# What are the odds of winning this bingo game?

I'll explain the game real quick, it's called Pick 8.

You get a sheet of paper, containing 3 rows of 8 boxes. You fill out each row with numbers 1-75, in any order, with no duplicates. The bingo caller then starts calling bingo numbers.

If you get a row filled during the first 20 numbers called, you win the jackpot, which is usually around \$7000. The first person to win after 20 balls are called wins \$500.

I'm interested in the jackpot. Mainly, how many sheets would I need to buy and fill out with an 8-number combination to ensure ~70% probability of winning in the first 20 numbers called? Each sheets costs \$3, so it's \$1 a row.

I started by calculating how many 8-number combinations there are in a pool of 75 (where order doesn't matter). Apparently that's called a binomial coefficient, and it comes out to ~16.8 billion combinations.

But I don't where to take it from there - how do I take into account the fact that I get 20 chances to get an 8-number combination correct? If I can figure out that, then I can multiply those odds by 3, and then multiply it even further to cover 70% of the possible combos, to see how many sheets I'd need to buy.

• I am a little confused about the mechanics of the game. You get blank rows and you fill them out with numbers? Then you cross them out as they are called? Who ensures that you filled in the numbers before the calling out began? – Thanassis Nov 25 '16 at 2:04
• Clarification needed: A single row cannot contain duplicate numbers, but the 3 rows in the same sheet can contain the same numbers, right? For example, is this valid? Row1: 1,2,3,4,5,6,7,8, Row2: 3,4,5,6,7,8,9,10, Row3: 1,10,20,30,40,50,60,70,75 – Thanassis Nov 25 '16 at 2:13
• Yes you're correct, and there's a carbon copy of each sheet that you turn in before it starts – Steven Nov 25 '16 at 2:16
• Yes, the 3 rows on the same sheet can contain the same numbers, your example is valid – Steven Nov 25 '16 at 2:16
• The answer will depend on the strategy used to fill the rows. – A.G. Nov 25 '16 at 3:25

For a single row, the probability of winning is the ratio of draws of 20 numbers out of 75 that include the 8 selected numbers in the row to the total number of draws of 20 numbers out of 75, i.e. $$p=\frac{75-8 \choose 20-8}{75 \choose 20}= \frac{1}{133\,929}.$$ If you play $n$ rows, each independently and randomly selected, then the probability of winning is $$1-(1-p)^n.$$ If you want that to be $70~\%$ you solve for $n$: $$1-(1-p)^n=.7 \rightarrow n=161\,246.$$ This is clearly an upper bound, as a better strategy should do better that selecting rows at random.
Edit: details on the computation of $p$. Once you have chosen your 8 numbers, the number of draws of 20 numbers out of 75 that will make you a winner is the number of ways to pick your 8 numbers exactly (just one way) and then to pick the remaining 12 numbers out of the remaining 67 numbers: $1\times{67\choose12}$. The total number of ways to pick 20 numbers remains ${75\choose20}$. The probability is the ratio of favourable draws to total draws, thus the above formula.
• Following your logic, you spend $161246$ to have a $70\%$ chance at $\$7000$. As you say, it can be improved by strategically choosing the rows, but it needs to improve a lot to be worthwhile. This is similar to lotteries worldwide. – Ross Millikan Nov 25 '16 at 3:59 • Can you please explain in more detail how you got the formula for the single row success? I cannot understand the current short explanation. I used a different (recursive) way to calculate this probability and our answers differ by about 20% (look at my answer). This could be due to rounding errors on my part, but I think it's more probable that either of us made a mistake. – Thanassis Nov 25 '16 at 6:41 • Your calculation is correct, but assumes that you are the only player. Your odds will reduce when there are other players, as someone else might win before you (or win with the same eightth number and split the prize). – Iadams Nov 25 '16 at 8:31 • @Iadams This is what I tought too, but the way the problem is posed suggests that there are as many jackpots as needed. If that were not the case it would of course have to be taken into account -- and strategy would be even more crucial. – A.G. Nov 25 '16 at 13:31 • @Thanassis See my last edit. – A.G. Nov 25 '16 at 13:40 I am sure there is an easier way to calculate this, but this is what I got: Update: Indeed A.G's answer does this in an easy way. I also found an easy way and perhaps more intuitive. We have$75 \choose 8$ways to pick$8$numbers out of$75$numbers. Out of these ways of picking there are$20 \choose 8$ways to have a jackpot winning pick. Hence the probability for a jackpot with one pick/row is: $$\frac{20 \choose 8}{75 \choose 8} = \frac{494}{66160995} \approx 7.467\cdot 10^{-6}$$ I also include my original way of calculating the probability. It's more complicated and requires a short program to do the calculation, but I think the general approach can be useful for other probability problems so I leave it here. Let$P_{N,k,m}$be the probability that you will match your$m$numbers when$k$numbers are called out, out of a possible$N$numbers. For your problem you want to calculate$P_{75,20,8}$We can calculate this recursively. Here's how: Let's say the first number out of the$20$is called. What are the possibilities? This number is either one of your$8$numbers or not. If it's not then this number gets out of the system and now you have the problem of getting 8 number matches, with 19 calls, out of 74 numbers. In short you want to calculate$P_{74,19,8}$. If, on the other hand, you do get a match, then again this number gets out of the system but now you have to match 7 numbers, with 19 calls, out of 74 numbers. In short you are after$P_{74,19,7}$. What's the probability of not getting a match on the first call? It's$\left(\frac{74}{75}\right)^8$. Edit: No this is wrong. If we have 8 distinct numbers and we want to match one randomly chosen number (out of 75 possible) the probability of not matching is$\frac{74}{75}\times\frac{73}{74}\times\dots \times\frac{67}{68} = \frac{67}{75}$. (It's interesting I made this mistake here because further down I ask essentially the same question -but from a different angle- and this time I got it right! I leave these notes here as a learning opportunity.) So we have: $$P_{75,20,8} = \frac{67}{75} \cdot P_{74,19,8} + \left(1- \frac{67}{75}\right)\cdot P_{74,19,7}$$ More generally: $$P_{N,k,m} = \frac{N-m}{N} \cdot P_{N-1,k-1,m} + \left(1- \frac{N-m}{N}\right)\cdot P_{N-1,k-1,m-1}$$ How does the recursion end? Consider what happens when we have the same number of calls as the numbers we have chosen, in other words we want to calculate$P_{N,k,k}$. This is simply the inverse of the binomial coefficient$N \choose k$. It's like we are asking "I have a specific combination of$k$numbers, what is the probability that this will be chosen out of$N$numbers?". $$P_{N,k,k} = \frac{1}{N \choose k}$$ Moreover we can directly calculate$P_{N,k,1}$. This means we have chosen just one number out of the$N$numbers and we want to find the probability to match this number when$k$different numbers are called. The probability of not matching is:$\frac{N-1}{N}\cdot \frac{N-2}{N-1}\cdot \frac{N-3}{N-2}\dots \frac{N-k}{N-k+1} = \frac{N-k}{N}$. Thus: $$P_{N,k,1} = 1- \frac{N-k}{N}$$ So now that we have our recursion-ending conditions we can calculate our probability. I wrote a short Python program to calculate this and the answer is: $$P_{75,20,8} \approx 7.467\cdot 10^{-6}$$ So you have about$7$chances in a million to match a single row. How many rows should you play in order to get a$70\%$probability of winning? If you choose rows randomly and independently then you want$n$rows where: $$1-(1-P_{75,20,8})^n = 0.7 \iff n = 161,246$$ This is non-optimal, because if you choose randomly you have a chance to choose the same row more than once, but given there are 16 billion combinations of possible rows, and you need around the ballpark of$150\,000$rows, the difference between the optimal strategy and just choosing randomly will not be significant. Update: Things are more complicated than avoiding duplicates. Bram28's answer explains how. We can also see it with one extreme example: Assume that$74$numbers are called instead of$20$. What would be our strategy then? We can choose one row at random and it will have a$\frac{67}{75}$probability of winning. What should our second guess/row be? Imagine that our first row does not win. This means that in our$8$chosen numbers there is the single number out of the$75$numbers that is not called (remember we are calling$74$numbers in this version of the game). If we keep most of our$8$numbers the same and we change only one, there is a$7/8$probability that we would keep the "bad" number and we will again not win. If, on the other hand, we change all numbers then we are guaranteed to get rid of the "bad" number and we are guaranteed to win! So the choice makes a big difference in this case. How about our case? It seems difficult to find an exact strategy but let's start our analysis. Let's assume that our first row did not win. What would be the best second row to choose? Consider all the ways our first row might not win. We can have$0,1,\dots 7$numbers matching. What is the probability of each case? Before giving this formula, consider any row (we do not know if it is winning or not) and we want to find the probability of exactly$i$matches ($i=0\dots 8$). For$i=0$this is simply$\frac{55}{75}\cdot \frac{54}{74}\cdot\frac{53}{73}\dots \frac{48}{68}$(i.e, probability of first number not matching, times probability of second number not matching, etc). How about if we have exactly one match? Let's say that the match is our second number (out of the eight we have picked). The probability for this to happen is$\frac{55}{75}\cdot \color{red}{\frac{20}{74}}\cdot\frac{54}{73}\dots \frac{49}{68}$. What if the matching number was the third, or any other number? There would be some rearrangement but all the numbers in the nominators and denominators would be the same. The result would be exactly the same. We just need to count all combinations we can have one number chosen out of 8. What if we have exactly two matches? Let's say the second and the seventh numbers. The probability for this is$\frac{55}{75}\cdot \color{red}{\frac{20}{74}}\cdot\frac{54}{73}\dots \color{red}{\frac{19}{69}}\cdot\frac{50}{68}$. What if they are some other two numbers matching? Notice that the denominators will be always the same. Notice also that the nominators will be the same, just arranged differently. The result will be the same, because we are multiplying. Again we have to account for all combinations. The general formula is: $$P(\text{matches}=i) = {8\choose i} \times \frac{\overbrace{20\times \dots \times(20-i+1)}^{i\;\text{terms}}\times \overbrace{55\times \dots \times(55-8+i+1)}^{8-i\;\text{terms}}}{75\times \dots \times 68} \\ = {8\choose i} \times \frac{\frac{20!}{(20-i)!}\times \frac{55!}{(55-8+i)!}}{\frac{75!}{(75-8)!}}$$ Note that for$i=8$this is yet another way to answer our original question (what is the probability of a row winning). Now let's find the formula for$i$exact matches given that the row is not a winning row (i.e. we know that the matches are less than$8$). So we know that one number out of our$8$is not matching. We can safely ignore that number (it does not matter where it is, or which number it is) and convert the problem to the equivalent unrestricted problem but with$7$numbers picked and a total pool of$74$numbers. This is the formula: $$P(\text{matches}=i| \text{matches}< 8) = {7\choose i} \times \frac{\frac{20!}{(20-i)!}\times \frac{54!}{(54-7+i)!}}{\frac{74!}{(74-7)!}}$$ Here are the approximate values for$i = 0,1,\dots 8$$$\begin{array}{c|c|c} \begin{array}{c} \text{Exact number}\\ \text{of matches} \end{array} & \begin{array}{c} \text{Probability for}\\ \text{any row} \end{array} & \begin{array}{c} \text{Probability for}\\ \text{non-winning row} \end{array} \\ \hline 0 & 0.07217 & 0.09841\\ 1 & 0.24056 & 0.28703\\ 2 & 0.32648 & 0.33390\\ 3 & 0.23506 & 0.20034\\ 4 & 0.09794 & 0.06678\\ 5 & 0.02411 & 0.01233\\ 6 & 0.00341 & 0.00116\\ 7 & 0.00025 & 0.00004\\ 8 & 7.467\cdot 10^{-6} & 0\\ \hline \text{Total} & 1 & 1 \end{array}$$ Here's the Python code that computed these values. We can also validate our calculations by checking that indeed the sums of probabilities are equal to$1$for unrestricted rows and for non-winning rows. (to be continued) I am not the best with probabilities, but I will give it a shot: You basically want to add up the chances of getting your 8 picks when exactly 8 are picked, when exactly 9 are picked, etc. until when exactly 20 are picked. That is: $$P(win \: jackpot) = \sum\limits_{i=0}^{12}P(win \: at \: 8+i )$$ Now, to win the jackpot at the moment the$8+i$-th ball is called means that 7 of your picks were called before then, and$i$balls were called that were not any of your picks, after which your 8th pick is called. What are the chances of this happening? Well, first of all we need to get those$7+i$initial numbers: 7 of your picks (and there are 8 ways of doing that), and$i$non-picks (and there you have$67 \choose i$options), and after that you need exactly that last number (with a chance of 1 out of$75-7-i$). So: $$P(win \: at \: 8+i)=8*\frac{67 \choose i}{75 \choose 7+i}*\frac{1}{68-i}$$ (Note that with i=0 you would get exactly your 1 in 16 billion chance) So, plug that into the earlier summation, and get out a really good calculator! And just to be clear: this would be the probability of a single row winning the jackpot. EDIT As A.G. points out, we can immediately calculate the chance of one line winning the jackpot: $$P(win \: jackpot) = \frac{\binom{67}{12}}{\binom{75}{20}}$$ This is because since the chosen collections have no order, this will cover all of the cases I handled separately above, i.e. where the winning 8th ball is the 8th ball drawn, or the 9th ball drawn, or ... or the 20th ball drawn. Both my proposed summation as well as this much simpler answer give a probability of: $$P(win \: jackpot) = \frac{494}{66160995} \approx \frac{1}{133929}$$ OK, but what about multiple lines? Here you need a good strategy. For example, for the second line, it would make sense to not have any number the same as what you have on the first line. To see this let's first consider the extreme case: the second line is exactly the first line. Well, then if the first line does not win the jackpot, then obviously the second line won't win the jackpot either. Having 7 numbers the same and 1 number different does not help much, since most of the 20-number draws that would make the second line a winner would make the first line a winner as well, and since the first line was not a winner, there is only s small chance that the 20-number draw we are dealing with with be one of the few draws that would make line 2 a winner but line 1 not. Much better, therefore, to pick 8 completely different numbers for line 2. Yes, there are still some 20-number draws that would rule out line 2 as well as line 1 as a winner (e.g. those that contain all 16 different numbers from those 2 lines), but there are clearly much fewer of those, i.e. the chances that the 20-number draw we are actually dealing with is one of those that makes line 2 a winner and line 1 not a winner are much greater now. Here is a simple example to illustrate this. Suppose we are dealing with only 5 numbers instead of 75, that we pick 2 instead of 8, and that 3 are drawn instead of 20. OK, first consider a equence of picks with lots of overlap: 12,13,14,15,23,24,25,34,35,45 The first pick (12) would be a winner with any of the following draws: 123,124,125. So, this has a$\frac{3}{10}$chance of winning (just as any other first pick would). What are the chances the second pick (13) wins? This would be when the first pick does not win (so the draw is not any of 123,124,125), but the second does, so this would be with draws 134 or 135, so the chance of the second pick winning is$\frac{2}{7}$. We can likewise go through the rest of the picks in this sequence, and when we just focus on the number of draws that would make the$i$-th pick a winner given that none of the earlier picks are, we get the following numbers: 3,2,1,0,2,1,0,1,0,0 (note that after 12,13,and 14 didn't win, 15 was certain not to win either!) Ok, now compare this with a strategy of avoiding overlap, e.g.: 12,34,15,23,45,... We already see an advantage of this strategy for the second pick, since it wins with draws 134,234,345, so it has a$\frac{3}{10}$chance of winning compard to the$\frac{2}{10}$for the second pick in the first sequence. Again just focusing on the number of draws that would give us a winner, we get: 3,3,2,1,1,0,0,0,0,0 (the last 5 don't matter!) So, whether in terms of how many lines you need to be guaranteed to win the jackpot, or whether how many picks we have to fill out in order to have an at least$x%$chance of winning the jackpot, the second strategy is better. Indeed, focusing on that 70%, the first strategy would take 5 picks, but the second strategy only 3. OK, so the general strategy is for your new lines to pick numbers that have not been picked before by earlier lines. OK, but after a while you will have to create lines that do overlap with earlier lines. Well, in that case the general advice is to keep the overlap at a minimum. What, in the end, is the optimal strategy though (in terms of minimizing the number of lines to get a winner or to have at least ~70% of winning), is not clear though. I suspect that is a real nasty proof! • The probability of wining with a single row is $$\frac{75-8 \choose 20-8}{75 \choose 20}= \frac{1}{133\,929}.$$ – A.G. Nov 25 '16 at 3:23 • @A.G. yeah, that makes sense ... That'll cover all possibilities of getting the winning 8th ball as the 8th ball called, or 9th ball, etc. Darn, so I did it way too complicated ... But I think still correct? – Bram28 Nov 25 '16 at 3:29 • Actually the$\sum_0^{12}$yields$\approx.59$. – A.G. Nov 25 '16 at 3:43 • Ugh. Like I said, I am not the best with probabilities .. What did I do wrong? – Bram28 Nov 25 '16 at 3:44 • Good example. I see it now, you are right. I was convinced that 1,5 has$0\$ probability of winning if 1,2,1,3,1,4 are chosen and not won. I also devised another (perhaps more straightforward) example of my own, and updated my answer with it. Minor correction: in your example the probability of the 2nd pick (1,3) winning (if the first pick 1,2 has not won), is 2/7, not 2/10. – Thanassis Nov 28 '16 at 14:45