Ask The Wizard #330

30%

First, shoot 90% of the cowboys in the leg.

Next, focus on the remaining 10% still standing and shoot them in the arm. You'll need to shoot 75% more in the arm, so let's take that number from those who have already been shot in the leg.

So, now we're at:

Leg only 15% (90% - 75%)
Arm only 10%
Both 75%
Neither 0%

Total leg: 90%
Total arm: 85%

Now, let’s address the gut injuries (80%). Target the 25% who have only one injury in the gut first. We still need to shoot 55% more, which we'll take from those who have sustained both types of injuries. So at this point, we have:

Leg & gut 15%
Arm & gut 10%
Leg & arm 20% (75% - 55%)
All three 55%
One injury 0%
Zero injuries 0%

Finally, let’s examine the 75% who suffered head injuries. Start by shooting the 45% who have exactly two injuries. After that, you still need to shoot 30%, which will come from the 55% who have all three other injuries. This leaves us with:

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%

Imagine there are 20 cowboys. This number is convenient because it allows for easy calculations with the probabilities involved, as everything is divisible by 5%, and 5% of 20 equals 1.

Arrange them in a line. Begin with the first cowboy on the left, shooting 90% of them, which amounts to 18 cowboys in the leg. Then, create a chart where the cowboy numbers are displayed along the top and their total injuries are listed along the left side, as shown.

Next, you need to shoot 85%, which is equivalent to 17 cowboys in the arm. Start shooting from the two cowboys who weren’t hit in the leg. Now you have 15 more to adjust. Return to the cowboy on the left and proceed down the line, targeting a total of 15 who have already been shot in the leg. Your injury chart should now look like this:

Following that, you should shoot 80%, or 16 cowboys in the gut. Start with the five cowboys that have only one injury. You have 11 more to go. Return to the first cowboy on the left and move down the row, targeting a total of 11 who have already been shot twice. Your injury chart should reflect this:

Next, you need to shoot 75%, or 15 cowboys in the head. Begin with the nine cowboys who have only two injuries. You’ll still need to shoot 6 more. Go back to the leftmost cowboy and move down the line, targeting a total of 6 who have been shot three times. Your injury chart will look as follows:

From this analysis, it is apparent that 6 cowboys have been shot four times, while 14 have been shot three times. Hence, the highest possible percentage of cowboys sustaining only three injuries stands at 14 out of 20, which equates to 70%.

In a general scenario, if the probabilities are labeled as a, b, c, and d, the maximum proportion of survivors can be expressed as 1 minus the sum of those probabilities, provided that the total sum is between 3 and 4.

Special thanks to CharliePatrick from the Wizard of Vegas forum for contributing this insightful solution.

The likelihood of all four players receiving a straight is calculated as (64/220) * (27/84) * (8/20) * 1, yielding a result of 216/5775, which simplifies to 72/1925, resulting in an approximate probability of 3.74%.

Deal to each player sequentially. The chance that the first player receives one of each rank is calculated as 4^3 divided by combin(12,3), or 64/220.

Assuming the first player has achieved a straight, there will be a remaining deck that consists of three cards per rank. The chance that the second player also receives one of each rank is found to be 3^3/combin(9,3), equating to 27/84.

If the first two players end up with a straight, the remaining deck now contains only two cards per rank. The probability of the third player getting one of each rank is 2^3 divided by combin(6,3), which calculates to 8/20.

If the first three players have all gotten a straight, just one card of each rank remains in the deck. Clearly, these three cards will also form a straight.

Therefore, the overall probability of all four players achieving a straight is (64/220) * (27/84) * (8/20) * 1, culminating to 216/5775, or approximately 72/1925, equivalent to about 3.74%.

89

1. If the frog can only jump one foot, there is evidently only one way to reach his target. Keep in mind that the frog must not overshoot the desired distance.
2. For a two-foot jump, the frog can choose between two distinct approaches: (1) two consecutive one-foot jumps, or (2) one two-foot jump.

In the case of a three-foot jump, the frog could be one foot or two feet away prior to his last leap. There is only one way to be two feet short (just one two-foot jump), while there are two ways to be one foot away (as explained in the previous step). Therefore, there are three different ways for the frog to reach three feet, confirmed by the sequences (1) 1+1+1, (2) 1+2, and (3) 2+1.

For a four-foot jump, the frog can land two or three feet away before making his final jump. From the previous analysis, there are 2 ways to be two feet short, and 3 ways to be one foot short. Consequently, there are 5 distinct methods to cover four feet, which can also be verified by the jumps (1) 1+1+1+1, (2) 1+1+2, (3) 1+2+1, (4) 2+1+1, (5) 2+2.

When aiming for a jump of 5 feet, the frog can be either 3 or 4 feet away from reaching the total distance on the last leap. Following the earlier steps, there are 3 pathways to be 2 feet shy and 5 pathways if only 1 foot remains, thereby resulting in a total of 3+5=8 distinct ways to reach five feet. This can be substantiated with the combinations (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.

Is a pattern becoming clear to you? It resembles the Fibonacci sequence. Using this same rationale, there are 89 different methods for the frog to precisely achieve a total of ten feet.