Random Number Poker Solution

Rules

Two participants are randomly assigned a number from a uniform distribution ranging from 0 to 1.
The first player has the option to retain their number or exchange it for a newly drawn random number.
The second player, being aware of the first player's choice, also has the option to keep their original number or switch to a new one.
Ultimately, the player with the highest number at the conclusion wins.

Questions

What would be the best approach for each player in this scenario?
If both players adopt the most effective strategy, what would be the chances of each player winning?

Answers

Player 1 should opt to switch if their number is below 0.567364; otherwise, they should stay with their current number.
If Player 1 decides to switch, Player 2 ought to switch too if their number is below 0.5; otherwise, they should maintain their original number.
In the case that Player 1 opts to stay, Player 2 should consider switching if their number is below 0.660951; otherwise, they should remain with their initial number.
Probability player 1 wins = 0.494333.
Probability player 2 wins = 0.505667.
If we assume each participant bets a single number, the expected value for Player 1 is calculated to be -0.011333.

Solution

It's clear that when Player 1 switches, Player 2 should also switch if their number is less than 0.5 and otherwise hold their number.

In other circumstances, Player 1 should keep their number as long as it exceeds a specified number, which we will denote as x.

If Player 1 decides to stay, Player 2 can reasonably deduce that Player 1 has a good number and should adopt an aggressive strategy, switching their number if it exceeds another specific value, y, upon Player 1's decision to stay.

To tackle problems like this, you need to determine these indifference points, x and y, by equating the expected values for choosing to stay versus switching.

In the following calculations, I will be determining the expected value from Player 1's perspective, under the assumption that both players wager one unit each.

Let’s solve for x first.

Expected value by standing = y*(2x-1) - (1-y)

Expected value by hitting = 0.5 * 0 + 0.25 * 0 + 0.25 * -1 = -0.25.

Next, we set these expected values equal to each other:

y*(2x-1) - (1-y) = -0.25
2xy - y - 1 + y =-0.25
2xy - 1 = -0.25
2xy = 0.75
xy = 3/8

Now, let's calculate the expected value for Player 2 if they have the value y and decide to stay after Player 1 chooses to stay:

(y-x)/(1-x) + (1-y)/(1-x) * -1 = (x-2y+1) / (x-1)
Now, let's calculate the expected value for Player 2 if they draw y after Player 1 has chosen to stay:

(1 / (1-x)) * [(1-x)^2 * 0 + x * (1-x) * -1] =
(1 / (1-x)) * [x^2 - x] =
x * (x-1) / -(x-1) =
-x

Next, we set these expected values equal to each other:

(x-2y+a) / (x-1) = -x
x^2 - 2y + 1 = 0
x^3 - 2xy + x = 0

Next, substitute 3/8 for xy.

x^3 + x - 0.75 = 0
4x^3 + 4x - 3 = 0.

At this stage, you can use a cubic equation solver to find that x equals 0.567364.

Knowing that the product of x and y equals 3/8, substitute the previously found value of x to determine y as 0.660951.

Afterward, it simply involves examining all possible combinations in which two to four numbers can appear to calculate the probability of victory for each player. This can be approached through geometric methods or calculus. Please forgive me for allowing the reader to tackle this part. Here are the results:

Probability player 1 wins = 0.494333.
Probability player 2 wins = 0.505667.
If we assume each participant bets a single number, the expected value for Player 1 is calculated to be -0.011333.

For those who require a precise formula for the solution:

Let z = (3/8 + (307/1728)^(1/2))^(1/3) 0.926962
Then x = z - 1/(3z) 0.567364
Then y = 3/(8x) 0.660951
Then expected value of player 1 = 3x/8 + y(y-1) -0.011333

Thanks to Joe Shipman for the problem.