Which gambling game is less disastrous?

Question

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

Answer 1

Keno:

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.

Answer 2

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".)