Fun craps question
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25 members have voted
February 21st, 2011 at 11:10:43 AM
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The following players have the given bankroll to play craps
Player A: $1000
Player B: $2000
Player C: $4000
Player D: $21,000
All players are told to make line bets only.(Player D must find a casino that permits 20X free odds)
Player A puts $25 on the pass line and plays until he is bankrupt or doubles his money.
Player B puts $25 on the pass line plus $25 free odds and plays until he is bankrupt or doubles his money.
Player C puts $25 on the pass line plus $75 free odds and plays until he is bankrupt or doubles his money.
Player D puts $25 on the pass line plus $500 free odds and plays until he is bankrupt or doubles his money.
The question is what is the probability that the player will double his money before he goes bankrupt? Hypothetically if he was playing a game with no house edge the answer would be 50%. Since there is a house edge the correct answer will be below 50%.
The choice of answers is in the poll . They are rounded off so that it is possible to pick one by intuition rather than calculation. Choose the answer closest to the correct answer.
Player A: $1000
Player B: $2000
Player C: $4000
Player D: $21,000
All players are told to make line bets only.(Player D must find a casino that permits 20X free odds)
Player A puts $25 on the pass line and plays until he is bankrupt or doubles his money.
Player B puts $25 on the pass line plus $25 free odds and plays until he is bankrupt or doubles his money.
Player C puts $25 on the pass line plus $75 free odds and plays until he is bankrupt or doubles his money.
Player D puts $25 on the pass line plus $500 free odds and plays until he is bankrupt or doubles his money.
The question is what is the probability that the player will double his money before he goes bankrupt? Hypothetically if he was playing a game with no house edge the answer would be 50%. Since there is a house edge the correct answer will be below 50%.
The choice of answers is in the poll . They are rounded off so that it is possible to pick one by intuition rather than calculation. Choose the answer closest to the correct answer.
February 21st, 2011 at 12:11:38 PM
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I have run these kind of sims in the past in WinCraps.
Mine were $5 bets with 5,10,20,40,50,100 units for the bankrolls and 0 up to 10x odds,20x and 100x odds.
I also had different win goals. 25%,50%,75%,100% for each.
I need to look for those results.
Wonder if the numbers would be the same as the $ per unit increases.
For fun, I re-ran with your $25 bets and bankrolls.
I agreed exactly with your first and last %s and was a little off with the middle ones.
I will not post my results until more vote.
FYI:
Player A: $1000 Average bet: $25
Player B: $2000 Average bet: $41.67
Player C: $4000 Average bet: $75
Player D: $21,000 Average bet: $358.33
Mine were $5 bets with 5,10,20,40,50,100 units for the bankrolls and 0 up to 10x odds,20x and 100x odds.
I also had different win goals. 25%,50%,75%,100% for each.
I need to look for those results.
Wonder if the numbers would be the same as the $ per unit increases.
For fun, I re-ran with your $25 bets and bankrolls.
I agreed exactly with your first and last %s and was a little off with the middle ones.
I will not post my results until more vote.
FYI:
Player A: $1000 Average bet: $25
Player B: $2000 Average bet: $41.67
Player C: $4000 Average bet: $75
Player D: $21,000 Average bet: $358.33
February 21st, 2011 at 12:19:44 PM
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Quote: pacomartinThe choice of answers is in the poll . They are rounded off so that it is possible to pick one by intuition rather than calculation. Choose the answer closest to the correct answer.
My intuition says this is a trick question and all players have the same probability of either doubling or busting regardless of the bet size. And given the varying bankrolls, they should even take similar amounts of time to reach either option.
Donald Trump is a fucking criminal
February 21st, 2011 at 2:26:45 PM
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Quote: guido111
I agreed exactly with your first and last %s and was a little off with the middle ones.
I will not post my results until more vote.
Thank you for not posting. The solutions are not exact in the poll, but you have to pick the best one.
I am interested in people's intuition. Should all four scenarios give you the same percentage? We all know that the house advantage for a single bet on the pass line without any odds is 1.4%. A single bet of the entire amount (if permitted by the casino) would result in 50.7% chance of the player winning double or nothing. What is reasonable for multiple smaller bets where you win some and lose some.
So far almost no one is voting, so people don't seem interested.
February 21st, 2011 at 5:22:03 PM
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I don't want to guess, so my approach would be to create simulations, just like guido111. But that takes TIME, and it sounds like you really don't want that kind of answer anyway... you want to know what people's gut reactions will be.
OK, my gut reaction tells me that player D will have the best resuts, followed by C, B, then A.
My gut tells me that the more free odds bets are made, the better the overall EV will be.
But my gut has a hook in it... I can feel it every time I approach a craps table!
OK, my gut reaction tells me that player D will have the best resuts, followed by C, B, then A.
My gut tells me that the more free odds bets are made, the better the overall EV will be.
But my gut has a hook in it... I can feel it every time I approach a craps table!
The difference between zero and the smallest possible number? It doesn't matter; once you cross that edge, it might as well be the difference between zero and 1.
The difference between infinity and reality? They are mutually exclusive.
February 21st, 2011 at 7:23:50 PM
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In the Wizard's Roulette problem (#116) on his web page, he works out a similar problem analytically instead of resorting to simulation.
February 21st, 2011 at 8:37:44 PM
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Quote: pacomartinIn the Wizard's Roulette problem (#116) on his web page, he works out a similar problem analytically instead of resorting to simulation.
There is a risk-of-ruin formula The Wizard has posted at WoO.
One must flat bet, meaning the bets must all be the same value, for the answers to be accurate and it works great in Excel.
https://wizardofodds.com/ask-the-wizard/roulette/general/
1/2 of the way down
Question:
"If I have a $200 bankroll that I don't mind losing, and keep playing $10 on one single number on European (single-zero) roulette, what are the probabilities of winning, $200, $500 or $1000? Assuming I'll stop after reaching the target. Thanks, great site, you wanted me to keep this short :) – Andy from Amsterdam"
Answer:
"With a bankroll of b units, winning goal of g units, probability of winning p, and probability of losing q your probability of success is ((q/p)^b-1)/((q/p)^(b+g)-1).
In this case b=20, p=18/37, q=19/37, and g=20, 50, and 100. So for a bankroll of $200 the probability is ((19/18)^20-1)/((19/18)^(40)-1) = 0.253252."
The Wizard gives a few more examples.
(edit:I added the ^ (power) in the above formulas for clarity here and () around b+g.)
The formula does not work (fails) if the bets are not the same, I have never been able to get even average bets to work, simulations are the best way to go if the bets are not the same.
Hint... So, one can use the above formula to calculate for Player A.
February 21st, 2011 at 8:55:39 PM
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Quote: pacomartinSo far almost no one is voting, so people don't seem interested.
Give it time. Some polls take a while to gather votes.
BTW I told you what I've gathered intuitively. Alas I can get a feel for which bet is better but not in a quantitative level. I've no idea how to vote.
Donald Trump is a fucking criminal
February 21st, 2011 at 9:55:37 PM
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Quote: guido111
Answer:
"With a bankroll of b units, winning goal of g units, probability of winning p, and probability of losing q your probability of success is
((q/p)^b-1)/((q/p)^b+g-1).
Slight error:
((q/p)^b-1)/((q/p)^(b+g)-1) -- you forgot the parenthesis around the "b+g".
February 21st, 2011 at 10:27:53 PM
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Quote: pacomartinSlight error:
((q/p)^b-1)/((q/p)^(b+g)-1) -- you forgot the parenthesis around the "b+g".
Thanks and corrected.
Was correct in my Excel.
I copied and pasted into notepad from the WoO site before here and lost the formatting!
when in a hurry, always double check. I would have found the mistake.
So, in a spreadsheet, like Excel, one can really see the relationship between flat betting and bet size, bankroll and win goal when considering risk-of-ruin..
February 22nd, 2011 at 12:28:24 AM
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Quote: guido111
FYI:
Player A: $1000 Average bet: $25
Player B: $2000 Average bet: $41.67
Player C: $4000 Average bet: $75
Player D: $21,000 Average bet: $358.33
Do you have the average number of rolls until resolution?
February 22nd, 2011 at 8:04:14 AM
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Quote: pacomartinDo you have the average number of rolls until resolution?
Yes I do.
Exactly. Just because Player A has an X% chance of doubling his bankroll, it is nice to see how long it takes.
One nice thing about simulations using WinCraps. It gives you statistics and the means to save them for future use. There are some statistics it does not save when you run multiple sessions (more than 1) but you can have any stat saved to a text file that can easily be imported into Excel.
I will post more results tonight, got to work today.
Here are some roll numbers for player A only. Our $25 flat bettor, no odds and a 20 unit bankroll and win goal.
# of Rolls
Mean: 4863
Median: 3772
Mode: 1626 to 1856
High: 25008
Low: 478
SD: 3846
Skew: 1.78 (tail to right)
I do not have the data in Excel so I can not show a histogram.
478-1000: 5.2%
1001-2000: 19.4%
2001-3000: 15.2%
3001+: 60.2%
Player B Mean: 5836.26
Player C Mean: 5866.78
Player D Mean: 5758.44
February 22nd, 2011 at 10:26:27 AM
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Quote: pacomartinThe choice of answers is in the poll . They are rounded off so that it is possible to pick one by intuition rather than calculation. Choose the answer closest to the correct answer.
I feel since everyone is betting 2.5% of their bankroll each bet, most times, that they all have the same chance to double their bankroll.
But, since the expected loss for each is the same but the variance for each is the big difference, then my mind says they all must have an unequal chance.
A,B,C,D 25%,33%,40%,48% is my choice
7 winner chicken dinner!
February 22nd, 2011 at 11:16:08 AM
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Concentrate on player A for moment
We have 19 votes now, so I assume most people who are interested have weighed in.
All possibilities for player A were chosen ranging from 5% to 45%. Since this is one of the lowest house edges in the casino, intuition should tell you that have a reasonable chance to double your bankroll. The most popular answers were 25% and 45%.
You can do this by simulation or with the following formula:
"With a bankroll of b units, winning goal of g units, probability of winning p, and probability of losing q your probability of success is
((q/p)^b-1)/((q/p)^(b+g)-1). The way to derive this formula is shown in problem #116 in the Wizard's math problem web site.
For example if you were betting $100 on the pass line (no odds) you would have $1000/$100=10 betting units
b=10,
g=10, (since you are doubling)
p=244/495,
q=251/495
The calculation for the pass line bet in craps is shown in many websites including the "Wizard of Odds".
The House Edge is q-p=7/495=1.4141%.
((q/p)^b-1)/((q/p)^(b+g)-1) = 43%
Would you expect the probability to go up or down with $25 bets instead of $100 bets?
Now can you choose the correct answer in the poll?
We have 19 votes now, so I assume most people who are interested have weighed in.
All possibilities for player A were chosen ranging from 5% to 45%. Since this is one of the lowest house edges in the casino, intuition should tell you that have a reasonable chance to double your bankroll. The most popular answers were 25% and 45%.
You can do this by simulation or with the following formula:
"With a bankroll of b units, winning goal of g units, probability of winning p, and probability of losing q your probability of success is
((q/p)^b-1)/((q/p)^(b+g)-1). The way to derive this formula is shown in problem #116 in the Wizard's math problem web site.
For example if you were betting $100 on the pass line (no odds) you would have $1000/$100=10 betting units
b=10,
g=10, (since you are doubling)
p=244/495,
q=251/495
The calculation for the pass line bet in craps is shown in many websites including the "Wizard of Odds".
The House Edge is q-p=7/495=1.4141%.
((q/p)^b-1)/((q/p)^(b+g)-1) = 43%
Would you expect the probability to go up or down with $25 bets instead of $100 bets?
Now can you choose the correct answer in the poll?
February 22nd, 2011 at 11:40:27 AM
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Quote: pacomartinConcentrate on player A for moment
We have 19 votes now, so I assume most people who are interested have weighed in.
All possibilities for player A were chosen ranging from 5% to 45%. Since this is one of the lowest house edges in the casino, intuition should tell you that have a reasonable chance to double your bankroll. The most popular answers were 25% and 45%.
You can do this by simulation or with the following formula:
"With a bankroll of b units, winning goal of g units, probability of winning p, and probability of losing q your probability of success is
((q/p)^b-1)/((q/p)^(b+g)-1). The way to derive this formula is shown in problem #116 in the Wizard's math problem web site.
For example if you were betting $100 on the pass line (no odds) you would have $1000/$100=10 betting units
b=10,
g=10, (since you are doubling)
p=244/495,
q=251/495
The calculation for the pass line bet in craps is shown in many websites including the "Wizard of Odds".
The House Edge is q-p=7/495=1.4141%.
((q/p)^b-1)/((q/p)^(b+g)-1) = 43%
Would you expect the probability to go up or down with $25 bets instead of $100 bets?
Now can you choose the correct answer in the poll?
Well, I think most smart gamblers realize the best chances you have to win the most is to bet the most for the fewest times.
Example: $1000 bankroll and bet $1000 on the pass gives you a 49.29% (244/495) chance of doubling your bankroll. May not be fun for most players.
Then as bet amounts go up you have a better chance of hitting a win goal.
Most craps players do not and never will play that way. Betting very high to double their bankroll.
They say they would rather do what it takes to play the longest.
So smaller bets would mean a lower chance to hit a win goal.
Again, most craps players want to Buy in for $100, bet the pass with double odds, place the 6&8 and wonder why they bust out in 20 minutes!
There appears to be a balance between bet size, bankroll and win goal and time at the table.
Thank goodness for computers.
Imagine Newton or Einstein, or for us gamblers, Fermat and Pascal having a PC!
7 winner chicken dinner!
February 22nd, 2011 at 6:17:52 PM
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Player A
If you are betting $25 on the pass line (no odds) you would have $1000/$25=40 betting units
b=40,
g=40, (since you are doubling)
p=244/495,
q=251/495
The calculation for the pass line bet in craps is shown in many websites including the "Wizard of Odds".
The House Edge is q-p=7/495=1.4141%.
((q/p)^b-1)/((q/p)^(b+g)-1) = 24.39%
So that narrows it down to two answers in the poll.
p=244/495*18/(18 +(5*m + 4*m + 3*m)
q=251/495*18/(18 +(5*m + 4*m + 3*m)
Where "m" is the free odds you are taking each time (assuming they are the same ). For this question m=0,1,3, and 20 for players A,B,C,D.
I would like to see the simulation results to see if they are the same as
25% 33% 40% and 48%.
If you are betting $25 on the pass line (no odds) you would have $1000/$25=40 betting units
b=40,
g=40, (since you are doubling)
p=244/495,
q=251/495
The calculation for the pass line bet in craps is shown in many websites including the "Wizard of Odds".
The House Edge is q-p=7/495=1.4141%.
((q/p)^b-1)/((q/p)^(b+g)-1) = 24.39%
So that narrows it down to two answers in the poll.
p=244/495*18/(18 +(5*m + 4*m + 3*m)
q=251/495*18/(18 +(5*m + 4*m + 3*m)
Where "m" is the free odds you are taking each time (assuming they are the same ). For this question m=0,1,3, and 20 for players A,B,C,D.
I would like to see the simulation results to see if they are the same as
25% 33% 40% and 48%.
February 22nd, 2011 at 9:46:03 PM
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I chose the first answer just by intuition. I know that with the odds bet you have fewer trials as well as a lower HA which gives you a better chance of doubling your bankroll. Even though your EV is the same, the variance is higher.
-----
You want the truth! You can't handle the truth!
February 23rd, 2011 at 11:21:06 AM
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Quote: pacomartin(1) I was curious how close that Guido's simulations came to the theoretical results (using the formulas outlined above) :
a)24.39%
b)33.65%
c)40.68%
d)48.03%.
Simulation results (100,000 sessions each):
player a: 24.281%
player b: 34.138%
player c: 41.756%
player d: 48.074%
>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>
player a
All bets are a single unit
------------------------------------
Simulation Results per Session
------------------------------------
Avg. No. games played . = 1446.94
Avg. No. games won . . = 713.18
Avg. No. games lost . . = 733.76
Avg. No. dice rolls . . = 4884.48
Avg. Total amount bet . = 1446.94
Bankroll was busted . . = 75.719% of the time ( 75719)
Win goal was met . . . = 24.281% of the time ( 24281)
Bankroll decreased . . = 75.719% of the time
Bankroll increased . . = 24.281% of the time
player b
All bets are a single unit
------------------------------------
Simulation Results per Session
------------------------------------
Avg. No. games played . = 1728.88
Avg. No. games won . . = 852.13
Avg. No. games lost . . = 876.74
Avg. No. dice rolls . . = 5836.26
Avg. Total amount bet . = 1728.88
Avg. amount bet on Odds = 1152.21
Bankroll was busted . . = 65.862% of the time ( 65862)
Win goal was met . . . = 34.138% of the time ( 34138)
Bankroll decreased . . = 65.862% of the time
Bankroll increased . . = 34.138% of the time
player c
All bets are a single unit
------------------------------------
Simulation Results per Session
------------------------------------
Avg. No. games played . = 1737.91
Avg. No. games won . . = 856.59
Avg. No. games lost . . = 881.32
Avg. No. dice rolls . . = 5866.78
Avg. Total amount bet . = 1737.91
Avg. amount bet on Odds = 3471.33
Bankroll was busted . . = 58.244% of the time ( 58244)
Win goal was met . . . = 41.756% of the time ( 41756)
Bankroll decreased . . = 58.244% of the time
Bankroll increased . . = 41.756% of the time
player d
All bets are a single unit
------------------------------------
Simulation Results per Session
------------------------------------
Avg. No. games played . = 1705.81
Avg. No. games won . . = 840.77
Avg. No. games lost . . = 865.04
Avg. No. dice rolls . . = 5758.44
Avg. Total amount bet . = 1705.81
Avg. amount bet on Odds = 22080.38
Bankroll was busted . . = 51.926% of the time ( 51926)
Win goal was met . . . = 48.074% of the time ( 48074)
Bankroll decreased . . = 51.926% of the time
Bankroll increased . . = 48.074% of the time
>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>
And for bets with different win goals, chances of success are:
Player A different win goals
1000 bankroll-$25 bets (40 bankroll units)
25 95.937% I thought it would be higher. 4% (1 in 25 ) chance of not even hitting a $25 win goal.
50 92.088%
100 84.974%
200 72.736%
250 67.450%
500 47.105%
1000 24.390%
1000 bankroll-$50 bets (20 bankroll units)
25 96.809%
50 93.773%
100 88.128%
200 78.292%
250 73.985%
500 56.926%
1000 36.223%
February 23rd, 2011 at 12:54:14 PM
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In terms of reaching a win goal, it's all about variance. That's why the results are slanted toward the higher odds.
Cheers,
Alan Shank
Woodland, CA
Cheers,
Alan Shank
Woodland, CA
Cheers,
Alan Shank
"How's that for a squabble, Pugh?" Peter Boyle as Mister Moon in "Yellowbeard"
February 23rd, 2011 at 2:01:56 PM
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Quote: goatcabinIn terms of reaching a win goal, it's all about variance. That's why the results are slanted toward the higher odds.
Cheers,
Alan Shank
Woodland, CA
pacomartin started out this thread "Player A puts $25 on the pass line and plays until he is bankrupt or doubles his money."
My years of playing craps, almost every player I had talked with wants to play for hours at the table and wants to end up doubling their bankroll.
I think we just do not know our chances of doing just that.
If player A was to win $500 instead of $1000 his chances go from 24.390% to 47.105%... almost doubles.
At a $250 win goal 67.450%, now he has the best of it, win percentage that is.
Just because a player can have a better session win rate does not guarantee an overall net win. So Alan is correct, as variance gives you a better chance to win, and the odds bet in craps gives you that chance without increasing the expected loss.
February 23rd, 2011 at 2:34:00 PM
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I am always amazed at web explanations that say something like Martingale works great for a while, but the longer you play it, the more you risk going bankrupt. It is a stupid statement, because as you can clearly see even flat betting has a good chance of getting you to a goal of increasing your bankroll by 25% or 50%.
(1) I was curious how close that Guido's simulations came to the theoretical results (using the formulas outlined above) :
a)24.39% b)33.65% c)40.68% d)48.03%.
(2) People's intuition was all over the map. From an extremely negative 5% probability that you will double your money for player A, to the very optimistic 45% for player A. Also people's intuition said that taking free odds has no effect. It confirms my guess that people don't really know what the effect of different house edge's in a game where you are winning and losing.
(2a) Many people felt intuitively that "free odds" did not increase the probability that you will double the bankroll.
(3) Baccarat has a slightly smaller house edge than pass line betting in craps, but not nearly as good as 20X odds in craps. Since the goal of the player is usually to accomplish some kind of goal like doubling your bankroll, I think that anyone who is betting $2K or less per roll ($5K at Rampart) is much better off at playing free odds at smaller casinos off the strip. But even playing 3,4,5X on the strip will still significantly increase the chance that you will double your bankroll than playing baccarat.
(4) If I drop my bet amount to $10 given the same bankrolls and doubling goal, the probabilities of doubling before bankruptcy go up dramatically.
5.58% 15.49% 28.03% 45.08%
(5) Think about the phrase grind joint which is the name of traditional budget casino with nothing to distract you from gaming, and the small house edge grinds away at the bankroll the longer you play. This thread illustrates how bad the grind can be. What happens if you are playing slots where the house edge is huge?
(1) I was curious how close that Guido's simulations came to the theoretical results (using the formulas outlined above) :
a)24.39% b)33.65% c)40.68% d)48.03%.
(2) People's intuition was all over the map. From an extremely negative 5% probability that you will double your money for player A, to the very optimistic 45% for player A. Also people's intuition said that taking free odds has no effect. It confirms my guess that people don't really know what the effect of different house edge's in a game where you are winning and losing.
(2a) Many people felt intuitively that "free odds" did not increase the probability that you will double the bankroll.
(3) Baccarat has a slightly smaller house edge than pass line betting in craps, but not nearly as good as 20X odds in craps. Since the goal of the player is usually to accomplish some kind of goal like doubling your bankroll, I think that anyone who is betting $2K or less per roll ($5K at Rampart) is much better off at playing free odds at smaller casinos off the strip. But even playing 3,4,5X on the strip will still significantly increase the chance that you will double your bankroll than playing baccarat.
(4) If I drop my bet amount to $10 given the same bankrolls and doubling goal, the probabilities of doubling before bankruptcy go up dramatically.
5.58% 15.49% 28.03% 45.08%
(5) Think about the phrase grind joint which is the name of traditional budget casino with nothing to distract you from gaming, and the small house edge grinds away at the bankroll the longer you play. This thread illustrates how bad the grind can be. What happens if you are playing slots where the house edge is huge?
February 23rd, 2011 at 2:45:04 PM
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Quote: pacomartin(2) People's intuition was all over the map. From an extremely negative 5% probability that you will double your money for player A, to the very optimistic 45% for player A.
Also people's intuition said that taking free odds has no effect.
It confirms my guess that people don't really know what the effect of different house edge's in a game where you are winning and losing.
For every one of us that do understand house edge and it's effect, there has to be 10,000 that have no clue or do not care or both.
"(1) I was curious how close that Guido's simulations came to the theoretical results (using the formulas outlined above) :
a)24.39% b)33.65% c)40.68% d)48.03%."
How did you come up with your formulas? They are interesting.
Thank you
February 23rd, 2011 at 3:38:57 PM
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The formulas are fairly standard math think, but they are all developed on the Wizard's websites. I tried to reference his pages as I was going along.
(1) The widely derived pass line result that there will be 244 wins for every 251 losses.
(2) The extension of cumulative house edge to include different "free odds" bets.
(3) The probability of achieving a percentage goal for flat betting given the goal, the number of betting units and the house edge.
(1) The widely derived pass line result that there will be 244 wins for every 251 losses.
(2) The extension of cumulative house edge to include different "free odds" bets.
(3) The probability of achieving a percentage goal for flat betting given the goal, the number of betting units and the house edge.
February 23rd, 2011 at 4:05:21 PM
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Quote: pacomartin
p=244/495*18/(18 +(5*m + 4*m + 3*m)
q=251/495*18/(18 +(5*m + 4*m + 3*m)
Where "m" is the free odds you are taking each time (assuming they are the same ). For this question m=0,1,3, and 20 for players A,B,C,D.
I would like to see the simulation results to see if they are the same as
25% 33% 40% and 48%.
How did you use the above formulas to arrive at the percentages? Or did you?
Player A p=244/495, q=251/495
Player B p=0.295757576 q=0.304242424
edit:
Thanks, I answered my own question
February 24th, 2011 at 2:07:09 PM
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p=244/495*18/(18 +(5*m + 4*m + 3*m))
q=251/495*18/(18 +(5*m + 4*m + 3*m))
I omitted a parenthesis by accident.
q=251/495*18/(18 +(5*m + 4*m + 3*m))
I omitted a parenthesis by accident.
February 11th, 2012 at 12:26:29 PM
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I got it!
Sally
Sally
I Heart Vi Hart
