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Ask The Wizard #358
On the latest culinary game show from the Food Network called Cranberries or Bust, contestants are faced with a decision between two doors: A and B. Behind one door is an endless supply of cranberry sauce, while behind the other, there is utterly nothing. And let's be honest, you're a big fan of cranberry sauce.
However, there's an interesting twist involved. The show’s host introduces a coin with two distinct sides labeled A and B, which correspond to the doors. The host reveals that this coin is biased towards the door that has the cranberry sauce behind it — though they do not disclose which door that is — meaning that this door will appear 60% of the time when the coin is flipped. For instance, if the cranberry sauce is located behind door A, the coin will show A 60% of the time and B the other 40%.
You have the opportunity to flip the coin twice before making your choice. If you strategize effectively, what are the odds of selecting the door with the cranberry sauce?
Bonus challenge: If instead of two flips you could flip the coin three, four, or even ten times, how would that affect your chances of correctly choosing the door with the cranberry sauce?
The scenario with a single flip is fairly straightforward. The coin boasts a 60% probability of landing on the door concealing the cranberry sauce. Therefore, your best strategy is to opt for the door indicated by the coin. This gives you a 60% probability of making the right choice.
Let's assume that behind door A is the cranberry sauce while door B is empty. In this case, the A side of the coin would have a likelihood of 60%. The strategy for the player should be to select the door that the coin lands on most frequently. In the event of a tie, either door can be chosen as there's no additional information to guide the decision.
Below are the possible results along with their associated probabilities. Combinations of A's and B's could appear in any sequence:
AA: 60%^2 = 36%
AB: 2*60%*40% = 48%
BB: 40%^2 = 16%
The player will succeed in choosing the right door if the coin shows A in both flips. If the result is one A and one B, they will lack useful information, leaving them with a 50/50 chance. If the output shows B twice, they will inevitably choose incorrectly.
Thus, in the case of two flips, the player will have a chance to correctly select the door calculated as 60% + 48%*(1/2) = 60%.
Assuming again that door A contains the cranberry sauce and door B has nothing, the A side of the coin carries a 60% chance of being flipped. The player’s strategy should revolve around selecting the door indicated by the coin the most.
Below are the possible results along with their associated probabilities. Combinations of A's and B's could appear in any sequence:
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 make the correct choice if the coin indicates A at least twice. Should it land on B two or more times, the player will choose the wrong door.
Therefore, when considering three flips, the player's odds of making the correct selection rise to 21.6% + 43.2% = 64.8%.
Let's assume that behind door A is the cranberry sauce while door B is empty. In this case, the A side of the coin would have a likelihood of 60%. The strategy for the player should be to select the door that the coin lands on most frequently. In the event of a tie, either door can be chosen as there's no additional information to guide the decision.
Below are the possible results along with their associated probabilities. Combinations of A's and B's could appear in any sequence:
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 select the correct door if the coin ends up on A a minimum of three times. If the result shows A twice and B twice, this will give them no useful insight, leaving them with a 50/50 chance. If B shows at least three times, they will pick the incorrect door.
So in the four flips scenario, the chances of choosing correctly amount to 12.96% + 34.56% + 34.56%*(1/2) = 64.80%.
The reasoning applied in the first four scenarios is applicable to all subsequent cases. Keep in mind that the number of ways to choose x from y items is computed as y!/(x! * (y-x)!).
This topic has been brought to attention on my forum at Wizard of Vegas .
In celebration of Thanksgiving, you find yourself seated at a circular table with 19 other mathematicians. Everyone at the table is eager for a serving of cranberry sauce, which is currently right in front of you.
You start by helping yourself. After that, rather than passing the sauce around the table in a clockwise manner, you decide to pass it randomly to the individual on your left or right. Each person then does the same, randomly delivering the sauce to the person next to them until everyone has had their share of the cranberry sauce.
Among the 20 seated at the table, who stands the highest chance of being the last person to receive the cranberry sauce?
Let’s refer to one of the mathematicians as G. For G to end up being last, two conditions must be satisfied:
- The cranberry sauce has to reach either of G's neighboring seats first.
- The sauce must travel 19 positions in the opposite direction without ever reaching G.
To be the last, the cranberry sauce inevitably needs to touch one of the neighbors. Hence, the probability of this happening is guaranteed at 100%.
As for the second part of the scenario, whatever its probability is, it applies equally across all participants. This ensures that anyone has an equal chance of being named last.
In case that explanation was unclear, Gialmere encountered this question over at fivethirtyeight.com. You can find it there. Just scroll down to the section that says 'Solution to last week’s Riddler Classic.' explain the solution Imagine a dart thrown at random towards a Gaussian curve. Let’s denote the dart's landing coordinates as (x,y). What would be the expected value of the absolute value of x?
This topic has been brought to attention on my forum at Wizard of Vegas .
If you want mathematical understanding to several decimal places, please refer to my
Here is my solution (PDF).
When someone from the general public is asked to name any card from a standard 52-card deck, which card do you think they are most likely to choose? Wiz Calculator .
Surprisingly enough, the ace of spades tends to be selected 24.59% of the time. Here’s a list of the top 5 cards:
The ace of spades, by far. According to Psychology of Magic Cards that were never selected in a sample size of 417 included the 5 of diamonds, the 6 of clubs, the 5 of clubs, the 6 of spades, and the 4 of spades.
- 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%
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