The Mathematics of Pricing Games on The Price is Right

[jlgarfield]

You know, sometimes, we like to think about the mathematics of the many pricing games of TPiR. For you math experts out there, this is your thread to post.

Here are two of my observations:

One Away: Nine times out of ten in this game, if only one number (out of five, or four in the 1980s) is correct, this usually means the odds are in your favor of winning the car, as the first one will be right, and the remaining ones are wrong.

Dice Game: The odds of guessing the price of the car exactly are 1-in-1296. It's only been done once.

[TimK2003]

You know, sometimes, we like to think about the mathematics of the many pricing games of TPiR. For you math experts out there, this is your thread to post.

Here are two of my observations:

One Away: Nine times out of ten in this game, if only one number (out of five, or four in the 1980s) is correct, this usually means the odds are in your favor of winning the car, as the first one will be right, and the remaining ones are wrong.

Dice Game: The odds of guessing rolling the price of the car exactly are 1-in-1296. It's only been done once.

FTFY

[TLEberle]

Strictly speaking, he’s not wrong, but that’s why you get to roll dice.

Interestingly, the odds of winning $25,000 in a single jab on Punch a Bunch are the same as they were to win $10,000 in the two-step game played the first handful of times.

[Jeremy Nelson]

Any Ten Chances player that knows the zero rule, and just blindly guesses at the other numbers still guarantees themselves at least two guesses at the car.

The optimal way to play Bonkers is to make a decision from the jump on whether the first number in the price is higher or lower than what's shown, and stick with it. This leaves three other paddles to switch around to make eight distinct combinations, which can be done in 30 seconds (unlike 16 if you were to mess with the first paddle as well)

Betting strategy would suggest to always take the prizes in Temptation, as you're taking a 1 in 16 chance to increase your winnings fivefold.

[MSTieScott]

One Away: Nine times out of ten in this game, if only one number (out of five, or four in the 1980s) is correct, this usually means the odds are in your favor of winning the car, as the first one will be right, and the remaining ones are wrong.

That isn't math. If the game were played randomly, then a one-horn scenario wouldn't result in anything-out-of-ten odds. And based on the approximately two times a one-horn scenario happens each season -- Dream Car Week aside -- a contestant who plays the game non-randomly has much, much better than a nine-out-of-ten chance of winning the car by leaving the first number alone.


It is mathematically impossible to not win the first prize in Ten Chances (assuming the host doesn't let you duplicate bids). The board is built in such a way that the first number jumble can't even completely sit behind the tenth slot.

I think I'm right about this, but someone please correct me if I'm mistaken: In Three Strikes, even if you were outright told the price of the car before the game began, you'd only have a 62.5% chance of pulling all five numbers and winning before you pull the three strikes.

[tvmitch]

The math behind the Pay The Rent game always fascinates me. Correct me if I'm wrong, but I believe it is, by far, the easiest game to win $10K on the show. However, usually only one combo of items/levels would get you the $100K.

Haven't been watching the show as much as I used to, but it seems like Pocket Change is proving difficult for contestants lately. I haven't seen the $2.00 card come up, but even if it did, that doesn't always guarantee a win.

Not sure if it fits in the discussion either, but the way the Cover Up game unfolds can be fun, especially if a contestant backs into a guaranteed win.

What's the probability to win the car if you play Let Em Roll with 1, 2, and 3 rolls? If the car is worth, say, $20K, what's the ideal exit fee for each roll?

[Mr. Armadillo]

I think I'm right about this, but someone please correct me if I'm mistaken: In Three Strikes, even if you were outright told the price of the car before the game began, you'd only have a 62.5% chance of pulling all five numbers and winning before you pull the three strikes.

Other way around - you lose 62.5% of the time if you know the price up front.  To win the game, you have to pull all five numbers before all three strikes, which mathematically works out the same as pulling out every ball except for one.  If the ball left in the bag is a number, you lose, and if it's a strike, you win.  Since 5 of the 8 balls are numbers, that's a 62.5% chance of losing.

What's the probability to win the car if you play Let Em Roll with 1, 2, and 3 rolls? If the car is worth, say, $20K, what's the ideal exit fee for each roll?

3.125% (1/32), 23.73% (243/1024), and 51.29% (16807/32768), respectively, assuming you never bail.

[MikeK]

The math behind the Pay The Rent game always fascinates me. Correct me if I'm wrong, but I believe it is, by far, the easiest game to win $10K on the show. However, usually only one combo of items/levels would get you the $100K.

Years ago, I created a rudimentary Pay the Rent analyzer with Excel.  I run the prices through it every time I see the game.  Usually, there are 2-3 possible solutions, but the number has as high as 7.

[Matt Ottinger]

The math behind the Pay The Rent game always fascinates me. Correct me if I'm wrong, but I believe it is, by far, the easiest game to win $10K on the show. However, usually only one combo of items/levels would get you the $100K.

Years ago, I created a rudimentary Pay the Rent analyzer with Excel.  I run the prices through it every time I see the game.  Usually, there are 2-3 possible solutions, but the number has as high as 7.

The original brilliance of that game was that it was counter-intuitive.  A player's instinct is to place the cheapest item at the bottom (necessary), then the next two cheapest on the floor above, the next two on the floor above that, and then the highest priced item in the attic (also necessary).  But (at least originally), the game was set so that the second and third highest priced items added together were more than the highest priced one.  So your middle four had to be carefully chosen (or lucky guesses).  That's why it was worth $100,000!

More recent stagings, at least the ones I've seen, have made it a lot easier to win by removing that trick in the middle, so that all you have to do is rank the items lowest to highest.

[Mr. Armadillo]

I didn't want to believe it, because that doesn't sound interesting at all, but I admittedly haven't seen it in forever, so I looked up the (last?) win on YouTube, and...yikes.  Yep, he just ranked them lowest to highest, and it wasn't close at any point. 

70c < ($1.89 + $3.49) < ($4.99 + $9.99) < $18.99

For comparison, I looked up a random playing from 2010.  These were the prices:

79c, $2.99, $3.39, $3.99, $4.99, $6.79

Good luck with that.

That combination has only one winning solution out of a possible one hundred and eighty, and even *if* you know what you're doing, and *if* you know the prices ahead of time, it'd take you five minutes to figure out which item goes on the bottom:

$3.99 < (79c + $4.99) < ($2.99 + $3.39) < $6.79

(Amazingly, the contestant actually placed them all correctly but bailed at $10,000.)

[Matt Ottinger]

For comparison, I looked up a random playing from 2010.  These were the prices:

79c, $2.99, $3.39, $3.99, $4.99, $6.79

Oh geez, I forgot about that.  In my description, I actually made it sound easier than it could be, because (when the game was nasty) the least expensive item was strategically key, and should NOT be placed at the bottom.

[Mr. Armadillo]

Okay, so that win I posted earlier is apparently an outlier since it was "rigged" for the season premiere.  Each of the 9 times the game's been played since then only had 2 or 3 winning combinations, and they all look a lot more...well, like I'd have expected them to look.

Typically, you want one of the two middle-priced prizes in the mailbox at the bottom, and (obviously) the most expensive prize has to be at the top.

In case you're curious, the most common winning combination is 4 < (2 + 3) < (1 + 5) < 6, which actually wins about two-thirds of the time.

[Mr. Armadillo]

Okay, one last tidbit for now.  The most recent playing, on Dec. 11, is actually fairly interesting.  The prices:

$1.39, $2.49, $3.49, $4.29, $6.99, $8.99

To win, you have to put the $8.99 item at the top (obviously), then El Cheapo with the $6.99 item on the $10,000 shelf, but then any combination of the other three items at the bottom works.

[MSTieScott]

Other way around - you lose 62.5% of the time if you know the price up front.

D'oh. Of course I screw up the easy part.


The difficulty of Pay the Rent changes based on whether the show wants it to be won. Using the information at tpirstats.com and looking at the five most recent complete seasons, the game has been won three times. Here are the number of solutions in each of those playings (mathematically, the game can have a maximum of fifteen correct solutions):

Season 47: Nine solutions (season premiere)
Season 46: Eleven solutions (Big Money Week)
Season 44: Eight solutions (Breast Cancer Awareness special)

There was also a season 46 playing with six solutions (summer beach party special) and a season 45 episode with five solutions (not a special episode).

Most of the other playings of the game over the past five seasons have had either three winning solutions or two winning solutions. There hasn't been a one-solution playing of the game since January 2014 (season 42). Over the past five seasons, no contestant has achieved a correct solution but quit before winning the grand prize.

[Mr. Armadillo]

(mathematically, the game can have a maximum of fifteen correct solutions):

Hadn't thought about it, but that makes sense.  There's 180 total possible combinations (6! divided by two for the second shelf being interchangeable, divided by two again for the third shelf), divided by 6 because the most expensive item trivially has to be on the top, divided by 2 again since the 30 remaining combinations can be divided into 15 pairs that are identical save for swapping the two middle shelves, and mathematically one of the two middle sums has to be more expensive than the other, so you can eliminate the combination in each pair with the more expensive pair on the lower shelf.