Game Card Probability
Yesterday a very helpful member of another forum answered my original question on this and that conversation came to a conclusion as a result. I have a second part that extends the problem a little.
This was the original post: Calculating Expected Increase In Prize Line Count
To summarise, the base problem of calculating the expected increase in prize lines from one turn using only number pools containing numbers was I think solved but I was having difficulty when introducing wildcard symbols. What the member was finally able to get through to me was that using my proposed solution I was overcounting the probability boost of the wildcard symbols because I was applying them to multiple incomplete lines originating from a particular cell and that in reality that wildcard would only be able to apply to one or the other number in the target column and therefore only one incomplete prize line containing the originating cell number. By making a pseudo choice of target prize line in my line calculations, the calculated probability matched my brute forced probability. So far so good again.
I went on in my investigation to increase the number of wildcards in the same column. My reasoning was that I would not be able to apply those wildcards to more incomplete lines but that the probability of achieving those lines to which it is applied would be increased because of the increased ratio of wildcards to numbers.
Here are my workings:
Number pools:
This time the brute forced combinations and prize line increases are as follows. There are now 45 (3x3x5) possible draw combinations.
147(1),148(0),149(1),14*(1),14*(1),157(1),158(0),159(1),15*(1),15*(1),167(1),168(0),169(1),16*(1),16*(1)247(0),248(1),249(0),24*(1),24*(1),257(0),258(1),259(0),25*(1),25*(1),267(0),268(1),269(0),26*(1),26*(1)347(1),348(0),349(1),34*(1),34*(1),357(1),358(0),359(1),35*(1),35*(1),367(1),368(0),369(1),36*(1),36*(1)
33 additional prize lines / 45 draw combinations = 0.73333 expected increase in complete prize lines.
This time we apply our solution in the same way as when there was only 1 wildcard on the draw but increase the probability boost for the boosted lines to reflect the ratio of wildcards to numbers in the pool.
The sum of these probabilities is 0.73332 which ~= 0.73333. Again if we boost the precision we’re going to be very close indeed.
Again, so far so good. So then I decided to move a step further and add wildcard symbols to multiple number pools. Below you can see I've added a wildcard to column 1's number pool and left one in column 3's number pool. This time I've changed the board configuration so only 2 numbers are checked off in advance. Here are my workings:
Number pools:
This time we have 48 (4x3x4) possible number pool combinations to brute force. (Using W instead of * as it isn't translated well by the editor)
147(2),148(1),149(2),14W(2),157(0),158(0),159(1),15W(1),167(0),168(0),169(1),16W(1) 247(1),248(2),249(1),24W(2),257(0),258(1),259(0),25W(1),267(0),268(1),269(0),26W(1) 347(2),348(1),349(2),34W(2),357(1),358(0),359(1),35W(1),367(1),368(0),369(1),36W(1) W47(2),W48(2),W49(2),W4W(2),W57(1),W58(1),W59(1),W5W(1),W67(1),W68(1),W69(1),W6W(1)
50 additional prize lines completed / 48 number pool combinations = 1.04167 expected additional lines completed by a single draw.
Now we apply our proposed solution:
The sum of these probabilities = 1.04166 ~= 1.04167. Again more precision will get these numbers very close together indeed.
So once again, things look to be so far so good. However, if I experiment moving the wildcard probability boost around to different incomplete lines, I get diverging values once again.
See my workings here (assuming the same board state and number pools as in the above example):
The above probabilities summed up give me 1.11041, which is way above my brute force calculation. I am hoping that this is because there is a rule for how wildcard probabilities must be applied that I am overlooking in my ignorance.
To be clear on how I'm adding the probability boost that a wildcard adds: If a number in pool 3 has a 1/4 chance of being drawn (1 instance of the number in the pool, 4 numbers and or wildcards in total), when I apply the wildcard to that incomplete prize line, I am boosting the probability of matching that number to 1/2 (1 instance of the number, 1 instance of the wildcard that I wish to use, 4 numbers and or wildcards in total). I am doing this in the same way for both columns when the boosts are applied.
I've also tried balancing out the probability boost across all numbers and also splitting the wildcards amongst incomplete lines according to the rule of not having boosted the probability for that number in any other incomplete prize line calculation but these methods also give me values that are varying degrees of divergence from the brute forced value.
I think I'm making an incorrect assumption somewhere or not applying the correct rule to placement of these wildcards somewhere that's causing my problem but I'm not sure what else to try.
Many thanks in advance for your time and any help!
This was the original post: Calculating Expected Increase In Prize Line Count
To summarise, the base problem of calculating the expected increase in prize lines from one turn using only number pools containing numbers was I think solved but I was having difficulty when introducing wildcard symbols. What the member was finally able to get through to me was that using my proposed solution I was overcounting the probability boost of the wildcard symbols because I was applying them to multiple incomplete lines originating from a particular cell and that in reality that wildcard would only be able to apply to one or the other number in the target column and therefore only one incomplete prize line containing the originating cell number. By making a pseudo choice of target prize line in my line calculations, the calculated probability matched my brute forced probability. So far so good again.
I went on in my investigation to increase the number of wildcards in the same column. My reasoning was that I would not be able to apply those wildcards to more incomplete lines but that the probability of achieving those lines to which it is applied would be increased because of the increased ratio of wildcards to numbers.
Here are my workings:
[1] [x] [7]
[2] [x] [8]
[3] [x] [9]
Number pools:
column 1: 1,2,3,*
column 2: 4,5,6
column 3: 7,8,9,*
This time the brute forced combinations and prize line increases are as follows. There are now 45 (3x3x5) possible draw combinations.
147(1),148(0),149(1),14*(1),14*(1),157(1),158(0),159(1),15*(1),15*(1),167(1),168(0),169(1),16*(1),16*(1)247(0),248(1),249(0),24*(1),24*(1),257(0),258(1),259(0),25*(1),25*(1),267(0),268(1),269(0),26*(1),26*(1)347(1),348(0),349(1),34*(1),34*(1),357(1),358(0),359(1),35*(1),35*(1),367(1),368(0),369(1),36*(1),36*(1)
33 additional prize lines / 45 draw combinations = 0.73333 expected increase in complete prize lines.
This time we apply our solution in the same way as when there was only 1 wildcard on the draw but increase the probability boost for the boosted lines to reflect the ratio of wildcards to numbers in the pool.
Line 1->4->7 = ⅓ * ⅗ = 0.20000
Line 2->5->8 = ⅓ * ⅗ = 0.20000
Line 3->6->9 = ⅓ * ⅗ = 0.20000
Line 1->2->3 = 0 (not possible to complete this line in a single draw)
Line 7->8->9 = 0 (not possible to complete this line in a single draw)
Line 1->5->9 = ⅓ * ⅕ = 0.06666
Line 3->5->7 = ⅓ * ⅕ = 0.06666
The sum of these probabilities is 0.73332 which ~= 0.73333. Again if we boost the precision we’re going to be very close indeed.
Again, so far so good. So then I decided to move a step further and add wildcard symbols to multiple number pools. Below you can see I've added a wildcard to column 1's number pool and left one in column 3's number pool. This time I've changed the board configuration so only 2 numbers are checked off in advance. Here are my workings:
[1] [4] [7]
[2] [x] [8]
[3] [x] [9]
Number pools:
column 1: 1,2,3,*
column 2: 4,5,6
column 3: 7,8,9,*
This time we have 48 (4x3x4) possible number pool combinations to brute force. (Using W instead of * as it isn't translated well by the editor)
147(2),148(1),149(2),14W(2),157(0),158(0),159(1),15W(1),167(0),168(0),169(1),16W(1) 247(1),248(2),249(1),24W(2),257(0),258(1),259(0),25W(1),267(0),268(1),269(0),26W(1) 347(2),348(1),349(2),34W(2),357(1),358(0),359(1),35W(1),367(1),368(0),369(1),36W(1) W47(2),W48(2),W49(2),W4W(2),W57(1),W58(1),W59(1),W5W(1),W67(1),W68(1),W69(1),W6W(1)
50 additional prize lines completed / 48 number pool combinations = 1.04167 expected additional lines completed by a single draw.
Now we apply our proposed solution:
Line 1->4->7 = ½ * ⅓ * ½ = 0.5 * 0.33333 * 0.5 = 0.08333 //both wildcards used to boost 1 + 7
Line 2->5->8 = ½ * ½ = 0.5 * 0.5 = 0.25 //both wildcards used to boost 2 + 8
Line 3->6->9 = ½ * ½ = 0.5 * 0.5 = 0.25 //both wildcards used to boost 3 + 9
Line 1->2->3 = 0
Line 4->5->6 = ⅓ = 0.33333
Line 7->8->9 = 0
Line 1->5->9 = ¼ * ¼ = 0.25 * 0.25 = 0.0625
Line 3->5->7 = ¼ * ¼ = 0.25 * 0.25 = 0.0625
The sum of these probabilities = 1.04166 ~= 1.04167. Again more precision will get these numbers very close together indeed.
So once again, things look to be so far so good. However, if I experiment moving the wildcard probability boost around to different incomplete lines, I get diverging values once again.
See my workings here (assuming the same board state and number pools as in the above example):
Line 1->4->7 = ¼ * ⅓ * ¼ = 0.25*0.33333*0.25 = 0.02083
Line 2->5->8 = ½ * ½ = 0.5 * 0.5 = 0.25 //both 2 + 8 boosted
Line 3->6->9 = ¼ * ¼ = 0.25 * 0.25 = 0.0625
Line 1->2->3 = 0
Line 4->5->6 = ⅓ = 0.33333
Line 7->8->9 = 0
Line 1->5->9 = ½ * ½ = 0.5 * 0.5 = 0.25 //both 1 + 9 boosted
Line 3->5->7 = ½ * ½ = 0.5 * 0.5 = 0.25 //both 3 + 7 boosted
The above probabilities summed up give me 1.11041, which is way above my brute force calculation. I am hoping that this is because there is a rule for how wildcard probabilities must be applied that I am overlooking in my ignorance.
To be clear on how I'm adding the probability boost that a wildcard adds: If a number in pool 3 has a 1/4 chance of being drawn (1 instance of the number in the pool, 4 numbers and or wildcards in total), when I apply the wildcard to that incomplete prize line, I am boosting the probability of matching that number to 1/2 (1 instance of the number, 1 instance of the wildcard that I wish to use, 4 numbers and or wildcards in total). I am doing this in the same way for both columns when the boosts are applied.
I've also tried balancing out the probability boost across all numbers and also splitting the wildcards amongst incomplete lines according to the rule of not having boosted the probability for that number in any other incomplete prize line calculation but these methods also give me values that are varying degrees of divergence from the brute forced value.
I think I'm making an incorrect assumption somewhere or not applying the correct rule to placement of these wildcards somewhere that's causing my problem but I'm not sure what else to try.
Many thanks in advance for your time and any help!
This post appears very similar to your part 2?
https://math.stackexchange.com/questions/4663341/calculating-prize-line-expectation-part-2
https://math.stackexchange.com/questions/4663341/calculating-prize-line-expectation-part-2
Original post by mqb2766
This post appears very similar to your part 2?
https://math.stackexchange.com/questions/4663341/calculating-prize-line-expectation-part-2
https://math.stackexchange.com/questions/4663341/calculating-prize-line-expectation-part-2
Yes it's the same. I couldn't get the help I needed.
Original post by daisy_day
Yes it's the same. I couldn't get the help I needed.
Given the length of the two posts, can you summarise what you were told and why it didnt help. By the looks of it, the guy offered to chat with you about the problem a couple of weeks ago. Is this a student project or ...?
(edited 1 year ago)
Thanks - yes he did offer to chat but when I tried to take him up on it he went silent. Other people said his answer was good and I even awarded him the bounty but unfortunately I didn't clearly follow his explanation. I'm looking for a rule or formula that I can follow in order to spread the probability boost of the wildcards across the incomplete prize lines (this will expand to many wildcards in much larger number pools). As I've said in the example, I appreciate that I must nominate prize lines to apply the boost to but I've tried spreading out the probability boost in other ways that seem logical to me but I get the incorrect calculation and I don't understand why one distribution is correct and another is incorrect.
Appreciate any help!
Appreciate any help!
Original post by daisy_day
Thanks - yes he did offer to chat but when I tried to take him up on it he went silent. Other people said his answer was good and I even awarded him the bounty but unfortunately I didn't clearly follow his explanation. I'm looking for a rule or formula that I can follow in order to spread the probability boost of the wildcards across the incomplete prize lines (this will expand to many wildcards in much larger number pools). As I've said in the example, I appreciate that I must nominate prize lines to apply the boost to but I've tried spreading out the probability boost in other ways that seem logical to me but I get the incorrect calculation and I don't understand why one distribution is correct and another is incorrect.
Appreciate any help!
Appreciate any help!
Im presuming youve understood the problem the other guy was pointing out namely that W5W has multiple meanings when youre trying to calculate the probabilities? Simply subtracting the 2*1/4*1/4 seems to account for this overcounting (note your probability sum is 1.1666 not 1.11041?) and this could be justified from triple counting the W5W case. You could probably scale/boost instead though I get the feeling that at some point youll have to properly address the fact that things are not mutually exclusive and you "can't" just add probabilities up.
(edited 1 year ago)
Original post by daisy_day
Yeah I was worried about the probabilities not being independent and someone assured me the linearity of expectation meant that that didn't matter. Would you go about this in a different way at all?
Youre really mixing up your terms here. Is this a student project or ... ?
Original post by daisy_day
I'm not a student no - I'm a programmer. It's not a project at all. It's just a problem that's been bothering me for years and a few weeks ago I decided to try to solve it. Sorry I'm mixing up terms!
How general are you imagining it to be, so is it restricted to your original bingo scenario 5*5 grid taken from 1..75, how many wild cards in the pool, ... Youve already stated it should be for any board state, so any number/combination of 'x's on the board. Are you trying to get some form of formula for the overall expected or for each prize line or ...?
In a previous post in this thread, you said you were looking for a formula to spread the -2*1/4*1/4 across the prize lines? In this case, youd simply do that to each of the 3 the *5* combinations as they have probabilities of 1/2*1/2 and are made from combinations like
159
w59
15w
w5w
where each combination has probability 1/4*1/4. You count the w5w part 3 times instead of choosing just 1 (or scaling by 1/3), hence the difference of 2*1/4*1/4 in the expected total.
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