# 3 card monte carlo variation

A friend wants to play a betting game with you. There are $3$ upside-down cards on the table $2$ black and $1$ red. Your job is to find the red card. For every dollar you bet he will give you $2$ to $1$ odds (i.e. you win $2$ dollars for every dollar you bet).

How do you solve this? What percentage of games would you have to win to come out on top?

What I have tried:
$\frac{1}{3}$ chance of winning $= 33.33%$
$2:1$ odds$= 66.66%$

$66.66 * 33.33 = 22.22$
Is this right? Or am I doing this wrong? Would you have to be correct more that $22%$ of the time.

• If you are correct 25% of the time and incorrect 75% of the time... lets look at an example of when that is true. If you play four games and win once and lose three times, how much is your net gain? You win $2$ for having won, but you lose $1$ for each time you lost... so if you won once in those four games you will have lost a dollar overall. So... no, $\frac{2}{9}$ is not correct. – JMoravitz Jul 29 '16 at 5:54
• A pet peeve, $\frac{3333}{9999}=\frac{1}{3}\neq 33.33\% = \frac{3333}{10000}$. Use $\approx$ to indicate "approximately equal" but only use equals signs for equality. – JMoravitz Jul 29 '16 at 5:56

Your question of "What percentage of games would you have to win to come out on top?"

If you happen to win $x\%$ of the games, then you will lose $(100-x)\%$ of the games.

For each game you win, you gain $\$2$. For each game you lose, you have a net loss of$\$1$.

The net gain per game on average with having won $x\%$ of the time will be $2x-1(100-x)$

You ask, what $x$ will make it so that this number is positive.

$2x-100+x>0$

$3x>100$

$x>\frac{100}{3}$

You will have needed to win greater than $\frac{100}{3}\%$ of the games, or worded another way, you will have needed to win greater than $\frac{1}{3}$ of the games to have a positive net outcome.

The question of how many games you are expected to win (assuming you guess randomly) and thus what your expected payout can be answered as it appears in Michael's answer. ($\frac{1}{3}$ chance to win, $\frac{2}{3}$ chance to lose, so your expected outcome is $2\cdot \frac{1}{3} - 1\cdot\frac{2}{3}=0$ and the game is fair)

Your expected value of the game is 0. In other words, the game is fair.

Consider this situation: the red card is the first card and the other cards are not. Option 1: You bet x and guess the first card. You win, which implies that you get x back plus 2 * x.

Option 2: You bet x dollars and guess the second card. You lose, which implies that you lose x dollars.

Option 3: Similar to Option 2 but you guess the third card instead.

Given that each option is equally likely and the winnings (and "losings") cancel out. The game is fair in this situation. Similar arguments apply to situations where the red card is in 2nd position or 3rd position. Thus the game is fair. (You should win 33.33% of the time.)

• This does not answer the OP's question of "What percentage of games would you have to win to come out on top?" – JMoravitz Jul 29 '16 at 5:59