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Direct: https://www.youtube.com/watch?v=S7nXtS3GwYQ

Here are the rules.

- There is a board containing tiles 1 to 30. Behind each number there is a C, A, R, or Car, distributed randomly. The distribution is as follows: C = 11, A = 11, R = 6, Car = 2
- The player immediately gets to pick two numbers.
- The player has an opportunity to win up to 3 more numbers playing a pricing game, which you can see in the video.
- The player wins a car if he can spell "C-A-R" (in any order) with the numbers he is given/earned. This can be done either with at least one each of the three letters in C-A-R, or at least one number with "CAR."

There is also a cash offer to surrender the game, but let's not confuse the issue with that, yet.

The question for now is what is the probability of winning the car given 2, 3, 4, and 5 tiles?

Quote:Joeman1 - [(28/30) * (27/29)] = 13.1%

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I agree.

How about 3, 4 and 5 tiles?

i - Pick 'CAR' at any time:

1 - [(28/30) * (27/29) * (26/28)] = 19.3%

ii - Pick one each of 'C', 'A', & 'R,' in any order. Since picking 'C' or 'A' are equally likely, the only variable is in which point you pick the 'R'. So...:

a. - Prob of spelling 'C-A-R' picking the 'R' first:

(6/30)*(22/29)*(11/28) = 5.96%

b. - Prob of spelling 'C-A-R' picking the 'R' second:

(22/30)*(6/29)*(11/28) = 5.96%

c. - Prob of spelling 'C-A-R' picking the 'R' third:

(22/30)*(11/29)*(6/28) = 5.96%

Summing up all probabilities, I get 37.2%

Note that since all 3 probabilities in "ii" are the same, I am probably missing a simpler solution.

That's it for me for now. I'll have to leave the 4 and 5 tile solutions to somebody else.

ETA: A little off topic, but FYI, there is a free streaming service called PlutoTV that has a channel that plays old Bob Barker TPiR episodes 24/7.

Quote:JoemanI get 37.2%

ETA: A little off topic, but FYI, there is a free streaming service called PlutoTV that has a channel that plays old Bob Barker TPiR episodes 24/7.

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I agree on the 37.2%.

I may check that out. I miss the classic three Barkers Beauties.

To answer the question you're all wondering -- I'm a Holly man.

Quote:JoemanThat's it for me for now. I'll have to leave the 4 and 5 tile solutions to somebody else.

For 4, there are C(30,4) = 27,405 combinations of tiles

Combinations with both CAR cards: C(28,2) = 378

Combinations with one CAR card: 2 x C(28,3) = 6552

Combinations of C,C,A,R: 55 x 11 x 6 = 3630

Combinations of C,A,A,R: 11 x 55 x 6 = 3630

Combinations of C,A,R,R: 11 x 11 x 15 = 1815

Total winning combinations = 16,005

Probability of winning = 1067 / 1827, or about 58.402%

For 5, there are C(30,5) = 142,506 combinations of tiles

Combinations with both CAR cards: C(28,3) = 3276

Combinations with one CAR card: 2 x C(28,4) = 40,950

Combinations of C,C,C,A,R: 165 x 11 x 6 = 10,890

Combinations of C,C,A,A,R: 55 x 55 x 6 = 18,150

Combinations of C,A,A,A,R: 11 x 165 x 6 = 10,890

Combinations of C,C,A,R,R: 55 x 11 x 15 = 9075

Combinations of C,A,A,R,R: 11 x 55 x 15 = 9075

Combinations of C,A,R,R,R: 11 x 11 x 20 = 2420

Total winning combinations = 104,726

Probability of winning = 52,363 / 71,253, or about 73.49%

I recall watching this game on a TPIR episode when I was 10 or so. Contestant had revealed 4/5 cards and they were all the same letter (let’s say C). Bob offers him whatever the surrender value was. Guy declines and goes for it, hoping to get 1 of the CAR cards. No other way to spell car. At the time I thought it was foolish but may not have been a bad decision, depending on the value of the car to him relative to the surrender value and number of unknown cards remaining.

Edit/conclusion: of course he got the “CAR” card. Otherwise doubt I’d remember this.

Quote:ThatDonGuy

For 4, there are C(30,4) = 27,405 combinations of tiles

Combinations with both CAR cards: C(28,2) = 378

Combinations with one CAR card: 2 x C(28,3) = 6552

Combinations of C,C,A,R: 55 x 11 x 6 = 3630

Combinations of C,A,A,R: 11 x 55 x 6 = 3630

Combinations of C,A,R,R: 11 x 11 x 15 = 1815

Total winning combinations = 16,005

Probability of winning = 1067 / 1827, or about 58.402%

For 5, there are C(30,5) = 142,506 combinations of tiles

Combinations with both CAR cards: C(28,3) = 3276

Combinations with one CAR card: 2 x C(28,4) = 40,950

Combinations of C,C,C,A,R: 165 x 11 x 6 = 10,890

Combinations of C,C,A,A,R: 55 x 55 x 6 = 18,150

Combinations of C,A,A,A,R: 11 x 165 x 6 = 10,890

Combinations of C,C,A,R,R: 55 x 11 x 15 = 9075

Combinations of C,A,A,R,R: 11 x 55 x 15 = 9075

Combinations of C,A,R,R,R: 11 x 11 x 20 = 2420

Total winning combinations = 104,726

Probability of winning = 52,363 / 71,253, or about 73.49%

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I agree.

Quote:TinManI know the surrender aspect is for later but this story occurred to me:

I recall watching this game on a TPIR episode when I was 10 or so. Contestant had revealed 4/5 cards and they were all the same letter (let’s say C). Bob offers him whatever the surrender value was. Guy declines and goes for it, hoping to get 1 of the CAR cards. No other way to spell car. At the time I thought it was foolish but may not have been a bad decision, depending on the value of the car to him relative to the surrender value and number of unknown cards remaining.

Edit/conclusion: of course he got the “CAR” card. Otherwise doubt I’d remember this.

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We can get to it now.

Regarding your story, I remember once on the Punchboard game, somebody got the second highest prize and had one punch to go. He foolishly went for it -- and got the top prize.

I find from the start, with two tiles only, the indifference point on the car value is $15,263.16.

In your case, the player obviously had a probability of 1/26 of winning of winning the car, for an indifference point of $26,000. I don't know your age, but I would imagine low-end cars were not going for as much back then.

Quote:WizardI may check that out. I miss the classic three Barkers Beauties.

To answer the question you're all wondering -- I'm a Holly man.

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Technically, neither Diane nor Holly are "classic"; the "classic" two are Janice, and Anitra Ford.

How familiar are you with the stories of how well the three of them got along with Bob?

There has only been one tie in a showcase in the 50-year history of the show, and even that was in the old weekly syndicated (if you are familiar with Match Game PM, think of it as "TPIR PM") version back when it was hosted by Dennis James.

Both Bob and Drew always include "If you are the winner" when they explain how someone can win both showcases. In the history of the show (although in the first year or two, they did not have the rule where you could win both showcases - including the first episode where somebody got a showcase exactly right), only one contestant was within the range needed to win both, but the other contestant was closer - and that was by one dollar.

There was a period of a few years when the show would not only not offer foreign cars, but would not offer GM cars either. The foreign cars were banned because the producers wanted to feature "American-made" cars, and GM was banned because Bob pretty much demanded it after GM would not sign something that said that it did not use animals for testing of any sort. Most are familiar with the sudden disappearance of furs from the show (and, in fact, Bob made it quite clear that no episodes with furs can air in repeats - and, somehow, his estate can still mandate this), but it's not quite as well-known that, at some point, the ban went to pretty much anything with meat inside of it. "Pepperoni Hot Pockets" were suddenly replaced by "Cheese & Broccoli Hot Pockets" on Grocery Game, for example.

Quote:avianrandyJust seen a lady on TPIR win a dodge Omni playing 3 strikes without pulling a single strike! Price had 4 numbers. Bob wanted to say something on her last pull but didn't want to hex her.Thenumbers were 4,6,7,9. Very lucky and she had I think 8 draws. I had never seen a contestant win with no strike before. Was glad to see her win.

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It is thought that, originally, it was easier somehow to tell the numbers from the strikes apart by feel. There is another school of thought that some players would "peek" just before the chip was pulled out of the bag, and if they saw the red of a strike, they would let go of it before pulling it out, but this is much less likely, although there was a brief period when the strikes were colored white as well as the numbered chips.

Eventually, the producers realized how hard it was to win the game, especially with 5-digit cars, which is why it is only played for really expensive cars.

Quote:avianrandyIt is a tough game to win. I remember a couple years ago they played it for a Ferrari. With 6 numbers.at one point,the contestant pulled out an8 and said it was the first number of the car.i distinctly remember drew Carey chuckling this was going to take a while. I don't think they have played it for a 6 figure vehicle since. I guess the producers do t want to run in to days of our lives

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I may have mentioned this before in this forum, but even if they tell you the price of the car in advance, you would still have only a 3/8 (with 5 digits - 1/3 with 6) chance of winning the car.

Oh, and pardon me for being pedantic (as usual), but Days of Our Lives is "on another network" - you're thinking of The Young & the Restless.

Quote:DeucekiesI'll take the easy one. The probability of winning with two tiles is (2/30)+(2/30)=(4/30) or 13.3%.

Quote:Joeman1 - [(28/30) * (27/29)] = 13.1%

I disagree!

Deucekies: Almost! But the second chance no longer has 30 options! Chances are slightly better at 2/29...

Get either of 2 "CAR" wild cards from the 30 numbers on your first guess. (2/30)

+ If that's wrong, there remain 2 winners in 29 options on your second guess. (2/29)

That's (2/30) + (2/29) = 13.6%

(I have yet to figure out 3, 4, and 5 which would include all sorts of combinations.)

EDIT: I withdraw my disagreement! Thank you for the quick response and explanation!

Normally with these kind of puzzles it's easier to use pr(any hit) = 1 - (pr(miss 1)*pr(miss 2)...).

btw if one used your logic for the chance of winning the car with 15 picks then it would be 2/30+2/29+...2/16 > 1!

Quote:DaveInSanFran@charliepatrick, thanks for the quick response! These combinations get tricky! I was simplifying the dependency between picks. (My logic error has led me into a math refresher. This stuff is oddly fascinating! I'll take a crack at 3, 4, and 5 picks later after learning more about "N choose K" problems.)

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Welcome to the forum.

I get the feeling you may like it here.

Quote:DieterThanks for the welcome!Quote:DaveInSanFranWelcome to the forum.

I get the feeling you may like it here.

I'm stepping through this problem interactively with an AI learning tool. It's fascinating to learn that odds change when given a choice to proceed after each reveal versus revealing all at once. It's gonna take a while to get my head around this!

Sorry for my ignorance! If this was directed to me, I don't understand the comment.Quote:DieterAnd welcome to "introductory effects of removal"...

Quote:DaveInSanFranSorry for my ignorance! If this was directed to me, I don't understand the comment.Quote:DieterAnd welcome to "introductory effects of removal"...

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"Effects of Removal" is how probabilities change as picks are made or cards are dealt.