Make em all
May 7th, 2024 at 3:43:06 AM
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The make em all bet in my local casino pays as follows:
~> small / tall : 30x
~> all : 150x
Yes, I am aware it’s a massive negative EV.
That being said, I have a more interesting concept to ask all you math wizards. Let’s assume I bet $50 a piece on each section (so $150 total, which is the max allowed).
1. Assuming there is only one number remaining and it’s a point number (ie 4,5,6,8,9,10) ~> I can hedge and guarantee a profit. For example say it’s a 8. So if an 8 hits, I would win 30x$50+150x$50 = $9,000. So laying the 8 for $6,000 would guarantee a profit of $3,000 OR $5,000-the juice (if a 7 were hit).
—> this is easy.
2. What if the only number remaining is a non point number? Is there a way to probabilistically generate a profit ? Ie positive EV?
3. What if there are 2 numbers?
4. What if there are n numbers? Ie when does the EV become positive? Because obviously a casino will not let you make the same bet with only 4 numbers remaining…
Some additional Qs:
1. How many rolls on average does it take to make the make em all when there is 1 number remaining (ie a 2,3,4 … 12)?
2. Same Q but with 2 and 3 and n
3. What’s the probability you make the make em all when there is 1 number remaining, 2, 3, n numbers ?
—
Some ideas to hedge can be as simple as betting 7s each roll once you are down to n numbers remaining? Or perhaps increasing your bet size on the points + the 7s… ? I just feel that with a combination of hops and / or place bets there is a probabilistic way of taking advantage of the fact there is only n numbers remaining.
Curious to hear people thoughts on the math behind it and how to think of this mathematically.
~> small / tall : 30x
~> all : 150x
Yes, I am aware it’s a massive negative EV.
That being said, I have a more interesting concept to ask all you math wizards. Let’s assume I bet $50 a piece on each section (so $150 total, which is the max allowed).
1. Assuming there is only one number remaining and it’s a point number (ie 4,5,6,8,9,10) ~> I can hedge and guarantee a profit. For example say it’s a 8. So if an 8 hits, I would win 30x$50+150x$50 = $9,000. So laying the 8 for $6,000 would guarantee a profit of $3,000 OR $5,000-the juice (if a 7 were hit).
—> this is easy.
2. What if the only number remaining is a non point number? Is there a way to probabilistically generate a profit ? Ie positive EV?
3. What if there are 2 numbers?
4. What if there are n numbers? Ie when does the EV become positive? Because obviously a casino will not let you make the same bet with only 4 numbers remaining…
Some additional Qs:
1. How many rolls on average does it take to make the make em all when there is 1 number remaining (ie a 2,3,4 … 12)?
2. Same Q but with 2 and 3 and n
3. What’s the probability you make the make em all when there is 1 number remaining, 2, 3, n numbers ?
—
Some ideas to hedge can be as simple as betting 7s each roll once you are down to n numbers remaining? Or perhaps increasing your bet size on the points + the 7s… ? I just feel that with a combination of hops and / or place bets there is a probabilistic way of taking advantage of the fact there is only n numbers remaining.
Curious to hear people thoughts on the math behind it and how to think of this mathematically.
May 7th, 2024 at 4:10:43 AM
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So, if you've placed $150 (50-50-50) of -EV bets, and bought at least $30 of amusement doing so...
Will placing a -EV "hedge" bet further increase the joy for you?
Welcome to the forum. Best of luck.
Will placing a -EV "hedge" bet further increase the joy for you?
Welcome to the forum. Best of luck.
May the cards fall in your favor.
