HF Habs: What are our real odds for Lafreniere? (Use this thread for Lafreniere talk)

[tinyzombies]

If the playoffs actually happen, we have a 50/50 chance to beat Pitt (or 35/65 depending on what you think the odds of the series are). If we lose, we then have a 12.5% chance in Phase II of the lottery. And right now, we have a 50/50 chance of there even being a playoffs. So what is 50/50 of 50/50 to attain 12.5%?

If the playoffs are cancelled, our odds improve dramatically, we simply have a 12.5% chance of drafting Lafreniere.

[spoiler: Zaide's answer is the correct one down below.]

Click to read more...

 

[LaP]

I'M not a mathematician but i'm a computer engineer and i've done a fair amount of math for my baccalaureate.

You can't calculate the chances for us to lose against the pens. You also can't calculate the chances for the season to be cancelled. Those parts of the equation can't be computed.

 

[tinyzombies]

 

[tinyzombies]

That's not the right formula because one proposition (us losing to Pitt) is dependent on the previous (that a playoff will take place).

 

[Frank JT]

The probability are that we will select at 9t or 16th.
Lafreniere is a bonus in that scenario.

 

[MXD]

I'd guesstimate @ 7.5%.

 

[tinyzombies]

I'd guesstimate @ 7.5%.

It is something like that at this point in time, maybe lower. If the playoffs are cancelled, it will nearly double and become the certain number 12.5%.

 

[CHfan1]

If the playoffs actually happen, we have a 50/50 chance to beat Pitt (or 35/65 depending on what you think the odds of the series are). If we lose, we then have a 12.5% chance in Phase II of the lottery. And right now, we have a 50/50 chance of there even being a playoffs. So what is 50/50 of 50/50 to attain 12.5%?

If the playoffs are cancelled, our odds improve dramatically, we simply have a 12.5% chance of drafting Lafreniere.

Hopefully my math is correct here but if the Habs have a 50/50 shot against the Pens and there is a 50/50 chance of the actual playoffs happening that means the Habs have a 25% chance of making the playoffs and losing their shot at Lafrenière and a 75% chance of being in the phase 2 lottery.

So 12.5*.75= 9.375% chance at Lafrenière.

 

[tinyzombies]

I was asking the question wrong. The question really is: What are the odds that a playoff will take place, we lose to Pittsburgh and then win Phase II of the lottery?



What is the formula for dependent probability?

If they are dependent, then P(A and B) = P(A)*P(B|A) which is the probability of A times the probability of "B happening if A has occurred," which is different than the "Probability of B if A has not occurred."

***

A-NHL playoffs will happen (let's say 25%)
B-Montreal loses to Pittsburgh (let's say 65%)

P(25%)*P(65%/25%) = 25%/2.6=‭‭9.62%

THEN

A-Playoffs happen and Montreal loses to Pittsburgh: 9.62%
B-We win Phase II of the lottery (12.5%)

P(9.62%)*P(12.5%/9.62%) = 9.62%/1.3=‭‭%7.4

***

If you change the odds:

A-NHL playoffs will happen (let's say 50%)
B-Montreal loses to Pittsburgh (let's say 60%)

P(50%)*P(60%/50%) = 50%/1.2=‭‭41.7%

THEN

A-Playoffs happen and Montreal loses to Pittsburgh: 41.7% chance of both of those things occurring
B-We win Phase II of the lottery (12.5%)

P(41.7%)*P(12.5%/41.7%) = 41.7%/0.30=‭‭12.51% (wut?)



***


If the playoffs don't take place: 12.5%

***

So, our odds are the same?

 

[Kairi Zaide]

Using your odds (50% there are playoffs, and 50% we beat Pittsburgh if there are playoffs)

P(Habs win 1st overall pick) = P(No playoffs)P(Win lottery) + P(Playoffs)P(Lose to Pittsburgh)P(Win lottery) = 0.5*0.125 + 0.5*0.5*0.125 = 0.09375 == 9.375%, as posted above

 

[MXD]

- Our odds to beat Pittsburgh are below 50%.
- What happens if the playoffs start but somehow don't go to the end?

 

[tinyzombies]

This brings me back to my school days. Ah, the memories:

Bergevin is traveling eastbound on a plane with $8,400,000 dollars. Sebastian Aho is traveling westbound on a plane with $0. How much will Bergevin offer Sammy Aho?

 

[ThaDevilGirl]

I never thought I'd find a math thread on HF but here we are.

 

[WeThreeKings]

- Our odds to beat Pittsburgh are below 50%.
- What happens if the playoffs start but somehow don't go to the end?

As long as the play in round finishes. They will go with the result of the lottery.

 

[MXD]

As long as the play in round finishes. They will go with the result of the lottery.

Thanks. To be honest, I think this (playoffs starts, the play-ins are completed, but the playoffs cannot go thru the end) is the likeliest result.

 

[WeThreeKings]

Thanks. To be honest, I think this (playoffs starts, the play-ins are completed, but the playoffs cannot go thru the end) is the likeliest result.

I just want to be eliminated by any means possible.

 

[Kairi Zaide]

I was asking the question wrong. The question really is: What are the odds that a playoff will take place, we lose to Pittsburgh and then win Phase II of the lottery?



What is the formula for dependent probability?

If they are dependent, then P(A and B) = P(A)*P(B|A) which is the probability of A times the probability of "B happening if A has occurred," which is different than the "Probability of B if A has not occurred."

***

A-NHL playoffs will happen (let's say 25%)
B-Montreal loses to Pittsburgh (let's say 65%)

P(25%)*P(65%/25%) = 25%/2.6=‭‭9.62%

THEN

A-Playoffs happen and Montreal loses to Pittsburgh: 9.62%
B-We win Phase II of the lottery (12.5%)

P(9.62%)*P(12.5%/9.62%) = 9.62%/1.3=‭‭%7.4

***

If you change the odds:

A-NHL playoffs will happen (let's say 50%)
B-Montreal loses to Pittsburgh (let's say 60%)

P(50%)*P(60%/50%) = 50%/1.2=‭‭41.7%

THEN

A-Playoffs happen and Montreal loses to Pittsburgh: 41.7% chance of both of those things occurring
B-We win Phase II of the lottery (12.5%)

P(41.7%)*P(12.5%/41.7%) = 41.7%/0.30=‭‭12.51% (wut?)



***


If the playoffs don't take place: 12.5%

***

So, our odds are the same?

I have to say, your math are completely off.

You do have the formula right :
P(B|A) = P(A∩B)/P(A) → P(A∩B) = P(A)*P(B|A)

However, P(B|A) is not P(B)/P(A), which is what you've been doing. P(B|A) is simply P(B|A), i.e. the probability that B occurs knowing that A has occurred.

You can illustrate all the scenarios with a probability tree (i did it with the 50-50 probabilities)

As you can see, the sum of the probabilities on the right is 100%. If you sum the desired scenarios (habs win the lottery - in green), then you get 9.375%.

You can change the odds, but the logic stays the same.

 

[SOLR]

I'M not a mathematician but i'm a computer engineer and i've done a fair amount of math for my baccalaureate.

You can't calculate the chances for us to lose against the pens. You also can't calculate the chances for the season to be cancelled. Those parts of the equation can't be computed.

Or someone would get rich at the stock market really quick.

 

[SOLR]

I have to say, your math are completely off.

You do have the formula right :
P(B|A) = P(A∩B)/P(A) → P(A∩B) = P(A)*P(B|A)

However, P(B|A) is not P(B)/P(A), which is what you've been doing. P(B|A) is simply P(B|A), i.e. the probability that B occurs knowing that A has occurred.

You can illustrate all the scenarios with a probability tree (i did it with the 50-50 probabilities)

As you can see, the sum of the probabilities on the right is 100%. If you sum the desired scenarios (habs win the lottery - in green), then you get 9.375%.

You can change the odds, but the logic stays the same.

All the 50% are bunkers - can't be computed, it's unknown.

 

[Kairi Zaide]

All the 50% are bunkers - can't be computed, it's unknown.

That is correct, hence why I mentioned using his odds.

That being said, they can be inferred (using analytics) or supposed. This is a valid approach when things are uncertain, as long as there is a rationale behind. We will never truly know the real odds because the event will only occur once with one of two outcomes.

The probability of Pittsburgh beating Montreal is, in a best of 5, as I suggested before, very likely to be between 55% and 60%. The best team in the NHL vs the worst team will only win the game (not series) at most 65-70% of the time, approximately, and this number gets closer to 50% the closer two teams are in the standings.

 

[DavePeak]

LaffyHabs: Your odds calculator! Just run the program and enter your odds for playoffs happening and Habs winning the play-in!

Edit: LaffyOrCup: You think our odds of getting Lafrenière are low? In this version, you can compare those to the odds of winning the Stanley Cup.

Using the latter, if there are 50% chances of playoffs going all the way and 50% of us winning the Play-in and then each round of the playoffs, we would have:

• 9.375% of getting Laffy (as was posted by other users previously)
• 1.5625% of winning the Stanley Cup.

If you change the odds of winning any series to 45%, then:

• 9.6785% of getting Laffy
• 0.9226% of winning the Stanley Cup.

Of course, as 26Mats said below, all those odds are based on pure speculation.

 

[ahmedou]

Odds = 1909% (IYKWIM)

 

[26Mats]

I'M not a mathematician but i'm a computer engineer and i've done a fair amount of math for my baccalaureate.

You can't calculate the chances for us to lose against the pens. You also can't calculate the chances for the season to be cancelled. Those parts of the equation can't be computed.

You can't compute the odds of beating the pens or of the season being canceled, you can only speculate what they are base don judgements and compute our odds based on those speculations.

So basically our odds of drafting Lafreniere can't be computed, they can only be speculated based on judgements.

 

[le_sean]

It’s 100% thanks to John Scott.

 

[LaP]

You can't compute the odds of beating the pens or of the season being canceled, you can only speculate what they are base don judgements and compute our odds based on those speculations.

So basically our odds of drafting Lafreniere can't be computed, they can only be speculated based on judgements.

There's so many variables this season though that it is very hard to make an educated guess.