Math Question for Casino Royale Night

I'm going to a "Casino Royale" night hosted by our local chamber of commerce.

They are having a "lootery" where the payouts are as follows. Plus every 10th number drawn $50, 50th $100

1st-$2000
2nd-$350
3rd-$250
4th-$200
5th-$150

What they do is they draw all the numbers down to the final three left in the drum. That is when they have a raffle for the 4th spot, in which you can buy an arms length of tickets for $20. Now there are four tickets in the drum. The fifth and final ticket is auctioned off to highest bidder.

Now this is where I have a question, for the maximum amount you should pay for the final ticket. Would you add up all five of the prizes and then divide by five? Or is it another way?

Never been to one of these, hopefully it will be fun. There will be alcohol and casino games, I like.

Obviously, you have a 1/5 chance of winning $2000, 1/5 of $350, 1/5 of $250, 1/5 of $200, and 1/5 of $150, so the EV is 400 + 75 + 50 + 40 + 30 = $595.

In other words, since your ticket has an equal chance of winning each prize, and zero chance of not winning, then yes, you add them up and divide by 5.

Quote: ThatDonGuy

Obviously, you have a 1/5 chance of winning $2000, 1/5 of $350, 1/5 of $250, 1/5 of $200, and 1/5 of $150, so the EV is 400 + 75 + 50 + 40 + 30 = $595.

In other words, since your ticket has an equal chance of winning each prize, and zero chance of not winning, then yes, you add them up and divide by 5.

Super! Thanks!

I suspect the value of the auctioned ticket is the average of the top four prizes plus the 5th, since you would be guaranteed at least 5th place. So my guess is $850.

Quote: Ayecarumba

I suspect the value of the auctioned ticket is the average of the top four prizes plus the 5th, since you would be guaranteed at least 5th place. So my guess is $850.

If you are using that approach, you must average the difference between all five prizes and the 5th prize, and then add the value of the 5th prize to that average.

It is easier to just average the five prizes. Either way, the mathematically fair price for the final ticket is $590.

Paying $500 for the final ticket has an 18% advantage.
Paying $525 for the final ticket has a 12.38% advantage.
Paying $550 for the final ticket has a 7.27% advantage.
Paying $575 for the final ticket has a 2.6% advantage.

Quote: JB

If you are using that approach, you must average the difference between all five prizes and the 5th prize, and then add the value of the 5th prize to that average.

It is easier to just average the five prizes. Either way, the mathematically fair price for the final ticket is $590.

Paying $500 for the final ticket has an 18% advantage.
Paying $525 for the final ticket has a 12.38% advantage.
Paying $550 for the final ticket has a 7.27% advantage.
Paying $575 for the final ticket has a 2.6% advantage.

I appreciate this, very helpful! I will keep you posted on what it went for, if I bought it, and or won.