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Ask the Wizard #295

I know of a promotion that pays a bonus after achieving a four of a kind in all 13 ranks. How many hands will this take on average?

AxelWolf

Let's look at the gold standard of video poker, 9-6 Jacks or Better to answer your question.

The first step is to modify my calculator to include a line item for all 13 four of a kinds. Here is that modified return table:

Modified Jacks or Better Return Table

Event Pays Combinations Probability Return
Royal flush 800 493,512,264 0.000025 0.019807
Straight flush 50 2,178,883,296 0.000109 0.005465
Four A 25 3,900,253,596 0.000196 0.004892
Four K 25 3,904,533,816 0.000196 0.004897
Four Q 25 3,898,370,196 0.000196 0.004889
Four J 25 3,886,872,684 0.000195 0.004875
Four 10 25 3,471,687,732 0.000174 0.004354
Four 9 25 3,503,226,684 0.000176 0.004394
Four 8 25 3,504,128,652 0.000176 0.004395
Four 7 25 3,504,825,252 0.000176 0.004396
Four 6 25 3,504,861,888 0.000176 0.004396
Four 5 25 3,504,895,944 0.000176 0.004396
Four 4 25 3,504,032,676 0.000176 0.004395
Four 3 25 3,503,177,148 0.000176 0.004394
Four 2 25 3,502,301,496 0.000176 0.004393
Full house 9 229,475,482,596 0.011512 0.103610
Flush 6 219,554,786,160 0.011015 0.066087
Straight 4 223,837,565,784 0.011229 0.044917
Three of a kind 3 1,484,003,070,324 0.074449 0.223346
Two pair 2 2,576,946,164,148 0.129279 0.258558
Jacks or better 1 4,277,372,890,968 0.214585 0.214585
Nothing 0 10,872,274,993,896 0.545435 0.000000
Total 19,933,230,517,200 1.000000 0.995439


The probability of getting any four of a kind is 0.002363.

The next question to be answered is how many four of a kinds will take on average to get all 13 kinds? To answer that question, I created my Expected Trials Calculator. To use it, enter the number of combinations of each four of a kind in the first 13 cells. The calculator will tell you that it will take an expected 41.532646 four of a kinds to get all 13 kinds.

So, the expected number of hands played to get all 13 four of a kinds is 41.341739/0.002363 = 17,580.

What are the odds of the Carolina Panthers going 16-0 in the regular season? Are either of these bets good?

Yes +425?
No -550?

Pinit2winit

I have a method to estimate the point spread for any given game that comes remarkably close to the actual spread, barring any major injuries, healings, suspensions, or things like that. Here is the formula for expected points scored by any given team:

[(Average offensive points) + (Average points allowed by opposing team)]/2 + (1.5 if playing at home, otherwise -1.5).

The point spread will be (Expected visiting team points) - (Expected home team points).

Let's look at the week 13 game against the Saints as an example. The Panthers are the visiting team. The Panthers have scored 32.3 offensive points per game on average this season. The Saints have given up 30.8 per game, on average. Using my formula, the Panthers can expect to score (32.3+30.8)/2 - 1.5 = 30.05 points.

Then, do the same for the Saints. They have scored 23.7 offensive points per game on average this season. The Panthers have given up 18.6 per game, on average. My formula yields (23.7 + 18.6)/2 + 1.5 = 22.65 points scored by the Saints.

Therefore, the Panthers can expect to win by 30.05 - 22.65 = 7.4 points. Next, use my Prop Bet Calculator to determine the probability of winning each game. My calculator will ask for the over/under on the game, but I find for who will win straight up the only thing that really matters is the spread. For the total just put in an NFL average this season of 46. You'll see that for a point spread of 7.4 the fair line of the home team winning is +271. That means the fair line on the Panthers is -271. That equates to a probability of winning of 271/371 = 73.05%.

Then just do that for the other four games and take the product. Or you can just use the table below.

Panthers Weeks 13 to 17

Week Opposing
Team
Location Expected
Panther
Points
Expected
Opponent
Points
Panthers
Winning
Margin
Panthers
Fair
Line
Probability
Win
13 Saints Away 30.05 22.65 7.4 -271 0.730458
14 Falcons Home 28.3 19.6 8.7 -323 0.763593
15 Giants Away 27.05 23.85 3.2 -154 0.606299
16 Falcons Away 25.3 22.6 2.7 -144 0.590164
17 Buccaneers Home 30.35 19.05 11.3 -458 0.820789


Taking the product of the probability column you get the probability of winning all five games, which is 0.163813. That corresponds to a fair line of +510. So, neither of those lines you quoted are good.

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

What is the probability of the ball landing in 1, 2, and 3 within 4, 5, 6, 7, 8, or 9 spins in roulette?

allinriverking

The general formula is:

Pr(Ball lands in 1) + Pr(Ball lands in 2) + Pr(Ball lands in 3) - Pr(Ball lands in 1 and 2) - Pr(Ball lands in 1 and 3) - Pr(Ball lands in 2 and 3) + Pr(Ball lands in 1, 2, and 3).

In double-zero roulette, for n number of spins, this comes to 3*(1-(37/38)^n)-3*(1-(36/38)^n)+(1-(35/38)^n).

The following table shows the probability of rolling all three numbers for various number of spins from 3 to 100 for single- and double-zero roulette.

Roulette Question

Spins Single
Zero
Double
Zero
3 0.000118 0.000109
4 0.000455 0.000420
5 0.001091 0.001009
6 0.002094 0.001939
7 0.003518 0.003261
8 0.005404 0.005016
9 0.007785 0.007234
10 0.010684 0.009937
15 0.033231 0.031066
20 0.068639 0.064476
25 0.114718 0.108254
30 0.168563 0.159750
35 0.227272 0.216265
40 0.288292 0.275379
45 0.349548 0.335089
50 0.409453 0.393835
55 0.466865 0.450467
60 0.521017 0.504191
65 0.571445 0.554501
70 0.617922 0.601122
75 0.660393 0.643951
80 0.698930 0.683016
85 0.733693 0.718435
90 0.764897 0.750386
95 0.792791 0.779086
100 0.817638 0.804773