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MATHEMATICS FOR THE PUNTER By the Turf Accountant

Last time I left you with the following questions:

 a) What is the chance of throwing either a 3 or a 6 with one throw of a die?

b) What is the chance of throwing either a 3 or a 6 with four throws of a die?

c) Using the subtraction rule what is the chance of drawing a club in one draw from a pack of cards?

d) What is the chance of throwing a 3 followed by a 6 in two consecutive throws of the die?

e) Is it the same as throwing a 1 followed by a 2 in two consecutive throws of the die?
 
 

The answers:

 (a) What is the chance of throwing either a 3 or a 6 with one throw of a die?

This is an application of the addition rule. "If two or more chances are mutually exclusive, the probability of making ONE OR OTHER of them is the sum of their separate probabilities." What then is the chance of throwing a three in one throw?

It is 1 in 6 or 1/6 or 16.667% or odds of 5 to 1 against. What then is the chance of throwing a six in one throw? Also 1 in 6 or 1/6 or 16.667% or odds of 5 to 1 against. From the addition rule then, the chance of throwing 6 OR 3, must be the 16.667% chance of the 6 plus the 16.667% chance of the 3.16.667 + 16.667 = 33.333% or two chances in six which is equivalent to one chance in three or odds of two to one against.

(b) What is the chance of throwing either a 3 or a 6 with four throws of a die?

For this answer we have to combine the addition rule as above with the formula c P = 1 - (LP / TP)

Now from the addition rule we know that the chance of throwing either a 6 or a 3 in one throw is 33.333% or .33333.

Now a winning chance of 33333 must be a losing chance of .66667; and from our formula then the chance of throwing a .66667 losing chance in four tries is: 4 (.6667) = .19753 [(.6667x.6667x.6667x.6667) = .19753] and a .19753 losing chance translates to a 1 - .19753 winning chance, or a .80247 winning chance. 80.247% chance of success = odds of four to one ON.

(c) What is the chance of drawing a club in one draw from a pack of cards?

The odds of drawing either a club, spade, diamond or heart are 100%. Therefore the subtraction rule says that the odds of drawing a club, are 100 - (the combined chances of drawing a spade, diamond or heart).

There are 13 cards of each suit in the 52 card pack. So that the chances of drawing a spade are 13 in 52 or 25%. Similarly the chances of a heart are 25% and the chances of a diamond are 25%.

Therefore, the chances of a club are ((100-(25+25+25))% = 25%.

(d)(i) What is the chance of throwing a 3 immediately followed by a 6 in two consecutive throws of a die?

This is an illustration of the use of the product or multiplication rule. "If two chances are mutually exclusive the chances of getting both together, or one immediately after the other, is the product of their respective probabilities."

What is the chance of getting the 3 on the first throw? It is 1 in 6 1/6.

What is the chance of getting the 6 on the second throw? It is 1 in 6 1/6. Then the chance of getting a 6 on the second throw, when we have already got our 3 on the first throw is: 1/6 * 1/6 = 1/36 = 2.77778% or odds of 35 to one against.
 
 

(d)(ii) Is the chance of throwing a 1 followed by a 2 the same as that of a 3 followed by a 6 in the question above. I trust most realize by now that the answer is YES. As each number on the die has the same 1/6 chance, the numbers selected are irrelevant, the 35/1 odds apply to 4 - 6; 2 - 2; 6 - 6; 1 - 6 etc etc.

(d)(iii) that then is the chance of throwing a 3 and a 6 in one cast of two dice? It is NOT 35/1 but 34/2 or 17/1! ??

Ever played backgammon?

There are 6 sides on each die, and there are two die, therefore the number of possible individual results are 6x6 = 36. We can conveniently tabulate the possibilities as follows: By the way this tabulation represents our first foray into the world of Statistics, I have described mathematics as the language of nature or science, statistics is the analysis or interpretation of this language.
 
 
 

Face Die 1 >	1	2	3	4	5	6
Face Die 2  1
2
3						x
4
5
6			x			*

FIGURE 1.

The 36 cells of the above grid represent the 36 possible results, the two X's represent the two results that fulfill our requirement.

This is a diagrammatic proof of our answer of 34/2 or 17/1.

But inductive reasoning, that is answers induced from observation are always dangerous, ‘logic’ can be flawed, mathematics are certainty. Therefore let us prove this observation mathematically.
 

• The possibility of throwing a three on dice A is 1/6,
• the possibility of throwing a six on dice B is also 1/6, therefore
• the possibility of throwing three on dice A and six on dice B in the same cast is 1/6 x 1/6 or 1/36;

that is we could have 6 on dice A and 3 on dice B, and this possibility has the same probability as the other, that is 1/36.

Applying the addition rule we have 2 chances for and 34 chances against or odds of 34/2 or 17/1.

But you may still ask why??.

Why is the probability of throwing the 6, followed by the 3, with two throws of the same die only 2.8% (35/1) when the probability of throwing 6 and 3 with one throw of two separate die is 5.6% (17/1)??.

Let me digress for a moment here, there is a very important point to illustrate.

 Notice that although the probability here has doubled from 2.8% to 5.6%, the odds have not quite doubled having gone from 35/1 to 17/1.

Do not be fooled into believing that the ratio of odds is the same as the ratio of probability. This can be more clearly illustrated with an example:

2/1 odds represent two chances against and one for, or a one in three probability or a probability of 33.3%.

4/1 odds represent four chances against and one for, or a one in five probability or a probability of 20%.

Back to the problem...

The difference in probabilities is because we have specified in the first case that the 6 is first and the 3 is second, had we instead said what is the probability of throwing a 3 or a 6 followed by a 6 or a three, that is getting two different numbers in two different ways, but not including pairs of threes or sixes, then the probability would be the same, as follows:

The probability of getting either a three or a six on the first throw would be 2/6 or 33.3%, but since on the second throw we no longer have the choice of either a 3 or a 6, but must instead throw the opposite of what has been thrown, then on the second throw we have only 1/6 probability or 16.7%, and our chance of getting this combination in one set of two throws is therefore 33.3% times 16.7% or 1/6 of 1/3, or 1/18 or 17/1.

What then is the probability of throwing a double six with one cast of two dice?

I hope that no-one said 17/1 !! All the backgammon players will tell you it is 35/1, as you can checkout in figure 1. where the only square that satisfies this criteria is marked with an *.

This is because when we want two different numbers, we have two chances on each die, when we want a ‘double’ we have only one choice on each die. One more for the road.

What is the probability then of throwing any ‘double’ in one cast of two dice ?. Is it 6/36 ? Is it 16.67% ? Is it 5/1 ? Is it 1/6 ?

Yes to all!!

These are only different ways of expressing the same probability.

Proof ?.

We need the addition rule here.

The probability of any one ‘double’ is as we proved above 1/36.

There are six separate possible ‘doubles’. So the possibility of a ‘double’ is 1/36+1/36+1/36+1/36+1/36+1/36= 6/36. We need to combine the product rule and the addition rule here. First the chance of a particular number is 1/6 therefore the product rule tells us the chance of a pair is 1/6 * 1/6 = 1/36 as explained in the answers to questions 4 and 5. Now since there are six possible pairs we use the addition rule to tell us the chance of getting any pair is the addition of the chances of getting each particular pair, i.e. 1/36+1/36+1/36+1/36+1/36+1/36 = 6/36 = 1/6 or odds of 5/1.

 Similarly if you had three dice the chances of a particular treble would be 1/6 * 1/6 * 1/6 = 1/216 or odds of 215/1.

 Whilst, using the addition rule the odds of ANY treble would be: 1/216+1/216 etc 6 times = 6/216 OR 1/36 or odds of 35/1.

Notice the odds of ANY treble are EXACTLY the same as a NOMINATED double with two dice.

The reason being that in the treble, if you treat the first die as the nominator.

i.e. when the first die is a 6 the chances of a pair of 6's with the other two dice; is exactly the same as nominating that you will throw a pair of 6s with two dice.

Think about that for quite a while, together with the difference between the addition rule and the product rule; and the circumstances in which you use one or the other, and I believe you should begin to "see" one of the "patterns" of probability theory.

Try to really understand how to arrive at the answers to the next set of questions before proceeding further, this grounding in probability is the foundation stone of the series on Profitable Punting Above we have seen that the chance of 3 followed by 6 in two casts of one die is 2.77778% or 1/36;

what then is the probability of throwing 1 & 2 with one cast of two dice?

If the answer is NOT 1/36 - then explain why is this so ? Notice I use the terms chance and probability as being interchangeable, think of them that way. If this question is a problem, refer back to Figure 1. How many combinations fulfil the criteria? 2 - right! How many do not fulfil the criteria - 34 - right! Therefore the probabilty is 2/36 or 1/18 or 17/1 against or 5.6%. Well so much for the laws of Probability.

But what is the purpose of all of this where am I heading?

I am heading to Profitable Punting something that took me forty years to achieve.

I have now written a book "Punting for Profit".

From which these articles are taken. My book concludes that Punting can indeed be Profitable.

The book examines "systems" that HAVE BEEN profitable.

Systems that are purchased seem to fall into two categories, those that never work in the first place, the majority, and those rare ones that show a profit for a time and then inevitably "fall into a hole".

These systems that perform for a while and then "go bad" can be sub-divided into two categories, those that merely strike the inevitable long run of outs, that then corrects itself, if one perseveres, and those that just lose slowly but very surely for ever more.

My studies show that the latter do not usually lose their original "strike rate", rather they retain their strike rate but the average return declines, turning a long term profit into a long term and permanent loss.

The reason for this is quite simple, the better a system is the more surely it will kill itself, if it is marketed. Because the more successful its the more people that buy and use it, and therefore the more people "on" its selections, and therefore the decreased returns,and therefore the inevitable losses.

How can this be solved?

Simply - by finding your own system, so that not only does the strike rate continue, but also the average return stays at the same level, and therefore the profitability continues.

But if we are to develop our own successful systems we must first understand statistics, and particularly their relationship to the laws of probability, so that we can analyse past performance, identify profitable statistics, and then coordinate this knowledge in a profitable and unique system.

• Start at any card.
• Note that the adjacent card in at least one direction is "Red", move to that card.
• Move left or right to the nearest ‘Black’ card.
• Move up or down to the nearest ‘Red’ card.
• Move diagonally to the nearest ‘Black’ card.
• Move down or right to the nearest ‘Red’ card.

So next time...

we continue on our introduction to understanding basic statistics, until then profitable Punting to all.

THE TURF ACCOUNTANT

EMAIL   ...... The Turf Accountant.
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