Price is Right

[Chelsea Thrasher]

[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.

[TheGameShowGuy]

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.

[Chuck Sutton]

[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??

[Fedya]

[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.)

[Unrealtor]

[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.

[Orion]

[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.

[Matt Ottinger]

[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.

[Steve Gavazzi]

[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.

[mcsittel]

[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!

[Chuck Sutton]

[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.

[Robert Hutchinson]

[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.)

[Jeremy Nelson]

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.

[Robert Hutchinson]

[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".

[gamed121683]

[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.

[mcsittel]

[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.