# Bell’s Casino Solution

2015-12-10T03:37:09Z

Bell’s Casino Problem

A new casino has opened up in town named “Bell’s Casino”. They are offering a coin game. The game works as follows.

The house will commit two coins on the table, oriented heads or tails each, and keep them covered. The player calls what the faces of the each of the coins are, either HH, HT, TH, or TT. The casino reveals the coins and if the player is correct, they win \$1, and otherwise they lose \$1.

Problem 1.
Prove that there is no strategy that can beat the casino.

Solution to problem 1.

Let pHH be the probability that the casino orients the coins as HH, and similarly for pHT, pTH, and pTT. We have pHH + pHT + pTH + pTT = 1. If the player calls the orientations XY, the they win \$1 with probability pXY, and lose \$1 with probability 1-pXY. This means the total expected value is 2*pXY - 1 dollars. The casino can minimize the players expected value by choosing each of the four possible orientations with equal proability of 25%. In the case the players expected value is \$-0.50 no matter what orientation they choose to call.

After opening the customers stop coming by to play this boring game, so to boost attendance the casino modifies the game as follows.

The house will commit two coins on the table, oriented heads or tails each, and keep them covered. The player calls what the faces of each of the two coins are, either HH, HT, TH, or TT. The casino reveals one coin, of the players choice. After seeing revealed coin, the player can elect to back out of the game and neither win nor lose, or keep going, and see the second coin. If the player’s call is correct, they win \$1, and otherwise they lose \$1.

Problem 2.
Prove that there is no strategy that can beat the casino.

Solution to problem 2.

Without loss of generality, suppose the player calls HH and they ask for the first coin to be revealed. With probability pTH + pTT the first coin is tails. At this point the player’s best option is to elect to back out of the game as it is now impossible for them to win. With probability pHH + pHT the first coin is heads. At this point the player can back out or continue. If the player always continues then they win \$1 with probability pHH and they lose \$1 with probability pHT. Their total expected value is pHH - pHT dollars. As before, the casino can choose each of the four possible orientations with equal proability of 25%. In the case the players expected value is \$0 if they chose to continue if the revealed coin is correct, and it is also \$0 if they chose to always back out of the game. No matter what strategy the player chooses, they cannot beat the casino as long as the casino chooses each of the four possible orientations with equal proability.

Even with the new, more fair, game, attendance at the casino starts dropping off again. The casino decides to offer a couples game.

The house will commit two coins on two tables, oriented heads or tails each, and keep them covered. The couple, together, calls what the the faces of each of the two coins are, either HH, HT, TH, or TT. Then, each player in the couple gets to see one coin each. Collectively they get to decide whether they are going to back out of the game or not by the following method. After seeing their revealed coin, each player will raise either a black flag or a red flag. If both players raise the different colour flags, the game ends and no one wins or loses. If both players raise the same colour flag, the game keeps going. If the couples original call was right, they win \$1, and otherwise, they lose \$1. To ensure that the couple cannot cheat, the two tables are places far enough apart such that each player’s decision on which flag to raise is space-like separated. Specifically the tables are placed 179 875 475 km apart and each player has 1 minute to decide which flag to raise otherwise a black flag will be raised on their behalf (or, more realistically, the tables are placed 400 m apart and each player has 100 nanoseconds to decide which flag to raise).

Problem 3.
Prove that there is no strategy for the couple that can beat the casino.

Solution to problem 3.

Alice and Bob enter the casino together. The casino covers two coins, and Alice and Bob call a random choice for the orientation of the coins. Without loss of generality let us assume they call HH.

Alice and Bob are separated by 179,875,475 km so they cannot communicate and simutaneously each of them see the value of one of the coins. If Alice sees heads on her coin, she can raise a black flag with probability a, and a red flag with probability 1-a. If she sees tails she can raise a black flag with probability b, and a red flag with probability 1-b. The pair (a, b) characterizes any strategy that Alice could employ. Similarly, if Bob sees heads on his coin can raise a black flag with probability c, and a red flag with probability 1-c, and if he sees tails he can raise a black flag with probability d, and a red flag with probability (1-d). So the quadruple (a,b,c,d) parameterizes all possible strategies that Alice and Bob can employ together.

Given the strategy parameterized by (a,b,c,d) they win \$1 if they raise the same colour flag if the coins are both heads. The probability of this happening is ac + (1-a)(1-c). They lose \$1 if they raise the same colour flag if any of the coins are tails. The probability of that happening is bc + (1-b)(1-c), ad + (1-a)(1-d), bd + (1-b)(1-d) if only the first coin is tails, or if the second coin is tails, or if both coins are tails, respectively. If the casino chooses the coins orientations with equal proability, their total expected winnings with their strategy is

E = (ac + (1-a)(1-c) - bc - (1-b)(1-c) - ad - (1-a)(1-d) - bd - (1-b)(1-d))/4.

Theorem. If 0 ≤ a ≤ 1, 0 ≤ b ≤ 1, 0 ≤ c ≤ 1, 0 ≤ d ≤ 1, and we let E = (ac + (1-a)(1-c) - bc - (1-b)(1-c) - ad - (1-a)(1-d) - bd - (1-b)(1-d))/4, then E ≤ 0.

Let G = bcd + (1-b)(1-c)(1-d) + abd + (1-a)(1-b)(1-d) + a(1-c)d + (1-a)c(1-d) + a(1-b)(1-c) + (1-a)bc. 0 ≤ G because G is the sum of products of non-negative numbers. But E = -G/4, so E ≤ 0. QED.

Thus no matter what strategy Alice and Bob use, they cannot expect to beat the casino. At least that is what the casino thinks.

Problem 4.
Devise a physical procedure that a couple can follow to beat the casino on average at this last game without cheating.

Solution to problem 4.

Alice and Bob enter the casino together. Each of them have one of a pair of maximally entangled qbits. For concreteness let say they are a pair of entangled photons, even though realistically they will use a different physical realization of their qbits.

The casino covers two coins, and Alice and Bob call a random choice for the orientation of the coins. Without loss of generality assume they call HH. Currently they only have a 25% chance of being right. In order to shift the odds in their favour they need to substantially increase their odds of dropping out of the game if they are wrong, but avoid dropping out if they are right.

Alice and Bob are separated by 179,875,475 km so they cannot communicate and simutaneously each of them see the value of one of the coins. Alice follows the following procedure. If she see an H, she measures the polarization of her qbit at an angle of 0°. If she sees a T, she measures the polarization of her qbit at an angle of -45°. In either case, if her measurement is positive she raises a black flag and if her measurement is negative she raises a red flag. In both cases she has a 50% chance of raising either flag.

Bob follows the following procedure. If he sees an H, he measures the polarization of his qbit at an angle of +22.5°. If he sees a T, he measures the polarization of his qbit at an angle of +67.5°. In either case, if his measurement is positive he raises a black flag and if his measurement is negative he raises a red flag. In either case he has a 50% chance of raising either flag.

Because their qbits are entangled, the flags they raise are correlated. The probability of Alice and Bob raising the same colour flag is cos²(θ), where θ is the difference between their measurement angles.

If both coins are heads, then the difference between their measurement angle is 22.5°, and the probability of them raising the same colour flag is cos²(22.5°) = (2 + √2)/4 ≅ 85.4%.

If one coin is heads and the other is tails, then the difference between their measurement angle is 67.5°, and the probability of them raising the same colour flag is cos²(67.5°) = (2 - √2)/4 ≅ 14.6%.

If both coins are tails, then the difference between their measurement angle is 112.5°, and the probability of them raising the same colour flag is cos²(112.5°) = (2 - √2)/4 ≅ 14.6%.

If the casino randomly selects the orientation of their coins with equal probability then Alice and Bob’s expected winnings is ((√2 + 2) - 3*(2 - √2))/16 ≅ \$0.103, which is a positive amount.

The casino cannot figure out how they keep losing money on this game and, soon, Bell’s Casino goes bankrupt.

This physical property that allows Alice and Bob to correlate their flags without communicating is called quantum pseudo-telepathy. What surprised me the most is that Bob measurements are not quite aligned with Alice’ measurments. I would have thought it was better that for Bob to measure at angles 0° and +45°.

If Bob measures at these angles then when the coins are HH, they are guarenteed to raise the same flag, and when the coins are TT, they are guarenteed to raise opposite flags, which is perfect. However, when one is tails and one is heads, then they have a 50% chance of raising the same flag. This makes their expected value 0.25*1 - 0.5*0.5 = \$0, and they do not come out ahead. But by having Bob rotate his measurements by a small amount, Bob can increase the chances of raising the different flags in case of HT or TH, while decreasing the chances of correctly staying or leaving the game in case of HH or TT. Because of the geometry of the circle, Bob increases the chances of raising different flags in case of HT or TH faster than he decreases the chances of raising the correct flags in case of HH or TT. If Bob makes measurements at angle φ and φ + 45°, then the optimal expected value occurs at φ = 22.5°.

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