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Question
in a game of pontoon dice, you continue throwing a single die until the sum of all your throws exceeds 21 (bust) or you decide to stop. you win plastic counters depending on the score you stop at. after four throws you have scored 17. should you throw once more? consider the chances of getting different scores and how much you will win or lose. what is best on average?
| stopping score | counters won or lost |
|---|---|
| 13 or 14 | win 1 |
| 15 or 16 | win 2 |
| 17 or 18 | win 3 |
| 19 | win 6 |
| 20 | win 8 |
| 21 | win 10 |
| over 21 | lose 4 |
Step1: Determine possible outcomes of the next throw
Current score is 17. When we throw the die once more, the possible scores we can get are \(17 + 1=18\), \(17 + 2 = 19\), \(17+3=20\), \(17 + 4=21\), \(17+5 = 22\) (bust), \(17+6=23\) (bust). Each outcome (rolling a 1, 2, 3, 4, 5, 6) has a probability of \(\frac{1}{6}\).
Step2: Find the counters for each outcome
- If we roll a 1: score = 18, counters won = 3
- If we roll a 2: score = 19, counters won = 6
- If we roll a 3: score = 20, counters won = 8
- If we roll a 4: score = 21, counters won = 10
- If we roll a 5: score = 22, counters lost = 4
- If we roll a 6: score = 23, counters lost = 4
Step3: Calculate the expected value
The expected value \(E\) is calculated as the sum of (probability of each outcome \(\times\) value of each outcome).
If we stop at 17, we win 3 counters. The expected value of throwing again (\(\approx3.17\)) is greater than 3. So we should throw once more.
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We should throw once more as the expected value of throwing again (\(\approx3.17\)) is greater than the 3 counters we win by stopping at 17.