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My roommate and I were discussing the stats around the Price is right game called Switcheroo.

The true goal of the game is to win the top prize, which is usually a car.

Let's simplify the game to the following:

Unknown to the player, there is a 5 digit master code made from the digits 1 - 5, non repeating (120 unique combinations). The game is played in 2 rounds.

First, a player picks a random 5 digit string. The goal is to get the first digit correct with the master code.

In the first round, after the player selects 5 digits, the host will tell them the quantity of digits they have correct, however not which specific digits.

After the host tells them, the player has the second round, where they can switch any of the numbers that they'd like. Again, the goal is to get the first digit correct.

We had a couple questions that we were discussing:

  1. Is it better to get 0 correct on the first round or 1 correct on the first round?
  2. In which cases should you switch the first digit vs keeping it?

My attempted answers: 1. If after the first round, the host tells you that you have 0 digits correct, you know you have to switch the first digit, and you have 4 remaining digits to choose from, so you have a 25% chance of winning.

If after the first round, the host tells you that you have 1 digit correct then: 20% of the time you have the first digit correct 80% of the time you have the first digit wrong.

If your strategy is to always change the first digit then:

  • When you have the first digit correct (20%), you will necessarily switch it to an incorrect number and lose with certainty.
  • When you have the first digit wrong (80%), you will have to keep one of the other digits constant (say the second digit), so you remove that digit AND whatever your first round guess was for the first digit, so you have three remaining digits to choose from and will win 1/3 of the time.

So your chances of winning if you have 0 digits right after the first round is 25% Your chances of winning if you have 1 digit right after the the first round is 80% * 33% = ~27%

So it is better to have 1 digit correct after the first round vs none.

It's been a long time since I've done any probability/combinatorics, so I'm still working on #2. I might write a program in R to brute force it. I'll probably do that to run simulations to check my work for #1.

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  • $\begingroup$ Draw the decision tree: it's not that large. $\endgroup$
    – whuber
    Jan 1, 2021 at 14:41

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