help me with math please

flyingwind

Well-Known Member
#1
a theoretical problem

I'm betting $100 per round on a game.
5% of the time I win.
95% lose.

When I win, I get 3X back. So I get $400, which is my $100 bet and the $300 winnings.

What is the casino edge on this horrendous game?

Instead of 3X, there is an X when I break even. How do I calculate that?
 

The Chaperone

Well-Known Member
#2
You risk $100x100 times which is 10k. You win 5 times out of 100 so 5 times you have $400 in front of you for a total of 2k. 95 times you are left with nothing for a total of 0. 2k+0=2k. 2k/100 trials=$20. You risk $100 to get an average of $20 back. House edge=(100-20)/100=80%
 

Canceler

Well-Known Member
#3
And you thought algebra was useless!

In your original scenario you win 100x, where x = 3, five percent of the time. Ninety-five percent of the time, you lose 100. So:

EV = (100x * .05) – (100 * .95) = 15 – 95 = -80

(Thankfully, this corresponds to The Chaperone’s numbers. Whew!)

To find out what value of x you need for a breakeven game, you set EV = 0, and then solve for x. So:

0 = (100x * .05) – (100 * .95)

Simplifying:

0 = 5x – 95

Adding 95 to both sides:

95 = 5x

Dividing both sides by 5:

19 = x

This makes sense because the 1 time out of 20 (5%) that you win gives you 1900, which balances the 19 times out of 20 that you lose 100.
 
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