Here is another puzzle from The Riddler.

There are 100 marbles in a bag. Each marble is either red, blue, or green. If three marbles were drawn from the bag, the probability would be 20% of getting one of each color. How many of each color marble are in the bag. Please note I specifically didn't say whether the marbles are drawn with or without replacement.

Gialmere

The quantity of red, blue, and green marbles are 21, 35, and 44, in any order. They are drawn without replacement from the bag.

Let's try to work it out under a "with replacement" assumption. Let r, b, and g be the number of red, blue and green marbles, respectively. Then, the probability of drawing one of each color would be 6*(r/100)*(b/100)*(g/100). Setting that equal to 0.2, we can say:

6*(r/100)*(b/100)*(g/100) = 0.2
6*r*b*g = 200000 ​

6 does not divide into 200,000 evenly. Thus, there are no integer solutions possible for r*b*g = 33333.333... So, we can eliminate the case of drawing with replacement

Next, let's try the "without replacement" assumption. In this case, the probability of drawing one of each color is r*b*g/combin(100,3) = 0.2. Trying to solve for that...

r*b*g/161700 = 0.2
r*b*g = 32340

The prime factorization of 32340 is 2*2*3*5*7*7*11.

We need to distribute those factors between r, b, and g, while maining r+b+g=100. For example, we could try:

r = 2*3*5 = 30
b = 2*11 = 22
g = 7*7 = 49

While these correctly use up all the prime factors, r+b+g = 101, so it's not a valid solution.

I'm afraid I had to write a brute force looping program to get an the solution of r, b, and g having values of 21, 35, and 44, in any order.

What is your analysis of the following way in craps to parlay a \$5 bet into \$1,200? Start with a \$5 bet on the 4. If that wins, you parlay the winnings on the 5. If that wins, parlay those winnings on the 6. You keep going, betting on the 8, 9, and then 10. You may assume the player adds \$1 after a wins on the 4 and 8, to keep the bets in round numbers.

John Cokos

The probability of winning on the 4 is 3/(3+6) = 3/9 = 1/3. A place bet on the 4 pays 9 to 5, so if that bet wins you'll have a total of \$9 + \$5 = \$14.

Next, the player adds \$1 to his bet for a total of \$15 on the 5. The probability of winning on the 5 is 4/(4+6) = 4/10 = 2/5. A place bet on the 5 pays 7 to 5, so if that bet wins you'll have a total of \$21 + \$15 = \$36. The probability of getting at least this far is (1/3)*(2/5) = 13.33%.

Next, the player bets \$36 on the 6. The probability of winning on the 6 is 5/(5+6) = 5/11. A place bet on the 6 pays 7 to 6, so if that bet wins you'll have a total of \$42 + \$36 = \$78. The probability of getting at least this far is (1/3)*(2/5)*(5/11) = 2/33 = 6.06%

Next, the player bets \$78 on the 8. The probability of winning on the 8 is 5/(5+6) = 5/11. A place bet on the 8 pays 7 to 6, so if that bet wins you'll have a total of \$91 + \$78 = \$169. The probability of getting at least this far is (1/3)*(2/5)*(5/11)^2 = 10/363 = 2.75%

Next, the player adds \$1 from his pocket to the \$169 and bets \$170 on the 9. The probability of winning on the 9 is 4/(4+6) = 2/5. A place bet on the 9 pays 7 to 5, so if that bet wins you'll have a total of \$238 + \$170 = \$408. The probability of getting at least this far is (1/3)*(2/5)^2*(5/11)^2 = 4/363 = 1.10%

Finally, we're ready to bet on the 10. With the lower house edge on the buy bet, let's assume the player bets that. You didn't specify whether the player must prepay the commission or only pay it on wins. Let's look at prepaying the commission first. Under that rule, the bet amount should be evenly divisible by \$21. Let's assume the player bets \$380 on the 10, prepays a 5% commission of \$19 and pockets the other \$9 from his \$408.

The probability of winning on the 4 is 3/(3+6) = 3/9 = 1/3. A winning \$380 bet will pay \$760 in winnings, for a total of \$760+\$380 = \$1,140. The probability of getting at least this far is (1/3)^2*(2/5)^2*(5/11)^2 = 4/1089 = 0.37% = 1 in 272.25.

Remembmer the player bet \$5+\$1+\$1 on the way, but pocketed \$9 after the win on a 9, for a net win of \$1,142. If we define the house edge as the expected loss to the original \$5 bet, then it would be \$1.06/\$5.00 = 21.16%.

Next, let's see what happens if the commission is paid on wins only on the 10. There, buy bets on the 10 should be evenly divisible by \$20. Let's assume the player pockets \$8 and bets the other \$400.

A winning \$400 bet will pay \$780 in winnings, for a total of \$780+\$400 = \$1,180.

Remembmer the player bet \$5+\$1+\$1 on the way, but pocketed \$8 after the win on a 9, for a net win of \$1,181. If we define the house edge as the expected loss to the original \$5 bet, then it would be \$0.92/\$5.00 = 18.44%.

So, we can't quite get to \$1,200 unless the player pulls more money out of his pocket after a win on 9, or somewhere else along the way. I can't endorse this strategy in terms of value, but it does seem like it would have a high fun and excitement factor.

At the Golden Lion casino in Panama City the dealer offers insurance with a 10-point card up. They pay the same 2 to 1, if the dealer gets a blackjack, as with an ace showing. They use six decks. What is the house edge when the dealer has a 10 up?

PG

Ouch! I saw this same awful and ignorant rule at the MGM in Macau.

The probability the hole card is an ace with a 10 up is (6*4)/(6*52-1) = 7.717%. The expected return is 0.077170×2 + 0.922830×-1 = -0.768489. In other words, the house edge is 76.85%.