Figuring the odds

If there are 52 balls in a hopper, where only 13 are drawn, what are the odds of correctly guessing 12 of the 13 correct. The Green Wizard

The probability is combin(13,12)*combin(52-13,13-12)/combin(52,13). This is 1:1,252,492,228.

This is assuming that you had 13 guesses, because you could possibly have 51 guesses and still guess 12 of the 13 correct.

For more info, lookup hypergeometric distribution on Google or Wikipedia.

Assuming drawing without replacement.

Dealing just 13 cards from a well shuffled standard deck of cards, where the object would be to get all the face cards, the four Kings, four Queens and the four jacks, If this is the answer is the above 1:1,252,492,228 and I'm reading it right thats one billion, 252 million, 494 thousand, 228 to one.

The next question is, if you have four boxes in a square that have five high value numbered balls in each box and 52 balls to draw from, with just 20 that have a high value and 32 that have a low value you can choose one box that would have more high valued balls in it than the others, what would be the odds of getting it correct, considering that if two, three or four boxes had the same amount you would lose, even if you bet all the boxes. An example would be if all four boxes had one ball each from the draw. Box #1 has these numbered balls in it 15,17,19,21.23 Box #2 has these numbered balls in it 16,18.20,22,24.
Box #3 has these numbered balls in it 25,27,29,31,33. Box #4 has these numbered balls in it 26,28,30,32,34
again only 13 or 25% of balls will be drawn out of th 52 ball hopper with all 20 high value balls in the boxes as illustrated above. the rest of the 32 balls are low valued with no box to put them in, because all balls are drawn from the same well mixed hopper the mathematical potential for high to be drawn is less than the low valued balls. One might think that the odds would be just 3-1, but I don't think so in this case. That might be so, if like in Keno, all the balls have the same value zero, only numbers to identify them. Thank you for your help, I hope this one is not to tough for you

Quote: thegreenwizard

The next question is, if you have four boxes in a square that have five high value numbered balls in each box and 52 balls to draw from, with just 20 that have a high value and 32 that have a low value you can choose one box that would have more high valued balls in it than the others, what would be the odds of getting it correct, considering that if two, three or four boxes had the same amount you would lose, even if you bet all the boxes. An example would be if all four boxes had one ball each from the draw. Box #1 has these numbered balls in it 15,17,19,21.23 Box #2 has these numbered balls in it 16,18.20,22,24.
Box #3 has these numbered balls in it 25,27,29,31,33. Box #4 has these numbered balls in it 26,28,30,32,34
again only 13 or 25% of balls will be drawn out of th 52 ball hopper with all 20 high value balls in the boxes as illustrated above. the rest of the 32 balls are low valued with no box to put them in, because all balls are drawn from the same well mixed hopper the mathematical potential for high to be drawn is less than the low valued balls. One might think that the odds would be just 3-1, but I don't think so in this case. That might be so, if like in Keno, all the balls have the same value zero, only numbers to identify them. Thank you for your help, I hope this one is not to tough for you

I get 1 in 6.584577833, what do you get?

Quote: CrystalMath

I get 1 in 6.584577833, what do you get?

a headache.

Quote: s2dbaker

a headache.

Maybe if the problem was rephrased:

There is a layout with four spaces - spades, hearts, diamonds, and clubs.
Deal 13 cards from a 52-card deck; ignore all 2s through 9s.
You win if the number of cards remaining in your suit is greater than the number of cards remaining in each of the other suits (counted separately). If there is a tie for the most, the house wins all bets.

The reason it's not 3-1 is because of the "the house wins all ties" rule.

Quote: thegreenwizard

if you have four boxes in a square that have five high value numbered balls in each box

you can choose one box that would have more high valued balls in it than the others, what would be the odds of getting it correct

The answer is zero since you state all boxes have the same amount.

Perhaps you could rewrite your question and state clearly what you want. As written it makes no sense--at least to me.

Quote: matilda

The answer is zero since you state all boxes have the same amount.

Perhaps you could rewrite your question and state clearly what you want. As written it makes no sense--at least to me.

You're right, the questions doesn't make much sense, but I think I know what he is trying to get at. I just want to know why and I want to know what the purpose of masking a card game with a keno type game.

After I answered the question, I thought of how to reword as a card question that makes more sense, so here is my rewording:

Player selects a suit. 13 cards are dealt from a deck. Count all of the cards for each suit that count toward a royal. If the player's suit has the highest number of royal cards, the player wins, but if the player is tied for the highest number of cards, then the player loses.

Quote: CrystalMath

You're right, the questions doesn't make much sense, but I think I know what he is trying to get at. I just want to know why and I want to know what the purpose of masking a card game with a keno type game.

After I answered the question, I thought of how to reword as a card question that makes more sense, so here is my rewording:

Player selects a suit. 13 cards are dealt from a deck. Count all of the cards for each suit that count toward a royal. If the player's suit has the highest number of royal cards, the player wins, but if the player is tied for the highest number of cards, then the player loses.

Alternatively, "the player wins iff their suit has a uniquely highest number of royal cards". When you solved, did you actually set up the multivariate hypergeometric terms and add them up or did you use another technique (like some software package that does it for you)? If you used software, what package?

Quote: MathExtremist

Alternatively, "the player wins iff their suit has a uniquely highest number of royal cards". When you solved, did you actually set up the multivariate hypergeometric terms and add them up or did you use another technique (like some software package that does it for you)? If you used software, what package?

I setup the multivariate hypergeometric terms and added them up in excel. To enumerate the terms, I wrote this macro in excel, and the rest was just the calculations. I have written Java code to analyze other similar games, but it was faster for this question to just do it in excel.

Sub enumerateTerms()
writerow = 0
For i = 0 To 5
For j = 0 To 5
For k = 0 To 5
For l = 0 To 5
writerow = writerow + 1
Cells(writerow, 1).Value = i
Cells(writerow, 2).Value = j
Cells(writerow, 3).Value = k
Cells(writerow, 4).Value = l
Next l
Next k
Next j
Next i

End Sub

Interesting -- I would have thought you'd filter in the loop by only the terms you care about. If you just want to get an enumeration of everything done in Excel, there's a faster way without code:
Column A is the index (1 .. N)
Column B =MOD(A, 6)
Column C =MOD(FLOOR(A/6,1), 6)
Column D =MOD(FLOOR(A/36,1), 6)
Column E =MOD(FLOOR(A/216,1), 6)
Then just fill-down all 5 columns until you get to 6^4 rows.

Edit: when I use this method, my results match yours exactly. 96439525155 wins.

Quote: MathExtremist

Interesting -- I would have thought you'd filter in the loop by only the terms you care about. If you just want to get an enumeration of everything done in Excel, there's a faster way without code:
Column A is the index (1 .. N)
Column B =MOD(A, 6)
Column C =MOD(FLOOR(A/6,1), 6)
Column D =MOD(FLOOR(A/36,1), 6)
Column E =MOD(FLOOR(A/216,1), 6)
Then just fill-down all 5 columns until you get to 6^4 rows.

Edit: when I use this method, my results match yours exactly. 96439525155 wins.

Actually, mine took me about 1 minute, which I'm OK with. I guess it just depends on what we are all comfortable with. The rest of the calculations took another minute or so. Sanity checking took about 5 minutes.

As for filtering in the loop, that would have just taken more time. Plus, my first sanity check is to see if all the individual probabilities add up to 1.

Quote: CrystalMath

Actually, mine took me about 1 minute, which I'm OK with. I guess it just depends on what we are all comfortable with. The rest of the calculations took another minute or so. Sanity checking took about 5 minutes.

As for filtering in the loop, that would have just taken more time. Plus, my first sanity check is to see if all the individual probabilities add up to 1.

OK. I just looked at it and I realized I could have filtered out any combination where the total hits are greater than 13.

I believe the 6.5 to one is accurate. Some are asking why the cross over from cards to balls. The original game of Crown Royal is played with cards. However its not a standard deck. I have the only patent in the world on a deck of playing cards, meaning you can play any game normally played with a standard deck. It is based on the binary system having a base of two. There are no numbers , suits or pictures. it made its entry on 7-7-77 at the North shore Club in Lake Tahoe with the permission of the Nevada Gaming Commission. Race Horse Keno is an extension of Crown Royal, its primary purpose is to offer a game that is simple, easy to learn, where all bets are based on the basis of a flip of a coin, thirteen times. Instead of a coin we use balls that create the same effect, that is win, lose or tie. related to the basis of all bets every where, in every game world wide. Examples are Banker, player and tie in Bacarat. Pass, don't pass, bar 12 in craps. Red, black and green or even, odd or zero on Roulette. Bank, player or tie in 21. The Yankees, Phillies or tie in baseball. Italy, France or tie in Soccer and so on. This basic binary principle having a base of two allows expansion to occur with added values, the number of coins available x the numberof flips identifies the odds and rules of any game. When the coins have the different values example nickles and dimes which are both coins having heads and tails, our drawing pot contains 20 dimes and 32 nickles. We only draw 13 out of 52 based on the first value a base of two, with a catalyst to make it work and a second value of either high or low that mutiples various mathematical proabilities or bets without numbers.

I have used numbers on balls to illustrate the second value that distinquishes high and low balls where there are only 20 high verses 32 low. The ticket layout is retangular 2 square wide and 26 square long, the 20 high values are located in the middle, 8 squares in from either end. This provides 16 spots on top, 20 spots in the middle and 16 spots on the bottom for a total of 52 spots. 25% of the 52 or 13 are drawn every race. From this explanation it is easier to understand that the four squares have 5 high valued balls in each. In the first question we are trying to get the from the deck or balls just 12 of 20 high that are in like all the picture cards Kings, Queens and Jacks of a standard deck. That I think has been addressed properly.

The second question then asks if those four boxes (see above) containing five high value cards each within the four squares in the middle of the ticket, you then can bet which one will get the most high valued cards. For example Box #1 has two, Box #2 has three. Box #3 & Box #4 have none. That a win for Box# 2.
it also tells you there were five high and eight low from the 13 drawn. My guess is that the 6.5 to one is proably right. I hope this clears up the question of cross over. I have need of someone to address the other odds and permutations required by the gaming authority. Thank you one an all

Is this it?

Dealing just 13 cards from a well shuffled standard deck of cards, where the object would be to get all the face cards, the four Kings, four Queens and the four jacks, If this is the answer is the above 1:1,252,492,228 and I'm reading it right that’s one billion, 252 million, 494 thousand, 228 to one.

There are two rows on the race card with 26 spots each
The left row has all the red balls and all the odd numbers
The left row has 16 red low balls #s 1, 3, 5, 7, 9, 11, 13, 15 on top.
#s 17, 19, 21, 23, (0), are all the high balls in the center.
#s 33, 35, 37, 39, 41, 43, 45, 47 are all low on the bottom
The right row has all the blue balls and all the even numbers
The right row has 16 blue low balls #s 2, 4, 6, 8, 10, 12, 14, 16, on top.
#s 18, 20, 22, 24, (00), (0000), 26, 28, 30, 32 are all high balls in the center
#s 34, 36, 38, 40, 42, 44, 46, 48 are all low on the bottom.

FOUR SQUARE BETS
Box #1 has these numbered balls in it 17, 19, 21, 23, (0), left upper box
Box #2 has these numbered balls in it 18, 20, 22, 24, (00), right upper box
Box #3 has these numbered balls in it (00), 25, 27, 29, 31 left lower box
Box #4 has these numbered balls in it (0000), 26, 28, 30, 32 left lower box

The next question is, the four boxes in the square have five high value numbered balls in each box as indicated above, 52 balls to draw from, with just 20 that have a high value and 32 that have a low value, if you can choose one box, that might have more high valued balls chosen, after the 13 ball draw, what would be the odds of getting it correct, considering that if two, three or four boxes had the same amount you would lose, even if you bet all the boxes. An example would be if all four boxes with one ball each from the draw.

Again only 13 or 25% of balls will be drawn out of the 52 ball hopper with all 20 high value balls in the boxes as illustrated above. the rest of the 32 balls are low valued with no box to put them in, because all balls are drawn from the same well mixed hopper the mathematical potential for high to be drawn is less than the low valued balls. One might think that the odds would be just 3-1, but I don't think so in this case. That might be so, if like in Keno, all the balls have the same value zero, and only numbers to identify them. Thank you for your help, I hope this one is not to tough for you

Dealing just 13 cards from a well shuffled standard deck of cards, where the object would be to get all the face cards, the four Kings, four Queens and the four jacks, If this is the answer is the above 1:1,252,492,228 and I'm reading it right that’s one billion, 252 million, 494 thousand, 228 to one.

There are two rows on the race card with 26 spots each
The left row has all the red balls and all the odd numbers
The left row has 16 red low balls #s 1, 3, 5, 7, 9, 11, 13, 15 on top.
#s 17, 19, 21, 23, (0), are all the high balls in the center.
#s 33, 35, 37, 39, 41, 43, 45, 47 are all low on the bottom
The right row has all the blue balls and all the even numbers
The right row has 16 blue low balls #s 2, 4, 6, 8, 10, 12, 14, 16, on top.
#s 18, 20, 22, 24, (00), (0000), 26, 28, 30, 32 are all high balls in the center
#s 34, 36, 38, 40, 42, 44, 46, 48 are all low on the bottom.

FOUR SQUARE BETS
Box #1 has these numbered balls in it 17, 19, 21, 23, (0), left upper box
Box #2 has these numbered balls in it 18, 20, 22, 24, (00), right upper box
Box #3 has these numbered balls in it (00), 25, 27, 29, 31 left lower box
Box #4 has these numbered balls in it (0000), 26, 28, 30, 32 left lower box

The next question is, the four boxes in the square have five high value numbered balls in each box as indicated above, 52 balls to draw from, with just 20 that have a high value and 32 that have a low value, if you can choose one box, that might have more high valued balls chosen, after the 13 ball draw, what would be the odds of getting it correct, considering that if two, three or four boxes had the same amount you would lose, even if you bet all the boxes. An example would be if all four boxes with one ball each from the draw.

Again only 13 or 25% of balls will be drawn out of the 52 ball hopper with all 20 high value balls in the boxes as illustrated above. the rest of the 32 balls are low valued with no box to put them in, because all balls are drawn from the same well mixed hopper the mathematical potential for high to be drawn is less than the low valued balls. One might think that the odds would be just 3-1, but I don't think so in this case. That might be so, if like in Keno, all the balls have the same value zero, and only numbers to identify them. Thank you for your help, I hope this one is not to tough for you