# Ask the Wizard #358

On the Food Network’s latest game show, Cranberries or Bust, you have a choice between two doors: A and B. One door has a lifetime supply of cranberry sauce behind it, while the other door has absolutely nothing behind it. And boy, do you love cranberry sauce.

Of course, there’s a twist. The host presents you with a coin with two sides, marked A and B, which correspond to each door. The host tells you that the coin is weighted in favor of the cranberry door — without telling you which door that is — and that door’s letter will turn up 60 percent of the time. For example, if the sauce is behind door A, then the coin will turn up A 60 percent of the time and B the remaining 40 percent of the time.

You can flip the coin twice, after which you must make your selection. Assuming you optimize your strategy, what are your chances of choosing the door with the cranberry sauce?

Extra credit: Instead of two flips, what if you are allowed three, four, ... ten flips? Now what are your chances of choosing the door with the cranberry sauce?

Gialmere

The one flip case is fairly trivial. The coin will have a 60% chance of landing on the door with the cranberry sauce. The player's strategy should be to pick whatever door the coin lands on. Thus he will have a 60% chance of picking correctly.

Let's say that door A has the cranberry sauce and door B has nothing. So, the A side of the coin will have a 60% chance. The player's strategy should be to pick whatever door the coin lands on the majority of the time. If it is a tie, the player may pick either door, as he has no useful information.

Here are the possible outcomes and their probability. Cases with a mixture of A's and B's may be in any order:

AA: 60%^2 = 36%

AB: 2*60%*40% = 48%

BB: 40%^2 = 16%

The player will correctly pick the right door if the coin lands on A both times. If it lands on A once and B once, he won't have any useful information and will have a 50/50 chance. If it lands on B both times, he will pick the wrong door.

So, in the two flip case, the player will have a 60% + 48%*(1/2) = 60% chance of picking the correct door.

Let's say that door A has the cranberry sauce and door B has nothing. So, the A side of the coin will have a 60% chance. The player's strategy should be to pick whatever door the coin lands on the majority of the time.

Here are the possible outcomes and their probability. Cases with a mixture of A's and B's may be in any order:

AAA: 60%^3 = 21.6%

AAB: 3*60%^2*40% = 43.2%

ABB: 3*60%^2*40% = 28.8%

BBB: 40%^3 = 6.4%

The player will correctly pick the right door if the coin lands on at least two times. If it lands on B two or more times, he will pick the wrong door.

So, in the three flip case, the player will have a 21.6% + 43.2% = 64.8% chance of picking the correct door.

Let's say that door A has the cranberry sauce and door B has nothing. So, the A side of the coin will have a 60% chance. The player's strategy should be to pick whatever door the coin lands on the majority of the time. If it is a tie, the player may pick either door, as he has no useful information.

Here are the possible outcomes and their probability. Cases with a mixture of A's and B's may be in any order:

AAAA: 60%^4 = 12.96%

AAAB: 4*60%^3*40% = 34.56%

AABB: 6*60^2*40%^2 = 34.56%

ABBB: 4*60%*40%^3 = 15.36%

BBBB: 40%^4 = 2.56%

The player will correctly pick the right door if the coin lands on A at least three times. If it lands on A twice and B twice, he won't have any useful information and will have a 50/50 chance. If it lands on B at least three times, he will pick the wrong door.

So, in the four flip case, the player will have a 12.96% + 34.56% + 34.56%*(1/2) = 64.80% chance of picking the correct door.

The logic for the first four cases will apply to all cases. Remember, the number of ways to choose x out of y items is y!/(x! * (y-x)!).

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

To celebrate Thanksgiving, you and 19 mathematicians are seated at a circular table. Everyone at the table would like a helping of cranberry sauce, which happens to be in front of you at the moment.

First, you serve yourself. Then, instead of passing the sauce around in a circle, you decide to pass it randomly to the person seated directly to your left or to your right. They then do the same, passing it randomly either to the person to their left or right. This continues until everyone has, at some point, received the cranberry sauce.

Of the 20 people in the circle, who has the greatest chance of being the last to receive the cranberry sauce?

Gialmere

Let's name one of the mathematicians G. In order for G to be last, two things must occur:

- The cranberries must first reach either neighbor of G.
- The cranberries must move 19 positions in the opposite direction without ever reaching G.

To be last, the cranberries must eventually reach either neighbor. So the probability of that is 100%.

Then, whatever the probability is for the second part is, it is the same for each person. Thus, each person has an equal probability of being last.

If that explanation wasn't clear. Gialmere got this problem from fivethirtyeight.com. Here they explain the solution. Scroll down to the part that says "Solution to last week’s Riddler Classic."

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

A dart is thrown randomly at the Gaussian curve. Let the location of the dart be (x,y). What is the expected value of the absolute value of x?

"Anonymous" .

Here is my solution (PDF).

For math to this math decimal places, please use my Wiz Calculator.

When a random person is asked to name any card from a 52-card deck, which card is the most likely they will pick?

"Anonymous" .

The ace of spades, by far. According to Psychology of Magic, the ace of spades is chosen 24.59% of the time. Here are the top 5:

- Ace of spades: 24.59%
- Queen of hearts: 13.71%
- Ace of hearts: 6.15%
- King of hearts: 5.91%
- Jack of spades: 4.26%

Never picked in their apparent 417 sample size were the 5 of diamonds, 6 of clubs, 5 of clubs, 6 of spades, and the 4 of spades.