 04:43 AM\']
[quote name=\'Fedya\' post=\'181374\' date=\'Mar 14 2008, 02:56 PM\']
Is it me, or has nobody selected Three Strikes yet? That one's not easy to win even if you know the exact price of the car.
[/quote]
Which might have something to do with why no one's selected it.
[/quote]
Actually, the original post did mention which games you'd least like to play for a car, as well as those you'd most like to play. True, the mention wasn't very prominent, but it was there.
(Obviously, I was mentioning the game I'd least like to play for a car.)
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[quote name=\'Chuck Sutton\' post=\'181361\' date=\'Mar 14 2008, 02:20 PM\']
[quote name=\'clemon79\' post=\'181357\' date=\'Mar 14 2008, 03:03 PM\']
[quote name=\'MikeK\' post=\'181356\' date=\'Mar 14 2008, 11:59 AM\']
Do you mean Double Prices? And how often is it played for a car, outside of the $1M Spectaculars?
[/quote]
No, he means Bonus Game. If you wild-ass guess, you should win 2 of the 4 screens, which means you have a 50/50 chance of getting the screen with the Bonus in it.
[/quote]
Yes the Bonus Game(or of couse the Shell Game) in the end is 50/50 even with wild ass guesses. If you want a longer explaination to get to the same answer; my formula for wild ass guesses at Plinko that got me in to so much trouble in the other thread works perfectly here.
There are 16 combinations of t/f guesses. 1 combination of wild ass guesses gets you a win. 4 combos get you a 3/4 chance 6 a 1/2 chance 4 a 1/4 and 1 combo no chance
That adds up to 8/16.
[/quote]
You have the right number, but the logic isn't right for Bonus Game. Since the prizes are attached to specific windows before the game starts, you have three prizes which increase your chance of winning by 0% and one (the one next to the bonus) which increases your chance of winning by 100%. So if you guess randomly on the small prizes, you get a probability of (3 x (1/2 x 0)) + (1 x (1/2 x 1)) = 1/2. It's a bit of a trick, really, based on how much information is available -- because we in the audience don't know which one has the bonus next to it, we have no choice but to value each of the small prizes equally, so it looks like winning more prizes increases the chances of winning the bonus.
However, your logic is right for Shell Game, because picking shells means that you truly do get a 1-in-4 chance for each prize you get right.
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[quote name=\'Matt Ottinger\' post=\'181315\' date=\'Mar 14 2008, 08:53 AM\']
Meanwhile, the probabilities in Cover Up fascinate me. Has any math type tried to tackle that one?
[/quote]
I did, a while back. Assuming that you know the first number and are guessing randomly for the rest, I found that you have a 31.9% chance of winning if you intentionally get the first number wrong on the first guess. If you're like most people and get the first number right on the first guess, it goes down slightly to 31.7%. Very good odds either way, especially by TPIR standards.
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[quote name=\'Orion\' post=\'181532\' date=\'Mar 15 2008, 05:13 PM\']I did, a while back. Assuming that you know the first number and are guessing randomly for the rest, I found that you have a 31.9% chance of winning if you intentionally get the first number wrong on the first guess. If you're like most people and get the first number right on the first guess, it goes down slightly to 31.7%. Very good odds either way, especially by TPIR standards.[/quote]
Interesting. I had always thought that missing the first number deliberately would be a useful strategy (since it gives you a lock on your second try, guaranteeing you a third unless you miss the first), but if you're right, it honestly doesn't look like it makes that much difference after all.
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[quote name=\'Matt Ottinger\' post=\'181536\' date=\'Mar 15 2008, 05:17 PM\'][quote name=\'Orion\' post=\'181532\' date=\'Mar 15 2008, 05:13 PM\']I did, a while back. Assuming that you know the first number and are guessing randomly for the rest, I found that you have a 31.9% chance of winning if you intentionally get the first number wrong on the first guess. If you're like most people and get the first number right on the first guess, it goes down slightly to 31.7%. Very good odds either way, especially by TPIR standards.[/quote]Interesting. I had always thought that missing the first number deliberately would be a useful strategy (since it gives you a lock on your second try, guaranteeing you a third unless you miss the first), but if you're right, it honestly doesn't look like it makes that much difference after all.[/quote]
I'm curious -- what if you factor in not getting the fourth number right until at least the third turn? Because that's usually what ends up happening.
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[quote name=\'Matt Ottinger\' post=\'181536\' date=\'Mar 15 2008, 04:17 PM\']
[quote name=\'Orion\' post=\'181532\' date=\'Mar 15 2008, 05:13 PM\']I did, a while back. Assuming that you know the first number and are guessing randomly for the rest, I found that you have a 31.9% chance of winning if you intentionally get the first number wrong on the first guess. If you're like most people and get the first number right on the first guess, it goes down slightly to 31.7%. Very good odds either way, especially by TPIR standards.[/quote]
Interesting. I had always thought that missing the first number deliberately would be a useful strategy (since it gives you a lock on your second try, guaranteeing you a third unless you miss the first), but if you're right, it honestly doesn't look like it makes that much difference after all.
[/quote]
It's funny-the exact stats for this game are a bear to calculate, but to simulate this is fairly simple.
I wrote a quick C++ program that assumes a random selection of a number on the board every pick. I got similar numbers to Orion... for multiple sets of 10,000,000 trials, P(winning) is 0.321. It's 0.322 if you always get the 1st number right, and 0.319 if you always get the first number wrong.
Given that you win the game:
P(win in 1 round) = 0.004
P(win in 2 rounds) = 0.130
P(win in 3 rounds) = 0.433
P(win in 4 rounds) = 0.364
P(win in 5 rounds) = 0.069
A quick Google search suggests someone recently asked a computer programming class to simulate this very game (since the game's description referred to Drew rather than Bob). They got 0.321 as well for P(win). Interesting stuff!
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[quote name=\'Unrealtor\' post=\'181489\' date=\'Mar 15 2008, 02:05 PM\']
You have the right number, but the logic isn't right for Bonus Game. Since the prizes are attached to specific windows before the game starts, you have three prizes which increase your chance of winning by 0% and one (the one next to the bonus) which increases your chance of winning by 100%. So if you guess randomly on the small prizes, you get a probability of (3 x (1/2 x 0)) + (1 x (1/2 x 1)) = 1/2. It's a bit of a trick, really, based on how much information is available -- because we in the audience don't know which one has the bonus next to it, we have no choice but to value each of the small prizes equally, so it looks like winning more prizes increases the chances of winning the bonus.
However, your logic is right for Shell Game, because picking shells means that you truly do get a 1-in-4 chance for each prize you get right.
[/quote]
In the end there are 64 possible combination of h/l answers and prizes windows. 32 of those combinations will get you the prize.
with no knwoledge of the correct prices 32/64 of the combinations will get you the prize. Make what you will of that fraction.
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[quote name=\'clemon79\' post=\'181357\' date=\'Mar 14 2008, 03:03 PM\']What about Let 'em Roll? Wild-assed guesses should get you one extra roll for a total of 2, so you have a 50/50 chance of getting a car on each die, and two chances to roll each one. In strict terms of probability, assuming a perfect spread of possible outcomes, those feel like pretty good odds.[/quote]
The odds you'll get a car on any one die, with two rolls, is 3/4 (1/2 plus 1/2 of the remainder). Raise that to the fifth power to find the odds of getting all five cars: 243/1024, or about 23.7%.
If you played the grocery part of the game randomly, that will be your expected outcome half of the time. One quarter of the time you'll get one roll, which is 1/2 to the fifth, or 1/32, or about 3.1%. And the last quarter of the time, you'll have three rolls, which is 7/8 to the fifth power -> 16807/32678 -> about 51.4%. Average those out: (3.1 + 23.7 + 23.7 + 51.4) / 4 = roughly 25.5%. Make the groceries gimmes, which they usually are, and you can of course call it 51.4%.
I believe the odds on 5 Price Tags, with random guessing, are a simple 40%, what with the probabilities all being "balanced" around getting two of the four small prizes right. And even with knowledge that they never have all 4 prizes as either True or False, I'm pretty sure there's no advantage to sticking with one guess all the way through. If they're split 3 and 1, you're just as likely to get 3 right as 1 right, which averages out to 2 right, and you're right back where you started.
Master Key with two keys is actually 70%: 40% (2/5 on the first pick) + 30% (2/4 on the second pick, or half of the remaining 60%). With one key, it's 40%, and with none it's . . . hmm, let's see . . . oh yeah, 0%. :) Averaging those out as I did above with Let 'Em Roll: (70+40+40+0) / 4 = 37.5% chance by picking randomly.
However, since we've discussed some games that are not regularly played for cars, I nominate Bullseye--IF you have a contestant who, in addition to not knowing any prices at all, is trying simply to maximize their chances. Just pick 1 of each of the three products, and you've given yourself a 60% chance to win. Start adding in "tricks" that don't necessarily involve pricing knowledge (for instance, I don't recall there ever being a grocery item that you needed only 1 of to get to $10-$12) and you can bump those odds up even further.
And again, phew. (Or Whew!, whichever you prefer.)
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Correct me if I'm wrong, but Three Strikes seems like the odds are pretty good as well. If we consider that each correctly placed number constitutes a "round", then you have a you have a 5 in 6 chance that you will pick a number. Each round, the probability slowly shifts in the house's favor, but it still only comes to a 1 in 2 chance that you will get the car, granted you haven't picked up 3 strikes by that time.
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[quote name=\'rollercoaster87\' post=\'181763\' date=\'Mar 17 2008, 05:10 PM\']Correct me if I'm wrong, but Three Strikes seems like the odds are pretty good as well. If we consider that each correctly placed number constitutes a "round", then you have a you have a 5 in 6 chance that you will pick a number.[/quote]
True--but you then have only a 1 in 5 chance of placing the number correctly.
Each round, the probability slowly shifts in the house's favor, but it still only comes to a 1 in 2 chance that you will get the car, granted you haven't picked up 3 strikes by that time.
That's a huuuuge "granted".
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[quote name=\'MikeK\' post=\'181375\' date=\'Mar 14 2008, 04:56 PM\']
[quote name=\'Steve Gavazzi\' post=\'181372\' date=\'Mar 14 2008, 04:38 PM\']
[quote name=\'MikeK\' post=\'181356\' date=\'Mar 14 2008, 02:59 PM\']I'll take Pick a Number, and I have to fill in the first digit.[/quote]
Oooohhhhhhhh, so sorry, Mike...that digit doesn't rise. :-)
[/quote]
Then give it some Viagra.
[/quote]
Would that be Numberwa--oh, never mind. We've killed this bit to the ground already, haven't we? Carry on.
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[quote name=\'rollercoaster87\' post=\'181763\' date=\'Mar 17 2008, 04:10 PM\']
Correct me if I'm wrong, but Three Strikes seems like the odds are pretty good as well. If we consider that each correctly placed number constitutes a "round", then you have a you have a 5 in 6 chance that you will pick a number. Each round, the probability slowly shifts in the house's favor, but it still only comes to a 1 in 2 chance that you will get the car, granted you haven't picked up 3 strikes by that time.
[/quote]
Since our network was unreachable at work today, I wrote another C++ program today to simulate 3 Strikes. The challenge in coding this up is keeping track of the previous guesses such that future guesses are conditioned upon past misses.
If you have *no* idea where any number goes, and randomly select each time, the probability of winning is 0.374. If you know where *every* number goes, and the only thing you have to do is pick each once while avoiding strikes, the probability of winning jumps to 0.624.
The probability of winning the 5-digit 3 Strikes in the older version with three strike chips in the bag is an abysmal 0.15.
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[quote name=\'mcsittel\' post=\'181820\' date=\'Mar 18 2008, 02:18 PM\']The probability of winning the 5-digit 3 Strikes in the older version with three strike chips in the bag is an abysmal 0.15.[/quote]
Did you take into account that in the old version a strike chip is taken out of the game when it's pulled?
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[quote name=\'clemon79\' post=\'181822\' date=\'Mar 18 2008, 04:22 PM\']
[quote name=\'mcsittel\' post=\'181820\' date=\'Mar 18 2008, 02:18 PM\']The probability of winning the 5-digit 3 Strikes in the older version with three strike chips in the bag is an abysmal 0.15.[/quote]
Did you take into account that in the old version a strike chip is taken out of the game when it's pulled?
[/quote]
Indeed...
if ( chip[pick] == 0 )
{
strikect++;
nchips--;
if ( strikect == 3 ) { cout << "Game over." << endl; }
}
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[quote name=\'MikeK\' post=\'181356\' date=\'Mar 14 2008, 01:59 PM\']I'll take Pick a Number, and I have to fill in the first digit.[/quote]
You just described the modern-day Balance Game.
Here's one for the people who have a lot of free time on their hands: Cliff Hangers. Here's what we know about the small prizes:
1) They're always between $10 and $50 in price. (They didn't use to be, but they are now.)
2) Each one is higher than the one before it.
That said, if three prizes are chosen completely at random among 41 prizes that have prices ranging from $10-$50 with no duplicates, how often can you keep [Drew]Yodelman[/Drew] safe if you just guess 20-30-40 every single time?
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[quote name=\'GS Warehouse\' post=\'181961\' date=\'Mar 19 2008, 06:30 PM\']
Here's one for the people who have a lot of free time on their hands: Cliff Hangers. Here's what we know about the small prizes:
1) They're always between $10 and $50 in price. (They didn't use to be, but they are now.)
2) Each one is higher than the one before it.
That said, if three prizes are chosen completely at random among 41 prizes that have prices ranging from $10-$50 with no duplicates, how often can you keep [Drew]Yodelman[/Drew] safe if you just guess 20-30-40 every single time?
[/quote]
I guess I resemble that remark, although it only took about 15 minutes to code up and run 10,000,000 trials.
The probability I came up with: 0.732.
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[quote name=\'GS Warehouse\' post=\'181961\' date=\'Mar 19 2008, 07:30 PM\']
[quote name=\'MikeK\' post=\'181356\' date=\'Mar 14 2008, 01:59 PM\']I'll take Pick a Number, and I have to fill in the first digit.[/quote]
You just described the modern-day Balance Game.[/quote]
But for a car, the first digit shouldn't be too hard to pick.
I really shouldn't have to explain this...
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[quote name=\'Robert Hutchinson\' post=\'181588\' date=\'Mar 15 2008, 10:08 PM\']However, since we've discussed some games that are not regularly played for cars, I nominate Bullseye--IF you have a contestant who, in addition to not knowing any prices at all, is trying simply to maximize their chances. Just pick 1 of each of the three products, and you've given yourself a 60% chance to win. Start adding in "tricks" that don't necessarily involve pricing knowledge (for instance, I don't recall there ever being a grocery item that you needed only 1 of to get to $10-$12) and you can bump those odds up even further.[/quote]
I believe the last playing of Bullseye had the $11.68 SlowMag in it.
Also, Dice Game isn't too hard to calculate, if you always go with the odds.
If you roll a 1 or a 6 on any die, you will be correct on that digit 100% of the time.
A 2 or 5 will succeed as long as the digit is not 1 or 6 (respectively), meaning you will be correct 83.3% of the time.
A 3 or 4, similarly, gives you 66.7%.
Thus, any given die will be correct (1/3)(1) + (1/3)(5/6) + (1/3)(2/3) = 5/6 of the time.
Which makes the probability of winning .8333^4, or 0.4823.
Oh, and Safe Crackers for a car is pretty sweet.