Does the house edge "compound" when pressing winning bets?
February 9th, 2010 at 8:37:50 AM
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Maybe I'm not asking that right. Maybe I should be asking about EV.
Say you place an inside bet at Roulette, for one unit. Assume that whenever you win, you press the bet to 5 units. Win again and it's 25 units, then 125, etc. If you lose, it's back to one unit.
Obviously, since you're always giving back part of the winnings, the EV is lowered.
But does that affect the house edge? If so, how can I calculate it?
Either way, is there a simple way to calculate the EV in this situation?
FYI: I tried to calculate it, but my brain hurts.
Thanks.
Say you place an inside bet at Roulette, for one unit. Assume that whenever you win, you press the bet to 5 units. Win again and it's 25 units, then 125, etc. If you lose, it's back to one unit.
Obviously, since you're always giving back part of the winnings, the EV is lowered.
But does that affect the house edge? If so, how can I calculate it?
Either way, is there a simple way to calculate the EV in this situation?
FYI: I tried to calculate it, but my brain hurts.
Thanks.
I invented a few casino games. Info:
http://www.DaveMillerGaming.com/ —————————————————————————————————————
Superstitions are silly, childish, irrational rituals, born out of fear of the unknown. But how much does it cost to knock on wood? 😁
February 9th, 2010 at 9:02:04 AM
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No, the house edge as a % of the total amount you bet does not change.
February 9th, 2010 at 9:02:22 AM
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I dont think that the amount bet makes a difference to the house edge.
The house edge is the constant advantage built in to the games, the size of the bet is completely unrelated to said bets chance of winning.
Thats the way I understand it.
The house edge is the constant advantage built in to the games, the size of the bet is completely unrelated to said bets chance of winning.
Thats the way I understand it.
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February 9th, 2010 at 11:40:04 AM
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Yeah, the more I think about it, the edge is a constant based upon a single bet.
In respect to the House Advantage, pressing a winner is nothing more than a new bet for a higher value. The fact that part of the wager comes from a prior win is irrelvant.
HOWEVER, pressing a bet does affect the EV, right?
Is there any reasonably simple way to calculate it?
In fact I'll make the example simpler:
Bet 1 unit on an inside number on Roulette. If it wins, press it to 5. If it wins again, press it to 25. Regardless if it wins again or not, next bet restart at 1.
I *think* that on occasions where you get the three in a row, the EV is the same as the HA. (or the reciprical). But what about those other times?
Can that be calculated?
In respect to the House Advantage, pressing a winner is nothing more than a new bet for a higher value. The fact that part of the wager comes from a prior win is irrelvant.
HOWEVER, pressing a bet does affect the EV, right?
Is there any reasonably simple way to calculate it?
In fact I'll make the example simpler:
Bet 1 unit on an inside number on Roulette. If it wins, press it to 5. If it wins again, press it to 25. Regardless if it wins again or not, next bet restart at 1.
I *think* that on occasions where you get the three in a row, the EV is the same as the HA. (or the reciprical). But what about those other times?
Can that be calculated?
I invented a few casino games. Info:
http://www.DaveMillerGaming.com/ —————————————————————————————————————
Superstitions are silly, childish, irrational rituals, born out of fear of the unknown. But how much does it cost to knock on wood? 😁
February 9th, 2010 at 11:45:48 AM
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One of the brains like JB, The Wizard, or Dorothy could probaly do it with their eyes closed. Me on the other hand, I wouldnt have a clue how to start.
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February 9th, 2010 at 11:50:52 AM
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DJ,
That is an interesting question, and the answer is somewhat complicated. Here is one way to look at it:
It depends upon your ultimate goal. Let's say you only have $1 and want to turn it into $30 or more. What is the best way to do this from the viewpoint of incuring least amount of "house edge" against you?
Everyone knows that both BJ and Craps beat Roulette hands down when it comes to the vig. Not even close. Or, is it?
If you place the $1 strait up, you have an opportunity to win $35 with only a vig of $.0526. So, with 00 Roulette it costs you $.0526 to get a shot at winning more than $30.
Now, let's look at BJ - assuming a .75% edge against the player.
You bet $1 and pay $.0075 for the privledge of getting a chance to return $2. If you win, you place $2 back in action to return $4. Remember, your goal is to end up with something over $30. You pay $.015 to make the $2 bet. You win, and then place $4 in action to return $8. This costs you $.03. Then, you place the $8 in action to return $16. This costs you $.06. Finally, you place $16 in action to return $32. This costs you $.12.
So, for the opportunity to turn $1 into over $30 with BJ you end up paying the house a total of vig of $.2325 (assuming you are a reasonably comptetent basic strategy player).
If your goal is to leave with over $30 or bust, Roulette is hands down your best play.
But, it's not really an apples to apples comparison. One of the benefits you are getting with the higher BJ vig is that you have the option of walking away with a lesser amount. This option is not given to you with the straight up Roulette bet.
Deciding which game is appropriate depends upon the ultimate goal of the player. If you want to walk into the casino with $1 and have the best chance of coming out with $2, BJ is clearly your best option. If your goal is to have the best chance to walk out with $30 or more, Roulette is your best option.
PS: The BJ option is actually much worse than I described because if you truly start with only $1 you don't have the capital to take advantage of DD or split opportunities, which comprise much of the player's advantage.
That is an interesting question, and the answer is somewhat complicated. Here is one way to look at it:
It depends upon your ultimate goal. Let's say you only have $1 and want to turn it into $30 or more. What is the best way to do this from the viewpoint of incuring least amount of "house edge" against you?
Everyone knows that both BJ and Craps beat Roulette hands down when it comes to the vig. Not even close. Or, is it?
If you place the $1 strait up, you have an opportunity to win $35 with only a vig of $.0526. So, with 00 Roulette it costs you $.0526 to get a shot at winning more than $30.
Now, let's look at BJ - assuming a .75% edge against the player.
You bet $1 and pay $.0075 for the privledge of getting a chance to return $2. If you win, you place $2 back in action to return $4. Remember, your goal is to end up with something over $30. You pay $.015 to make the $2 bet. You win, and then place $4 in action to return $8. This costs you $.03. Then, you place the $8 in action to return $16. This costs you $.06. Finally, you place $16 in action to return $32. This costs you $.12.
So, for the opportunity to turn $1 into over $30 with BJ you end up paying the house a total of vig of $.2325 (assuming you are a reasonably comptetent basic strategy player).
If your goal is to leave with over $30 or bust, Roulette is hands down your best play.
But, it's not really an apples to apples comparison. One of the benefits you are getting with the higher BJ vig is that you have the option of walking away with a lesser amount. This option is not given to you with the straight up Roulette bet.
Deciding which game is appropriate depends upon the ultimate goal of the player. If you want to walk into the casino with $1 and have the best chance of coming out with $2, BJ is clearly your best option. If your goal is to have the best chance to walk out with $30 or more, Roulette is your best option.
PS: The BJ option is actually much worse than I described because if you truly start with only $1 you don't have the capital to take advantage of DD or split opportunities, which comprise much of the player's advantage.
Prediction is very difficult, especially about the future. - Niels Bohr
February 9th, 2010 at 12:12:53 PM
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Quote: DJTeddyBearYeah, the more I think about it, the edge is a constant based upon a single bet.
In respect to the House Advantage, pressing a winner is nothing more than a new bet for a higher value. The fact that part of the wager comes from a prior win is irrelvant.
HOWEVER, pressing a bet does affect the EV, right?
Is there any reasonably simple way to calculate it?
In fact I'll make the example simpler:
Bet 1 unit on an inside number on Roulette. If it wins, press it to 5. If it wins again, press it to 25. Regardless if it wins again or not, next bet restart at 1.
I *think* that on occasions where you get the three in a row, the EV is the same as the HA. (or the reciprical). But what about those other times?
Can that be calculated?
Scenario 1: You bet $1 on an inside number, one time. Stop win or lose.
P(win) = 1/37 (single zero roulette, double is obviously 1/38, but I'm using single here because you should never play double zero roulette)
Payoff for a win = $36
Payoff for a lose = $0
EV = 36/37 (HA = 2.70%)
Scenario 2: You bet $1 on an inside number. Stop if you lose. Bet $5 on next spin if you win.
P(win then lose) = 1/37*36/37 = 36/1369
Payoff = 36-5 = $31
P(win then win) = 1/37*1/37 = 1/1369
Payoff = 31 + 180 = $211
EV = 31*36/1369 + 211/1369 = 1327/1369
To show that the HA remains unchanged, we could have approached the EV calc for scenario 2 using the HA for a single spin (1/37 = 2.70% calculated in Scenario 1): You are betting $1 37/37 times and $5 1/37 times.
So EV = 1 -($1 * 1/37 + $5 * 1/37 * (1/37)) = 1327/1369 and therefore the HA remains unchanged!
February 9th, 2010 at 1:12:50 PM
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CardShark -
Your example with only two steps in the system, actually helped the brain pain go away - except the math is coming out just the opposite of what I would have expected.
Bet one unit, unless I've got one win in a row, then bet 5. If I've got two or more in a row, bet one unit. Therefore, I only concern myself with the 1369 combinations of two spins of a single zero wheel.
OK. I start with a bankroll of 1369 and keep the winnings seperate, although pressure comes from the prior win.
In those 1369 spins, statistically, there will be 36 instances of getting one in a row, and one instance of getting two in a row.
To keep the math simple, the bets are paid, "and down", I.E. it pays 36.
36 single wins of 32 (I'm pressing 4) = 36 * 32 = 1152 pocketed.
1 double win = 32 + 5 * 36 = 212 pocketed.
Total pocketed = 1364
Loss = bankroll - pocketed winnings = 1369 - 1364 = 5
It only cost five units? Out of 1369?
That indicates the HA = 5 / 1369 = 0.0037
That can't be right, can it?
Your example with only two steps in the system, actually helped the brain pain go away - except the math is coming out just the opposite of what I would have expected.
Bet one unit, unless I've got one win in a row, then bet 5. If I've got two or more in a row, bet one unit. Therefore, I only concern myself with the 1369 combinations of two spins of a single zero wheel.
OK. I start with a bankroll of 1369 and keep the winnings seperate, although pressure comes from the prior win.
In those 1369 spins, statistically, there will be 36 instances of getting one in a row, and one instance of getting two in a row.
To keep the math simple, the bets are paid, "and down", I.E. it pays 36.
36 single wins of 32 (I'm pressing 4) = 36 * 32 = 1152 pocketed.
1 double win = 32 + 5 * 36 = 212 pocketed.
Total pocketed = 1364
Loss = bankroll - pocketed winnings = 1369 - 1364 = 5
It only cost five units? Out of 1369?
That indicates the HA = 5 / 1369 = 0.0037
That can't be right, can it?
I invented a few casino games. Info:
http://www.DaveMillerGaming.com/ —————————————————————————————————————
Superstitions are silly, childish, irrational rituals, born out of fear of the unknown. But how much does it cost to knock on wood? 😁
February 9th, 2010 at 1:16:02 PM
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Quote: DJTeddyBearThat indicates the HA = 5 / 1369 = 0.0037
1327/1369 = 0.9693Quote: cardsharkEV = 31*36/1369 + 211/1369 = 1327/1369
36/37 = 0.9729
0.9729 - 0.9693 = 0.0037
Maybe my math is right, but I'm still missing the 'why'.
I invented a few casino games. Info:
http://www.DaveMillerGaming.com/ —————————————————————————————————————
Superstitions are silly, childish, irrational rituals, born out of fear of the unknown. But how much does it cost to knock on wood? 😁
February 9th, 2010 at 1:33:50 PM
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Quote: DJTeddyBearThat indicates the HA = 5 / 1369 = 0.0037
Careful! This is not true! From the wizard: "The house edge is defined as the ratio of the average loss to the initial bet." I think you are mixing up EV with HA. In scenario 2, we have 2 bets being resolved one out of every 37 spins, so you can't use the EV directly to get the HA per spin.
If you look back at my scenario 2, I was able to derive the same EV using the HA of 1/37 (2.7%), therefore, the HA remains unchanged.
In terms of cost, you're system will cost you an extra $5/1369 to play, than scenario 1 (one single $1 bet). This is just because you are butting up against the HA of 2.7% twice once every 37 rounds.
February 9th, 2010 at 2:01:35 PM
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So let's just sum up...
The house advantage is always 2.7% of the total amount wagered.
The expected value of a normal $1 bet is $0.9730. The expected value of a $1 bet using your system is $0.9693.
The house advantage is always 2.7% of the total amount wagered.
The expected value of a normal $1 bet is $0.9730. The expected value of a $1 bet using your system is $0.9693.
February 9th, 2010 at 2:15:52 PM
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That's it exactly!
February 9th, 2010 at 7:18:07 PM
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Ok, maybe I'm confusing HA and EV.Quote: DJTeddyBear36 single wins of 32 (I'm pressing 4) = 36 * 32 = 1152 pocketed.
1 double win = 32 + 5 * 36 = 212 pocketed.
Total pocketed = 1364
Loss = bankroll - pocketed winnings = 1369 - 1364 = 5
It only cost five units? Out of 1369?
That indicates the HA = 5 / 1369 = 0.0037
That can't be right, can it?
But can I really risk 1369 and lose only 5?
When I started this thread, I assumed that the 'system' would cost me MORE than the HA. Or at best, the same as the HA.
This seems to suggest that I'm actually beating the HA.
While that would be great, I refuse to believe that I've trimmed my expected loss almost to zero.
Sigh... I gotta sleep on this...
I invented a few casino games. Info:
http://www.DaveMillerGaming.com/ —————————————————————————————————————
Superstitions are silly, childish, irrational rituals, born out of fear of the unknown. But how much does it cost to knock on wood? 😁
February 9th, 2010 at 7:32:15 PM
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Quote: DJTeddyBear36 single wins of 32 (I'm pressing 4) = 36 * 32 = 1152 pocketed.
1 double win = 32 + 5 * 36 = 212 pocketed.
Total pocketed = 1364
Loss = bankroll - pocketed winnings = 1369 - 1364 = 5
It only cost five units? Out of 1369?
That indicates the HA = 5 / 1369 = 0.0037
That can't be right, can it?
I think this is what you're missing...
When you win at roulette, the dealer gives you $35 + the $1 stays on the square. If you press it $4, then you only pocket $31 each time, not $32.
So your net is $31 * 37 + $180 = $1327. $1327/1369 = an expected loss of about $42 or 3.07%. This is more than the 2.7% HA because you are actually betting 1369 times @ $1 + 37 times @ $5 = $1554 total bet. $42 loss / $1554 = 2.7%
If you press your bets, you will lose more, but only because you are actually betting more! The betting strategy does not alter the house edge.
February 10th, 2010 at 5:38:00 AM
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I took that into consideration. Note in my original post "And Down".Quote: jeremykayI think this is what you're missing...
When you win at roulette, the dealer gives you $35 + the $1 stays on the square. If you press it $4, then you only pocket $31 each time, not $32.
If I leave the $1 up and press $4, when I'm done I'll have $37 left in my bankroll, so the net is the same.
I invented a few casino games. Info:
http://www.DaveMillerGaming.com/ —————————————————————————————————————
Superstitions are silly, childish, irrational rituals, born out of fear of the unknown. But how much does it cost to knock on wood? 😁
February 10th, 2010 at 5:55:51 AM
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HA and EV are two very different things, but yes, the HA changes based on your compounding system.
Let's take a good example: the hard ways press in Craps.
Place $1 on the hard 6 and 8. If it hits press it for the full amount. Then take it down if it hits again.
The HA on the $1 bet is ($9 x 1/11 - 1 x 10/11) = -1/11 = 9.09091%
The HA on the $10 bet is the same.
But because you are grouping the bet, all the time, together, the HA can be combined to give a different result.
The HA on the "System" is 1/11 x (90 x 1/11 - 10/11) - 10/11 = -.24793 = 24.79%.
Let's do the same "system": betting Red in "00" Roulette. You press your bet if you win, twice, taking a maximum of 3 units (1+2) if it hits twice.
The HA on the "System" is 18/38 x (3 x 18/38 - 20/38) - 20/38 = -.10249 = 10.25%.
The "system" is a rule that you put on yourself and it does change your HA.
Let's take a good example: the hard ways press in Craps.
Place $1 on the hard 6 and 8. If it hits press it for the full amount. Then take it down if it hits again.
The HA on the $1 bet is ($9 x 1/11 - 1 x 10/11) = -1/11 = 9.09091%
The HA on the $10 bet is the same.
But because you are grouping the bet, all the time, together, the HA can be combined to give a different result.
The HA on the "System" is 1/11 x (90 x 1/11 - 10/11) - 10/11 = -.24793 = 24.79%.
Let's do the same "system": betting Red in "00" Roulette. You press your bet if you win, twice, taking a maximum of 3 units (1+2) if it hits twice.
The HA on the "System" is 18/38 x (3 x 18/38 - 20/38) - 20/38 = -.10249 = 10.25%.
The "system" is a rule that you put on yourself and it does change your HA.
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You want the truth! You can't handle the truth!
February 10th, 2010 at 6:12:13 AM
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Quote: DJTeddyBearI took that into consideration. Note in my original post "And Down".Quote: jeremykayI think this is what you're missing...
When you win at roulette, the dealer gives you $35 + the $1 stays on the square. If you press it $4, then you only pocket $31 each time, not $32.
If I leave the $1 up and press $4, when I'm done I'll have $37 left in my bankroll, so the net is the same.
I'm sorry, but that's just not correct. You only get paid 35 to 1 for a win at roulette. If you win and pick up all your bets, you only have $36, not $37!
February 10th, 2010 at 6:45:24 AM
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BoyMimbo -
I gotta rattle that around, but it seems like you're confirming what I was suspecting.
(BTW: Love your handle.)
I gotta rattle that around, but it seems like you're confirming what I was suspecting.
(BTW: Love your handle.)
35 for each of the single wins, plus 2 for that double win.Quote: jeremykayYou only get paid 35 to 1 for a win at roulette. If you win and pick up all your bets, you only have $36, not $37!
I invented a few casino games. Info:
http://www.DaveMillerGaming.com/ —————————————————————————————————————
Superstitions are silly, childish, irrational rituals, born out of fear of the unknown. But how much does it cost to knock on wood? 😁
February 10th, 2010 at 7:42:49 AM
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Quote: DJTeddyBear35 for each of the single wins, plus 2 for that double win.
I'm not sure what you mean by this. If you place a $1 bet and win, you get paid $35, so you have $36 total. If you then place your $5 bet, you only have $31 left. If you lose this next bet, you only have $31 in your pocket (this will happen 36 times in the simulation). If your $5 bet wins, you get paid $175, plus you have your $5 bet and the $31 in your pocket = $180 + $31 = $211 (this will happen once). So you get paid $31 * 36 + $180 + $31 = $1327. If you don't believe me, go ahead and try it out in a free online game.
February 10th, 2010 at 7:50:34 AM
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Boymimbo, the cleverness of your "handle" escapes me. What am I missing?
the next time Dame Fortune toys with your heart, your soul and your wallet, raise your glass and praise her thus: “Thanks for nothing, you cold-hearted, evil, damnable, nefarious, low-life, malicious monster from Hell!” She is, after all, stone deaf. ... Arnold Snyder
February 10th, 2010 at 8:01:57 AM
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Quote: boymimboHA and EV are two very different things, but yes, the HA changes based on your compounding system.
Let's take a good example: the hard ways press in Craps.
Place $1 on the hard 6 and 8. If it hits press it for the full amount. Then take it down if it hits again.
The HA on the $1 bet is ($9 x 1/11 - 1 x 10/11) = -1/11 = 9.09091%
The HA on the $10 bet is the same.
But because you are grouping the bet, all the time, together, the HA can be combined to give a different result.
The HA on the "System" is 1/11 x (90 x 1/11 - 10/11) - 10/11 = -.24793 = 24.79%.
Let's do the same "system": betting Red in "00" Roulette. You press your bet if you win, twice, taking a maximum of 3 units (1+2) if it hits twice.
The HA on the "System" is 18/38 x (3 x 18/38 - 20/38) - 20/38 = -.10249 = 10.25%.
The "system" is a rule that you put on yourself and it does change your HA.
I completely disagree. The house advantage never changes... it is always the expected loss per amount bet. When you press your bets, you are not just betting the $1... every 11 times, you're betting $1 + $10 = $11. So for every 11 rolls, you are actually betting $21. Your expected loss may be 24.79% of your original $1 bet, but the house advantage on the total amount bet is no different! The probability of winning your press is 1/121, which nets you $100. You put out a total of $121 to get that $100 = loss of $21. However, your total amount wagered was $231 ($1*121 + $10*11). $21 loss / $231 = 9.09% HA.
February 10th, 2010 at 8:47:34 AM
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Hmmm... let's think about this. The Casino's HA never changes on a bet, that is true. The HARD 6 and 8 always carry the 9.0909% HA.
But the two possible outcomes of this system is to lose the original one unit or to win 89 units. (1 -> 10 units -> 90 units). The odds of winning 89 units is 1/121. The odds of losing 1 unit is 120/121. Therefore the "system" has a house advantage of 25.62% (89 / 121 - 120 / 121) because the better refuses to take the choice of taking the bet down and thus must ride the rules that he set for himself.
Imagine if the casino offered the same bet which basically said, "put $1 down, and if it hits twice before a seven is rolled, we'll give you $90. Otherwise you lose the $1" You would go and calculate the house advantage as:
89/121 - 120/121 = 25.62%.
My point is that the player is setting the rule for him/herself in defining the "system". You can calculate the house advantage on the "system". The fact that there is $10 being bet for part of the time has no effect on the house advantage because it doesn't factor into the final result. The final result can either be a win of $89 or a loss of $1. You calculate the HA on that. The house edge on the bet itself is indeed 9.0909%, but because the player lets it ride, the average bet size changes (I am guessing to 2.812 units) bit that has no bearing on the HA.
But the two possible outcomes of this system is to lose the original one unit or to win 89 units. (1 -> 10 units -> 90 units). The odds of winning 89 units is 1/121. The odds of losing 1 unit is 120/121. Therefore the "system" has a house advantage of 25.62% (89 / 121 - 120 / 121) because the better refuses to take the choice of taking the bet down and thus must ride the rules that he set for himself.
Imagine if the casino offered the same bet which basically said, "put $1 down, and if it hits twice before a seven is rolled, we'll give you $90. Otherwise you lose the $1" You would go and calculate the house advantage as:
89/121 - 120/121 = 25.62%.
My point is that the player is setting the rule for him/herself in defining the "system". You can calculate the house advantage on the "system". The fact that there is $10 being bet for part of the time has no effect on the house advantage because it doesn't factor into the final result. The final result can either be a win of $89 or a loss of $1. You calculate the HA on that. The house edge on the bet itself is indeed 9.0909%, but because the player lets it ride, the average bet size changes (I am guessing to 2.812 units) bit that has no bearing on the HA.
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You want the truth! You can't handle the truth!
February 10th, 2010 at 9:48:33 AM
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I do understand what you're saying and perhaps it's just difference of symantics... I prefer to only think of the expected value changing while the house edge is constant.
Side note... you should check your math. While the payout on the second bet is $90, you also pick up your original $10. So the win is $100 not $90. 99/121 - 120/121 = 17.36%. This is the same number as just squaring your expected value. With a 9.09% house edge, your expected value is 0.9091. 0.9091^2 = 0.8265. That's an expected loss of 17.35% (difference due to rounding) if you press.
Side note... you should check your math. While the payout on the second bet is $90, you also pick up your original $10. So the win is $100 not $90. 99/121 - 120/121 = 17.36%. This is the same number as just squaring your expected value. With a 9.09% house edge, your expected value is 0.9091. 0.9091^2 = 0.8265. That's an expected loss of 17.35% (difference due to rounding) if you press.
February 10th, 2010 at 10:53:18 AM
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Bingo. I stand corrected.
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You want the truth! You can't handle the truth!
February 10th, 2010 at 11:41:17 AM
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Quote: DJTeddyBearQuote: jeremykayI think this is what you're missing...
When you win at roulette, the dealer gives you $35 + the $1 stays on the square. If you press it $4, then you only pocket $31 each time, not $32.
I took that into consideration. Note in my original post "And Down".
I'm re-reading your post... Do you mean that on the second spin, you will add another $1 from your bankroll? If this is the case, your bankroll needs to be $1369 + 37 = $1406. So you win 32*37 + 180 = $1364, but it cost you $1406, which is stil a $42 loss.
February 10th, 2010 at 1:24:02 PM
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Yeah, I use $1 from the bankroll.Quote: jeremykayI'm re-reading your post... Do you mean that on the second spin, you will add another $1 from your bankroll? If this is the case, your bankroll needs to be $1369 + 37 = $1406. So you win 32*37 + 180 = $1364, but it cost you $1406, which is stil a $42 loss.
There are only 37 * 37 = 1369 combinations of two spins on a single zero wheel. I'll be using $1 from the bankroll for EVERY spin, whether the prior spin was a winner or not. Therefore a bankroll of $1369 is all I need.
When I win, I get the $35 win, and take the bet down. I bet $1 from the bankroll, plus $4 from the win, leaving $32 to into my pocket. I do that 37 times. 32 * 37 = 1184.
In addition, one time that $5 bet wins. I take the $5 down, and along with the $175 win, put $180 in my pocket.
My total win is 1184 + 180 = 1364.
That's only $5 less than I started with!
I invented a few casino games. Info:
http://www.DaveMillerGaming.com/ —————————————————————————————————————
Superstitions are silly, childish, irrational rituals, born out of fear of the unknown. But how much does it cost to knock on wood? 😁
February 10th, 2010 at 1:58:18 PM
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Quote: DJTeddyBearYeah, I use $1 from the bankroll.
There are only 37 * 37 = 1369 combinations of two spins on a single zero wheel. I'll be using $1 from the bankroll for EVERY spin, whether the prior spin was a winner or not. Therefore a bankroll of $1369 is all I need.
When I win, I get the $35 win, and take the bet down. I bet $1 from the bankroll, plus $4 from the win, leaving $32 to into my pocket. I do that 37 times. 32 * 37 = 1184.
In addition, one time that $5 bet wins. I take the $5 down, and along with the $175 win, put $180 in my pocket.
My total win is 1184 + 180 = 1364.
That's only $5 less than I started with!
You are correct that there are 1369 combinations of two spins. However, you have to bet $1 on each spin + $5 for each of the 37 spins that win.
Think about it this way... You need to bet a number 1369 times on average for it to come up a winner 37 of those times. Then you need to bet $5 each of those 37 extra times on average to come up a winner again. The total number of bets will actually be 1369+37. (You can either bet $1 1369 times and $5 37 additional times or you can bet $1 1332 times and $6 37 times.)
Lets say spin number 259 is a winner. For spin number 260 you need to place a $1 bet to start a new series, plus $5 to try for the "second spin" win. So everytime you win, you are essentially taking $2 out of your bankroll for that round, not just one.
Does this make sense?
February 10th, 2010 at 2:06:15 PM
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No. It's $1 every spin (from the bankroll) + $4 (pressure from the win) for the spins that follow the a single win, for a TOTAL of $5 bet on that second spin.Quote: jeremykayyou have to bet $1 on each spin + $5 for each of the 37 spins that win.
I invented a few casino games. Info:
http://www.DaveMillerGaming.com/ —————————————————————————————————————
Superstitions are silly, childish, irrational rituals, born out of fear of the unknown. But how much does it cost to knock on wood? 😁
February 10th, 2010 at 3:00:21 PM
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Quote: DJTeddyBearNo. It's $1 every spin (from the bankroll) + $4 (pressure from the win) for the spins that follow the a single win, for a TOTAL of $5 bet on that second spin.
That's fine, but the second spin of $5 doesn't count toward the average 1369 spins necessary to make this work. You need to bet $1 an average of 1369 times PLUS $5 an average of 37 times (whether they occur simultaneously is irrelevant).
I'm not trying to argue with you or anything. You seem to be skeptical of your calculation of a $5 loss (and for good reason). There are just some flaws in the logic that drives your numbers, which I'm doing my best to point out. However, I'm not a math teacher and not very good at explaining these things! Perhaps someone else can chime in... otherwise you'll just have to take my word for it!
February 10th, 2010 at 3:22:50 PM
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Maybe this will help... even though there are 1369 combinations of 2 spins, their are 2738 individual spins that make those up. You need to play all of the spin #1's, but most of the spin #2's you'll be sitting out (except for 37 of them, which you'll be playing at $5 each). Perhaps this will help you understand why you need to have a bankroll of $1369 + $37.
February 10th, 2010 at 3:37:12 PM
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Alright... You won't be able to argue with this. I made an excel sheet with every one of the 1369 combinations and the amounts bet and won from your bankroll on each one.
roulette.xlsx (97KB)
roulette.xls (273KB)
roulette.xlsx (97KB)
roulette.xls (273KB)
Last edited by: jeremykay on Feb 11, 2010
February 10th, 2010 at 5:29:19 PM
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Whoa!Quote: jeremykayMaybe this will help... even though there are 1369 combinations of 2 spins, their are 2738 individual spins that make those up. You need to play all of the spin #1's, but most of the spin #2's you'll be sitting out (except for 37 of them, which you'll be playing at $5 each). Perhaps this will help you understand why you need to have a bankroll of $1369 + $37.
Every spin is followed by another spin. 1369 combinations can be accomplished by 1370 spins. (I guess that means the math is off by $1)
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FYI: .xlsx is a new file format. Old versions of Excel can't open it. New versions can, AND can save it in the old format if selected.
I invented a few casino games. Info:
http://www.DaveMillerGaming.com/ —————————————————————————————————————
Superstitions are silly, childish, irrational rituals, born out of fear of the unknown. But how much does it cost to knock on wood? 😁
February 10th, 2010 at 6:23:50 PM
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Hey, I'm just trying to play by your "rules". When you win one, you bet only $5 the next spin.
Yes, you're right... xlsx was a new file format in 2007. Welcome to 2010!
If you haven't upgraded in a while, you can download the compatibility pack for older versions of Office here: Microsoft Office Compatibility Pack for Word, Excel, and PowerPoint 2007 File Formats
Once you take a look at the file, let me know where I went wrong.
Yes, you're right... xlsx was a new file format in 2007. Welcome to 2010!
If you haven't upgraded in a while, you can download the compatibility pack for older versions of Office here: Microsoft Office Compatibility Pack for Word, Excel, and PowerPoint 2007 File Formats
Once you take a look at the file, let me know where I went wrong.
February 10th, 2010 at 6:34:58 PM
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Quote: DJTeddyBearEvery spin is followed by another spin. 1369 combinations can be accomplished by 1370 spins.
Nope! You need two spins for each combination. That's why it's called a combination...
February 11th, 2010 at 5:51:06 AM
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Man, this is straining my brain muscle.
I'm starting to think that you're right. I'm not convinced, but I'm not so stubborn about it either.
Here's part of where I went wrong:
As I pointed out, since every spin is the first of two spins, as well as the second of two spins, I assumed that with 1369 combinations, only 1369 (ok 1370) spins are required to hit all combinations.
I even came up with my own XL spreadsheet to show them.
Except it didn't show them all.
There were 37 combinations missing, as well as 37 duplicates. No matter what I did, I couldn't fix that. I could add 37 spins to get the missing combinations, but I could not get away from the duplicates.
I'm going to wrap my head around this a little more.
Later today, I'll share my own XL documents.
Stay tuned....
I'm starting to think that you're right. I'm not convinced, but I'm not so stubborn about it either.
Here's part of where I went wrong:
As I pointed out, since every spin is the first of two spins, as well as the second of two spins, I assumed that with 1369 combinations, only 1369 (ok 1370) spins are required to hit all combinations.
I even came up with my own XL spreadsheet to show them.
Except it didn't show them all.
There were 37 combinations missing, as well as 37 duplicates. No matter what I did, I couldn't fix that. I could add 37 spins to get the missing combinations, but I could not get away from the duplicates.
I'm going to wrap my head around this a little more.
Later today, I'll share my own XL documents.
Stay tuned....
I invented a few casino games. Info:
http://www.DaveMillerGaming.com/ —————————————————————————————————————
Superstitions are silly, childish, irrational rituals, born out of fear of the unknown. But how much does it cost to knock on wood? 😁
February 11th, 2010 at 5:55:25 AM
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On a side note....
Here in my office, I have the new version, but when I downloaded it, the file I got was roulette.zip
I opened the zip file and found nothing usable.
Then I remembered something from somewhere.
Change the .zip to the correct extension (.xlsx). I then opened the file without problems.
Note: At home, I have an older version of XL, so I didn't try to download it.Quote: jeremykayI made an excel sheet....
roulette.xlsx
Here in my office, I have the new version, but when I downloaded it, the file I got was roulette.zip
I opened the zip file and found nothing usable.
Then I remembered something from somewhere.
Change the .zip to the correct extension (.xlsx). I then opened the file without problems.
I invented a few casino games. Info:
http://www.DaveMillerGaming.com/ —————————————————————————————————————
Superstitions are silly, childish, irrational rituals, born out of fear of the unknown. But how much does it cost to knock on wood? 😁
February 11th, 2010 at 8:03:51 AM
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Ok, ok... Here's the "old" excel version: roulette.xls
February 11th, 2010 at 10:11:28 AM
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Hmmmm.... The .xls version downloaded and opened fine.
I think the problem I had was caused by Internet Explorer. Whatever.
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I finally wrapped my head around this to the point where I agree with you. I imagine it was frustrating for you to convince me, but it was equally frustrating to me since I believed I was right.
Although duplication is introduced, it takes 1407 consecutive spins so that all 1369 two spin combinations can be found somewhere in the sequence. Part of my temporary argument was that you said that you need 1369*2 spins. I of course argued that that would result in EVERY combination appearing twice. Truth is, as I later found out, it will result in many duplicates, but also many triplicates, as well as many single instances of the unique two spin combinations.
Here's my own Excel spreadsheet that shows how I cam up with the 1407 spins.
Admin note: removed link to www.djteddybear.com/stuff/roulette_1406_chart.xls
It actually requires 1406 spins, plus the first of the next itteration to resolve the last spin.
Since that means a bankroll of $1406, matches the number you have been insisting, I started working with your spreadsheet. I made a couple minor changes: I added a variable for the amount being pressed. I also calculated the Player Expectation (Is that the right term?)
Here's my tweaked copy of your spreadsheet:
Admin note: removed link to www.djteddybear.com/stuff/roulette.xls
As you can see by playing with the value pressed, when you press zero, the player expectation is the same as the HA. It gets worse as the value pressed increases.
FYI: That's what I was trying to calculate / prove when I started this thread: That pressing a bet makes things worse. I mean, I knew it, but wanted the math to prove it.
Thanks.
I think the problem I had was caused by Internet Explorer. Whatever.
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I finally wrapped my head around this to the point where I agree with you. I imagine it was frustrating for you to convince me, but it was equally frustrating to me since I believed I was right.
Although duplication is introduced, it takes 1407 consecutive spins so that all 1369 two spin combinations can be found somewhere in the sequence. Part of my temporary argument was that you said that you need 1369*2 spins. I of course argued that that would result in EVERY combination appearing twice. Truth is, as I later found out, it will result in many duplicates, but also many triplicates, as well as many single instances of the unique two spin combinations.
Here's my own Excel spreadsheet that shows how I cam up with the 1407 spins.
Admin note: removed link to www.djteddybear.com/stuff/roulette_1406_chart.xls
It actually requires 1406 spins, plus the first of the next itteration to resolve the last spin.
Since that means a bankroll of $1406, matches the number you have been insisting, I started working with your spreadsheet. I made a couple minor changes: I added a variable for the amount being pressed. I also calculated the Player Expectation (Is that the right term?)
Here's my tweaked copy of your spreadsheet:
Admin note: removed link to www.djteddybear.com/stuff/roulette.xls
As you can see by playing with the value pressed, when you press zero, the player expectation is the same as the HA. It gets worse as the value pressed increases.
FYI: That's what I was trying to calculate / prove when I started this thread: That pressing a bet makes things worse. I mean, I knew it, but wanted the math to prove it.
Thanks.
I invented a few casino games. Info:
http://www.DaveMillerGaming.com/ —————————————————————————————————————
Superstitions are silly, childish, irrational rituals, born out of fear of the unknown. But how much does it cost to knock on wood? 😁
February 11th, 2010 at 12:31:04 PM
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I'm glad we got to the bottom of this! By the way, you can successfully do this with a bankroll of only $1369 by only using pressure for the second bet. This gives you a more realistic representation of the "player expectation" %. I made a modification that shows this here: roulette2.xls
February 12th, 2010 at 6:04:31 AM
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Indeed, the problem I encountered with the .xlsx file was due to Internet Explorer.
At home, on my Mac, if I clicked on the .xlsx file, a page opened with gibberish. If I downloaded it instead, I got the correct .xlsx file - which I was easily able to email to myself and open on my PC in my office.
Home, and in the office, had no problems with the .xls version. On both platforms, they downloaded and opened in Excel without issues.
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When the 'pressure' variable is changed to zero, the player expectation equals the house advantage, as expected.
When the 'pressure' is changed to 36, the player expectation equals the house advantage squared.
I.E.: -1 + (36/37)^2 = -5.3324%
Of course, any pressure inbetween results in a PE that is somewhere inbetween.
FYI: This is what I originally meant when I asked if the edge gets compounded by pressing. OK. Not the edge, but the PE gets compounded.
BTW: Are there other names for this value? Is that the same as the 'hold'? Yeah, I realize that the house 'hold' normally refers to a table's take at the end of the day... But is it the 'hold' for this one bet? Or is there a different term?
At home, on my Mac, if I clicked on the .xlsx file, a page opened with gibberish. If I downloaded it instead, I got the correct .xlsx file - which I was easily able to email to myself and open on my PC in my office.
Home, and in the office, had no problems with the .xls version. On both platforms, they downloaded and opened in Excel without issues.
-----
When the 'pressure' variable is changed to zero, the player expectation equals the house advantage, as expected.
When the 'pressure' is changed to 36, the player expectation equals the house advantage squared.
I.E.: -1 + (36/37)^2 = -5.3324%
Of course, any pressure inbetween results in a PE that is somewhere inbetween.
FYI: This is what I originally meant when I asked if the edge gets compounded by pressing. OK. Not the edge, but the PE gets compounded.
BTW: Are there other names for this value? Is that the same as the 'hold'? Yeah, I realize that the house 'hold' normally refers to a table's take at the end of the day... But is it the 'hold' for this one bet? Or is there a different term?
I invented a few casino games. Info:
http://www.DaveMillerGaming.com/ —————————————————————————————————————
Superstitions are silly, childish, irrational rituals, born out of fear of the unknown. But how much does it cost to knock on wood? 😁
February 12th, 2010 at 6:48:20 AM
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By the way, my original question was:
But my last post gave me an idea about the formula:
-1 + (36/37) ^ ( (36+pressure) / 36 )
This formula matches the PE in roulette2.xls when the pressure is 0 or 36. It's off for all other values.
OK. "EV" might not be the correct term.Quote: DJTeddyBearEither way, is there a simple way to calculate the EV in this situation.
But my last post gave me an idea about the formula:
Based on that, it seems that the formula should be:Quote: DJTeddyBearWhen the 'pressure' is changed to 36, the player expectation equals the house advantage squared.
I.E.: -1 + (36/37)^2 = -5.3324%
-1 + (36/37) ^ ( (36+pressure) / 36 )
This formula matches the PE in roulette2.xls when the pressure is 0 or 36. It's off for all other values.
I invented a few casino games. Info:
http://www.DaveMillerGaming.com/ —————————————————————————————————————
Superstitions are silly, childish, irrational rituals, born out of fear of the unknown. But how much does it cost to knock on wood? 😁
