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Calculate Possible Losing Bets In a Row With Odds Calculator.

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Before we start here are some of the facts that we all know very well:

1) We have the same chance of winning on every bet.

2) Past outcome(s) does not influence future outcome(s.) Both are independent of each other.

The odds calculator is just to get an idea of the odds we are up against. We all are very much aware than anything can happen while gambling.

Most of the time as gamblers we tend to underestimate the odds that we are up against and end up losing. Hope this help in better understanding the odds and making better decisions,

For those who love to martingale on auto bet and get surprised when they come across the most unlikely losing streak. Hopefully they do not make the same mistake again by betting without stop limits as unexpected long losing streak can happen anytime.

Check it out: Odds Calculator

* Calculations are done on probability and actual result(s) might be less or more.

If you have any suggestions or doubts feel free to comment below :)

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6 minutes ago, williamshennie9 said:

Thanks for sharing and creating this! Just a reminder that the chatbot also has the !odds command which effectively does the same thing :)

Yes, its almost the same. The plus point is that with this you can calculate your odds much faster. :)

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1 hour ago, williamshennie9 said:

Thanks for sharing and creating this! Just a reminder that the chatbot also has the !odds command which effectively does the same thing :)

But isnt chatbot odd feature only calculates for dice? Im pretty sure it is, cause it doesnt really give the option for any other game. xD

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On 1/20/2019 at 6:52 PM, KiXxnTRiXx said:

But isnt chatbot odd feature only calculates for dice? Im pretty sure it is, cause it doesnt really give the option for any other game. xD

It is for any game :) You just type !odds [nr of games] [% chance of winning]. For example, !odds 15 49.5 = Odds of losing 15 bets in a row at 49.5000%: Once every 28224.70 bets.

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oh nice.. i did not know that... but really doesnt matter.. cause possible losing bets in a row calculator will never be accurate or help really.. cause even the impossible (shouldnt be possible) repeated streams of reds every minute will still happen... theres been plenty of times (every.. single...day...) that makes me go "wtf" and ask myself "how is this happening",. "how is this justified to happen over and over getting endless streak of reds without even hitting anything in between".. the kind of streaks that you would expect if youve hit big or won alot.. but ya no, you dont need to win or hit anything it turns out.. its still shows up repeatedly to bust you... o.O 

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This would be useful when i am hunting some multipliers so i will bookmark this Thanks for the the calculator :) 

Edit : Umm is the site still up or down ? I click on the Link and it says that link cannot be reached. 

Edited by Kate

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14 hours ago, Kate said:

This would be useful when i am hunting some multipliers so i will bookmark this Thanks for the the calculator :) 

Edit : Umm is the site still up or down ? I click on the Link and it says that link cannot be reached. 

Your welcome, Its still up might be some issue at your end :)

 

13 hours ago, FotisNt said:

Well, this is actually useful to me, same like each bot you've made! Thanks for helping out:). By the way, I think that it can't calculate higher amounts than 1000 on both fields, except if it's only me. Great job though, keep it going!

The max winning chance is less than 100. Its impossible to have more than or 100 % winning chance. :) 

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50 minutes ago, amarcrypto said:

Your welcome, Its still up might be some issue at your end :)

 

The max winning chance is less than 100. Its impossible to have more than or 100 % winning chance. :) 

Umm i tried opening from another browser ( Firefox ) it works over there for me but not in google chrome seems weird . Will try to reinstall chrome and try again and thanks :) 

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4 hours ago, amarcrypto said:

Your welcome, Its still up might be some issue at your end :)

 

The max winning chance is less than 100. Its impossible to have more than or 100 % winning chance. :) 

Oh, yeah you're right , I ment the mines calculator, sorry.

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On 1/20/2019 at 4:12 PM, williamshennie9 said:

Thanks for sharing and creating this! Just a reminder that the chatbot also has the !odds command which effectively does the same thing :)

I spent some time looking at the site and trying different numbers, because at first, I didn't understand at all what the output meant. Then I realised it is attempting to take the inverse of what the "!odds" command does.

The !odds command is very simple. It tells you, for example, that the odds of losing 10 bets in a row with 50% win chance is 1 in 1024 bets. But what does this mean? It's a simple sentence, but I think fully understanding it is much harder than it first appears.

Does this simply mean, for every 1024 times I spin the dice, with a 50% win chance on each one, that I am expected to hit 10 losses in a row, on average once? That's what I think @amarcrypto's calculator is saying. And I am not sure that this is correct. This is very hard to understand, so bear with me as I attempt to flesh out one small detail at a time.

I ran a simulator for seeing the odds of getting 10 losses in a row (and then aborting if you get the 10 losses in a row, a la martingale with the "kiss of death" at 10). For 50% win chance, doing 1024 bets, the odds of seeing 10 reds in a row was 0.392876 (3,928,761 times out of 10 million trials).

I ran this a second time, and got again 0.392685, which is very close to the previous number. Okay, so our average odds of hitting the 10 reds in a row with 1024 bets with 50% win chance each is about ~39.3%. This does not fit with this being the "average number of bets to see 10 reds in a row". For this to fit, the odds would be around ~63%*.

So, "1024" is not the correct number to this problem of, "how many rolls do I make before I'm expected to see the 10 reds in a row and 'die' with my martingale strategy?" (hypothetical question).

So what exactly is this "!odds" command telling us? If it's not telling us that after 1024 rolls, we're expected to see the 10 losses in a row (which as shown above by the simulation results, is not true), then what is it telling us? Here's what it's telling us. If we have a game called "The Big Ten Spins", which lets us spin 10 times at 50% win chance each, and to win the big prize, we have to get 10 losses in a row, in a single round of the game, then how many times on average do we need to play "The Big Ten Spins" in order to get the big prize, and see our 10 reds in a row? THIS is what the !odds command is telling us, we need to play "The Big Ten Spins" game on average, 1024 times. This is NOT the same as spinning continuously 1024 rolls, and there is a subtle but important difference there, which I hope you can note.

So this got me thinking, is it as simple as multipling the 10 spins by 1024 times? So let's simulate it.

For 10,240 spins, the odds of seeing 10 reds in a row (and aborting, a la martingale), is 0.993386 (over 10 million simulated trials). So this is not the "average number" either, as we're looking for odds close to 0.63 (see below for explanation). This is more like, if you've done 10,240 rolls, you are "nearly guaranteed" to have seen the 10 reds in a row and 'died' with a martingale strategy. The odds of not having seen it by then, according to my simulation results, is 0.006614 or 0.6614%. This should be compared to a "On average it takes 100 rolls to get a 1% win chance roll, but if you've done the 100 rolls, you are most certainly NOT guaranteed to have seen it. It's just that on average, you should see it every 100 rolls, not that you WILL see it every 100 rolls."

I hope you're still with me, because we're nearly through.

Just as a test, because I have no idea how far to extend this idea, and I don't really know how to do it, I doubled the 1,024 spins to 2,048 and ran the simulation again. This time, the odds of seeing the 10 reds in a row (and dying with our martingale strategy) was found to be 0.632579 by simulation. This matches our expectation of about 63% for the "average number"! Therefore, I would say that the actual number of rolls before you're expected to see the 10 reds in a row for 50% win chance is not 1,024, but actually 2,048. If I understood this concept better, I might see how to extend this for any situation (do you just double it? I don't think so, but I don't know - feel free to try it if you've read this far! And share your results!)

I hope this brings the tiniest bit of clarity to what seems like a simple problem, but actually is not so simple.

 

* Where did I get this 63% from? I don't fully understand the phenomenon, but if we look at "1 in X" situations, a very interesting pattern emerges.

For a "1 in 2" bets situation, the odds of getting it after 2 bets = 1 - (0.5 * 0.5) = 0.75
"1 in 3", odds of getting it after 3 bets = 1 - (2/3 * 2/3 * 2/3) = 0.703704
"1 in 4", odds of getting it after 4 bets = 1 - (3/4 * 3/4 * 3/4 * 3/4) = 0.683594
"1 in 5", ... 0.67232
1 in 10: 0.651322
1 in 100: 0.633968
1 in 1000: 0.632305
1 in 1000000: 0.632121

So you can see the pattern, which is that for a "1 in X bets" problem, after X bets, your odds of getting it are roughly: 1 - (1/e) or 0.632121, as long as X is at least 100 or so.

Edited by Pirnitho | fixed a typo

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