Ask the Wizard #330
In an old western saloon, an argument over a game of cards escalated into all the cowboys in the vicinity drawing their guns and shooting at each other.
When the smoke finally cleared, 90% of the cowboys had been shot in the leg, 85% had been shot in the arm, 80% had been shot in the gut, and 75% had been shot in the head. Surprisingly, only the cowboys who received all four types of wounds died in the big gunfight.
What is the smallest possible percentage of cowboys who ended up being laid to rest?
First, shoot 90% of the cowboys in the leg.
Next, shoot the 10% left standing in the arm. You have 75% more to shoot in the arm, so take them from those already shot in the leg.
So, now we're at:
Leg only 15% (90% - 75%)
Arm only 10%
Total leg: 90%
Total arm: 85%
Next, let's go onto the gut injuries (80%). Shoot the 25% with just one injury in the gut. We have 80%-25% = 55% more people to shoot. We will take that 55% from the pool of people with both injuries. So now we're at:
Leg & gut 15%
Arm & gut 10%
Leg & arm 20% (75% - 55%)
All three 55%
One injury 0%
Zero injuries 0%
Finally, consider the 75% with head injuries. First, shoot the 45% with exactly two injuries. We have 30% more to go, so get them from the 55% with all three injuries. That leaves:
Head, Leg & gut 15%
Head, Arm & gut 10%
Head, Leg & arm 20%
Leg, arm & gut: 25% (55% - 30%)
All four 30%
Zero injuries 0%
One injury 0%
Two injuries 0%
Let there be 20 cowboys. We choose this number because all the probabilities involved are evenly divisible by 5% and 5% of 20 is 1.
Line them up in a row. Then, starting with the left, shoot 90% of them, which comes to 18, in the leg. Then make a diagram with the cowboy number along the top row and the total injuries of each along the left column, as follows.
Next, you'll need to shoot 85%, or 17 in the arm. Start with the two cowboys not shot in the leg. You have 15 more to go. Go back to the cowboy on the left and move down the row, shooting a total of 15 already shot in the leg. Your injury card should look like this:
Next, you'll need to shoot 80%, or 16 in the gut. Start with the FIVE cowboys with one injury only. You have 11 more to go. Go back to the cowboy on the left and move down the row, shooting a total of 11 already shot twice. Your injury card should look like this:
Next, you'll need to shoot 75%, or 15 in the head. Start with the nine cowboys shot twice only. You have 6 more to go. Go back to the cowboy on the left and move down the row, shooting a total of 6 already shot three times. Your injury card should look like this:
As you can see, 6 cowboys have been shot four times and 14 three times. Thus the maximum percentage that can sustain three injuries only is 14/20 = 70%.
For the general case, if the four probabilities are a, b, c, and d, then the maximum ratio that can live is 1-(a+b+c+d), as long as a+b+c+d >=3 and a+b+c+d <=4.
I would like to thank and credit Wizard of Vegas forum member CharliePatrick for this solution.
This question is asked and discussed in my forum, beginning with this post.
A casino dealer is working on a new Three Card Poker variant. She takes all the face cards from a standard deck and thoroughly shuffles them. She then deals 3 cards to Player #1, three cards to Player #2, three cards to Player #3, and the final three cards to Player #4. What is the probability that all four hands will contain a straight (J-Q-K of any suit)?
Deal to one player at a time. The proability the first player gets one of each rank is 4^3/combin(12,3) = 64/220.
Assuming the first player got a straight, there is a depleted deck of three of each rank left. The proability the second player gets one of each rank is 3^3/combin(9,3) = 27/84.
Assuming the first two players got a straight, there is a depleted deck of two of each rank left. The proability the third player gets one of each rank is 2^3/combin(6,3) = 8/20.
Assuming the first three players got a straight, there is a depleted deck of one of each rank left. Those three cards obviously form a straight.
Thus, the probability all four players get a straight is (64/220)*(27/84)*(8/20)*1 = 216/5775 = 72/1925 = 3.74%.
This question is asked and discussed in my forum at Wizard of Vegas, beginning with this post.
Let's assume that we are basing each system on the Player bet in baccarat. Let's also assume that we have a bankroll equal to 50x our bankroll with Oscar's Grind and the Labouchere. Let's make it 53x for the Fibonacci, which is the sum of the Fibonacci numbers 1,2,3,5,8,13, and 21.
Here is the probability of success of each:
- Labouchere: 97.53%
- Oscar's Grind: 97.69%
- Fibonacci: 97.93%
You may wonder why they are different if I keep saying that "all betting systems are equally worthless." The reason is that I qualify that statement with "Measured by total money lost to total money bet." The Fibonacci has the greatest probability of success because the player bets less, on average. The other two involve a greater average amout bet, which is more opportunity to grind down the player's bankroll. The Labouchere, with the lowest probability of success, has the highest amount bet, which let's the player enjoy the experience longer. Overall, here is the ratio of the average amount bet to winning goal of each:
- Labouchere: 20.95
- Oscar's Grind: 14.56
- Fibonacci: 9.59
All things considered, your choice of betting system should depend on why you're playing. If you want to maximize your probability of success, the Fibonacci is bet. If you want to play longer and bet more, the Lobouchere is best.
Since they are all based on the same bet, the ratio of money lost to money bet will always get closer to 1.235%, the house edge on the Player bet, the more you play, regardless of what system you may use.
- If the frog needs to jump one foot only, there is obviously one way only. Remember, the frog can't overshoot his goal.
- If the frog needs to jump two feet, there are two ways this can be done -- (1) 1 foot and 1 foot, or (2) 2 feet.
If the frog needs to jump three feet, he can be either one foot or two feet away before the final jump. There is one way to be two feet away, as shown from step 1, and two ways to be one foot away, as shown from step 2. Thus, there are three ways to jump three feet away. This is also easily verified as (1) 1+1+1, (2) 1+2, (3) 2+1.
If the frog needs to jump four feet, he can be either two feet or three feet away before the final jump. There are 2 ways to be 2 feet away, as shown from step 2, and 3 ways to be 1 foot away, as shown from step 3. Thus, there are 5 ways to jump four feet away. This is also easily verified as (1) 1+1+1+1, (2) 1+1+2, (3) 1+2+1, (4) 2+1+1, (5) 2+2.
If the frog needs to jump 5 feet, then he can be either 3 feet or 4 feet away before the final jump. There are 3 ways to be 2 feet away, as shown from step 3, and 5 ways to be 1 foot away, as shown from step 4. Thus, there are 3+5=8 ways to jump five feet away. This is also easily verified as (1) 1+1+1+1+1, (2) 1+1+1+2, (3) 1+1+2+1, (4) 1+2+1+1, (5) 2+1+1+1, (6) 2+2+1, (7) 2+1+2, (8) 1+2+2.Do you start to see a pattern? It is the Fibonacci sequence. Continuing with the same logic, there are 89 ways the frog can jump exactly a total of ten feet.