hml48
hml48
  • Threads: 4
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Joined: Dec 7, 2009
December 11th, 2009 at 12:12:11 AM permalink
Here is what I did please tell me if the results seem reasonable. (And give me a name for it.)
I simulated 20,000 sessions of craps. Each session consisted of up to 500 rolls with the following parameters. I started each with $10,000 [in simulated money]. If I hit my target of 15,000 or my floor of 2,500 I got out on the next losing roll. I bet $5 on the pass line and 500 on the odds bet at every opportunity to do so. [These are the only 2 bets I understand.]

Here are the results in descending order of gain
38.4% of the sessions hit the $15k target; that is the original $10k and $5k in winnings.
14.0% of the sessions produced some positive return but less than the target.
2.2% lost $500 or less
2.5% lost more than $500 but not more than $1,000
4.6% lost more than $1,000 but less than $2,000
38.1% lost more than $2,000 including the 1.5% whose bankrolls dipped below $2,500.

The average ending bankroll was 9,974.98 reflecting a small house advantage. Still a good evenings entertainment was free of cost to 52.4% of the sessions and quite profitable for the rather large winning group. And perhaps losers 500 or less might simulate the attitude of money well spent. But for the rest.. Well, there is nothing here to get me to go to a casino. I enjoyed the programing exercise and the education.

Thanks

HML
pacomartin
pacomartin
  • Threads: 649
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Joined: Jan 14, 2010
January 16th, 2010 at 5:36:58 PM permalink
Quote: hml48


I simulated 20,000 sessions of craps. The average ending bankroll was 9,974.98 reflecting a small house advantage.
HML




The formula for pass line bets is (-7 / 495) / [ 1 + ((5x + 4y + 3z) / 18) ] where the permissable maximum is x times odds on the 6 and 8, y times on the 5 and 9, and z times on the 4 and 10.

Pretty much the standard of (x,y,z) is (3,4,5).
If (x,y,z) is (0,0,0) or no free odds the probability is you will lose 7 out of 495 bets or 1.41%.
If (x,y,z) is (3,4,5) or maximum free odds the probability is you will 0.398%.

But you were betting 100X odds, so the expected value is that you would lose
If (x,y,z) is (100,100,100) or maximum free odds the probability is you will -0.021%.

So your expected value is you would only lose $2.09. However 20,000 sessions is not very many to get any statistical reasonable simulation. Typically you do billions of sessions.

The rest of the results would theoretically be nearly even because the house edge is nearly zero.
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