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+--- Thread: Yet Another Probability Problem (/thread-6877.html)
Yet Another Probability Problem - AltruismAndCake - 06-28-2015
You start with $1000. You get to play a game of double or nothing for 10 rounds, betting any amount you want. You also magically know that you will win exactly 5 rounds and lose 5 rounds, but you don't know what order the wins and loses will happen in. What's your best strategy for profiting guaranteed?
RE: Yet Another Probability Problem - PabloDons - 06-29-2015
if I lose even once, I have nothing to double, so I would say flip a coin on the first one, but then the chances to win the next will be lower cuz you will win 4 times and lose 5 times.
I say flip a coin on the first one, then go for that
RE: Yet Another Probability Problem - Apuly - 06-29-2015
Constantly bet everything, until you've won 5 times. Then just bit 1$.
RE: Yet Another Probability Problem - AltruismAndCake - 06-29-2015
@Pablo You don't have to bet all your money all the time. You can bet half of it ($500) and then that amount will be double or nothing, not all your money. Sorry if I didn't make it clear. Also you don't choose heads or tails, you just win or lose, no matter what the bet is.
@Apuly what if the first round is a loss? You would be out of the game on the first go xD. Also, $0 is an option in this game if you know you are losing.
There is a strategy for guaranteed money after the 10 rounds, with no cheating and no coin flipping. If you need hints I can give you some
I know how to make a (very) small profit within these rules, I just want to see if anyone knows a better strategy
RE: Yet Another Probability Problem - qwerasd205 - 06-29-2015
for the first bet you bet $500, and then as long as you lost your most recent bet you bet half of your total money. if you win a bet bet 1/4 your total money.
RE: Yet Another Probability Problem - qwerasd205 - 06-29-2015
Another strat I thought of is start at 1/4 and then every time you lose double your bet and when you win go back down to 1/4
RE: Yet Another Probability Problem - qwerasd205 - 06-29-2015
wait that's an awful idea.
RE: Yet Another Probability Problem - qwerasd205 - 06-29-2015
Always bet half your money until EITHER:
you have lost 5 times:
Go all in! you can't lose and you'll get back up to 1000!
OR
you have one 5 times:
bet $0 for the rest of the game.
RE: Yet Another Probability Problem - AltruismAndCake - 06-29-2015
(06-29-2015, 02:25 AM)qwerasd205 Wrote: for the first bet you bet $500, and then as long as you lost your most recent bet you bet half of your total money. if you win a bet bet 1/4 your total money.Assuming you do what you did on post #4, and bet everything after all the loses, and $0 after all the wins, this is a pretty good strategy. If you lose 5 times in a row, you can recover all your money back, but no gain. If you win 5 times in a row, you get around $3662 as your total. If you get LWLWLWLWLW you get a total of $1201, without applying the strategy for the final win. So it would actually earn even more than $1201. For other patterns though, it's a bit weaker, like LLWWLWLWLW will make a total of $667. Again though, if we take into account the final guaranteed win, you would do better. All in all a good strategy, just no guaranteed gains.
(06-29-2015, 02:30 AM)qwerasd205 Wrote: Another strat I thought of is start at 1/4 and then every time you lose double your bet and when you win go back down to 1/4With this strategy you run out of money a bit quick if you lose a few, perhaps a smaller fraction that than starting with a fourth? You were getting really close to my strategy here
(06-29-2015, 02:33 AM)qwerasd205 Wrote: Always bet half your money until EITHER:Second part of the strategy is sound, but betting half of your total money each bet is actually one of a few terrible strategies for this game. Without the second part, you are actually guaranteed to lose money with this strategy! The amount you lose is 25% every two rounds you participate in, so $1000 * 0.75^5= $237, if you don't include the second part. The second part of this post gives a major advantage, but I think it's a better fit for one of your other strategies
you have lost 5 times:
Go all in! you can't lose and you'll get back up to 1000!
OR
you have one 5 times:
bet $0 for the rest of the game.
RE: Yet Another Probability Problem - Apuly - 06-29-2015
(06-29-2015, 02:11 AM)AltruismAndCake Wrote: @Apuly what if the first round is a loss? You would be out of the game on the first go xD. Also, $0 is an option in this game if you know you are losing.
It's double or nothing. If I lose, I just go double or nothing.
It has never been stated that I can't go outside my budget.
RE: Yet Another Probability Problem - GISED_Link - 06-29-2015
Okay, let me trying something :
bet = p^k * c -> where p is the probability to win, k a constant and c the current fortune you play with. (p ^ k means p power k)
p = "nb of possible win" / "nb turn to play"
The idea : If you are sure to win (if you have lost 5 times), you will bet all your fortune. But it's really effective only if you have more than 1/2 of the start fortune !
You bet according to the probability to win. But the biggest idea, is the k factor. This factor is a "security" factor. if k > 1, it reduces the bet (p ^ k, p is always <=1). With this, you have always enough fortune to bet when you are "sure" to win. When you are sure to win (p = 1), you bet all your money and you double it
.Please look at the joined excel file, you will test by yourself my method.
- the best k factor, in my opinion, is e (2.718)
RE: Yet Another Probability Problem - jxu - 07-04-2015
Bet $1 until you have 5 losses then bet all your money and win
idk
RE: Yet Another Probability Problem - AltruismAndCake - 07-04-2015
Any objection to me declaring GISED_Link the winner? I don't have excel but I calculated some values with k=2 and it seems to work very VERY well. It would probably do even better if I used e (2.718) like he said. AFAIK his formula guarantees winnings, and is in one convenient formula as well.
RE: Yet Another Probability Problem - jxu - 07-04-2015
I object because mine is better probably
RE: Yet Another Probability Problem - slugdude - 07-04-2015
I object because I know I can calculate this.
However I an way too lazy for that, therefore you should declare me the winner because I CAN do it.
Because that's how it works. Totally.
RE: Yet Another Probability Problem - Apuly - 07-04-2015
I can crack quantum physics. I am physically capable of doing so. My brain is able to think of the solution on its own.
Therefor you most make me the winner of life and all that has ever existed.
You must make me a God amongst man.
Because I can.
RE: Yet Another Probability Problem - tokumei - 07-05-2015
Make me unix god. nao.
RE: Yet Another Probability Problem - Apuly - 07-05-2015
ur an god nao.
RE: Yet Another Probability Problem - AltruismAndCake - 07-06-2015
(07-04-2015, 11:01 PM)Apuly Wrote: I can crack quantum physics. *snip* Because I can.I though quantum physics says that you can and can't at the same time
(07-04-2015, 11:01 PM)Apuly Wrote: Therefor you most make me the winner of life and all that has ever existed.Is that how you spell therfore? Or are you so powerful, that you can change the spelling of a word at your whim?
Getting sidetracked ftw!
RE: Yet Another Probability Problem - Apuly - 07-06-2015
That's the power you get when you crack quantum physics.
RE: Yet Another Probability Problem - GISED_Link - 07-07-2015
I will try to approximate the best k value with a simple programm. I will try all the possibilities and then make an average of each best k value.
Anyway, there is maybe a better way to do it, but I think my formula is simple to use so it's maybe the better way to do it.