Dice don't lie
Tonight's idea
Roll two dice and add them. Which total should you bet on? Most people guess that every total from 2 to 12 is equally likely, or that the middle ones are "a bit" more likely. The truth is lopsided: 7 is six times as likely as 2.
The reason is that a total is not an outcome. An outcome is what the two dice actually show, red die and blue die, and there are \(6 \times 6 = 36\) of those, each exactly as likely as the others. A total of 2 happens only one way (1 and 1). A total of 7 happens six ways: 1+6, 2+5, 3+4, 4+3, 5+2, 6+1. Here is the whole table, with the sevens shaded:
| + | 1 | 2 | 3 | 4 | 5 | 6 |
|---|---|---|---|---|---|---|
| 1 | 2 | 3 | 4 | 5 | 6 | 7 |
| 2 | 3 | 4 | 5 | 6 | 7 | 8 |
| 3 | 4 | 5 | 6 | 7 | 8 | 9 |
| 4 | 5 | 6 | 7 | 8 | 9 | 10 |
| 5 | 6 | 7 | 8 | 9 | 10 | 11 |
| 6 | 7 | 8 | 9 | 10 | 11 | 12 |
Probability is then just counting: the number of outcomes you want, divided by 36. \(P(\text{total } 7) = \tfrac{6}{36} = \tfrac16\). \(P(\text{total } 2) = \tfrac{1}{36}\). The table does all the thinking.
One more idea for tonight: the expected value of a game is what you would win per play, on average, if you played it thousands of times. Multiply each prize by its probability and add. If you win $5 on a 7 and lose $1 on anything else, the expected value is \(\tfrac{6}{36}(5) + \tfrac{30}{36}(-1) = 0\). Fair, exactly. Casinos make sure their games come out a little below zero for you.
Watch one: at least one six
Roll two dice. What is the probability that at least one of them shows a 6?
Count in the table of outcomes (not totals). The red die is 6 in six outcomes. The blue die is 6 in six outcomes. But 6-and-6 has been counted twice, so the number of outcomes with at least one six is \(6 + 6 - 1 = 11\). \(P = \tfrac{11}{36}\), a bit under one third.
A faster way for "at least one" questions: count the opposite. No six on either die is \(5 \times 5 = 25\) outcomes, so at least one six is \(36 - 25 = 11\). This trick, probability of the opposite, is the most useful move in all of probability. Remember it for the stretch.
Your turn
- Using the table, find the probability of each total from 2 to 12. Check your eleven answers add up to 1.
- What is the probability of rolling doubles? Of a total of 10 or more? Of a total of 6 or 8?
- Your friend offers a bet: you win if the total is 6 or 8, they win if it is 7. Who has the better side? By how much?
- Now multiply the dice instead of adding. What is the probability the product is even? (Hint: count the outcomes where it is odd.)
- A game costs $2 to play. You roll two dice and win $10 if the total is 2 or 12, nothing otherwise. Find the expected value per play. Would you play?
- Roll two dice and keep the higher of the two numbers. Make a table like the one above showing the higher number in each cell. Which value is most likely, and what is its probability?
Try the dice demo at mathdetours.com/demos/dice.html: it rolls thousands of times in a second and draws the results next to the probabilities you just computed.
Stretch: the gambler who asked a mathematician
In the 1650s a French gambler, the Chevalier de Méré, had a profitable bet: he would get at least one 6 in four rolls of a die. He then switched to a bet he reckoned was equally good: at least one double-six in 24 rolls of two dice. He started losing money and wrote to Blaise Pascal to ask why. Pascal's reply helped invent probability theory.
Use the "probability of the opposite" trick. The chance of no six in one roll is \(\tfrac56\). In four independent rolls it is \(\left(\tfrac56\right)^4\). So \(P(\text{at least one six in 4}) = 1 - \left(\tfrac56\right)^4\). Compute it. Then do the same for double-six: the chance of no double-six in one roll of two dice is \(\tfrac{35}{36}\), and he gets 24 tries. Which bet is above a half and which is below?
Answers
- Totals 2 through 12 have 1, 2, 3, 4, 5, 6, 5, 4, 3, 2, 1 outcomes, so the probabilities are \(\tfrac1{36}, \tfrac2{36}, \tfrac3{36}, \tfrac4{36}, \tfrac5{36}, \tfrac6{36}, \tfrac5{36}, \tfrac4{36}, \tfrac3{36}, \tfrac2{36}, \tfrac1{36}\). The counts add to 36, so the probabilities add to 1.
- Doubles: 6 outcomes, \(\tfrac{6}{36} = \tfrac16\). Total 10 or more: 3 + 2 + 1 = 6 outcomes, \(\tfrac16\). Total 6 or 8: 5 + 5 = 10 outcomes, \(\tfrac{10}{36} = \tfrac{5}{18}\).
- You win on 10 outcomes, your friend on 6. You have the better side, by 10 to 6. On the other 20 outcomes nobody wins; roll again.
- The product is odd only when both dice are odd: \(3 \times 3 = 9\) outcomes. So the product is even in 27 outcomes, \(P = \tfrac{27}{36} = \tfrac34\).
- Totals 2 or 12: 2 outcomes out of 36. Expected winnings \(\tfrac{2}{36} \times 10 = \tfrac{20}{36} \approx 0.56\) dollars, against a cost of $2. Expected value about \(-\$1.44\) per play. Do not play.
- The higher number is 1 in 1 cell, 2 in 3 cells, 3 in 5, 4 in 7, 5 in 9, 6 in 11 (the pattern is the odd numbers). Most likely is 6 with probability \(\tfrac{11}{36}\), the same count as "at least one six", because they are the same event.
Stretch. \(1 - \left(\tfrac56\right)^4 = 1 - 0.482 = 0.518\), just above a half, so the first bet wins slightly more often than it loses. \(1 - \left(\tfrac{35}{36}\right)^{24} = 1 - 0.509 = 0.491\), just below a half. De Méré's reasoning was that \(4/6 = 24/36\), so the bets should be "the same". Probabilities of repeated tries do not scale like that; you have to multiply the chances of missing. Over many evenings a 1.8 percent edge became a 0.9 percent deficit, and he noticed.