What is my Average Bet

Quick math question (from a non-math person). I want to calculate my average bet for standard deviation purposes.

If I bet the pass line for $25 with full odds (3x, 4x, 5x) and I am always coming, what is my average bet?

I am guessing it would be ($125 * 3) + $25 come bet = $400 average bet

Help please! Thank you.

Quote: gambler

and I am always coming

That's what SHE said. (Sorry, couldn't resist!)

Seriously, though, I think goatcabin has a formula for this so hopefully he will chime in here and help you out.

The always coming part makes it hard. If I were inclined to answer the question, I would run a simulation.

I am not sure what you mean by average. Do you mean the average amount of money on the table per dice throw, or per series of rolls that ends in either a 7 out or making your point (including an unknown number of wins/losses on the come out roll)? If you go by dice then you will have some rolls with only a $25 bet on the table (but even that is variable since some come out rolls will have a come bet with odds on them since you made your point).

It's also not clear if you have three come bets, and you make your point, do you add another three come bets on your next roll, or do you replace only those come bets that you have won.

What I suspect you really want to know is given a rate of play (say 100 dice rolls per hour) what is the amount you will have gambled per hour.

There is a three decade old picture hanging in Binion's honoring what was claimed to be a world record 34 passes in a row. It is more customary today to count the number of throws of a dice to include the variable of how many wins and losses were made on the come out roll. We can hypothesize that 34 passes was approximately (34*3.7) throws of the dice. However, the current world record throw of 154 dice throws is probably over 40 passes.

for comp purposes, $100

I had asked a question previously regarding standard deviation and bankroll. The excellent answer that PapaChubby gave me was:

Standard Deviation = Average Bet x sqrt(bets/hour x hours played)

So my follow up question is "What is my average bet?"

If I bet $25 pass line, with full odds (average of 4x) with a come bet each and every time with full odds, (yes Teddy, always coming. That's what she said.) how much should I calculate my average bet to be?

Quote: appistappis

for comp purposes, $100

It's funny Appistappis, you would think that for comp purposes my average bet would be $100. I agree with you completely. However, perhaps it is because of my large buy in, my friendly personality, or the fact that I tip well, my average bet playing like this is between $100 and $300 depending on the pit boss. On rare occations even higher then that.

I always ask how they rated me right before I leave the table, and sometimes my jaw drops slightly with the inflated ratings.

Quote: gambler

I had asked a question previously regarding standard deviation and bankroll. The excellent answer that PapaChubby gave me was:

Standard Deviation = Average Bet x sqrt(bets/hour x hours played)

So my follow up question is "What is my average bet?"

If I bet $25 pass line, with full odds (average of 4x) with a come bet each and every time with full odds, (yes Teddy, always coming. That's what she said.) how much should I calculate my average bet to be?

For this purpose, each pass or come bet is a new bet. Your average bet is:

$25 x 12/36 + $100 x 6/36 + $125 x 8/36 + $150 x 10/36 = $94.44

Now the question is not how many dice rolls per hour, but how many resolved bets per hour.

Also, because you may lose multiple bets due to a seven out, the bets are not independent, which is an assumption of the formulas I gave you earlier. The correlation between multiple bets should probably be taken into account somehow. This is probably the reason that I have sucked at craps for so many years!

Quote: gambler

Quick math question (from a non-math person). I want to calculate my average bet for standard deviation purposes.

If I bet the pass line for $25 with full odds (3x, 4x, 5x) and I am always coming, what is my average bet?

I am guessing it would be ($125 * 3) + $25 come bet = $400 average bet

Help please! Thank you.

If you are making a come bet every chance you get, then the average amount you have on the line is going to be 3.44 * $25 = $86. This is based on a WinCraps simulation I did a while back. Here is an excerpt from my post about that study:

Quote: goatcabin, prior

Just for fun, I created a WinCraps autobet file that takes as input the maximum number of come bets to allow at a time. I set it up to stop betting at 200 rolls, wait for any bets to be resolved and end the session. I ran 2000 sessions for each input 1-6. I had discovered before that, although a come bet is made only 2/3 of the time, when the passline comeout establishes a point, the number of come bets resolved is more like 80% of the number of passline bets. This is because sometimes the passline bet gets resolved and there's still a come bet outstanding. On the next pass comeout, the come bet may be resolved and another come bet put up.

max comes	1	2	3	4	5	6
come/pass	.82	1.45	1.91	2.21	2.38	2.44

The numbers are fairly close to .8 + .8^2 + .8^3 + ... .8^n.

So, basically, if you make a come bet on every non-comeout roll, your average total of flat bets will be close to 3 1/2 times your basic amount. So, if you are a $10 passline bettor, and you make 60 bets, your total bet handle would be close to $2060, yielding an expected loss of about $29. That's without taking any odds, of course. If you took odds, the proportions would remain the same, assuming you always worked your come odds. Taking the odds would not increase the expected loss, just the volatility, risk of ruin, chance of coming out ahead and chance of a big win.

So, you may have as little as $25 or as much as $150 on the table in flat bets, but averaging about $86. Now if you are taking 3, 4, 5X odds on your passline bet, that means another $100 2/3 of the time, so add $67 = $153. Are you taking odds on those come bets, too? That's going to add about .67 * 86 * 4 = ~$230. So, that would be about $383. However, you should not combine the flat bets and odds to figure the standard deviation, because the SD on the odds is a lot higher, averaging about 1.22 times the amount of the bet, compared to, essentially, the amount of the bet for the flat part.

After that other post, I did a simulation of $20 pass, double odds, always $20 come bet, double odds. No more bets made after 200 rolls, wait for everything to resolve, $1500 bankroll. The average number of bets resolved was 319 (flat and odds), the average net result was -$55 and the SD was $1037.

Another way to get at this is to look at the ev and SD for $25 flat, 3, 4, 5X odds. For one bet, the ev is -$.3535 and the SD is $122.88, average bet $94.44. To extrapolate this, just multiply the ev by the number of bets and the SD by the square root of the number of bets. For example, for 60 bets, the ev is -$21.21 and the SD is $951.85.

Maybe I will run a sim for you.
Cheers,
Alan Shank

OK, I ran a WinCraps sim for $25 pass, 3, 4, 5X odds and $25 come, same odds, always coming.
Each session stopped betting after 200 rolls, waited for all bets to resolve. I set the bankroll at $4000, but there were some busts, so not all sessions reached 200 rolls, but only about 8% ended early. The average number of pass bets resolved was 57.9, 141.5 come bets, total bets 332, including all odds bets.

The mean net outcome was -$86, and the SD was $2408.

I ran it again with an $8000 bankroll, so there wouldn't be any busts. The standard deviation was $2398. An average of 203 flat bets were resolved.

If we just did 203 passline bets at $25 w/3, 4, 5X odds, we'd expect the SD to be sqrt(203) * 122.88 = $1752, so you are getting more variance by making pass and come bets.
Cheers,
Alan Shank