# Ask the Wizard #325

"Anonymous" .

A farmer plants 5 apple seeds. Each day, each seed will have a 1/3 chance of sprouting. What is the average time until all five trees sprout?

Let's work out way backwards. If there is one seed left that hasn't sprouted, it will take on average 1/p days to sprout, where p is the probability of sprouting any given day. Since p = 1/3, it will take on average 3 days to sprout. Let's call that t_{1} = 3.

What if there are two seeds left? There is a p^{2} = 1/9 chance both will sprout in the next day and we're done. The chance one will sprout the next day is 2×p×q, where q is the chance of not sprouting. Thus, the chance of one seed sprouting is 2×(1/3)(2/3) = 4/9. The chance of neither seed sprouting is q^{2} = (2/3)^{2} = 4/9. Let's call the expected number of days with two seeds t_{2}.

t_{2} = 1 + (4/9)×t_{1} + (4/9)t_{2}

t_{2} = (1 - (4/9)) = 1 + (4/9)×t_{1}

t_{2} = (1 + (4/9)×3) / (1 - (4/9))

t_{2} = (21/9) / (5/9)

t_{2} = (21/9) × (9/5) = 21/5 = 4.2

What if there are three seeds left? There is a p^{3} = 1/27 chance all will sprout in the next day and we're done. The chance one will sprout the next day is 3×p×q^{2} = 3×(1/3)(2/3)^{2} = 12/27. The chance two will sprout the next day is 3×p^{2}×q = 3×(1/3)^{2}×(2/3) = 6/27. The chance of no seeds sprouting is q^{3} = (2/3)^{3} = 8/27. Let's call the expected number of days with three seeds t_{3}.

t_{3} = 1 + (6/27)t_{1} + (12/27)×t_{2} + (8/27)×t_{3}

t_{3} = 1 + (6/27)×3 + (12/27)×4.2 + (8/27)×t_{3}

t_{3} × (1 - 8/27) = (1 + 18/27 + 28/15)

t_{3} = (1 + 18/27 + 28/15) / (1 - 8/27) = 477/95 = apx. 5.02105263

What if there are four seeds left? There is a p^{4} = 1/81 chance all four will sprout in the next day and we're done. The chance one will sprout the next day is 4×p×q^{3} = 4×(1/3)(2/3)^{3} = 32/81. The chance two will sprout the next day is combin(4,2)×p^{2}×q^{2} = 6×(1/3)^{2}×(2/3)^{2} = 24/81. The chance three will sprout the next day is combin(4,3)×p^{3}×q = 4×(1/3)^{3}×(2/3) = 8/81. The chance of no seeds sprouting is q^{4} = (2/3)^{4} = 16/81. Let's call the expected number of days with three seeds t_{4}.

t_{4} = 1 + (8/81)×t_{1} + (24/81)×t_{2} + (32/81)×t_{3} + (16/81)×t_{4}

t_{4} = 1 + (8/81)×3 + (24/81)×4.2 + (32/81)×5.02105263 + (16/81)×t_{4}

t_{4} = (1 + (8/81)×3 + (24/81)×4.2 + (32/81)×5.02105263) / (1 - (16/81))

t_{4} = apx. 5.638056680161943319838056680.

What if there are all five seeds left? There is a p^{5} = 1/243 chance all 5 will sprout in the next day and we're done. The chance one will sprout the next day is 5×p×q^{4} = 5×(1/3)(2/3)^{4} = 80/243. The chance two will sprout the next day is combin(5,2)×p^{2}×q^{3} = 10×(1/3)^{2}×(2/3)^{3} = 80/243. The chance three will sprout the next day is combin(5,3)×p^{3}×q = 10×(1/3)^{3}×(2/3)^{2} = 40/243. The chance four will sprout the next day is combin(5,4)×p^{4}×q = 5×(1/3)^{4}×(2/3) = 10/243. The chance of no seeds sprouting is q^{5} = (2/3)^{5} = 32/243. Let's call the expected number of days with three seeds t_{5}.

t_{5} = 1 + (10/243)×t_{1} + (40/243)×t_{2} + (80/81)×t_{3} + (80/243)×t_{4} + (32/243)×t_{5}

t_{5} = (1 + (10/243)×t_{1} + (40/243)×t_{2} + (80/81)×t_{3} + (80/243)×t_{4}) / (1 - (32/243))

t_{5} = (1 + (10/243)×3 + (40/243)×4.2 + (80/243)×(477/95) + (80/243)×5.63805668) / (1 - (32/243))

t_{5} = apx. 6.131415853.

This problem is adapted from a similar problem by Presh Talwalkar of Mind Your Decisions.

There is a teepee with radius 1 meters and a slant height (distance from any point on bottom edge to tip) of 4 meters. You wish to stake a rope to anywhere on the base of the teepee, wrap it entirely around the teepee once somehow, and tie the other end to the stake where you started.

- What is the minimum length of rope needed?
- Assuming the minimum distance is used, how close does any point on the rope get to the tip?

"Anonymous" .

- What is the minimum length of rope needed? A: sqrt(2)*4 = apx. 5.6569.
- Assuming the minimum distance is used, how close does any point on the rope get to the tip? A: 2*sqrt(2) = apx. 2.828427125.

Assume the bottom of the teepee is bare ground. In other words, the teepee is just the walls and not the base. Since the radius was 1, the diameter of the base of the teepee is 2*pi.

Cut the teepee from any point on the base to the tip and lay the material flat.

The curved part of this slice spape will still be 2*pi. Since the slant height was 4, if this slice were extended to a full circle, the radius would be 8*pi. Thus, this slice is 1/4 of a circle.

With sides of length 4 and angle of 90-degrees, the hypotenuse of the triangle between the three points has length sqrt(2)*4 = apx. 5.6569. If you reassembled the teepee, this distance would be the length of the rope.

Laying the slice flat again, using the Pythagorean formula, it is easy to see the closest point from the hypotenuse to the tip of the teepee is 2*sqrt(2) = apx. 2.828427125.

This question is raised and discussed in my forum at Wizard of Vegas.

I was playing 10-play video poker and held a pair after the deal. All ten hands then improved to a four of a kind on the draw. What are the odds?

"Anonymous" .

The probability of a pair improving to a four of a kind is 45/COMBIN(47,3) = apx. 0.002775208.

The probability of that happening in ten out of ten hands is (0.002775208)^{10} = apx. 1 in 36,901,531,632,979,700,000,000,000.

That probability is like buying three independent and random Powerball tickets and winning on ALL three of them.

The explanation is that this is NOT a normal video poker game, with natural probabilities, in which each card had an equal chance to be drawn out of the remaining cards in the deck. No, this is what is called a "VLT," or Video Lottery Terminal. In such games, the outcome is predestined, regardless of how the player pays his hand. It is like a scratch card lottery ticket, but the outcome is displayed to the player like a game of video poker. You might ask what would happen if the player held all five cards. Then a genie would have come along to change some of the cards or the player would have won a bonus to get the final win to 2,500 credits.

This question is raised and discussed in my forum at Wizard of Vegas.