Hi laguy,
You wrote:
1. It's currently 5:15 - at what time (to a few decimal places of a minute) will the the minute hand meet the hour hand?
Let HH denote the position of the hour-hand between 5 and 6 o'clock. So, HH=5+x, where 0<x<1.
Let MH denote the position of the minute-hand. Then MH=12x. So, the condition of minute-hand meeting hour-hand is the same as MH=HH; i.e., 5+x=12x.
There is another way of doing this without using an equation, but instead a series.
You wrote:
2. You have a regular six-sided die, and you play the following game: Roll the die and you get $1 per number that comes up (ie roll a 1 get $1, roll a 6 get $6). If you are happy with your outcome after one roll, you can stop. If you would like to roll again you may do so up to two more times (so a max of 3 rolls). You will get whatever dollar amount you rolled last, so if you rolled 1-1-6, you would get $6. You can quit at any time, and have a max of 3 rolls. What is your strategy, and using that, what is the expected value for the game?
Nice problem. There are couple of ways I am aware of for doing this problem. Here's one:
First do the problem when the player has a max of only two rolls. Let's say if a number b or less comes up, then the player rolls again (for the second and last time). Otherwise the player sticks with b. Obviously b must be less than or equal to the expected winning on the second roll, which is 7/2. So, for b <=3, the player must roll again, otherwise must keep his winning.
Now going to the case of a max of three rolls. Let's say the first roll comes up a. If a is less than the expected amount that can be won by rolling again, then the player must roll again. Otherwise, the player must stick with a. But what is the expected amount that can be won if rolled again. By the previous paragraph, either the second roll (with a probability of 1/2) is less than or equal to three, in which case a third roll with an expected amount of 7/2 is in order, or the second roll (with a probability of 1/2) is greater than or equal to 4, in which case the expected amount is (4+5+6)/3.
So, the expected amount won by rolling again after the firs roll is
(1/2)(7/2)+(1/2)(4+5+6)/3=17/4=4.25
So, if a is less than or equal to 4.25; i.e., 1,2,3, or 4, then the player must roll again for a second time. If the first roll comes up a 5 or 6, then the player should stick with it.
After this it is easy to compute the expected amount of winning.
Another way to do this problem is to write a random variable strategy S in terms of three iid random variables X, Y Z, representing the three possible rolls, followed by computing expectation, and concluding with a little (calculus) partial differentiating for maximization, etc.
You wrote:
3. 50 coins in hand problem seen earlier, add up to $1, drop one, what is probability it was a penny
This is a partition problem, a good one, though. Will leave it at that.
You wrote:
4. You have 5 quarters on the table in front of you: four fair (regular two-sided coins) and one double-sided (both sides are heads). You pick one of them up at random, flip it five times, and get heads each time. Given this information, what is the probability that you picked up the double-sided quarter?
There are two different random variables or processes involved, one is the selection of the coin, the second the tossing of that coin. So, if we let c denote the random coin selection, and T(c,i) denote the i-th tossing of the coin c, BC biased double-sided coin, FC fair coin, then the question can be stated as follows:
x=P{c=BC | T(c,i)=H for i=1, 2, ..., 5} = ?
x=P{c=BC, T(c,i)=H for i=1, 2, ..., 5}/P{T(c,i)=H for i=1, 2, ..., 5}
=P{T(c,i)=H for i=1, 2, ..., 5 | c=BC}*P{c=BC}/P{T(c,i)=H for i=1, 2, ..., 5}
where the denominator is
P{T(c,i)=H for i=1, 2, ..., 5}=P{T(c,i)=H for i=1, 2, ..., 5 | c=BC}*P{c=BC} +
P{T(c,i)=H for i=1, 2, ..., 5 | c=FC}*P{c=FC}.
The rest involves plugging in the numbers.
There is another less mathematical way of doing this problem.
Let's differentiate one 'kind' of head from another. There are two ways that a head can come up, either from a fair coin or from a biased coin.
Let's say with a probability of p a fair coin was chosen, and with a probability of q=1-p, a biased coin was chosen. (A tree diagram will help a lot here!)
If a fair coin was chosen, then, let's say with a probability of r we get n many heads in a row (here n=5 and r=1/(2^n) ). So, flipping a randomly chosen coin will result in heads due to fair coin with a probability of p*r.
If a biased coin was chosen, then with a probability of s we get n heads in a row (here s=1). So, flipping a randomly chosen coin will result in heads due to biased coin with a probability of q*s.
Between the two 'kinds' of heads, we'd want to know with what probability they come from the biased coins:
FC=p*r
BC=q*s
So, the answer is x=q*s/(q*s + p*r). Nice. Isn't it? Cheers!
