# Easier way of solving betting probability problem?

You have 2 blue and 2 red balls, in a box. You draw one out at a time (without replacement) and guess the color of the ball; you receive a dollar if you are correct. What is the dollar amount you would pay to play this game?

I think this is somewhat straightforward to calculate the maximum expected payoff using conditional expectation with the optimal strategy, which I think is to guess randomly ($$P(correct) = \frac{1}{2}$$), then to guess opposite ($$P(correct) = \frac{2}{3}$$), then to guess randomly again ($$P(correct) = \frac{1}{2}$$). I thought about conditioning on the event that you either guess correctly or don't at each turn, but it's still a relatively tedious calculation since I would have to condition for each turn.

Is there an insight that I'm missing that can make this easier/quicker to solve?

I see only three patterns that the balls can be in, and taking that they are drawn and seen one by one and that your next guess is the color with the higher probability, the scenario that emerges for each of the three patterns is as given below. A true guess $$T$$ gives a dollar and $$F$$ does not.

$$RRBB$$
$$TFTT$$
$$FFTT\quad$$2.5 dollars for this pattern

$$RBRB$$
$$TTTT$$
$$FTTT$$
$$FTFT\quad 3$$ dollars for this pattern

$$RBBR$$
$$TTTT$$
$$FTTT$$
$$FTFT\quad 3$$ dollars for this pattern

Since each pattern is equiprobable, game is worth $$\frac{8.5}{3}$$ dollars

If you win a dollar every time you guess correct, the expected payout is $$2.8\bar 3$$.

If you win a dollar just for the very end, if you made all correct guesses, it's just 1/6.

In both cases, there are 6 routes you can go down.

• I guess the question wasn't too clear about that
– Vons
Mar 28, 2021 at 6:12
• How did you get $.8\bar 3$ part Mar 28, 2021 at 6:31
• My calculation was $1/6(3+4+3+2+3+2)$
– Vons
Mar 28, 2021 at 6:38