about gamble on black/red
As we know, most of the slot game has option to gamble the winning bet. The most common gamble is to guess the black or red poker card which randomly drawn form a deck. I want to figure out the payback on the gamble. So if I win $1 and I gamble it, I've got 50% chance of winning or losing the game. So my expectation is $0*50% + $2*50% = 100%, so the return to player for gamble is 100%?
Today, my friend told me he saw a slot in Casino which support gamble 50% of the winning or 200% of the winning. I wonder how's the return to player change when with those option. Let's say I won $1, if I gamble 50%, so even I lose the gamble, I still have 50% of wager back to me, so I estimate the math as
$0.5*50% + ($0.5 + $0.5*2)*50% = 1
where the first 50% is the chance of losing the gamble (black/red), the first $0.5 is the amount I kept excluding the wager, the second $0.5 is the part I kept excluding the wager, the last $0.5 is my wager and the following "2" is the odd of winning the game, I am quite surprise that it gives me 100% return again. Is my math correct or I made some mistakes?
For gamble 200%, it is quite confusing me. At the first glance, I should get 100% return to player because no matter how much you bet, when calculating the return to player, we always consider the unit wager, so bet $1 or $2 will not change the return to player. But if I work out the math the following way, I see something different.
-$1*50% + $2*2*50% = 1.5
The first -$1 is for the extra wager needed for gamble 200%. In gamble 200%, I need to bet my original $1 as well as $1 extra from my own pocket. The frist 50% stands for chance of losing the game. The $2 is the total bet due to the gamble 200% feature, the following "2" is when winning, pay double. So I got 1.5 (or 150%) return to player? I just don't understand why it is over 100%?
Quote: konglifyHi there,
As we know, most of the slot game has option to gamble the winning bet. The most common gamble is to guess the black or red poker card which randomly drawn form a deck. I want to figure out the payback on the gamble. So if I win $1 and I gamble it, I've got 50% chance of winning or losing the game. So my expectation is $0*50% + $2*50% = 100%, so the return to player for gamble is 100%?
I believe the formula should be:
(1 ×.50) + (-1 ×.50) = 0
In your formula you had nothing wagered on one side.
Quote: mrcleanI believe the formula should be:
(1 ¡Ñ.50) + (-1 ¡Ñ.50) = 0
In your formula you had nothing wagered on one side.
Thanks. But I am confusing. So the formula should be
return to player = P(lost)*PAY(LOST) + P(win)*PAY(WIN)
In this case, P(lost)=P(win)=0.5, if you win $1 from the game, you are allowed to gamble $2 ($1 from your win and $1 from your own pocket), and if you the gamble, you will get total $4 back, but if you lose, you lose all $2. So should it be
return to player = $4*0.5 =200% ?
If using your formula, the return to player is ZERO, so does it mean gamble will gain 0%? I did the simulation with computer for the case (wager 1, and pay $2 if win), playing the game for couple million times with probability of winning or losing is 50%, each time just increaset he wager by 1, if win, record the total payout. I got the return to player as 100%.
Quote: konglifyToday, my friend told me he saw a slot in Casino which support gamble 50% of the winning or 200% of the winning. I wonder how's the return to player change when with those option. Let's say I won $1, if I gamble 50%, so even I lose the gamble, I still have 50% of wager back to me, so I estimate the math as
$0.5*50% + ($0.5 + $0.5*2)*50% = 1
where the first 50% is the chance of losing the gamble (black/red), the first $0.5 is the amount I kept excluding the wager, the second $0.5 is the part I kept excluding the wager, the last $0.5 is my wager and the following "2" is the odd of winning the game, I am quite surprise that it gives me 100% return again. Is my math correct or I made some mistakes?
For gamble 200%, it is quite confusing me. At the first glance, I should get 100% return to player because no matter how much you bet, when calculating the return to player, we always consider the unit wager, so bet $1 or $2 will not change the return to player. But if I work out the math the following way, I see something different.
-$1*50% + $2*2*50% = 1.5
The first -$1 is for the extra wager needed for gamble 200%. In gamble 200%, I need to bet my original $1 as well as $1 extra from my own pocket. The frist 50% stands for chance of losing the game. The $2 is the total bet due to the gamble 200% feature, the following "2" is when winning, pay double. So I got 1.5 (or 150%) return to player? I just don't understand why it is over 100%?
If I understand what you mean by "100% return", it is correct; you are expected to get back the 1 that you won on the spin. The expected value of the bonus bet is zero, regardless of how much is bet; you have 50% chance of winning what you bet, and 50% chance of losing what you bet.
In the 200% calculation, you forgot to subtract the "extra 1 from your own pocket" from the win.
Let's say you win 1, and you bet 2 on the red/black bet; if you win that bet, your total winnings = 4, but your profit is only 4 - 1 = +3.
The calculation should be -1 x 0.5 + 3 x 0.5 = 1.
Quote: ThatDonGuyIf I understand what you mean by "100% return", it is correct; you are expected to get back the 1 that you won on the spin. The expected value of the bonus bet is zero, regardless of how much is bet; you have 50% chance of winning what you bet, and 50% chance of losing what you bet.
In the 200% calculation, you forgot to subtract the "extra 1 from your own pocket" from the win.
Let's say you win 1, and you bet 2 on the red/black bet; if you win that bet, your total winnings = 4, but your profit is only 4 - 1 = +3.
The calculation should be -1 x 0.5 + 3 x 0.5 = 1.
That's very clear and helpful. I learn something. Thanks a lot.
Quote: ThatDonGuyIf I understand what you mean by "100% return", it is correct; you are expected to get back the 1 that you won on the spin. The expected value of the bonus bet is zero, regardless of how much is bet; you have 50% chance of winning what you bet, and 50% chance of losing what you bet.
In the 200% calculation, you forgot to subtract the "extra 1 from your own pocket" from the win.
Let's say you win 1, and you bet 2 on the red/black bet; if you win that bet, your total winnings = 4, but your profit is only 4 - 1 = +3.
The calculation should be -1 x 0.5 + 3 x 0.5 = 1.
I am back with some questions, sorry for that. I read your reply again, so based on the statement "The expected value of the bonus bet is zero, regardless of how much is bet;" does it mean a fair gamble should always return the ZERO expectation, that is no gain no loss?
I have the following questions
1) in the book of math for gaming that I found online, it said the expectation is given as
P(lost)*(lost amount) + P(win)*(net win amount)
In the regular gamble for red/black card, P(lost)=P(win)=0.5, and says I bet $1 each gamble, so if I lost the gamble, I lost $1, if I win, I win double ($1x2=$2). So the expectation is
0.5*(-$1) + 0.5*($1x2-$1) = 0
Is that correct? So what is this expectation value related to the so called return-to-player? I ever read the book about return-to-player, it calculate that term by multiplying the winning probability with the winning amount (including the bet) so for this gamble, we have 0.5*($1x2) = 1 (or 100%). That's why I am confusing the calculations in this threads. Any way to associate the return-to-player to expectation?
2) So get back to the 200% gamble based on the expectation calculation, let's assume the pay is still 2x if I win and the probability won't change P(lost)=P(win)=0.5, but this time, I beg $2 ($1 from the spin and $1 from my pocket). If I lost the gamble, I lost all $2. But if I win, I win ($2x2-$1) so the expectation should be
0.5*(-$2) + 0.5*($2x2-$1) = $0.5
It is confusing me that you wrote it in -$1x0.5 + $3x0.5 = $1, in which it means that if you lost you lost $1 only? So which one is correct?
Also , if I get the positive expectation (like my $0.5 or your $1) what does it mean? Does it mean I will make profit for long time if I play 200% gamble comparing to the regular gamble?
3) So if I want to make the 200% gamble "fair", should I change the payout so to make the expectation zero? Thanks
Quote: konglifyI am back with some questions, sorry for that. I read your reply again, so based on the statement "The expected value of the bonus bet is zero, regardless of how much is bet;" does it mean a fair gamble should always return the ZERO expectation, that is no gain no loss?
I have the following questions
1) in the book of math for gaming that I found online, it said the expectation is given as
P(lost)*(lost amount) + P(win)*(net win amount)
In the regular gamble for red/black card, P(lost)=P(win)=0.5, and says I bet $1 each gamble, so if I lost the gamble, I lost $1, if I win, I win double ($1x2=$2). So the expectation is
0.5*(-$1) + 0.5*($1x2-$1) = 0
Is that correct? So what is this expectation value related to the so called return-to-player? I ever read the book about return-to-player, it calculate that term by multiplying the winning probability with the winning amount (including the bet) so for this gamble, we have 0.5*($1x2) = 1 (or 100%). That's why I am confusing the calculations in this threads. Any way to associate the return-to-player to expectation?
If I understand the terms correctly, "expected value" is the amount you are "expected to" win/lose in relation to how much you bet. In a 50/50 game, you are just as likely to win as you are to lose, and since the amount you win is the same as the amount you lose, the expected value is zero.
"Return to player" means what percent of the amount that you bet is returned as winnings. I think it is usually used when talking about video poker or slot machines. A video poker machine with a "99% return" means that for every $100 that you put into it, it will pay out $99. A 50/50 game has "100% return."
Quote: konglify2) So get back to the 200% gamble based on the expectation calculation, let's assume the pay is still 2x if I win and the probability won't change P(lost)=P(win)=0.5, but this time, I beg $2 ($1 from the spin and $1 from my pocket). If I lost the gamble, I lost all $2. But if I win, I win ($2x2-$1) so the expectation should be
0.5*(-$2) + 0.5*($2x2-$1) = $0.5
It is confusing me that you wrote it in -$1x0.5 + $3x0.5 = $1, in which it means that if you lost you lost $1 only? So which one is correct?
Your actual bet is 2, not 1 (never mind where the money came from; you are betting 2), so the expectation for just that bet is 0.5 * (-2) + 0.5 * (2 * 2 - 2) = 0.
Quote: konglifyAlso , if I get the positive expectation (like my $0.5 or your $1) what does it mean? Does it mean I will make profit for long time if I play 200% gamble comparing to the regular gamble?
No, because the numbers from the "200% gamble" weren't taking both bets into account, but had assumed that the first bet had won. The 1 that I came up with was not an expectation of the two bets combined, but was what you were expected to have after your second bet if you won 1 on your first bet.
There are three things that can happen with a "bet 2 if you win your first bet" strategy:
(a) 1/2 of the time. you lose your first bet; your total is -1
(b) 1/2 * 1/2 = 1/4 of the time, you win your first bet but lose your second; your total is 1 + (-2) = -1
(c) 1/2 * 1/2 = 1/4 of the time, you win both bets; your total is 1 + 2 = 3.
The total expected value = 1/2 * (-1) + 1/4 * (*-1) + 1/4 * 3 = 0.
Quote: konglify3) So if I want to make the 200% gamble "fair", should I change the payout so to make the expectation zero?
The total expectation of the "200% bet" by itself is already zero.
Quote: ThatDonGuyQuote: konglifyI am back with some questions, sorry for that. I read your reply again, so based on the statement "The expected value of the bonus bet is zero, regardless of how much is bet;" does it mean a fair gamble should always return the ZERO expectation, that is no gain no loss?
I have the following questions
1) in the book of math for gaming that I found online, it said the expectation is given as
P(lost)*(lost amount) + P(win)*(net win amount)
In the regular gamble for red/black card, P(lost)=P(win)=0.5, and says I bet $1 each gamble, so if I lost the gamble, I lost $1, if I win, I win double ($1x2=$2). So the expectation is
0.5*(-$1) + 0.5*($1x2-$1) = 0
Is that correct? So what is this expectation value related to the so called return-to-player? I ever read the book about return-to-player, it calculate that term by multiplying the winning probability with the winning amount (including the bet) so for this gamble, we have 0.5*($1x2) = 1 (or 100%). That's why I am confusing the calculations in this threads. Any way to associate the return-to-player to expectation?
If I understand the terms correctly, "expected value" is the amount you are "expected to" win/lose in relation to how much you bet. In a 50/50 game, you are just as likely to win as you are to lose, and since the amount you win is the same as the amount you lose, the expected value is zero.
"Return to player" means what percent of the amount that you bet is returned as winnings. I think it is usually used when talking about video poker or slot machines. A video poker machine with a "99% return" means that for every $100 that you put into it, it will pay out $99. A 50/50 game has "100% return."
Your actual bet is 2, not 1 (never mind where the money came from; you are betting 2), so the expectation for just that bet is 0.5 * (-2) + 0.5 * (2 * 2 - 2) = 0.
No, because the numbers from the "200% gamble" weren't taking both bets into account, but had assumed that the first bet had won. The 1 that I came up with was not an expectation of the two bets combined, but was what you were expected to have after your second bet if you won 1 on your first bet.
There are three things that can happen with a "bet 2 if you win your first bet" strategy:
(a) 1/2 of the time. you lose your first bet; your total is -1
(b) 1/2 * 1/2 = 1/4 of the time, you win your first bet but lose your second; your total is 1 + (-2) = -1
(c) 1/2 * 1/2 = 1/4 of the time, you win both bets; your total is 1 + 2 = 3.
The total expected value = 1/2 * (-1) + 1/4 * (*-1) + 1/4 * 3 = 0.
The total expectation of the "200% bet" by itself is already zero.
Thanks. I need some times to digest your replay. But from the last sentence, does it mean no matter if there is 200% gamble, 100% gamble or x% gamble, for black/red gamble, it always have zero expectation because -x*0.5 + (2x-x)*0.5 = 0? That is to say, if I play a game in casino and I have option to choose 200% gamble, 100% gamble or x% gamble, it won't make any difference to my profit for long time, is that correct?
Quote: konglifyThanks. I need some times to digest your replay. But from the last sentence, does it mean no matter if there is 200% gamble, 100% gamble or x% gamble, for black/red gamble, it always have zero expectation because -x*0.5 + (2x-x)*0.5 = 0? That is to say, if I play a game in casino and I have option to choose 200% gamble, 100% gamble or x% gamble, it won't make any difference to my profit for long time, is that correct?
Correct - if the expectation is zero.
If the expectation is negative, which is true for just about every bet in the casino, then the more you bet, the more you are expected to lose.
