Pointspread payouts question for Chin or Mattty

[ProlinePlayer]

Sometimes when dealing with problems regarding probabilities in complex situations, I've found that often things are clearer/simpler if one just changes one's viewpoint or perspective.

We've been trying to work out the problem from the player's view. But how about if iwe look at the situation from the other (OLG) side instead. Instead of looking at our payout and our expected loss per game - let's see what it's like for them - what is their expected gain and payout per game.

I'll use the 10 team parlay here to explain
All they need to know is
-there are 1024 possible outcomes
-there is over those outcomes an expected net payout of 600

my question is does it matter at all how that 600 breaks down? It could be one payment or it could be any possible combination of payments. In the end all that matters is that for every 1024 taken in 600 is paid out leaving a gain of 424 units (or +41.40625%). The 424 and the 41% will remain the same no matter how you change the distribution of the payouts.

Now, back to my previous post. I would argue that from the +41% we can now derive their ev per game and all the other values in between. Since the two values are linked ( the ev per parlay and the ev per game), the only way to alter one is to alter the other. And changing the distribution of the payout does not do that.

We've now reached at point where we know that any distribution of the money is exactly the same as a single payout
so ...
in the end we can come back to this : 600 ^ (1/10)


What do you think?

PLP

[MattyKGB]

Now I hate to admit this but it just is not obvious to me that this is the same thing.

Not sure what you mean here. The same thing as what?

[MattyKGB]

I'll use the 10 team parlay here to explain
All they need to know is
-there are 1024 possible outcomes
-there is over those outcomes an expected net payout of 600

my question is does it matter at all how that 600 breaks down? It could be one payment or it could be any possible combination of payments. In the end all that matters is that for every 1024 taken in 600 is paid out leaving a gain of 424 units (or +41.40625%). The 424 and the 41% will remain the same no matter how you change the distribution of the payouts.

Now, back to my previous post. I would argue that from the +41% we can now derive their ev per game and all the other values in between. Since the two values are linked ( the ev per parlay and the ev per game), the only way to alter one is to alter the other. And changing the distribution of the payout does not do that.

We've now reached at point where we know that any distribution of the money is exactly the same as a single payout
so ...
in the end we can come back to this : 600 ^ (1/10)


What do you think?

PLP

Everything you are saying is correct, under the assumption that all 1024 outcomes are equally likely, i.e. that each game is a 50-50 proposition. But is that really a good assumption? It depends on what you're going to do with the outcome.

For a 10 game ticket, your method gives 1.895899. If you take 1/1.895899 = 0.527454 and plug it into cell E4, you will see in cell C9 that the EV of this parlay is -3.47%. So, how can 1.895899 be the "single game equivalent odds" if you take 10 games that are each break-even @ those odds, parlay them together and end up with a -EV ticket?

[ProlinePlayer]

Everything you are saying is correct, under the assumption that all 1024 outcomes are equally likely, i.e. that each game is a 50-50 proposition. But is that really a good assumption? It depends on what you're going to do with the outcome.

But haven't we been using the assumption of .5 all along? I mean when we do the 5 teamer and calculate the odds per game the formula 20 ^ (1/5) implies a probability of .5.

For a 10 game ticket, your method gives 1.895899. If you take 1/1.895899 = 0.527454 and plug it into cell E4, you will see in cell C9 that the EV of this parlay is -3.47%. So, how can 1.895899 be the "single game equivalent odds" if you take 10 games that are each break-even @ those odds, parlay them together and end up with a -EV ticket?

yes. this is a problem
I'll have to rethink things because of this little detail

PLP

[MattyKGB]

For single tiered payout structures like the 5 gamer, I guess you could say that we are making the same 0.5 assumption but we don't need it - it doesn't matter because all of the approaches would give the same result.

The reason it only matters for multi-tiered structures is that we need to make some kind of statement about the relative probabilities / frequencies between tiers, e.g. that the 20 gets paid out 10x as often as the 400. That's what is driving the difference between the results of the various approaches.

[ProlinePlayer]

Now I hate to admit this but it just is not obvious to me that this is the same thing.

Not sure what you mean here. The same thing as what?

Not sure what I mean either now !?!?!

stupid sometimes

PLP

[ProlinePlayer]

just so I'm sure I have this clear .....

Would you not say that solution 1 and 3 are really the same thing. The fair payout per game and break even win probability are just different ways of expressing the same value. Altering one directly alters the other. In both cases you're looking for the value of theses variables that brings the ev to 0.

PLP

[MattyKGB]

They're related but not the same thing.

Method 1 finds the break-even point and converts it to odds.

Method 3 takes the win probs as an input and uses them to convert a "complex parlay" with a multi-tiered structure into an equivalent "simple parlay" with a single-tiered structure.

If you know the underlying win probabilities (like you do in parlaymaker) then method 3 makes more sense. If you don't, then method 1 makes more sense.

[ProlinePlayer]

MattyKGB,

Just wanted to thank you for taking the time to do this and explain it so well for myself and any others who are challenged mathematically.

Good not only to get the right answers but also extend one's abilities a bit.



PLP