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I've been trying for a few hours now to wrap my brain around, what at first, seemed like a simple concept. I've tried so many different things and ways of constructing scenarios so I can figure out how to calculate this, but I just can't seem to get it right

Here's the scenario:

Imagine there are 25 different TYPES of "prizes" each worth a monetary value. The total quantity of all the prizes is infinite.

Type. Value

 1. $0.5
 2. $0.5
 3. $1
 4. $1
 5. $1
 6. $1
 7. $2
 8. $2
 9. $2
 10. $4
 11. $5
 12. $5
 13. $7
 14. $8
 15. $10
 16. $10
 17. $12
 18. $15
 19. $15
 20. $20
 21. $20
 22. $20
 23. $25
 24. $30
 25. $100

Now let's say that you ran a "casino" and someone could pay $2 in return for one of the prizes. What winning-percentage would the "casino owner" have to set for each type of prize so that in the long run, the buyer would walk away even? The buyer should lose as much money as they gain.

In other words, if the buyer has lost \$98 due to lots of bad luck, the \$100 prize should be the next prize to be won.

This is also taking place in a "perfect world"... The smallest prizes should have the largest percentages.


At first, I just thought about how many \$2 payments it would take to make your money back for each value (50 payments to win the \$100 prize). So that means a 2% chance for the \$100 prize? But when I did that for all the rest, it was clear to me that was the wrong approach since it was a 100% chance for the \$2 prize.

How do I calculate the percentage chance to win each prize so the buyer walks away with the "same" amount of money after a gazillion plays? Thanks for your time. :)

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  • $\begingroup$ You want to find probabilities $p_i$ corresponding to each prize value $x_i$ such that $\sum \limits_{i=1}^{25} p_i x_i = 2$ and $\sum \limits_{i=1}^{25} p_i = 1$. $\endgroup$ – Rammus May 28 '15 at 5:38
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    $\begingroup$ Hint: You want the expected prize to be $2. The only constraint on our problem is that the sum of the prize probabilities equal 1 and all are non-negative. There is no one unique solution. $\endgroup$ – user237392 May 28 '15 at 5:40
  • $\begingroup$ Dang. Well this is for a real situation, not theoretical, so I will eventually need to make up some percentages. Hmmmm. $\endgroup$ – mrg95 May 28 '15 at 6:10
  • $\begingroup$ I have a quite satisfying solution, but I need to know if it's possible to win no prize. If so, I'll post an answer right now. $\endgroup$ – Masclins May 28 '15 at 8:02
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I'll post this answer assuming that it's possible to win no prize, which seems reasonable.

Of course, there's no unique answer to this question, since for example, you could simply always give back the $8^{th}$ prize. Still, I took an approach that tries to go around this problem of non-uniqueness.

Let's use for that the concept of RTP. We can force the game to make every prize has the same RTP and that all together the game has an RTP of 100%. What I mean by every prize having the same RTP is that each price, taking into account how often it appears, contribute the same to the total RTP. So for each prize the formula $prize\times P(prize)$ remains constant. That means that $RTP=4\%$.

Now let's find what should be $P(prize_n)$ where $prize_n$ is winning the $n^{th}$ prize.

$$P(prize_n)=2\times\frac{4\%}{prize\ value} $$

This gives the following table of probabilities

Type Value Probability Frecuency

 1.  $0.5  0.16        6.25
 2.  $0.5  0.16        6.25
 3.  $1    0.08        12.5
 4.  $1    0.08        12.5
 5.  $1    0.08        12.5
 6.  $1    0.08        12.5
 7.  $2    0.04        25
 8.  $2    0.04        25
 9.  $2    0.04        25
 10. $4    0.02        50
 11. $5    0.016       62.5
 12. $5    0.016       62.5
 13. $7    0.01143*    87.5
 14. $8    0.01        100
 15. $10   0.008       125
 16. $10   0.008       125
 17. $12   0.00667*    150
 18. $15   0.00533*    187.5
 19. $15   0.00533*    187.5
 20. $20   0.004       250
 21. $20   0.004       250
 22. $20   0.004       250
 23. $25   0.0032      312.5
 24. $30   0.00267*    375
 25. $100  0.0008      1250

Probabilities with * had been rounded so if you have to implement it, I'd recommend you to use frecuencies ($\frac{1}{probability}$), that are not rounded. That's why I added the Frecuency column as well. Frecuencies are telling you that you should see that prize once for every time you'd played that number of games. So, you should see one prize 7 every 25 prize (that's the expected value), or after 100 games you should see 16 prizes 1.


The solution for your problem if you always win a prize requires some computation.

Let $p_i$ be the probability of wining prize $i$. Let $v_i$ the value of the przie $i$. You have the following system.

$$ \sum_{i=1}^{25} p_i=1\\ \sum_{i=1}^{25} p_iv_i=2\\ p_i>p_{i+1}\ \ \forall i\in [1,24] $$

Last condition wouldn't be needed but it's for making it more beautiful, every prize is more probable the smaller it is.

Here you have 26 equations (actually 2 equations and 24 inequations) that can solve through computer.

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  • $\begingroup$ Wow! This is great. Makes much more sense to me now. I understand the formula. Thanks so much :) However, they do win a prize every time... So should I just count a "miss" as giving out a $2 prize? Or would it have to be worth nothing? $\endgroup$ – mrg95 May 28 '15 at 23:53
  • $\begingroup$ Thanks. You should include a prize $0$ or $26$ that gives back $0$ and that is win when none of the others are won. That would mean that you have a probability to win it equal to $1-\sum_{n=1}^{25} P(prize_n)$ $\endgroup$ – Masclins May 29 '15 at 5:15
  • $\begingroup$ Wait. How come the total percentage when you add it up is only 83%? $\endgroup$ – mrg95 May 31 '15 at 1:39
  • $\begingroup$ The remaining is "no-prize". Or the new prize = 0. $\endgroup$ – Masclins May 31 '15 at 2:36
  • $\begingroup$ Is there any way to eliminate 0 as an option? So that way the max you could ever lose would be $1.50? $\endgroup$ – mrg95 May 31 '15 at 3:19

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