From Keith Wynrow comes a clever little puzzle that stumped a few folks on social media:
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 or four flips? Now what are your chances of choosing the door with the cranberry sauce?
If both coins are the same, assume they are both A. If both coins are different, half the time you will guess A correctly.
This yields
$$EV(2) = (1)\big(1(0.6^2)\big) + (0.5)\big(2(0.6)(0.4)\big) + (0)\big(1(0.4^2)\big) = 0.6$$
In general, assume the coin that shows up more often is A. If there is an even number of coin flips and both appear half the time, half the time you will guess A correctly.
$$EV(3) = (1)\big(1(0.6^3)\big) + (1)\big(3(0.6^2)(0.4)\big) = 0.648$$
$$EV(4) = (1)\big(1(0.6^4)\big) + (1)\big(4(0.6^3)(0.4)\big) + (0.5)\big(6(0.6^2)(0.4^2)\big) = 0.648$$