Expected Total Payout of the Die Game

You roll a fair six-sided die repeatedly. Each roll pays its face value. If the roll is 4, 5, or 6, you roll again. If the roll is 1, 2, or 3, the game stops immediately. We want the expected total payout.

1) Outcome-by-outcome view

Let E be the expected total payout from the start of the game. After one roll, six things can happen:

1
Payout = 1
STOP
2
Payout = 2
STOP
3
Payout = 3
STOP
4
Payout = 4 + E
ROLL AGAIN
5
Payout = 5 + E
ROLL AGAIN
6
Payout = 6 + E
ROLL AGAIN
E = (1/6)[1 + 2 + 3 + (4 + E) + (5 + E) + (6 + E)]
Combine constants: 1 + 2 + 3 + 4 + 5 + 6 = 21
There are three continuing outcomes (4, 5, 6), so: (E + E + E) = 3E
Therefore:
E = (1/6)(21 + 3E) = 21/6 + 3E/6 = 3.5 + 0.5E
Move 0.5E to the left:
E - 0.5E = 3.5 ⟹ 0.5E = 3.5 ⟹ E = 7
Final Answer
Expected total payout = 7

2) Intuition / recursive view

Every roll gives you some money right away, and sometimes it restarts the same game.

Average value of one roll
3.5
Probability of continuing
1/2
E = (average from current roll) + (chance to continue) × (same expectation again)
E = 3.5 + (1/2)E

Because with probability 1/2 (rolling 4, 5, or 6), you are back in exactly the same situation as before — so the expected future payout is still E.

Game flow
  • Roll 1, 2, or 3 → collect that amount and stop
  • Roll 4, 5, or 6 → collect that amount and restart the whole problem

3) Simulation

Click the button to simulate many games and compare with the theoretical answer.

No simulation yet.
The simulated average should get closer to 7 as the number of games increases.