Calculate the payout of this Dice Game

Hey, I created a dice game and am having a bit of trouble calculating the odds for it. I think I know how to calculate each individual "round" of the game but have no idea how to calculate the total payout if all or only some of the rounds are chosen to be played.

The game starts with the gambler betting on what number won't be rolled when an x sided dice is rolled. Say a person bets 1 unit

Round 1:
6 sided dice
If their number isn't rolled they now have two units.
If their number is rolled they lose their bet and owe -5x their current units (-5*1-1)

Payout for this round is simple, 5/6*2-1/6*-5 = 0.83

If the player wins, they have the option to risk their money in round 2.

Optional Round 2:
A 5 sided dice is rolled,
If number isn't rolled again, they are now at 5 units (2.5*previous round's unit total)
If their number is rolled they lose their bet and owe -5*1 their previous round's unit total (-5*2-1)

Optional Round 3:
4 sides
Not rolled again, they are now at 15 units (3*previous rounds total)
Rolled they lose their bet and lose -5*1 the previous round's unit total (-5*5-1)

Optional Round 4:
3 sides
Not rolled they, are now at 52.5 units (3.5*previous round's total units)
Rolled they lose their bet and lose -5*1 the previous round's unit total (-5*15-1)

An example game: Gambler bets 1 unit on a "1" not being rolled,
First round: 6 sided dice is rolled, result is a 3. Player currently has 2*1=2 units of winnings. The player decides to risk their winnings.
Second Round: 5 sided dice is rolled, result is 3. Player currently has 2.5*2=5 units of winnings. They decide to risk again.
Third Round: 4 sided dice rolled, result is 1. Player now loses any winnings, their original, and lose -5*5-1 = -26

So my question is, how would I go about calculating the payout if the player decided to risk their winnings on every round? What if they stopped after winning round 2, etc? Really appreciate the help, hopefully my explanation of the games rules are clear; if not I have no problem clearing up any confusing parts :).

I don't think I'm clear on the rules (1.5:1 doesn't pay 2.5* anything), but the conditional part is pretty easy. Just multiply the odds of success in the round by the odds of reaching the round. It's very similar to the house edge calculation that you've already done for round 1.

Updated it to make it a bit more clear how much money is won/lost each round.

Also if you wouldn't mind giving me an example of how you'd calculate the payout if only two rounds were offered and the player decided to risk their money on the second round whenever given the opportunity? Having a bit of trouble understanding the formula.

There's a 1/6 chance that the player loses in round 1. If the player wins in round 1, there's a 1/5 chance that they lose in round 2, and a 4/5 chance that they win in round 2.
Assuming that the player always chooses to go to round 2 when they can:
(1/6)*1*-5=-.8334
(5/6)*(4/5)*2*2.5=3.334
(5/6)*(1/5)*2*-5=-1.667

So I'm getting -.8334+3.334-1.667=.8334.

Then round 3 would add

(5/6)*(4/5)*(3/4)*5*3=
(5/6)*(4/5)*(1/4)*5*-5=

and so on

*Edited to make formulas have items in consistent order and to correct a misunderstanding I had about a round 1 loss

Your math on round 1 is off.

You lose 6 units if you lose, at least from what OP wrote: "Lose initial bet AND owe 5x wager"
And on rounds 2,3, and 4 we lose 5x our previous winnings plus the initial wager (if I am reading it right)

5/6*2 - 1/6*6 = 0.66666

Now if players play Round 2:
4/5*3 - 1/5*11 = 0.2
The probability of playing round 2 is 5/6 = 0.8333333
This gives a weighted return of 1/6 = 0.1666666

We are multiplying only by 3 in the "win" because 3 units is all we gained in level 2, we always have to have 2 units to play level 2, and we lose 11 (5*2 - 1) if we roll the number in Level 2.

Now if players play Round 3, we will gain 10 additional units if we win and lose (5*5 + 1) 26 units if we lose
3/4*10 - 1/4*26 = 1 = 1
The probability of playing Round 3 is: 5/6*4/5 = 2/3
This gives a weighted return of 2/3 = 0.666666

Now if players play Round 4 we win 37.5 units if we win and lose (5*15+1) 76 units if we lose.
2/3*37.5 - 1/3*76 = -1/3
The fact players lose money means they should not play Round 4.
If they do play Round 4, however, the weighted probability is:
5/6*4/5*3/4 = 1/2
Weighted return is then: 1/2*(-1/3) = -1/6

If you are forced to play all four levels, the return is 1.3333 bets, or 0.3333 bets more than the original wager.

Playing optimally and stopping at Step 3, the return is 1.5 bets, or 0.5 bets more than the original wager.

So this is a bad game for a casino. :P

Appreciate the help from both of you! I finally understand how i'd go about calculating this. I've altered it so only 3 rounds are optional and losing in the second/third round requires the person to pay -6x their bet which would make the weighted return -1/6 for rounds 2/3 :).

Quote: Play4fun

Appreciate the help from both of you! I finally understand how i'd go about calculating this. I've altered it so only 3 rounds are optional and losing in the second/third round requires the person to pay -6x their bet which would make the weighted return -1/6 for rounds 2/3 :).

Are you sure you want it set up that way (I'm not sure of your goals)? The way I calculated it (out of convenience) is based on return. So a total return of 1.0 is a fair game.

I'm trying to make the game profitable for the house. The way I changed it was keeping round 1 the same while making it so you lose -6x your current winnings and your bet during round two and three. Example round two: 4/5*3-1/5*-13. If I calculated it correctly that means rounds 2 and 3 will have a weighted average of -1/6. Also just double checking but a -1/6 weighted return would mean the house edge is 1/6%?

Is there any game in any casino where a player "owes" more than his bet if he loses?

What happens if you bet the 1 and don't have the 5?

Quote: Play4fun

I'm trying to make the game profitable for the house. The way I changed it was keeping round 1 the same while making it so you lose -6x your current winnings and your bet during round two and three. Example round two: 4/5*3-1/5*-13. If I calculated it correctly that means rounds 2 and 3 will have a weighted average of -1/6. Also just double checking but a -1/6 weighted return would mean the house edge is 1/6%?

No, -1/6 weighted return means you lose 1/6th of a betting unit in that step.

Quote: QuadDeuces

Is there any game in any casino where a player "owes" more than his bet if he loses?

What happens if you bet the 1 and don't have the 5?

Not to my knowledge. So the current setup is quite flawed.