538 Riddler
This season, on the way to winning her fourth WNBA championship in her 17-year career, Sue Bird made approximately 50 percent of her field goal attempts. Suppose she and Seattle Storm teammate Breanna Stewart are interested in testing whether Bird has a “hot hand” — that is, if her chances of making a basket depend on whether or not her previous shot went in. Bird happens to know that her chances of making any given shot is always 50 percent, independent of her shooting history, but she agrees to perform an experiment.
In each trial of the experiment, Bird will take three shots, while Stewart will record which shots Bird made or missed. Stewart will then look at all the trials that had at least one shot that was immediately preceded by a made shot. She will randomly pick one of these trials and then randomly pick a shot that was preceded by a made shot. (If there was only one such shot to pick from, she will choose that shot.)
What is the probability that Bird made the shot that Stewart picked?
Now that LeBron James and Anthony Davis have restored the Los Angeles Lakers to glory with their recent victory in the NBA Finals, suppose they decide to play a game of sudden-death, one-on-one basketball. They’ll flip a coin to see which of them has first possession, and whoever makes the first basket wins the game.
Both players have a 50 percent chance of making any shot they take. However, Davis is the superior rebounder and will always rebound any shot that either of them misses. Every time Davis rebounds the ball, he dribbles back to the three-point line before attempting another shot.
Before each of Davis’s shot attempts, James has a probability p of stealing the ball and regaining possession before Davis can get the shot off. What value of p makes this an evenly matched game of one-on-one, so that both players have an equal chance of winning before the coin is flipped?
Quote: Ace2....What is the probability that Bird made the shot that Stewart picked?
There are eight ways, all equally likely, of having three shots; think binary they are YYY YYN YNY YNN NYY NYN NNY NNN.
Of these the last two do not have a shot following a success. Therefore there are six left.
According to the rules stated one of those six are picked at random.
Some of those six only have one possible pick (5) YNY, (4) YNN, (3) NYY, (2) NYN, and there is only one possible result (5) N (4) N (3) Y (2) N. each with Pr=1/6.
The other two possibilties there are two possible picks (7) YYY (6) YYN giving (7) Y (7) Y (6) Y (6) N, each with Pr=1/12 (since 1/6 to pick the trial, and 1/2 whether to pick the second or third shot).
So the combinations of successes are 7Y (1/12) 7Y (1/12) 6Y (1/12) 3Y (1/6) = 5/12
And the combinations of failures are 6N (1/12) 5N (1/6) 4N (1/6) 2\N (1/6) = 7/12.
So the chances that the Shot picked was a success = 5/12.
The trick is not to look at the Y's and N's and just count them up as they have different probabilities.
(7) Y Y Y - Y Y (either Y could be picked)
(6) Y Y N - Y N (either Y or N could be picked)
(5) Y N Y - N - (only N can be picked)
(4) Y N N - N - (only N can be picked)
(3) N Y Y - - Y (only Y can be picked)
(2) N Y N - - N (only N can be picked)
(1) N N Y - - - (this trial doesn't qualify as no throw was after a success)
(0) N N N - - - (this trial doesn't qualify as no throw was after a success)
Quote: Ace2And the Riddler
Now that LeBron James and Anthony Davis have restored the Los Angeles Lakers to glory with their recent victory in the NBA Finals, suppose they decide to play a game of sudden-death, one-on-one basketball. They’ll flip a coin to see which of them has first possession, and whoever makes the first basket wins the game.
Both players have a 50 percent chance of making any shot they take. However, Davis is the superior rebounder and will always rebound any shot that either of them misses. Every time Davis rebounds the ball, he dribbles back to the three-point line before attempting another shot.
Before each of Davis’s shot attempts, James has a probability p of stealing the ball and regaining possession before Davis can get the shot off. What value of p makes this an evenly matched game of one-on-one, so that both players have an equal chance of winning before the coin is flipped?
LeBron has a 25% chance of winning the game on the first shot (probability 1/2 of winning the toss and 1/2 of making the first shot)
If he doesn't, then each "situation" begins with Davis having the ball (either because he won the coin toss, or somebody missed a shot)
In each situation:
LeBron steals the ball and makes his shot with probability p/2
Davis makes his shot with probability (1-p)/2
If neither player makes a shot, the situation resets itself, so the probability that LeBron wins once the situations begin
is p / (p + (1 - p)) = p
The overall probability that LeBron wins is 1/4 + 3/4 p
This is 1/2 when 3/4 p = 1/4 -> p = 1/3
That’s what I got as wellQuote: ThatDonGuy
LeBron has a 25% chance of winning the game on the first shot (probability 1/2 of winning the toss and 1/2 of making the first shot)
If he doesn't, then each "situation" begins with Davis having the ball (either because he won the coin toss, or somebody missed a shot)
In each situation:
LeBron steals the ball and makes his shot with probability p/2
Davis makes his shot with probability (1-p)/2
If neither player makes a shot, the situation resets itself, so the probability that LeBron wins once the situations begin
is p / (p + (1 - p)) = p
The overall probability that LeBron wins is 1/4 + 3/4 p
This is 1/2 when 3/4 p = 1/4 -> p = 1/3
For any value of p between zero and one, this procedure will always return heads half the time and tails half the time. This is pretty remarkable! But there’s a downside to von Neumann’s approach — you don’t know how long (i.e., how many flips) the simulation will last.
Suppose I want to simulate a fair coin in at most three flips. For which values of p is this possible?
Extra credit: Suppose I want to simulate a fair coin in at most N flips. For how many values of p is this possible?”
My comment: I’m not sure I understand. The expected number of flips to get a head/tail or tail/head is 3 for a fair coin and goes up from there. I don’t see how it could be <= 3 for any value of p between 0 and 1 besides 0.5. And that’s the expected value... the problem makes it seem like you only get 3 flips.
Quote: Ace2“Mathematician John von Neumann is credited with figuring out how to take a biased coin (whose probability of coming up heads is p, not necessarily equal to 0.5) and “simulate” a fair coin. Simply flip the coin twice. If it comes up heads both times or tails both times, then flip it twice again. Eventually, you’ll get two different flips — either a heads and then a tails, or a tails and then a heads, with each of these two cases equally likely. Once you get two different flips, you can call the second of those flips the outcome of your “simulation.”
For any value of p between zero and one, this procedure will always return heads half the time and tails half the time. This is pretty remarkable! But there’s a downside to von Neumann’s approach — you don’t know how long (i.e., how many flips) the simulation will last.
Suppose I want to simulate a fair coin in at most three flips. For which values of p is this possible?
Extra credit: Suppose I want to simulate a fair coin in at most N flips. For how many values of p is this possible?”
My comment: I’m not sure I understand. The expected number of flips to get a head/tail or tail/head is 3 for a fair coin and goes up from there. I don’t see how it could be <= 3 for any value of p between 0 and 1 besides 0.5. And that’s the expected value... the problem makes it seem like you only get 3 flips.
Ace, here’s my take. Assume you know the p going in. For which p can you simulate a fair coin flip in at most three flips.
I can think of a couple of p where you can but haven’t thought of a systematic way to group them.
For example if p=0.5 then it takes one flip. That flip is fair.
Also if p=0.5^2 then you can do it in two flips. If it’s tails twice that happens 50% of the time. Anything other than tails twice counts as heads. Similarly can do p=0.5^3 in three flips.
Quote: Ace2Suppose I want to simulate a fair coin in at most three flips. For which values of p is this possible?
The way I see it is, there are eight ways to toss a coin three times: one has three heads, three have two heads and a tail, three have one head and two tails, and one has three tails.
There are 15 ways to select one or more of the possible results, but one is choosing all four, which has probability 1.
Each of the four ways of choosing three results has a "matching" way of choosing one. For example, if three heads has probability 1/2, then "not three heads", which is the sum of two heads/one tail, one head/two tails, and three tails, is also 1/2. Similarly, each of the three ways of choosing three heads and one other result has a mathcing way of choosing the other two.
This means only seven sets of results need to be considered - let p be the probability that heads is tossed:
1. HHH: p^3 = 1/2
2. HHT: 3 p^2 (1 - p) = 3 p^2 - 3 p^3 = 1/2
3. HTT: 3 p (1 - p)^2 = 3 p^3 - 6 p^2 + 3 p = 1/2
4. TTT: (1 - p)^3 = 1 - 3 p + 3 p^2 - p^3 = 1/2
5. HHH + HHT: p^3 + 3 p^2 (1 - p) = 3 p^2 - 2 p^3 = 1/2
6. HHH + HTT: p^3 + 3 p (1 - p)^2 = 4 p^3 - 6 p^2 + 3 p = 1/2
7. HHH + TTT: p^3 + (1 - p)^3 = 1 - 3 p + 3 p^2 = 1/2
All values of p, 0 < p < 1, that are solutions to any of these seven equations will work for three tosses.
Quote: ThatDonGuyThe way I see it is, there are eight ways to toss a coin three times: one has three heads, three have two heads and a tail, three have one head and two tails, and one has three tails.
There are 15 ways to select one or more of the possible results, but one is choosing all four, which has probability 1.
Each of the four ways of choosing three results has a "matching" way of choosing one. For example, if three heads has probability 1/2, then "not three heads", which is the sum of two heads/one tail, one head/two tails, and three tails, is also 1/2. Similarly, each of the three ways of choosing three heads and one other result has a mathcing way of choosing the other two.
This means only seven sets of results need to be considered - let p be the probability that heads is tossed:
1. HHH: p^3 = 1/2
2. HHT: 3 p^2 (1 - p) = 3 p^2 - 3 p^3 = 1/2
3. HTT: 3 p (1 - p)^2 = 3 p^3 - 6 p^2 + 3 p = 1/2
4. TTT: (1 - p)^3 = 1 - 3 p + 3 p^2 - p^3 = 1/2
5. HHH + HHT: p^3 + 3 p^2 (1 - p) = 3 p^2 - 2 p^3 = 1/2
6. HHH + HTT: p^3 + 3 p (1 - p)^2 = 4 p^3 - 6 p^2 + 3 p = 1/2
7. HHH + TTT: p^3 + (1 - p)^3 = 1 - 3 p + 3 p^2 = 1/2
All values of p, 0 < p < 1, that are solutions to any of these seven equations will work for three tosses.
Nice. I agree with this for three flips, though the question was at most three flips so need to add the 6 ways you can do it in one or two flips.
