## Dino V2

Self-supervised learning from Meta claimed to beat state of the art without fine tuning.

## gpt4 on probability problems (part 3)

### A, B, and C are to fight a three-cornered pistol duel. All know that A’s chance of hitting his target is 0.3, C’s is 0.5, and B never misses. They are to fire at their choice of target in succession in the order A, B, C, cyclically (but a hit man loses further turns and is no longer shot at) until only one man is left unhit. What should A’s strategy be?

1. If A shoots and hits either B or C, then the remaining two players will fight a regular duel. Since B never misses, B would win against either A or C. So, eliminating one player is not advantageous for A.
2. If A intentionally misses his first shot, then B and C will have their turns. B knows that C is a more dangerous opponent because C’s probability of hitting the target is 0.5, while A’s is only 0.3. Therefore, B should target C, and if B hits C, the duel becomes a one-on-one match between A and B.

In the one-on-one duel, A has the first shot against B. A has a 0.3 chance of hitting B and surviving the duel. If A misses, B will shoot and win since B never misses. Therefore, A’s overall chance of winning the duel is 0.3 when he intentionally misses his first shot, which is better than the alternative of targeting either B or C in the beginning.

### coupon in cereal boxes are numbered 1 to 5, and a set of one of each is required for a price. with one coupon per box, how many boxes on the average are required to make a complete set?

Let’s break down the process of collecting a complete set of coupons:

1. The first coupon: Since you don’t have any coupons yet, you are guaranteed to get a unique coupon in the first box. The expected number of boxes to get the first coupon is 1.
2. The second coupon: The probability of getting a new coupon is 4/5 (since there are 4 new coupons you don’t have yet). The expected number of boxes to get the second coupon is 1 / (4/5) = 5/4.
3. The third coupon: The probability of getting a new coupon is 3/5 (since there are 3 new coupons you don’t have yet). The expected number of boxes to get the third coupon is 1 / (3/5) = 5/3.
4. The fourth coupon: The probability of getting a new coupon is 2/5 (since there are 2 new coupons you don’t have yet). The expected number of boxes to get the fourth coupon is 1 / (2/5) = 5/2.
5. The fifth coupon: The probability of getting the last coupon is 1/5 (since there is only 1 new coupon you don’t have yet). The expected number of boxes to get the fifth coupon is 1 / (1/5) = 5.

To find the expected number of boxes needed to complete the set, we add the expected number of boxes for each unique coupon:

Expected number of boxes = 1 + 5/4 + 5/3 + 5/2 + 5 = 1 + 1.25 + 1.67 + 2.5 + 5 ≈ 11.42

On average, you would need to buy approximately 11.42 boxes to collect a complete set of coupons numbered 1 to 5.

### A 1-meter stick is randomly broken into 4 pieces, what is the expected length of the shortest piece?

Let’s represent the break points on the stick as X1, X2, and X3, which are random variables uniformly distributed between 0 and 1. We can arrange these break points in ascending order, such that X1 <= X2 <= X3. Now, the stick is divided into 4 pieces with lengths Y1 = X1, Y2 = X2 – X1, Y3 = X3 – X2, and Y4 = 1 – X3.

We are interested in the expected value of the minimum length, which is given by min(Y1, Y2, Y3, Y4). To calculate the expected value of the shortest piece, we can find the expected value of each piece and then average them. Since the breaking points are uniformly distributed, the expected value of each piece is symmetric.

For the first piece, Y1 = X1, the expected value is:

E(Y1) = E(X1) = 1/4

For the second piece, Y2 = X2 – X1, the expected value is:

E(Y2) = E(X2) – E(X1) = 1/2 – 1/4 = 1/4

Similarly, for the third piece, Y3 = X3 – X2, the expected value is:

E(Y3) = E(X3) – E(X2) = 3/4 – 1/2 = 1/4

And for the fourth piece, Y4 = 1 – X3, the expected value is:

E(Y4) = 1 – E(X3) = 1 – 3/4 = 1/4

Now we can average the expected values of each piece to find the expected value of the shortest piece:

E(min(Y1, Y2, Y3, Y4)) = (E(Y1) + E(Y2) + E(Y3) + E(Y4)) / 4

E(min(Y1, Y2, Y3, Y4)) = (1/4 + 1/4 + 1/4 + 1/4) / 4

E(min(Y1, Y2, Y3, Y4)) = 1/4

So, the expected length of the shortest piece when a 1-meter stick is randomly broken into 4 pieces is 0.25 meters or 25 cm.

The answer is incorrect and I asked a similar question earlier and it is still wrong.