average bet conondrum

[sagefr0g]

there is a game that has two play types, type A and type B. the types differ with respect to the minimum amounts than can be bet, otherwise they are the same game. type A having a higher bet requirement and type B having a lower bet requirement.

for any given session there are 56 potential type A plays and 133 potential type B plays, a grand total of 189 potential plays for the overall game, for a given session.

type A plays afford an average amount bet = $20.00

type B plays afford an average amount bet = $3.00

session data is collected for the number of plays of type A and type B realized.

the (overall for data collected) average amount bet for those number of plays is calculated by multiplying the number of plays (for a given play type) by the above average amount bet for the play type.

the sum of all the plays for a given session is calculated by adding the number of plays recorded for type A and type B.

the average bet per play for a given session is calculated by summing the (overall) average amount bet for each play type and dividing by the total number of plays.

see example image #1

we can calculate the average number of plays for the three sessions, by first adding together the number of plays for type A and type B for each session , session 1, 6.00 + 8.00 = 14.00, session 2, 4.00 + 6.00 = 10.00, session 3, 0.00 + 15.00 = 15.00 . then by adding the total number of plays together of each session and dividing by the total number of sessions. (14.00 + 10.00 + 15.00)/3 = 13.00 . so on average it appears (from the data) that one gets 13 plays/session.

now again from the data in image 1:

the average bet per play for a session 1 is, ($120.00 +$24.00)/14.00 = $10.29 . for session 2, ($80.00 + $18.00)/10.00 = $9.80 . for session 3, ($0.00 + $45.00)/15.00 = $3.00 .

calculating the average of the avg bet/play for the three sessions in total we come up with ($10.29 + $9.80 + $3.00)/3 = $7.70 . so on average it appears that one has an average bet [edit] per play per [end edit] session of $7.70 .

so, but it should be possible to calculate the average bet per [edit] play per [end edit] session and the average number of plays per session by another method.

consider example image # 2

one way to calculate the average bet for the three sessions of data is as follows:

first find the average of the number of plays for the three sessions of data in image 1 for the type A data. that would be avg # realized type A plays = (6.00 + 4.00 + 0.00)/3 = 3.33 . next find the average of the number of plays for the three sessions of data in image 1 for the type B data. that would be avg # realized type B plays = (8.00 + 6.00 + 15.00)/3 = 9.67 .

now we can find the actual avg bet of type A plays realized and the actual avg bet of type B plays realized. this would be done by multiplying avg # realized type A plays by the average amount bet of $20.00 that is afforded type A plays (mentioned above). and then multiplying avg # realized type B plays by the average amount bet of $3.00 that is afforded type B plays (also mentioned above).

so 3.33 * $20.00 = $66.60 actual avg bet of type A plays

and 9.67 * $3.00 = $29.01 actual avg bet of type B plays .

now we can calculate the avg bet per play by dividing the sum of the type A and type B actual avg bets by the total number of plays realized.

so ($66.60 + $29.01)/13.00 = $7.35 the average bet per play (for the data collected).

the question is which result for average bet per play is the proper one, $7.35 or $7.70 (for the data collected) ?

i tend to think the method that comes up with $7.35 is the more proper approach.

as an aside if the data collected has certain types of symmetry for the number of type A and type B plays then both methods agree.

 

[London Colin]

The short answer is, yes, $7.35.

The longer answer is that this whole question only arises because you are thinking about sessions. Sessions are meaningless.

All you need to do is divide the total bet by the total number of plays -
120 + 80 + 0 + 24 + 18+ 45 = 287.
287/39 = 7.36

This both gives you the thing you are trying to calculate, the average per play, and avoids any rounding errors that come from dealing with lots of intermediate results.

 

[sagefr0g]

The short answer is, yes, $7.35.

The longer answer is that this whole question only arises because you are thinking about sessions. Sessions are meaningless.

All you need to do is divide the total bet by the total number of plays -
120 + 80 + 0 + 24 + 18+ 45 = 287.
287/39 = 7.36

This both gives you the thing you are trying to calculate, the average per play, and avoids any rounding errors that come from dealing with lots of intermediate results.

agree, thank you.

apparently, as a youth, i didn’t pay enough attention whilst being educated about arithmetic matters such as rounding errors. i did at least notice that something awry was happening, and even had a half witted awareness that some sort of rounding error was occurring. but until your succinct response i doubt i’d ever have come to grips with what was the problem there. oh, and i do get your point about sessions.

i must say, it’s kind of scary that one can stumble into such errors of computation. like gronbog stated in reference to ever running into a real life two envelope like problem, “You would probably smell something fishy.” that said, i now don’t think this was a two envelope like problem, but instead a problem that was non-intuitive in nature. again, i think back to my grade school math teacher’s question, “right answer, but did you check your math?”i may never have realized the rounding error if i'd not attempted to check the math by attempting another method to arrive at the answer, and landing up smelling something fishy.

 

[London Colin]

I'm not sure, but it sounds like you may have misunderstood.

The problem with your first method is not that it may give rounding errors. Both your methods have the potential to give rounding errors if you drop decimal places in the intermediate results you are using. Using your second method, you got 7.35, whereas I got 7.36, which is why I mentioned rounding errors.

Your first method is just plain wrong, because it gives equal weight to each session, regardless of the number of plays in the sessions.

 

[sagefr0g]

I'm not sure, but it sounds like you may have misunderstood.

i very much appreciate your time and effort.
again, you are correct, i did not fully understand. but, i think i'm at least getting into the right ball park, lol.

The problem with your first method is not that it may give rounding errors. Both your methods have the potential to give rounding errors if you drop decimal places in the intermediate results you are using. Using your second method, you got 7.35, whereas I got 7.36, which is why I mentioned rounding errors.

yes, i recognize the fact that i came up with 7.35 (using my hand calculation) and that it apparently has to do with dropping decimal places with respect to using intermediate results. i fail to see where i dropped any decimal places, but i'll look it over again. strangely, if you look at the result in image #2, excel did not have that problem and came up with 7.36.

Your first method is just plain wrong, because it gives equal weight to each session, regardless of the number of plays in the sessions.

i was unaware of that and i don't fully understand that, but i'll mull it over and try to understand. i was indeed just thinking that method had a greater rounding issue. but apparently as you say it has a weighting issue because it gives equal weight to each session, regardless of the number of plays in the sessions.. this is something i'm not fully understanding but like i say, i'll mull it over.
does this weighting issue, perhaps have something to do with the fact that as i stated in the original post, "as an aside if the data collected has certain types of symmetry for the number of type A and type B plays then both methods agree."?

 

[sagefr0g]

....

Your first method is just plain wrong, because it gives equal weight to each session, regardless of the number of plays in the sessions.

i do notice that if the sum of the number of type A plays and the sum of the number of type B plays for all the sessions is the same (or equal) then and seemingly only then the first method and the second method agree (give the same result). so i guess the essence of that scenario rectifies the equal weighting problem (at least and only for such cases)?
now, i'm wondering why the avg amnt's bet don't need to be weighted the same. [edit: rhetorical statement[end edit]

 

[sagefr0g]

........

The problem with your first method is not that it may give rounding errors. Both your methods have the potential to give rounding errors if you drop decimal places in the intermediate results you are using. Using your second method, you got 7.35, whereas I got 7.36, which is why I mentioned rounding errors.

......

ok, leaving more decimal places, I was able to get 7.36 and yes i'm understanding that having intermediate results leaves one open for such errors.
edit: as an aside, apparently rounding issues can be a big deal, I read one account where a patriot missile wouldn't work properly because of it. end edit