Ask The Wizard #319

The following table shows the Vegas Insider money lines on both teams for each game. The column for the fair money line on the road time splits the juice in half between the two teams. The probability column shows the probability of the road team visiting, based on that fair line.

Taking the product of the road team probability of winning in every game, we get 0.00422, which rounds to 1 in 237.

A home team Martingale to win $100 would have resulted in a loss of $28,081.06.

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

First, I'll go over the number of permutations. This means that not only do the numbers matter but also their order on the card. There are permut(15,5) = 15!/(15-5)! = 15*14*13*12*11 = 360,360 possible permutations for the B, I, G, and O columns. For the N column, the number of permutations is permut(14,4) = 15!/(15-4)! = 15*14*13*12 = 32,760. Thus, the total number of permutations of bingo cards is 360,360⁴ × 32,760 = 552446474061128648601600000.

Second, I'll go over the number of combinations. This means that the numbers matter but not their order on the card. There are combin(15,5) = 15!/(5!*(15-5)!) = (15*14*13*12*11)/(1*2*3*4*5) = 3,003 possible combinations for the B, I, G, and O columns. For the N column, the number of permutations is combin(14,4) = 15!/(4!*(15-4)!) = (15*14*13*12)/(1*2*3*4) = 1,365. Thus, the total number of permutations of bingo cards is 3,003⁴ × 1,365 = 111007923832370565.

On the show, Sheldon asks himself how may UNIQUE bingo cards exist. Based on the later incorrect formulas, I assume he means permutations. In other words, two cards with the same numbers but in different positions would both be unique.

The image above shows Sheldon's formula for the B, I, G, and O columns. He initially gets the formula right at 5! × combin(15,5). However, he incorrectly simplifies it to 15!/(15!-5)!. The second exclamation point shouldn't be there. It should read 15!/(15-10)!. However, he then gets back to the correct answer at 360,360.

We have exactly the same problem with the N column. The formula should be 15!/(15-4)!, not 15!/(15!-4)!. The second exclamation point ruins it.

The ironic thing is later in the Episode, Sheldon become obsesses with errors in the chronology of the Lord of the Rings, much as I'm obsessing with this.

First, let's determine the number of combinations of player and board cards where this could happen. There are obviously four suits. Then there are combin(13,4)=715 ways you can choose four out of 13 cards of the given suit.

Second, one way this could happen is with three cards of that same suit the players have on the board and the other two among the other 39 cards. There are combin(9,3)=84 ways the board can have three out of the 9 cards left of the chosen suit. Then there are combin(39,2)=741 ways to pick two more cards out of the other 39 in the other three suits. So, there are 84*741=62,244 ways with three of the suit in question on the board.

Third, another way this could happen is with four cards of that same suit the players have on the board and the other one among the other 39 cards. There are combin(9,4)=126 ways the board can have four out of the 9 cards left of the chosen suit. Then there are 39 ways to pick one more card out of the other 39 in the other three suits. However, not all of these will result in both player's using both hole cards. For that condition to be met, the lowest card in the suit in question must be on the board. The probability of that, out of the 8 cards of that suit in play, is 4/8 = 1/2. So, there are 126*39*(1/2)=2,457 ways with four of the suit in question on the board.

Fourth, the last way this could happen is with five cards of that same suit the players have on the board. There are combin(9,5)=126 ways the board can have five out of the 9 cards left of the chosen suit. However, not all of these will result in both player's using both hole cards. For that condition to be met, the lowest two cards in the suit in question must be on the board. The probability of that, out of the 9 cards of that suit in play, is combin(5,2)/combin(9,2) = 10/36 = 5/18. So, there are 126*(5/18)=35 ways with four of the suit in question on the board.

So, the number of combinations where this will happen is 715*(62,244 + 2,457 + 35) = 46,286,240.

The total number of combinations of ways to pick four cards for the player hole cards out of 52 and then 5 more out of the 48 left on the board is combin(52,4)*combin(48,5) = 463,563,500,400.

Thus, the probability is 46,286,240 / 463,563,500,400 = 0.000399395 = 1 in 2,504.

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

To analyze such a bet, I would first estimate the number of points each team will score. I do this using simple algebra with the point spread and the over/under. For example, consider the first game between the Bengals and Ravens. The Ravens are favored by 12 and the over/under is 48. Let:

b = points scored by Bengals
r = points scored by Ravens

b+12=r
b+r=48

To rearrange the first equation: b-4=-12. Then add that equation to b+r=48 and you get 2b=36, so b=18. If the Bengals are expected to score 18 points, then the Ravens are expected to score 18+12=30.

Once we have estimated total points we can get to estimated touchdowns. I do this by subtracting six field goal points from each team and then dividing the rest by 7.

The total number of touchdowns expected to be scored among these teams is 29.57. Next, divide estimated touchdowns for each team by that total. That will give an estimated probability that team will score the first touchdown. Then determine the expected value given that probability and what the bet pays.

As you can see in the table, I perceive a positive expected value on two teams only. The Redskins (yes, I call them that) at a 0.48% advantage and the Bengals at a 21.7% advantage. The edge on the Redskins is too small, but I would definitely bet the Bengals.

P.S. The Bengals DID score the first touchdown that day!

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