The probability of winning "Lucky Seven" game in "The Price is Right"?

Question

I saw this YouTube video about a TV show called "The Price is Right" where a kid got insanely lucky.

In this they play the game "Lucky \$even" which is the first game they play. Details about the game:

Number of player(s): $1$
Goal: To guess correctly the $5$-digit price of the car to win it.
At start: Player is handed $\$7$. First digit of the price is already shown.
Gameplay: Remaining $4$ digits are shown one by one but before that, for each digit, the player has to guess it (from $0-9$). The difference between guess and actual digit is what they pay from $\$7$.
Win/Lose: Player wins if by the time the last digit is shown (and the difference collected), at least $\$1$ is still left with the player. Otherwise, the player loses (the car).

Example run of the game (from the video):

Price: $2\ \_\ \_\ \_\ \_$
Amount with player: $\$7$
Guess $\bf1$: $6$
Price: $2 1\ \_\ \_\ \_$
Amount with player: $\$2$
Guess $2$: $7$
Price: $2 1 8\ \_\ \_$
Amount with player: $\$1$
Guess $\bf3$: $9$
Price: $2 1 8 9\ \_$
Amount with player: $\$1$
Guess $\bf4$: $5$
Price: $2 1 8 9 5$
Amount with player: $\$1$
Game ends.
Status: Player wins the car!

Question:

What is the probability of winning such a game?

My attempt:

The more I thought about this problem the complex(er) it got. I couldn't think of a smarter way than to make a billion (metaphorical) cases. Such as in the following manner:
First the possibilities of the actual price digits may be from $0000$ to $9999$. From here we see that after $4999$ symmetricity will come into play so the problem is reduced to half. But from here on, I feel that, I'll have to, for each number, and for each digit within that number, calculate the probability separately. Thus, the billion cases.
I tried to make use of stars and bars but failed to do so.

How to overcome the problem? Or what is a better way to solve the problem?

Sidekick: Is there any strategy, whatsoever, in this game to maximize the chances of winning?

Answer 1

The best digit to guess is the one which minimizes the expected absolute value of the difference between the guess and the true digit.

This guess has a name: the median.

One you've figured out a sensible distribution of possible values for the prize (and how one does that is outside the scope of this question and indeed of mathematics itself, beyond that some sort of normal-ish distribution is probably reasonable), for each digit, find the weight of each possible remaining guess and select the median of those.

If you have no knowledge whatsoever of the possible price, you could decide that it's uniformly distributed, in which case a guess of 4 or 5 for each digit will perform best, with an expected money loss of \$2.50 per digit. Using this strategy, you will win the prize $12.53\%$ of the time.

Guessing completely randomly, the expected money loss is \$3.30 per digit. Using this strategy, you will win the prize $0.73\%$ of the time. (by the way, I calculated this in python completely naively and it took a matter of seconds, so it's not that much work if you have a computer)

Answer 2

Short answer:

The probability is very close to $\frac18$.

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My full answer has three sections:

1. Distribution of the digits
2. Winning strategy
3. Probability to win

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DISTRIBUTION

It seems that you assume that the numbers $0000$ to $9999$ are evenly distributed, which means that the probability for each quadruple of $4$ digits is equal. In this case, the probability that the first digit is $0$ is also equal to the probability that it is $9$. But you have no guarantee that this is the case.

However, if the price of the car is a real price that had to be paid to a seller, then this is guaranteed not to be the case. There is Benford's law, also known as the law of the first digit:

Probability of the first digit of the price

Benford analyzed the first digit of numbers he found in all sorts of tables and collections of numbers. Originally, he discovered this law in logarithm tables, but then also in lengths of rivers, areas of lakes, heights of buildings, population figures of cities, physical and mathematical constants, molecular weights, house numbers of celebrities, etc.

If these numbers were evenly distributed, then $11.11\%$ ($\frac19$) of all first digits would be the digit $1$ and you would find just as many nines. (The digit $0$ cannot be the first digit, so there are only $9$ possible digits that can be the first).

However, Benford found that about $30\%$ of all first digits are the digit $1$, and less than $5\%$ of all numbers start with the digit $9$. You can check this yourself by looking at the prices of any goods in any online store. $3$ out of $10$ start with the digit $1$, and only about $1$ out of $20$ start with the digit $9$.

Here is the law:

$$ P(D_1=d)=\log_{10}\left(1+{\frac{1}{d}}\right) $$

"The probability that digit $1$ ($D_1$) equals $d$ is the decadic logarithm of $1$ plus $1$ over $d$."

The result is these probabilities:

$d$	$P(D_1=d)$
$1$	$30.1\%$
$2$	$17.6\%$
$3$	$12.5\%$
$4$	$9.7\%$
$5$	$7.9\%$
$6$	$6.7\%$
$7$	$5.8\%$
$8$	$5.1\%$
$9$	$4.6\%$

^{Table 1}

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Probability of the second digit of the price = first digit to be guessed

But we already know what the first digit is. It is told to us by the host of the game. The first digit we have to guess is the second digit of the prize. How can Benford's law help us here?

We can combine the first two digits into a super digit, which can have values from $01$ to $99$, and we need to change the base of the logarithm from $10$ to $100$. We do not need the results for the super digits $01$ to $09$. Their sum is exactly $\frac12$, so the sum of the remainder is also $\frac12$. Thus, if we sum the probabilities of the super digits $10$ through $19$ and divide by $\frac12$ (i.e., multiply by $2$), we get exactly $30.1\%$, which is exactly the probability of the "normal" first digit $1$.

To get the dependent probability of the second digit, if we know the first digit, we just have to multiply the probability of the super digit by $2$ and divide by the result of the corresponding "normal" first digit.

Then we get a new table. Each column represents the first digit (i.e. the digit that the host told us) and each row represents the second digit of the prize, which is the first digit we have to guess:

$D_2$	$1$	$2$	$3$	$4$	$5$	$6$	$7$	$8$	$9$
$0$	$13.8\%$	$12.0\%$	$11.4\%$	$11.1\%$	$10.9\%$	$10.7\%$	$10.6\%$	$10.5\%$	$10.5\%$
$1$	$12.6\%$	$11.5\%$	$11.0\%$	$10.8\%$	$10.7\%$	$10.5\%$	$10.5\%$	$10.4\%$	$10.4\%$
$2$	$11.5\%$	$11.0\%$	$10.7\%$	$10.5\%$	$10.4\%$	$10.4\%$	$10.3\%$	$10.3\%$	$10.3\%$
$3$	$10.7\%$	$10.5\%$	$10.4\%$	$10.3\%$	$10.3\%$	$10.2\%$	$10.2\%$	$10.2\%$	$10.2\%$
$4$	$10.0\%$	$10.1\%$	$10.1\%$	$10.1\%$	$10.1\%$	$10.1\%$	$10.1\%$	$10.0\%$	$10.0\%$
$5$	$9.3\%$	$9.7\%$	$9.8\%$	$9.8\%$	$9.9\%$	$9.9\%$	$9.9\%$	$9.9\%$	$9.9\%$
$6$	$8.7\%$	$9.3\%$	$9.5\%$	$9.6\%$	$9.7\%$	$9.8\%$	$9.8\%$	$9.8\%$	$9.8\%$
$7$	$8.2\%$	$9.0\%$	$9.3\%$	$9.4\%$	$9.5\%$	$9.6\%$	$9.7\%$	$9.7\%$	$9.7\%$
$8$	$7.8\%$	$8.7\%$	$9.0\%$	$9.2\%$	$9.4\%$	$9.5\%$	$9.5\%$	$9.6\%$	$9.6\%$
$9$	$7.4\%$	$8.4\%$	$8.8\%$	$9.1\%$	$9.2\%$	$9.3\%$	$9.4\%$	$9.5\%$	$9.5\%$

^{Table 2}

(Note, that the 10 probabilities of each column add to exactly $100\%$.)

Using a similar technique, you can calculate the probability of the digits $0$ to $9$ being the third digit of the prize if you already know the first two digits. However, you will soon find that the distribution of the later digits very quickly resembles a uniform distribution, so it is not really worth doing this complicated calculation for the remaining digits.

But let me make it clear that this distribution is just a guess. You can try to find out the real distribution by analyzing the games that already have been played. Maybe the numbers have a completely different distribution. And everything that follows in my answer depends on this distribution.

But without knowing better, I assume a distribution according to Benford's law.

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STRATEGY

The probability of winning strongly depends on the strategy you use during the game. If you always bet that the next digit is a $9$, your chances are lower than if you always bet on a $4$. So before we can calculate a probability of winning, we need to determine a strategy. In game theory, we often try to find the best strategy, that is, the strategy with the highest estimated profit.

This game is a two-person game (you against the host), where the host made his choice at the beginning of the game by setting the price of the car (without telling you) and then revealing the first digit. Then you make exactly $4$ moves, and the host's reaction is only to reveal the previously determined number, so the host doesn't really react to your move. (The digits you are shown do not depend on your guess). Under these conditions, we can also pretend that the game is a one-person game (only you play).

In the example you gave, the first digit of the prize was $2$. If we guess that the next digit will be $0$, then we are correct with a probability of $12.0\%$ and do not have to pay anything (you can read the $12.0\%$ from column $2$, row $0$ in table 2.). But if the next digit is $1$, we pay $1$, and the probability of that is $11.5\%$. If the next digit is $2$, we have to pay $2$ dollars, and the probability of this is $11.0\%$, so the expected loss in this case is $0.22$ dollars ($22$ cents). We can add up all these expected losses and get in total the expected amount we have to pay if we guess $0$. Then we do the same for guessing $1$ and so on and in the end we have this table:

The number shown by the host is 2:

Guess	expected loss (Dollar)
$0$	$4.17$
$1$	$3.41$
$2$	$2.88$
$3$	$2.57$
$4$	Minimum $\rightarrow 2.47$
$5$	$2.57$
$6$	$2.86$
$7$	$3.34$
$8$	$4.00$
$9$	$4.83$

^{Table 3}

If the first digit of the price is $1$, we get this table:

Guess	expected loss (Dollar)
$0$	$3.93$
$1$	$3.21$
$2$	$2.73$
$3$	$2.49$
$4$	Minimum $\rightarrow 2.46$
$5$	$2.63$
$6$	$2.99$
$7$	$3.52$
$8$	$4.22$
$9$	$5.07$

^{Table 4}

Even if the first digit were any other number, the tip $4$ would always be the best choice. Only if the numbers are not distributed according to Benford's law, but are evenly distributed, then both the guess on $4$ and the guess on $5$ will bring the same expected minimum amount, exactly $2.5$ dollars.

And this pattern repeats for all the other digits you have to guess: The best strategy is always to guess $4$.

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PROBABILITY TO WIN

Just to make it clear once again: Our strategy is to always guess that the next digit will be $4$.

Let's calculate this task with equally distributed digits first, because that's easier.

There are $10000$ different combinations of digits. You pay nothing ($0$ Dollar) if this combination is $4444$. The probability of this is $\frac{1}{10000}=0.01\%$.

If the $4$ digits you had to guess are in fact $4443, 4445, 4434, 4454, 4344, 4544, 3444$ or $5444$, you have to pay $1$ dollar. That's $8$ combinations, so the chance of paying $1$ Dollar is $\frac{8}{10000}=0.08\%$.

There are formulas to calculate this, but I did it in a spreadsheet with $10000$ rows and just counted the results.

The chance of paying exactly $2$ dollars is $0.32\%$ and the chance of paying $2$ or less is therefore $0.01\%+0.08\%+0.32\%=0.41\%$.

If I understand the rules correctly, if you pay $6$ or less, you win. If you pay $7$ or more, you lose. The chance to pay $6$ or less is

$$ 12.53\%. $$

This is approximately $\frac{1}{8}$, so every $8$^th game is won if the numbers are evenly distributed.

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If the numbers are distributed according to Benford's law, you can calculate the probability of each combination of $4$ digits individually using this law. Note that these probabilities depend on the first digit of the price of the car. The rest of the calculation is the same as shown before.

But the influence of the first digit is very small. It turns out that all probabilities are between $12.59\%$ and $12.69\%$. So if you know that the numbers are distributed according to Benford, the probability increases a little bit compared to equally distributed numbers, but not much.