# Ask the Wizard #339

I saw a Super Bowl 55 bet on whether the game would end in a unique NFL combination of scores that had never happened before in NFL history, called a Scorigami. The lines were:

Yes: +1100

No: -1400

What do you make the odds?

Actuarial

Good question! Fortunately, there is NFL Scorigami that tells us the count of every score combination in NFL history.

I'm sure the frequentists out there will hate my answer, but I had to make some assumptions to get a probability of an event that has never happened.

First, to get an individual team score, I looked at historical NFL games. In particular, the games between 1994 and 2018. The reason I chose 1994 is that is the year the two-point conversion rule started, which should result in smoothing out the individual team score distribution a little. I ended in 2018 because that was the upper end of the data I had available. Here is that distribution.

### Individual NFL Team Scores 1994-2018

Points | Count | Probability |
---|---|---|

0 | 170 | 0.013490 |

1 | 0 | 0.000000 |

2 | 2 | 0.000159 |

3 | 303 | 0.024044 |

4 | 0 | 0.000000 |

5 | 5 | 0.000397 |

6 | 267 | 0.021187 |

7 | 420 | 0.033328 |

8 | 29 | 0.002301 |

9 | 188 | 0.014918 |

10 | 706 | 0.056023 |

11 | 32 | 0.002539 |

12 | 123 | 0.009760 |

13 | 646 | 0.051262 |

14 | 530 | 0.042057 |

15 | 128 | 0.010157 |

16 | 434 | 0.034439 |

17 | 892 | 0.070782 |

18 | 91 | 0.007221 |

19 | 282 | 0.022377 |

20 | 860 | 0.068243 |

21 | 511 | 0.040549 |

22 | 189 | 0.014998 |

23 | 548 | 0.043485 |

24 | 821 | 0.065148 |

25 | 118 | 0.009364 |

26 | 267 | 0.021187 |

27 | 673 | 0.053404 |

28 | 382 | 0.030313 |

29 | 131 | 0.010395 |

30 | 336 | 0.026662 |

31 | 578 | 0.045866 |

32 | 61 | 0.004841 |

33 | 146 | 0.011585 |

34 | 394 | 0.031265 |

35 | 200 | 0.015870 |

36 | 71 | 0.005634 |

37 | 163 | 0.012934 |

38 | 265 | 0.021028 |

39 | 30 | 0.002381 |

40 | 50 | 0.003968 |

41 | 146 | 0.011585 |

42 | 78 | 0.006189 |

43 | 25 | 0.001984 |

44 | 58 | 0.004602 |

45 | 85 | 0.006745 |

46 | 7 | 0.000555 |

47 | 16 | 0.001270 |

48 | 47 | 0.003730 |

49 | 35 | 0.002777 |

50 | 5 | 0.000397 |

51 | 15 | 0.001190 |

52 | 14 | 0.001111 |

53 | 1 | 0.000079 |

54 | 4 | 0.000317 |

55 | 6 | 0.000476 |

56 | 6 | 0.000476 |

57 | 2 | 0.000159 |

58 | 3 | 0.000238 |

59 | 5 | 0.000397 |

60 | 0 | 0.000000 |

61 | 0 | 0.000000 |

62 | 2 | 0.000159 |

Total | 12602 | 1.000000 |

Not that it matters, but the mean team score is 21.60165.

Second, for every score x-y, that has never occured, I calculated the probability as 2×prob(x)×prob(y). Why multiply by two? Because a score of x-y can happen two ways. For example, Super Bowl 55 could end in a result of Kansas City x -- Tampa Bay y, or Kansas City y -- Tampa Bay x. A Super Bowl may not end in a tie, so we don't need to concern ourselves with x-x scores. If we did, we wouldn't multiply by 2.

For example, a score of 11-15 has never happened. I put the probability of an 11 at 0.002539 and 15 at 0.010157. This would make the probability of an 11-15 score 2×0.002539×0.010157 = 0.0000515835.

Doing this for every score that has never happened results in a total probability of 0.0179251. The fair line for a bet on that would be +5479, or about 55 to 1. So to lay only 11 to 1 is a great bet! I wish I had access to it.

I admit this assigns a zero to the possibility of either team getting one point, which has never happened, but could. Yes, there is such a thing as a one-point safety. The probability of either team getting a score of one I feel is extremely negligible.

In all reality, the over/under in Super Bowl 55 was 56.5. Such a high scoring game would increase the probability of a Scorigami. If forced to make an estimate, I'd put it at 2%, for a fair line of 49 to 1.

This question is asked and discussed in my forum at Wizard of Vegas.

What is the probability of getting a sum of 53 on a roll of 15 dice?

gordonm888

There is an easy way to get such answers in a spreadsheet. To illustrate, consider an alternate question, what is the probability of getting a total of 20 with eight dice?

For the "1 Die" column there is obviously one way to roll each total from 1 to 6.

For every cell for two or more dice, go one cell to the left and then add the six cells above that cell. It shold be obvious why this works. Copy and paste this formula through the cell for eight dice and a total of 20.

You can see that cell has a total of 36,688. There are 8^{6} = 262,144 ways to roll eight six-sided dice. This makes the answer for the probability of a total of 20 with eight dice 36688 / 262,144 = 0.139954.

Using the same logic, the probability of a total of 53 with 20 dice is 0.059511.

### Dice Totals

Total | 1 Die | 2 Dice | 3 Dice | 4 Dice | 5 Dice | 6 Dice | 7 Dice | 8 Dice |
---|---|---|---|---|---|---|---|---|

0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 |

0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 |

0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 |

0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 |

0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 |

1 | 1 | 0 | 0 | 0 | 0 | 0 | 0 | 0 |

2 | 1 | 1 | 0 | 0 | 0 | 0 | 0 | 0 |

3 | 1 | 2 | 1 | 0 | 0 | 0 | 0 | 0 |

4 | 1 | 3 | 3 | 1 | 0 | 0 | 0 | 0 |

5 | 1 | 4 | 6 | 4 | 1 | 0 | 0 | 0 |

6 | 1 | 5 | 10 | 10 | 5 | 1 | 0 | 0 |

7 | 6 | 15 | 20 | 15 | 6 | 1 | 0 | |

8 | 5 | 21 | 35 | 35 | 21 | 7 | 1 | |

9 | 4 | 25 | 56 | 70 | 56 | 28 | 8 | |

10 | 3 | 27 | 80 | 126 | 126 | 84 | 36 | |

11 | 2 | 27 | 104 | 205 | 252 | 210 | 120 | |

12 | 1 | 25 | 125 | 305 | 456 | 462 | 330 | |

13 | 21 | 140 | 420 | 756 | 917 | 792 | ||

14 | 15 | 146 | 540 | 1161 | 1667 | 1708 | ||

15 | 10 | 140 | 651 | 1666 | 2807 | 3368 | ||

16 | 6 | 125 | 735 | 2247 | 4417 | 6147 | ||

17 | 3 | 104 | 780 | 2856 | 6538 | 10480 | ||

18 | 1 | 80 | 780 | 3431 | 9142 | 16808 | ||

19 | 56 | 735 | 3906 | 12117 | 25488 | |||

20 | 35 | 651 | 4221 | 15267 | 36688 |

This question is asked and discussed in my forum at Wizard of Vegas.

You are a pyrotechnician in charge of the nightly firework display at an amusement park. You've received some new style rockets from Europe and are testing one in order to time it to your show's music soundtrack.

The firework rocket is fired vertically upwards with a constant acceleration of 4 ms^-2 until the chemical fuel expires. Its ascent is then slowed by gravity until it reaches a maximum height of 138 meters where it detonates.

Assuming no air resistance and taking acceleration due to gravity is 9.8 meters per second per second, how long does it take the rocket to reach its maximum height?

Gialmere

Let:

t = time since rocket fuel runs out.

r = time rocket fuel lasted.

I'm going to express acceleration in terms of an upward direction. So, the acceleration after the rocket fuels burns out is -9.8.

As a reminder, the integral of acceleration is velocity and the integral of velocity is location. Let's make location relative to the ground.

When the rocket is first launched, we're given that the acceleration is 4.

Taking the integral, the velocity of the rocket after r seconds equals 4r.

Taking the integral of the velocity gives us the location of the rocket after r seconds of 2r^{2}.

Now let's look at what happens after the rocket fuel burns out.

We're given that acceleration due to gravity is -9.8.

The velocity due to gravity at time t is -9.8t. However, it also has upward velocity of 4r from the rocket.

Let v(t) = velocity at time t

v(t) = -9.8t + 4r

The rocket will achieve a maximum height when v(t) = 0. Let's solve for that.

v(t) = 0 = -9.8t + 4r

4r = 9.8t

t = 40/98 r = 20r/49.

In other words, whatever time the rocket fuel lasted, the rocket will keep traveling up for 20/49 of that time.

We are also given the distance traveled at the maximum altitude achieved is 138.

Let's take the integral of v(t) to get the formula for distance traveled, which we'll call d(t).

d(t) = -4.9t^{2} + 4rt + c, were c is a constant of integration.

As we already showed, the rocket traveled 2r^{2} by the time the fuel burned out, so that must be the constant of integration. That gives us:

d(t) = -4.9t^{2} + 4rt + 2r^{2}

We know the maximum altitude of 138 was reached at time 20r/49 So let's plug t=20r/49 into the equation to solve for r:

d((20r/49) = -4.9((20r/49)^{2} + 4r(20r/49) + 2r^{2} = 138

r^{2}*(-1960/2401 + 80/49 + 2) = 138

r^{2} = 49

r = 7

So, the rocket fuel lasted for seven seconds.

We already know the rocket kept going up for 20/49 of that time, which is 140/49 = apx. 2.8571 seconds.

Thus, the time from launch to maximum velocity is 7 + 140/49 = 483/49 = apx. 9.8571 seconds

This question is asked and discussed in my forum at Wizard of Vegas.