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Ask The Wizard #359
Here is another puzzle from The Riddler .
Imagine you have a bag filled with 100 marbles, each painted either red, blue, or green. If you randomly select three marbles from this bag, there's a 20% chance of pulling one marble of each color. Can you determine the quantity of each colored marble in the bag? Keep in mind that I haven't specified if the marbles are drawn one after another or all at once.
Let's analyze the scenario assuming we replace each marble after drawing. Denote r, b, and g as the counts of red, blue, and green marbles. The likelihood of drawing one marble of each color would be calculated as 6*(r/100)*(b/100)*(g/100). Setting this value to equal 0.2 gives us the equation to solve:
6*(r/100)*(b/100)*(g/100) = 0.2
6*r*b*g = 200000
The number 6 does not divide 200,000 into whole numbers evenly, indicating that integer values for r*b*g = 33333.333.. do not exist. Thus, we can rule out the case of drawing with replacement.
Now, let’s switch to the assumption where we do not replace the marbles drawn. In this scenario, the probability of drawing one of each color changes to r*b*g/combin(100,3) = 0.2. Let's attempt to derive a solution using this model.
r*b*g/161700 = 0.2
r*b*g = 32340
The prime factorization of 32340 is 2*2*3*5*7*7*11.
We have to appropriately allocate our factors among r, b, and g while ensuring that the equation r+b+g=100 holds. For instance, we might consider the following combination:
r = 2*3*5 = 30
b = 2*11 = 22
g = 7*7 = 49
This attempt succeeds in utilizing all the prime factors accurately, but the sum r+b+g = 101 does not comply with our original constraint, making it an invalid solution.
Unfortunately, I had to resort to a brute force looping algorithm to correctly derive the values of r, b, and g, which turn out to be 21, 35, and 44 in some order.
What are your thoughts on this approach in craps to transform a $5 wager into $1,200? It starts with a $5 bet on the number 4. If that succeeds, you place the winnings on 5. If that bet also wins, you continue the pattern by betting on 6, 8, 9, and finally 10. Assume the player adds $1 to the bet after each win on the 4 and 8 to round off the numbers.
The likelihood of winning on the 4 is calculated as 3/(3+6) = 3/9 = 1/3. A successful place bet on the 4 pays out at 9 to 5, so a winning bet means you'll have $9 plus your original $5, bringing the total to $14.
Next, the player places a $15 wager by adding $1 to the previous winnings for the 5. The odds of winning on the 5 are 4/(4+6) = 4/10 = 2/5. The payout for a winning bet on the 5 is 7 to 5, which means a total of $36 after winning this round. The probability of achieving this outcome is (1/3)*(2/5) = about 13.33%.
Following that, the player advances with a $36 bet on the 6. The chances of winning here are determined as 5/(5+6) = 5/11. Winning this bet, which pays 7 to 6, gives a total of $78 following the win. The likelihood of reaching this point is (1/3)*(2/5)*(5/11) = approximately 6.06%.
The player then bets $78 on the 8. The odds for the 8 are likewise 5/(5+6) = 5/11. If successful, this bet also pays 7 to 6, raising the total to $169. The probability of advancing this far is calculated to be (1/3)*(2/5)*(5/11)^2, making it about 2.75%.
Next, the player contributes an additional $1 from their own funds to reach a $170 bet on the 9. The winning odds here are 4/(4+6) = 2/5, with a payout of 7 to 5. Should they succeed, the total amounts to $408 after securing that win, bringing the total probability to (1/3)*(2/5)^2*(5/11)^2, roughly 1.10%.
At the final step, they are prepared to bet on the 10. Given a lower house edge on the buy bet, let’s assume this is the route taken. You did not clarify whether the player pays the commission ahead of time or only when winning. We’ll first assume the commission is prepaid. Under these rules, the betting amount should ideally be divisible by $21. Let's presume the player wagered $380 on the 10, covering a 5% commission of $19 upfront while pocketing the remaining $9 from their $408.
Calculating the probability of winning on the 4 gives us the same odds, 3/(3+6) = 3/9 = 1/3. Winning a $380 bet means an earning of $760, totaling $760 + $380 = $1,140. The combined probability of achieving this point is (1/3)^2*(2/5)^2*(5/11)^2, which evaluates to about 0.37% or 1 in 272.25.
It's important to remember the player wagered $5+$1+$1 throughout the process, ultimately pocketing $9 after claiming a win on the 9, leading to a net gain of $1,142. If we determine the house edge as the expected loss compared to the original $5 bet, that equates to $1.06 or 21.16%.
Now, let’s examine the scenario where the commission is only paid on winning the 10. In this case, buy bets on the 10 should be divisible by $20, presuming the player keeps $8 and bets the remaining $400.
A winning $400 bet would yield $780 in winnings, making a grand total of $780 + $400 = $1,180.
Don't forget, the player had initially wagered $5+$1+$1, pocketing $8 after winning on the 9, resulting in a final net win of $1,181. Defining the house edge from the expected loss based on the original $5 bet, this results in $0.92 or 18.44%.
Achieving the target of $1,200 seems elusive unless the player contributes more money after winning on the 9 or at another point. Although I can't officially advocate for this strategy in terms of profitability, it indeed appears to promise a thrill and enjoyable experience.
At the Golden Lion casino located in Panama City, the dealer offers insurance against a 10-point card being shown. The payout for insurance remains consistent at 2 to 1 if a blackjack is achieved by the dealer with an ace visible. This play involves six decks. What would be the house edge if the dealer shows a 10?
Oh dear! I encountered this alarming and misguided rule at the MGM in Macau as well.
The likelihood that the dealer’s hidden card is an ace when a 10 is showing is calculated as (6*4)/(6*52-1) = 7.717%. Consequently, the expected return from this scenario would be 0.077170*2 + 0.922830*-1 = -0.768489, denoting that the house edge reaches a staggering 76.85%.