Find the most optimal strategy to increase your money

Question

There is a race consisting of 10 individual races in a row. You are able to bet on the racecars for a chance to multiply your money. The potential payout and chances of winning per racer are listed below.

Blue

- 30% Chance of winning

- 2x Payout on winning bids

Red

- 25% Chance of winning

- 2.5x Payout on winning bids

Purple

- 18% Chance of winning

- 3x Payout on winning bids

Yellow

- 10% Chance of winning

- 5x Payout on winning bids

Black

- 8% Chance of winning

- 7x Payout on winning bids

Green

- 5% Chance of winning

- 10x Payout on winning bids

Orange

- 3% Chance of winning

- 15x Payout on winning bids

Pink

- 1% Chance of winning

- 50x Payout on winning bids

At the end of the race, bets placed on the top 3 winning racecars will be paid back to the player depending on the sheep's placement

1st Place: 100% Bet Payout

2nd Place: 40% Bet Payout

3rd Place: 10% Bet Payout

All other bets placed on non top 3 place racecar will be lost

You can safely assume that if one is likely to win but doesn't win then they are statistically most likely to come second and so on

During the 10 rounds, what is the most efficient bets to increase your money?

Answer 1

What you are asking for is two fold.

1. What is the expected value or the likely return on investment (ROI) given some probability of winning?
2. What races have a positive expected ROI?

The expected value, or average ROI is defined as a sum:

ROI=∑P(event i)*(Return/Payout from event i)

for this problem

ROI=P(1st)*Payout(1st)+P(2nd)*Payout(2nd)+P(3rd)*Payout(3rd) + P('losing')*Payout('losing')

The payout for losing is negative, since that's the cost of the bet, which also varies depending on the racer & other factors.

More over, we know the probability of coming in first, and the probability of not coming in first, but we don't know the probability of coming fourth or after.

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Based on the information given, the most obvious interpretation is that there are four possible situations:

1. Racer gets first place with a known probability
2. Racer gets second place with an unknown probability
3. Racer gets third place with an unknown probability
4. Racer doesn't win first place with a known probability

So to answer your overall question: Insufficient information.

What is the probability of getting first, second, and third place?

The information provided only gives the chance of winning, which is presumably the chance of getting 1st place. To answer your question, we need the additional information.

You say that they are likely to come in second or third if they don't come in first, but at what probability? For example blue has a 30% chance of getting first and a 70% of not getting first. What is the probability that, given they don't come in first, they come in second?

If I try to use Bayes' Theorem for blue to help clarify we immediately run into a problem:

A: The probability that the racer comes in second (unknown=x)

B: The probability that the racer doesn't come in first (70%)

P(A|B) = P(B|A) * P(A) / P(B) = P(B|A) * x / .7

Aside from the variable x, which prevents us from calculating P(A|B), the only other point of clarification is P(B|A), or the probability of not coming in first given that the racer came in second. Since the racer cannot have two finishing places, the probability here is 1. That is to say, if the racer comes in first, then they cannot have come in second; if the racer comes in second, they cannot have come in first. So we can simplify the formula to:

P(A|B) = x / .7 = 10x / 7

So to calculate the probability that the racer comes in second we either need to know the conditional probability of coming in second given that they don't come in first, or just be told the probability of coming in second.

This same exact problem prevents us from calculating the probability of third place, and the probability of coming in fourth or after and completely losing the bet.

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So I reiterate, Insufficient Information.