# Expected gain

2 players put one dollar into a pot. They decide to throw a pair of dice alternately. The first one who throws a total of 5 on both dice wins the pot. How much should the player who starts add to the pot to make it a fair game?

Is the answer = 1 / 8 one eighth?

How to think the equation?

### 2 個解答

• 最愛解答

P(total of 5 in 1 throw) = 4/36 = 1/9 = p

Let player A starts first.

P(A wins the pot in 1st try) = p = 1/9

P(B wins the pot in 1st try) = (1-p)p = 8/81

P(nobody wins the pot in 1st try) = 1-p-(1-p)p = (1-p)^2 = 64/81

P(A wins the pot) = 1/9 + 64/81 * 1/9 + (64/81)^2 * 1/9 + ... = 1/9 [1 + 64/81 + (64*81)^2+...]

P(B wins the pot) = 8/81 + 64*81 * 8/81 + (64/81)^2 * 8/81 + ... = 8/81 [1 + 64/81 + (64/81)^2 + ... ]

Let x = [1 + 64/81 + (64/81)^2 + ... ]

P(A wins the pot) + P(B wins the pot) = 1

1/9 x + 8/81 x = 1

9/81 x + 8/81 x = 1

17/81 x = 1

x = 81/17

P(A wins the pot) = 9/17

P(B wins the pot) = 8/17

Let \$y be the first player&#39;s extra amount of the bet.

Expect gain of A = Expect gain of B = zero

9/17 * 1 - 8/17 * (1 + y) = 8/17 * (1 + y) - 9/17 * 1 = 0

8 * (1 + y) - 9 = 0

1 + y = 9/8

y = 1/8

Therefore, player A needs to put an extra 1/8 dollar to make the game fair.

資料來源： my mathematical knowledge
• haha I saw you mention my name in some post 2 months ago, then answer you la :P

Ivan and Jessica is playing, Ivan play first. Ivan add \$x to the pot. so if Ivan win he will gain \$1 and if Jessica win she will gain \$1+x.

Probability to throw 5 is 4/36 = 1/9

Probability for Ivan to win is: (first time 5) + (not 5, not 5, is 5) + (not 5, not 5, not 5, not 5, is 5) +......... = 1/9 + 8/9*8/9*1/9 + (8/9)^4 * 1/9 +........

= 1/9 Sum (64/81)^n

= 1/9 * (1/ (1- 64/81)) = 9/17

So Ivan has probability 9/17 to win, and so Jessica has probability 8/17 to win.

To be a fair game, the expected gain for both people should be the same:

9/17 (1) = 8/17(1+x)

so solve for x, Ivan need to add \$1/8 more.

資料來源： I wont play this game though :P