Easy math puzzles

Let's say that you start with 15 ml of milk and 250 ml of coffee. You move one teaspoon (5 ml) of milk into the coffee mug.

The ratio of coffee to total volume becomes 250/255 which is about 98%. Then, you take one teaspoon (5ml) of the 98% coffee + 2% milk mixture and dump it into the milk container. This teaspoon is 98% coffee and 2% milk, so .98*5ml coffee = 4.9 ml and 0.1 ml milk.

The milk container had 10 ml of milk after a teaspoon of it was removed. After adding back 5 ml, its volume is again 15 ml, and most of that, but not all, that is returned (98%) is coffee.

The amount of coffee in the milk cup is 4.901961 ml. The amount of milk in the coffee cup is 5 ml.

Quote: KevinAA

Let's say that you start with 15 ml of milk and 250 ml of coffee. You move one teaspoon (5 ml) of milk into the coffee mug.

The ratio of coffee to total volume becomes 250/255 which is about 98%. Then, you take one teaspoon (5ml) of the 98% coffee + 2% milk mixture and dump it into the milk container. This teaspoon is 98% coffee and 2% milk, so .98*5ml coffee = 4.9 ml and 0.1 ml milk.

The milk container had 10 ml of milk after a teaspoon of it was removed. After adding back 5 ml, its volume is again 15 ml, and most of that, but not all, that is returned (98%) is coffee.

The amount of coffee in the milk cup is 4.901961 ml. The amount of milk in the coffee cup is 5 ml.
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I'm not sure where your mistake is, but here are the amounts and ratios after each movement.

After first movement

Cup	Coffee	Milk	Total	Ratio Coffee	Ratio Milk
Coffee	250	5	255	0.980392	0.019608
Milk	0	10	10	0.000000	1.000000

After second movement

Cup	Coffee	Milk	Total	Ratio Coffee	Ratio Milk
Coffee	245.0980392	4.901960784	250	0.980392	0.019608
Milk	4.901960784	10.09803922	15	0.326797	0.673203

Note the milk in the coffee cup = coffee in the milk cup = 4.901960784.

We at least agree there is 4.901960784 ml of coffee in the milk cup. That had to displace the same amount of milk that used to be there. Where else could it have gone but the coffee cup?

OK, I get it now. That was a good one!

You flip a fair coin 50 times. What is the probability of never seeing three or more heads in a row? A numeric answer to six significant digits will suffice.

Exact answer should be the 52nd tribonacci number divided by 2^50. There is a closed form formula for tribonaccis but it’s much longer than the Fibonacci formula.

Very easy to Markov the answer though I consider that brute forcing it

Quote: Wizard

I've been working on this on my own. I found this video, by one of my favorite YouTube channels, to be very useful in finding the nth term in the Fibonacci sequence. From there, I can see how to adapt the logic for the dice question.

Direct: https://www.youtube.com/watch?v=ITSbuT9ojOw

By the way, this was hardly an Easy math puzzle.
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This was about the 500th puzzle on this thread that wasn’t “easy”

Quote: Ace2

Exact answer should be the 52nd tribonacci number divided by 2^50. There is a closed form formula for tribonaccis but it’s much longer than the Fibonacci formula.
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It does not help that determining a closed form for a recursive sequence that depends on the three previous terms requires finding the roots to a cubic equation.

I'm still looking for a simple Markov Chain answer, in part to verify my own answer.