I’m pretty useless when it comes to probabilities, so could someone solve the following bingo game for me? Thanks.
There are 75 balls, of which 37 are randomly selected. What are the odds that a player with a random bingo card could match 12 numbers?
Once drawn, the balls are not replaced.
Well, unless I made an arithmetic mistake somewhere, the probability of getting (at least)12 numbers on the card is 56.65606354571%. I don’t have time right now to tell where that came from, but I’ll get back to that later, if you want.
It’s gotta be very close to half, doesn’t it? There are twenty-four numbers on a bingo card, so the other twelve numbers non-match the remaining 38 numbers.
The only reason it’s more than half is because the twelve-twelve split works for both sides.
Are we to assume that the balls are chosen without replacement (as in a Bingo game)?
um,
You guys have to be wrong, because Yahoo! http://games.yahoo.com runs a Jackpot bingo contest, where they run this game and give away a prize worth ~ $150. And from experience, I can tell you that the odds are a lot higher than 1:2.
The odds of what are a lot higher than 1:2?
I haven’t played Yahoo Bingo, but I have done other offline and online bingo games.
The difference between the scenarios is that in a real Bingo game, only the first player to win actually wins. In the scenario given, there was no such condition. There were no other people.
In a real Bingo game, I belive your odds of winning are simply 1/N where N is the number of competing players. After all, the #of balls and squares and draws is completely irrelevant as long as all players are under the same set of rules. (Note: Actual winnings are less because of the house “rake”, and reduced winnings through ties.)
So I would ask Seth to let us know why he wants this number. I don’t doubt the 56% figure, but I suspect it’s the answer to the wrong question.
This is a dependent event; each outcome depends on the preceding event (since a particular ball is being removed without replacement each time).
Therefore, the probability of the first ball being one of your numbers is 12/75. If that ball is indeed one of your numbers, the probability of the second ball being one of them too is 11/74 (since one number has been “picked” and that ball removed). If the first ball had not been one of your numbers, the probability for the second ball would have been 12/74.
My stats book here says the “total” probability of dependent events is calculated by multiplying each individual outcome, so someone can knock themselves out here…!
I think that’s only true if you can only pick 12 balls – the question was to pick 37, and the odds of the correct 12 of your 24 numbers (also having been selected randomly from the 75?) being among those 37.
Followup question:
The 24 numbers on your card aren’t random are they? I thought each group of 5 (for for the ‘N’ column) were selected from a strata of width 15. That is,
1-15 B numbers
16-30 I "
31-45 N "
46-60 G "
61-75 O "
Can anyone confirm that?
This is a little more complicated then all that, because the numbers are distributed on the card a very specific way, i.e. 1-15 ar under ‘B’ etc.
This is important, although I don’t have the time to do the calculation.
The exact way to answer this is to first consider the case of requiring that the first 12 balls drawn are all on the card, and the subsequent 25 all are not. For that case, the probability is:
p = P(25,12) / P(75,12) * P(50,37-12) / P(75-12,37-12)
Then you can multiply this by the number of combinations of 37 things taken 12 at a time. Note: I think there are 25 total numbers on a bingo card, and they’re arranged so that five are from the group [1…15], five are from [16…30], etc., but I don’t think this affects the probability.
So the overall probability for exactly n matches is:
p = C(37,n) * P(25,n) / P(75,n) * P(50,37-n) / P(75-n,37-n)
With n=12, the probability is 19.1%. Now this is for exactly 12 matches, and if you’re asking for at least 12, then you’d need to add the probability for 13, 14, 15, 16, etc., up to 37. Fortunately the numbers get small pretty quickly and you can neglect many of them. But for at least 12, I get an answer right around 65.8%.
I see your logic <b>Curt</b>, but something must be wrong. Unless the drawing is rigged, that is. But there are typically no winners (I played for hours yesterday), so the numbers have to be a little wacky.
To the people who argue about the B-I-N-G-O derivation, this would not affect the probability because the chance of drawing one number is identical to any other number.
To the people who say that there has to be a winner every game, this is wrong as well. They draw 37 balls; if/when there is no winner, they reset the game.
And again, they are giving away a prize that is worth ~$150, and through experience I can tell you that the odds are slim. I was just wondering how slim they really are.
The B-I-N-G-O configuration absolutely affects the outcome. Think of it this way. Let me just alter the problem to this, select 30 match 11. Now one of the potentail draws is balls 1-30 inclusive. But given the configuration of a BINGO card the probability that you match 11 is zero. If it was just a random selection of numbers on your card then there would be some probability of 11 matches.
sethdallob wrote:
Then I must not understand the setup. I’ve just run a computer simulation one million times, and I get 18.84% for exactly 12, and 68.84% for at least 12. I did it this way: I selected a 25 different numbers from the set [1…75] and assigned them to be the “card.” Then I pick 37 different numbers from the set [1…75], and see how many out of that set match a number on the card. Is this how it works?
And Tretiak, I just ran the simulation again, but used random numbers like on a bingo card (five from [1…15], five from [16…30], etc.) and got the same answer. That was expected, though, because nothing in the setup assumes what those numbers are, it just assumes that the card has 25 different numbers.
Well, here’s what I did. Remember that the middle of the card is “Free Space”, so there are 24 numbers on the card.
So you’ve got a card (doesn’t matter how the numbers are distributed, since the numbers are being drawn randomly), let’s count the number of ways you get exactly n numbers right.
n of the numbers must come from the card: C(24,n).
The remaining 37-n numbers drawn must not come from the card (there are 51 numbers not on the card): C(51,37-n).
The product of these two give you the number of ways you get exactly n numbers right. Divide this by the number of different ways 37 numbers can be drawn from 75 numbers, and you get the probability of getting exactly n numbers right:
( C(24,n) * C(51,37-n) ) / C(75,37)
Plug in n=12 to get the probability the card has exactly 12 numbers right.
Sum from n=12 to n=24 to get the probabilty the card has at least 12 numbers right. That’s where I got the 56…% in my first post, but as I said, I can’t guarantee I didn’t screw up the calculation somewhere.
Well, I took a look at the actual game, and it appears that the goal of this form of Yahoo Bingo is to match a specific pattern on the bingo card. Like a diamond or a heart for instance. To win you have to match exactly those 12 numbers. The rest of the numbers on the card don’t really matter at all. The normal bingo rules don’t really apply.
So I think the real question is: what are the odds of picking all 12 specific random numbers (out of 75 possible choices) when given only 37 draws?
[sub](This might be the same question as the OP – I’ve managed to confuse myself badly)[/sub]
Boy, this has become a lot more involved than I thought it would be. The pattern, the letter (B-I-N-G-O) shouldn’t make a difference, because all the numbers have the same opportunity to be picked.
And through playing it (over 100 times X ~ 900 players at any given time and no winners, I can definitely tell you that the odds of winning are less than 56%. I must not be describing the situation correctly. Let me try one more time:
It’s bingo. 75 numbers. B = 1-15, I = 16-30, N = 31-45, G = 46-60, O = 61-75. A 5x5 matrix with random numbers within the above constraints are picked, with the middle square (in the N column) “free.” 37 numbers are drawn, without replacement, from the lot of 75. The winner is the first player able to match a given pattern of 12 numbers on their card (in the pattern of a “Z”, an “A”, etc.) What are the odds of winning this game?
CurtC, is that software that you used shareware-able? I’d be interested in playing around with it.
Cabbage, I’ll back you up on the 56.67% probability. I worked it out that way before seeing your description. The “odds” for this are about 17 : 13 in favor.
CurtC, I think your figure is from considering 25 numbers on a card instead of 24. When I use 25 I get your figure (pretty close).
The BINGO distribution of numbers doesn’t matter. This isn’t going to turn into another “East side -West side taxicab” great debate, is it?
Now, then, if the rules say you have to match a specific pattern on a card - that’s a different thing. Not only do you have to have a card with those numbers, but they also have to line up in that particular way on your card. That’s gotta be a lot tougher.
You guys are doing the probability backward. The OP says that given the numbers picked what are the odds you have a “matching” card. That is not thae same as what you are doing, which is given a particular card what are the odds you have a match given a random selection.