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Bob Dancer writes a video poker column for beginners to experts. He also writes a column with Jeffrey Compton, "Player's Edge", featuring information on promotions at various Las Vegas casinos. Player's Edge is published each Friday in the Neon section of the Las Vegas Review-Journal. Click here to send Bob Dancer an e-mail.

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Sep 23, 2008

What's the Difference? Redux

In last week's column we had a supposedly fair coin that turned up tails 19 times in a row. I argued that was sufficient to doubt the fairness assumption. We then drew to AAA 50 times in a row without collecting the fourth ace. I argued that was NOT sufficient to doubt the fairness assumption. So how can 19 be "enough" while 50 be "not enough?" When you read my answer, most of you will agree it's not a difficult concept. At the same time, I believe that many of you are not ready to answer the question now without reading on first. If so, the column will be a success. My major goal, after all, is to teach a significant percentage of my readers something useful they didn't know before.

What we need to consider is the probability for success. Let's start with the aces. When you draw two cards to trips, you should get quads two times in 47. If that seems like a strange number, consider that we started with 52 cards in the deck and we initially drew 5 cards, leaving 47 remaining. For convenience we'll say we were dealt the hand we identified last week, Ah As Ac 6d 6s. Of the 47 remaining cards in the deck, when we toss the sixes we don't care whether the fourth ace ends up in the fourth position or the fifth. That gives us two chances out of 47, which comes out to be 0.0425531914893617, which I will shorten to 4.255% for convenience.

If I have a 4.255% success rate each time, then that means I have a 95.745% failure rate. To figure out the chances of failing 50 times in a row, we multiply (0.95745) * (0.95745) . . . * (0.95745) a total of 50 times. There are lots of ways to do this (I'll demonstrate the way I do it shortly) but the answer comes out to be 0.1137, which is 11.37%, which is one chance in 8.8.

On the coin flip, it's supposed to be a 50-50 chance for success. To get 19 heads in a row we multiply (0.50) times itself a total of 19 times. This gives us 0.0000019073486328, which comes out to one chance in 524,288.

So I'm not really comparing the numbers 50 and 19, but rather one chance in 8.8 (which I see as not so rare) with one chance in 524,288 (which I consider very rare indeed.) So it isn't how long the lucky (or unlucky) streak is, it is how long it is compared to what the probability is.

Now let's look at how we can easily calculate these numbers, using Excel, which is a program found on most computers these days. The formula I am going to use is called the Binomial Distribution. Without defining it, for purposes of this article I'll just say that it's the appropriate one to use for questions like we have here.

Check out the following table:

1	A	B	C	D	E
2
3	Number of	Chance for	Chance for	Cumulative	Inverse
4	Successes	1 Hit	Total Hits	Chances	of Total Hits
5
6	0	0.0426	0.1137	0.1137	8.80
7	1	0.0426	0.2527	0.3663	3.96
8	2	0.0426	0.2751	0.6415	3.63
9	3	0.0426	0.1956	0.8371	5.11
10	4	0.0426	0.1022	0.9392	9.79
11	5	0.0426	0.0418	0.9810	23.94
12	6	0.0426	0.0139	0.9949	71.82
13	7	0.0426	0.0039	0.9988	257.07
14	8	0.0426	0.0009	0.9998	1076.10

To the far left we have line numbers. The next column, A, represents the number of successes, which on this chart go from zero to eight.

Column B represents 2/47, which I've been calling 4.255% in the text. To place it there I placed the cursor on "cell" B6, entered =2/47 on the formula line, and hit enter. This gave me 0.0426 in area B6. (I set up my cells to have 4 digits to the right of the decimal. This is done for convenience. The computer handles 16 digits to the right of the decimal whether you choose to display them or not.)

Column C represents the chances for having a specific number of hits. Cell C6 represents 0 hits, cell C7 represents 1 hit, cell C8 represents 2 hits, etc. I put the cursor on C6 and on the formula line I enter =BINOMDIST(A6,50,B6,FALSE) and then hit enter. Cell A6 contains a zero (which means the number of hits), the 50 (i.e. the second position in the formula) refers to the number of chances, cell B6 contains the probability of success each time, and FALSE is a required parameter to make this work. When we do this, the answer we come up with is 0.1137, which means this happens 11.37% of the time.

Column D represents the chances for that many hits or fewer, and the formula is =BINOMDIST(A6,50,B6,TRUE)

Column E is simply 1 divided by whatever is in Column C. I put the cursor on E6 and enter on the formula line =1/C6

Once I do this for line 6, I select all of the information from B6 through E6, copy it using CNTL C, and then paste it (using CNTL V) from B7 through B14. When I do this, the chart looks like what we see above.

Let's look at line 10, which corresponds to exactly 4 hits --- meaning we get four quads in 50 tries. C10 tells us we have a 10.22% chance of getting four hits. D10 tells us we have a 93.92% chance of getting four or fewer hits (i.e. 0, 1, 2, 3, or 4), and E10 tells us that 10.22% corresponds to one chance in 9.79.

Once you've duplicated this table, I recommend you save it on your computer for next time under some name like Binomial Distribution Worksheet. It's quite handy and can be used to answer a number of questions. In a column in the near future I'll address some of the questions it can answer, but you'll need to have created this on your own computer before you can get much benefit out of that column.