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I am a math developer for a gaming technology and would like to discuss the following problem:

We have a standard bingo card, with five rows, five columns, and a free space in the center. Column 1 contains 5 integers chosen at random from the set 1 thru 15. Column 2 ... 16-30. Column 3 contains 4 integers ... 31 thru 45. Column 4 contains 5 integers ... 46 thru 60. Column 5 ... 61 thru 75. This game is a slot machine version of bingo, where only one bingo card can used per game, and every card eventually achieves a bingo.

The Draw # is defined as the number of balls drawn when the player FIRST arrives at bingo. For instance if a player hits bingo after 4 numbers have been called, then the Draw # is 4.

If a number is called such that the player has this number on their card, then we call this a daub. Important note: A player can have only so many daubs before a bingo is reached.

The Pay Group # is defined to be the ones digit of the integer that results when the four "corner" numbers are summed together. For instance, if the four corner numbers are 6,12, 68, 71, then the pay group is 7. Important note: Column 1 can only contain integers contained in the set 1 thru 15, Column 5 ... 61 thru 75.

The pay table used to calculate winnings, if any, is based on three factors. Factor #1) The Draw #, Factor #2) The number of daubs, Factor #3) The Pay Group #. For instance if the Draw # is 9, number of daubs is 4, and the paygroup is 7, then the player receives a 2 to 1 return on their bet. Important Note- The pay-table details are not important to this problem, the 3 factors used in determining pay are important.

Here's the problem: I need to calculate the probability of the occurrence of each winning hand. The probabilities of the losing hands are not as important. Now, which events are winners has already been determined, and I can supply the list of winning events, if necessary. For instance, I need to calculate the probabilities of the events,

5 draws, 4 daubs, Paygroup: 0 7 draws, 6 daubs, Paygroup: 0 11 draws, 9 daubs, Paygroup: 0 . . .

Any ideas on how to solve this?

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A first observation: The paygroup is independent of the rest. As for the draw/daub probabilities, I see nothing better than to enumerate all daub patterns with one of the daubs markes as last and such that the pattern has at least one bingo but no bingo if the last daub is removed. There are only $24\times 2^{23}$ patterns you have to investigate. –  Hagen von Eitzen Sep 16 '12 at 12:24
    
yeah only 24*2^23 –  pagal Sep 16 '12 at 12:28
1  
Well you do have a computer, don't you? –  Hagen von Eitzen Sep 16 '12 at 13:13
    
Never mind. Maybe it's more important that I have a computer ;) –  Hagen von Eitzen Sep 16 '12 at 14:05

1 Answer 1

Assuming that "Bingo" means complete row or column or diagonal, I computed the following numbers (count of "winning pattern") by computer: $$S[0] = 0\\ S[1] = 0\\ S[2] = 0\\ S[3] = 0\\ S[4] = 16\\ S[5] = 360\\ S[6] = 3800\\ S[7] = 25080\\ S[8] = 116040\\ S[9] = 399048\\ S[10] = 1053120\\ S[11] = 2167368\\ S[12] = 3493152\\ S[13] = 4380480\\ S[14] = 4197720\\ S[15] = 2975160\\ S[16] = 1478304\\ S[17] = 472392\\ S[18] = 84312\\ S[19] = 6552\\ S[20] = 120\\ S[21] = 0\\ S[22] = 0\\ S[23] = 0\\ S[24] = 0 $$ With these numbers, the probability that the first BINGO is obtained with the $n$th daub equals $$\frac{S[n]}{n\cdot{24\choose n}},$$ provided we can assume that in all situation, the next daub is equally likely to occur among all non-daub positions. This assumption is justified in spite of the not completely random distribution of numbers among columns.

Then the probability that the first BINGO is achieved with $n$ daubs and $m$ draws is $$p(n,m) = S[n]\cdot\frac1{75}\cdot\frac{75-24\choose m-n}{74\choose m-1} $$ For each winning pattern, the last number drawn must be the last daub of the pattern and the remaining $m-n$ numbers not determined by the winning pattern must be taken from the $75-24$ losing numbers, while in principle there are 74 numbers to chose the $m-1$ first draws (i.e. without the last draw). For example, $p(4,5)=\frac{136}{14382825}$.

Note (to quote D.E.Knuth): I have only proved the above calculations correct, not tested them - so there may be errors.

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