The math has been posted several times long ago. You are better off believing and practicing the 5 Count. Spend your practice times improving your Advantage play. 😀 😀

Dice Shark,

You asked a good question. I will explain the answer in two ways. First I will give you another example that has the same criteria you list and show how a similar premise can be proven mathematically despite having the two facts that you listed. Second I will give you a link to an article that shows how Dr. Catlin proved the 5-count mathematically.

In my first example I will discuss tossing a fair coin with heads on one side and tails on the other. I think you will agree that the odds of throwing a head is 1 in 2. Thus a coin toss satisfies both of your facts.

1. odds of a head is 1 in 2. Each toss of the coin is an independent trial – the past or future has no effect.

Now if I toss the coin twice in a row I can expect one of the four following results: HH, TH, HT, TT

If I were to toss the coin twice in a row 1,000 times I can expect to see two heads (HH) approximately 250 times.

All I have done is used both facts to determine the probability of throwing 2 heads in a row.

The five count does the same thing for craps. It determines the probability of a random roller sevening out before completing the five count. The five count is using the probabilities of tossing 2 fair die by a random roller. But because it is a bit more complicated than my simple coin toss example, Dr. Don Catlin set up a computer simulation to do the calculation.

Dr. Catlin did a computer simulation of 200 million random shooters placing 6 and 8 for $6 each. One player, called the "Bet-All" player, obviously bet on every shooter who rolled the dice. The 5-Counter only placed the 6 and 8 after the shooter made it through the 5-Count. The 5-Count eliminated approximately 57 percent of the random shooters. During the period of play the 5-count shooter lost about 57% less than the "Bet All" player.

Here is the link to the entire article where you can read about the proof of the 5-count: http://www.goldentouchcraps.com/proof.shtml

Skinny, thank you for your continued contributions to the GTC message board. We do miss you.

Rose and Doc

Skinny as always a great job!

Just to add:

Using the 5-count will not give you an edge over the game unless you throw in comp time value as Dr Catlin, (God rest his soul) points out. There isn’t a mathematical system that can beat the game of craps. BUT – and the is BIG BUT – what the 5-count does is eliminate shooters so your money is only at risk 47% of the time! And that is a huge thing! it gives us a way of not betting on everyone and if we get lucky and a random roller has a big roll we only missed out on 5 of their throws.

The only way to beat the game is this:
1. Learn the controlled throw
2. calculate your edge when you are throwing – and
3. ONLY BET INTO YOUR EDGE – (as NF5 just posted in his great article)

Dom

Dom,
You posted

"…The only way to beat the game is this:
1. Learn the controlled throw
2. calculate your edge when you are throwing – and
3. ONLY BET INTO YOUR EDGE – (as NF5 just posted in his great article) "

Do you mean that we calculate our edge WHEN we are throwing at the casino’s table
and that edge is not our calculated edge done at home?

7×7

I still not know why the 5 Count has so many detractors. Just think how much money you lose on early 7 outs,I remember times when I hit my loss limits on the first 3 shooters having very short hands and not hitting a 6 or 8 . Also any great hand continues after the 5 Count,

DaveG

I have another article I once wrote on this topic I think you may find interesting. Upon rereading it I find it is a bit long. But I hope you feel it is worthwhile based on the information it gives.

http://www.goldentouchcraps.com/article.php?p=skinny0001.shtml

thanks for the explanation and links.

next queston: wth an SSR 0f 7.0 and using the hardway set what would be the probability of tossing a 6 on the first toss? What would the probability of repeating
the 6? standard random would be 13.88%

"Dice Shark" wrote: thanks for the explanation and links.

next queston: wth an SSR 0f 7.0 and using the hardway set what would be the probability of tossing a 6 on the first toss? What would the probability of repeating
the 6? standard random would be 13.88%

The probability of tossing a 6 on the first toss would be: 5.14/36

The probability of tossing two sixes in a row would be: 26.42/1296

If you establish the 6 as your point the probability of making your point would be 1/2.

With an SRR of 7.0 your chance of throwing a 6 or 8 is the same as that of throwing a 7.