To clarify I acknowledge that both gambling systems will produce a loss long term. The question is, if my calculations are correct regarding them, and if one system is more disastrous than the other.

First game is Keno. I select 6 numbers (to win) out of 80 and 20 numbers are drawn. I assume that each number has a 0,25 chance (for simplicity) and the payouts are as following:

    *6 numbers won = 1600
    *5 numbers won = 50
    *4 numbers won = 7
    *3 numbers won = 1

I will come back to keno in a bit. Moving forward to sports betting where I select 6 events on 4.00 odds assuming no bookmaker edge (for simplicity) and 0,25 chance for each event to win. I gamble the 6 events in the combinations of 6fold,5fold,4fold and 3fold resulting in a total of 42 bets. Assuming 1 euro per bet the total is 42 euros.

Now back to Keno assuming a x42 multiplier the winnings change to:

    *6 numbers won = 1600*42=67200
    *5 numbers won = 50*42=2100
    *4 numbers won = 7*42=294
    *3 numbers won = 1*42=42

While the results of sports betting would be:

    *6 events won = 15360
    *5 events won = 2944
    *4 events won = 512
    *3 events won = 64

The question is which of the two gambling methods is less disastrous than the other and why?

To answer my own question Keno will have a payout of 0,48 and sports betting a payout of 0,22 but I am not sure if I calculated it correct

EDIT: How sports betting payouts are calculated:

    *Assuming you win a three fold the payouts are 4*4*4 = 64
    *Assuming you win a four fold the payouts are 1 four fold + 4 three folds = 4*4*4*4 + 4*64= 256+256=512
    *Assuming you win a five fold the payouts are 1 five fold + 5 four folds + 10 three folds = 4*4*4*4*4 + 5*(4*4*4*4) + 10*64 = 1024 + 1280 + 640 = 2944
  • $\begingroup$ What do you mean by "disastrous"? $\endgroup$ Commented May 17, 2014 at 20:37
  • $\begingroup$ I mean both gambling ways will provide a loss long term. which way (keno or sports betting) will provide less or more of a loss $\endgroup$ Commented May 17, 2014 at 20:39
  • $\begingroup$ That really depends on how fast you play. It wouldn't be fair to compare, say, one slot machine pull to one lottery ticket, since playing the slots involves pulling a lot faster than you're likely to buy lottery tickets. $\endgroup$ Commented May 17, 2014 at 20:42
  • $\begingroup$ Assuming 1 ticket of keno, as I described costs 42 currency units and 1 ticket of sports betting costs 42 currency units as well AND you gamble both 100000 times that would make the frequency exactly the same $\endgroup$ Commented May 17, 2014 at 20:45
  • $\begingroup$ I get also 0.48 for Keno assuming you approximated the chnaces for having $n$ numbers right as $0.25^n$, which is quite different from my my exact calculation (0.69). Can't say anything about the other since I dont understand what "3fold" means and how the chances are calculated. Please write more detailed what you are doing. $\endgroup$
    – Simon S
    Commented May 17, 2014 at 21:19

2 Answers 2



The chance to get $n$ numbers right is $\cfrac{\binom{74}{20-n} * \binom{6}{n}} {\binom{80}{20}}$

Therefore the expected value on a single bet is

$\cfrac{\binom{74}{14}} {\binom{80}{20}}*1600+\cfrac{\binom{74}{15} * \binom{6}{5}} {\binom{80}{20}}*50+\cfrac{\binom{74}{16} * \binom{6}{4}} {\binom{80}{20}}*7+\cfrac{\binom{74}{17} * \binom{6}{3}} {\binom{80}{20}} \approx 0.69$

Sport betting:

The chance to win $n$ events is $\binom6{n}*0.25^n*0.75^{6-n}$

Therefore the expected value on a 42 units bet is

$\binom660.25^6*15360+\binom65*0.25^5*0.75*2944+\binom64 *0.25^4*0.75^2*512+ \binom63*0.25^3*0.75^3*64 = 42$

Therefore you dont lose and win anything.

  • $\begingroup$ I will try and understand the math tommorow. If indeed you don't lose or win anything that must be because I assumed no bookmaker's edge. By bookmaker's edge I mean the bookmaker offering a 0,50 chance event at 1,90 price (instead of 2,00). The difference is his profit. $\endgroup$ Commented May 17, 2014 at 22:44
  • $\begingroup$ first of all I hate that I don't know how to format my text! What I dont understand in sports betting case, is how you calculate the chance to win. I thought the chance to win was just 0,25^n. $\endgroup$ Commented May 17, 2014 at 22:50
  • $\begingroup$ MathJax help: meta.math.stackexchange.com/questions/5020/… $\endgroup$
    – Simon S
    Commented May 17, 2014 at 23:04
  • $\begingroup$ Sport betting is a binomial distribution: en.wikipedia.org/wiki/Binomial_distribution $\endgroup$
    – Simon S
    Commented May 17, 2014 at 23:07

So you're simply playing to lose? That is impossible to believe! If that were really the case, you would be asking us for the fastest possible way to lose money.

Therefore you are playing for something else. It may be the pleasure of sitting in a luxurious casino, in which case you might want to maximise your time at the table. Or it may be that you are hoping to get lucky and win some set amount with which you can pay for your mother's life-saving surgery, in which case your aim would be to maximise your chance of winning this sum before going broke.

But until you tell us what you are playing for, your question is impossible to answer. (Unless the answer is "don't play".)

  • $\begingroup$ What drives "sane" gamblers is the chance to win a "respectable" amount of money with low wagers, before the loss of all these low wagers add and exceed that "respectable" amount of money. According to the above Keno would be a better choice in the examples above. However there should be (and there is) a way to gamble in sports betting, because keno is boring and sports betting is fascinating, with the same aspiration and payout in mind. $\endgroup$ Commented May 17, 2014 at 22:35

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